3 Marker

Physics 3-Marker Previous Year Questions (PYQs) - Class 11 & 12

Class 11 Physics

Chapter 1: Units and Measurements

Question: Why is measurement essential in physics? Explain with two examples.

Solution: Measurement allows quantification of physical phenomena. For example, measuring velocity helps analyze motion, and measuring temperature helps study thermodynamics.

Question: Distinguish between fundamental and derived quantities with examples.

Solution: Fundamental quantities are independent (e.g., length, mass). Derived quantities depend on fundamental ones (e.g., speed = distance/time).

Question: What are the characteristics of a good unit of measurement?

Solution: A good unit should be universally accepted, invariable, reproducible, and easy to use.

Question: Define physical quantity. How are physical quantities classified?

Solution: A physical quantity is a measurable property. Classified as scalar or vector; and fundamental or derived.

Question: Explain the difference between scalar and vector quantities with examples.

Solution: Scalars have magnitude only (e.g., mass). Vectors have magnitude and direction (e.g., velocity).

Question: Why is it important to have standard units in scientific measurements?

Solution: Standard units ensure consistency, accuracy, and universal understanding in scientific communication.

Question: Name two instruments used for measuring length and state their least counts.

Solution: Vernier caliper (least count = 0.1 mm), Screw gauge (least count = 0.01 mm).

Question: What is meant by direct and indirect measurement? Give one example of each.

Solution: Direct: using instruments (e.g., ruler for length). Indirect: calculated (e.g., density = mass/volume).

Question: How does measurement help in validating scientific laws?

Solution: Measurement provides data to test hypotheses and verify predictions of scientific laws.

Question: Why are multiple measurements taken in experiments?

Solution: To reduce random errors and improve reliability through averaging.

Question: Name the four commonly used systems of units and mention one unit from each.

Solution: CGS (centimetre), FPS (foot), MKS (metre), SI (metre).

Question: What are the advantages of the SI system over other systems of units?

Solution: SI system is coherent, universally accepted, and based on fundamental constants.

Question: List the seven base quantities in SI and their units.

Solution: Length (m), Mass (kg), Time (s), Current (A), Temperature (K), Amount (mol), Luminous Intensity (cd).

Question: Define the SI unit of electric current. How is it realized practically?

Solution: Ampere (A) is defined via the force between two parallel conductors carrying current.

Question: Differentiate between base units and derived units with examples.

Solution: Base units are fundamental (e.g., metre). Derived units are combinations (e.g., m/s).

Question: What is meant by a coherent system of units?

Solution: In a coherent system, derived units are directly obtained from base units without conversion factors.

Question: Define mole. Why is it considered a base quantity?

Solution: Mole is the amount of substance containing 6.022 × 10²³ entities. It’s fundamental in chemistry.

Question: What is the SI unit of pressure? Derive it from base units.

Solution: Pascal (Pa) = N/m² = kg·m⁻¹·s⁻².

Question: How is the SI unit of temperature defined?

Solution: Kelvin is defined based on the triple point of water (273.16 K).

Question: Why is the SI system preferred in scientific work?

Solution: It provides consistency, precision, and global standardization.

Question: Define significant figures. How are they useful in scientific measurements?

Solution: Significant figures reflect precision and help convey the reliability of measurements.

Question: How many significant figures are there in 0.00450? Explain your answer.

Solution: Three significant figures: 4, 5, and the trailing zero after decimal.

Question: What is meant by absolute error and percentage error?

Solution: Absolute error = |measured - true|. Percentage error = (absolute error / true value) × 100.

Question: A quantity is measured as 5.2 ± 0.1 cm. Calculate the percentage error.

Solution: Percentage error = (0.1 / 5.2) × 100 ≈ 1.92%.

Question: State the rules for rounding off significant figures.

Solution: If digit after rounding place is ≥5, increase preceding digit by 1; if <5, retain it.

Question: How is uncertainty calculated in multiplication and division of quantities?

Solution: Add percentage errors of individual quantities to get total percentage error.

Question: Define least count. How does it affect the accuracy of measurement?

Solution: Least count is the smallest measurable value. It sets the minimum uncertainty.

Question: Two resistances are measured as 5 ± 0.2 Ω and 10 ± 0.1 Ω. Find the total resistance in series with error.

Solution: Total resistance = 15 Ω, Error = 0.2 + 0.1 = 0.3 Ω.

Question: What is the difference between accuracy and precision? Give one example of each.

Solution: Accuracy is closeness to true value; precision is repeatability. Measuring 9.8 m/s² is accurate; repeated values like 9.76, 9.77 are precise.

Question: Why is it important to report uncertainty in scientific results?

Solution: It reflects the reliability of data and helps in comparing results meaningfully.

Question: What is dimensional analysis? Mention one use.

Solution: Dimensional analysis involves using dimensions to check equations or convert units. It helps verify correctness of physical equations.

Question: Derive the formula for time period of a simple pendulum using dimensional analysis.

Solution: T ∝ √(l/g); Dimensions: [T] = √([L]/[LT⁻²]) = [T]

Question: What are the limitations of dimensional analysis?

Solution: Cannot determine dimensionless constants; fails for trigonometric, exponential, and logarithmic functions.

Question: How can dimensional analysis help in converting units?

Solution: By expressing quantities in terms of base units, it allows conversion between systems (e.g., CGS to SI).

Question: Use dimensional analysis to check the correctness of the formula: v = u + at

Solution: Dimensions of v, u, and at are all [LT⁻¹]; hence, equation is dimensionally correct.

Question: What is the dimensional formula of energy?

Solution: Energy = Force × Distance = [MLT⁻²] × [L] = [ML²T⁻²]

Question: Find the dimensions of Planck’s constant.

Solution: E = hν ⇒ h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]

Question: What is the dimensional formula of pressure?

Solution: Pressure = Force/Area = [MLT⁻²]/[L²] = [ML⁻¹T⁻²]

Question: How can dimensional analysis help derive relationships between physical quantities?

Solution: By assuming dependence on variables and equating dimensions, one can derive formulas (e.g., T ∝ √(l/g)).

Question: What is the dimensional formula of gravitational constant G?

Solution: F = Gm₁m₂/r² ⇒ G = Fr²/m² = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²]

Chapter 2: Motion in a Straight Line

1. Define inertial and non-inertial frames of reference with one example each.

Solution:
Inertial frame: A frame where Newton’s first law holds, i.e., an object remains at rest or in uniform motion unless acted upon by a force. Example: A stationary laboratory on Earth.
Non-inertial frame: A frame where Newton’s first law does not hold, and fictitious forces appear. Example: An accelerating bus, where passengers feel a push due to acceleration.
[1 mark for definitions, 1 mark for examples, 1 mark for clarity]

2. Explain why laws of physics are the same in all inertial frames.

Solution:
Inertial frames move with constant velocity relative to each other. Newton’s laws remain unchanged because there is no acceleration in these frames, ensuring consistent physical behavior. For example, a ball dropped in a stationary room or a train moving at constant speed falls vertically with the same acceleration (g = 9.8 m/s²).
[1 mark for explanation, 1 mark for example, 1 mark for clarity]

3. Differentiate between absolute and relative motion with an example.

Solution:
Absolute motion: Motion described relative to a fixed frame, like Earth. Example: A car moving at 20 m/s relative to the ground.
Relative motion: Motion described relative to another moving object. Example: A car moving at 20 m/s relative to another car moving at 10 m/s appears to move at 10 m/s.
[1 mark for definitions, 1 mark for examples, 1 mark for clarity]

4. How does motion appear different in stationary and moving frames?

Solution:
In a stationary frame (e.g., ground), a ball thrown upward follows a vertical path. In a moving frame (e.g., a train moving at constant velocity), the ball appears to follow a curved path due to the train’s motion relative to the ground. This difference arises because the moving frame adds its velocity to the ball’s motion.
[1 mark for explanation, 1 mark for example, 1 mark for clarity]

5. What is a Galilean transformation? Explain with an example.

Solution:
Galilean transformation relates coordinates of an event between two inertial frames moving relative to each other. If frame S’ moves at velocity v relative to frame S, then x’ = x - vt. Example: A car at x = 10 m in frame S (ground) at t = 2 s, with S’ (train) moving at v = 3 m/s, is at x’ = 10 - 3×2 = 4 m in S’.
[1 mark for definition, 1 mark for equation, 1 mark for example]

6. Explain reference frame using a train passenger example.

Solution:
A reference frame is a coordinate system used to describe motion. For a passenger in a train moving at constant speed, objects inside (e.g., a book) appear stationary (train’s frame). On the ground frame, the book moves with the train’s velocity. The choice of frame changes the observed motion.
[1 mark for definition, 1 mark for example, 1 mark for clarity]

7. Why is Earth an approximate inertial frame?

Solution:
Earth is an approximate inertial frame because its rotation and orbital motion cause small accelerations, which are negligible for most experiments. For example, in a lab, a pendulum’s motion follows Newton’s laws closely, ignoring Earth’s slight rotation effects.
[1 mark for explanation, 1 mark for example, 1 mark for clarity]

8. Explain fictitious forces in non-inertial frames with an example.

Solution:
Fictitious forces appear in non-inertial frames due to the frame’s acceleration. Example: In an accelerating car, passengers feel pushed backward (fictitious force) because the car’s frame is non-inertial, unlike a stationary frame where no such force is observed.
[1 mark for definition, 1 mark for example, 1 mark for clarity]

9. How does frame choice affect projectile motion description?

Solution:
In a ground frame, a projectile follows a parabolic path due to gravity. In a frame moving with the projectile’s horizontal velocity, it appears to fall vertically, as the horizontal component is canceled. The frame choice alters the observed path.
[1 mark for explanation, 1 mark for example, 1 mark for clarity]

10. Compare laboratory and center of mass frames in collisions.

Solution:
Laboratory frame: Motion is measured relative to the ground, e.g., two cars colliding have individual velocities. Center of mass frame: The total momentum is zero, simplifying analysis, e.g., cars approach each other symmetrically. The latter simplifies calculations.
[1 mark for lab frame, 1 mark for CM frame, 1 mark for comparison]

1. Define displacement, distance, and path length.

Solution:
Displacement: Shortest distance from initial to final position, a vector (e.g., 10 m east).
Distance: Total length of the path traveled, a scalar (e.g., 20 m).
Path length: Same as distance, total length covered.
[1 mark each for displacement, distance, path length]

2. Differentiate scalar and vector quantities with examples.

Solution:
Scalar: Has only magnitude, e.g., distance (50 m), speed (20 m/s).
Vector: Has magnitude and direction, e.g., displacement (50 m east), velocity (20 m/s east).
Scalars are directionless, vectors include direction.
[1 mark for definitions, 1 mark for examples, 1 mark for clarity]

3. A particle moves 50 m east, then 30 m west. Find displacement and distance.

Solution:
Displacement = final position - initial position = 50 m east - 30 m west = 20 m east.
Distance = total path = 50 m + 30 m = 80 m.
[1.5 marks for displacement, 1.5 marks for distance]

4. Describe position vs. time graph for rectilinear motion.

Solution:
Position vs. time graph shows position (x) on y-axis and time (t) on x-axis. For constant velocity, it’s a straight line (x = vt). For accelerated motion, it’s a parabola (x = (1/2)at²). Slope gives velocity.
[1 mark for description, 1 mark for graph types, 1 mark for slope]

5. What is one-dimensional motion? Give examples.

Solution:
One-dimensional motion occurs along a straight line with one coordinate (x). Examples: (i) A car moving on a straight road, (ii) An elevator moving vertically up or down.
[1 mark for definition, 1 mark for examples, 1 mark for clarity]

6. How to determine direction of motion using position?

Solution:
Direction is determined by the sign of displacement. If final position x₂ > initial position x₁, motion is in the positive direction (e.g., right). If x₂ < x₁, motion is in the negative direction (e.g., left).
[1 mark for method, 1 mark for explanation, 1 mark for example]

7. A car travels 100 km in 2 hours. Find average speed in m/s.

Solution:
Average speed = total distance / total time.
Distance = 100 km = 100,000 m, Time = 2 h = 2 × 3600 = 7200 s.
Average speed = 100,000 / 7200 = 13.89 m/s.
[1 mark for formula, 1 mark for conversion, 1 mark for calculation]

8. Discuss sign convention for displacement.

Solution:
Displacement is a vector. Positive sign is assigned for motion in the chosen positive direction (e.g., right or up, +x). Negative sign is used for motion in the opposite direction (e.g., left or down, -x). Example: Moving 5 m right is +5 m, left is -5 m.
[1 mark for convention, 1 mark for explanation, 1 mark for example]

9. How does straight-line motion differ from motion in a plane?

Solution:
Straight-line motion is one-dimensional, along a single axis (e.g., train on tracks). Motion in a plane is two-dimensional, involving two axes (x, y), e.g., a projectile’s parabolic path. The former uses one coordinate, the latter uses two.
[1 mark for definitions, 1 mark for examples, 1 mark for clarity]

10. Find time for a train to cover 200 m from rest with a = 2 m/s².

Solution:
Use s = ut + (1/2)at², where u = 0, s = 200 m, a = 2 m/s².
200 = (1/2) × 2 × t² = t².
t² = 200, t = √200 = 14.14 s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define velocity as the derivative of position.

Solution:
Velocity is the rate of change of position with respect to time. Mathematically, v = dx/dt, where x is position and t is time. Example: If x = 2t², v = d(2t²)/dt = 4t m/s.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. How is acceleration obtained by differentiating velocity?

Solution:
Acceleration is the rate of change of velocity with respect to time. Mathematically, a = dv/dt. Example: If v = 3t + 2, then a = d(3t + 2)/dt = 3 m/s².
[1 mark for definition, 1 mark for formula, 1 mark for example]

3. Find velocity from x = 3t² + 2t + 1 using differentiation.

Solution:
Velocity v = dx/dt. Given x = 3t² + 2t + 1.
v = d(3t² + 2t + 1)/dt = 6t + 2 m/s.
[1 mark for formula, 1 mark for differentiation, 1 mark for answer]

4. Explain integration to find displacement from velocity.

Solution:
Integration is the reverse of differentiation. Displacement s = ∫v dt, the area under the velocity-time curve. Example: For v = 2t, s = ∫(2t) dt = t² + c (c = 0 at t = 0).
[1 mark for definition, 1 mark for formula, 1 mark for example]

5. Derive position from constant acceleration using integration.

Solution:
Given a = constant, a = dv/dt, so v = ∫a dt = at + c₁ (c₁ = u at t = 0).
v = u + at. Now, v = dx/dt, so x = ∫(u + at) dt = ut + (1/2)at² + c₂ (c₂ = 0).
Thus, x = ut + (1/2)at².
[1 mark for velocity, 1 mark for position, 1 mark for derivation]

6. Find acceleration if v = 4t³ - 5t + 2.

Solution:
Acceleration a = dv/dt. Given v = 4t³ - 5t + 2.
a = d(4t³ - 5t + 2)/dt = 12t² - 5 m/s².
[1 mark for formula, 1 mark for differentiation, 1 mark for answer]

7. Explain the physical meaning of dx/dt.

Solution:
dx/dt represents the instantaneous velocity, the rate at which position changes with time. Example: For x = t², dx/dt = 2t, meaning velocity increases linearly with time.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

8. Integrate v = 2t + 3 to find displacement from t=0 to t=5 s.

Solution:
Displacement s = ∫v dt = ∫(2t + 3) dt = t² + 3t + c.
From t=0 to t=5: s = (5² + 3×5) - (0) = 25 + 15 = 40 m.
[1 mark for integration, 1 mark for limits, 1 mark for answer]

9. Find instantaneous velocity for x = t⁴ - 3t² + 7 at t=2 s.

Solution:
Velocity v = dx/dt. Given x = t⁴ - 3t² + 7.
v = d(t⁴ - 3t² + 7)/dt = 4t³ - 6t. At t=2: v = 4(2³) - 6(2) = 32 - 12 = 20 m/s.
[1 mark for differentiation, 1 mark for substitution, 1 mark for answer]

10. Derive s = ut + (1/2)at² using integration.

Solution:
Given v = u + at. Velocity v = ds/dt.
ds = (u + at) dt. Integrate: s = ∫(u + at) dt = ut + (1/2)at² + c.
At t=0, s=0, so c=0. Thus, s = ut + (1/2)at².
[1 mark for setup, 1 mark for integration, 1 mark for answer]

1. Define uniform motion with two examples.

Solution:
Uniform motion: Motion with constant velocity (constant speed and direction).
Examples: (i) A car moving at 60 km/h on a straight road, (ii) A satellite in a circular orbit with constant speed.
[1 mark for definition, 1 mark for examples, 1 mark for clarity]

2. Differentiate uniform and non-uniform motion with graphs.

Solution:
Uniform motion: Constant velocity, v-t graph is a horizontal line, x-t graph is a straight line.
Non-uniform motion: Velocity changes, v-t graph is sloped/curved, x-t graph is curved.
Example: A falling ball has a sloped v-t graph due to acceleration.
[1 mark for uniform, 1 mark for non-uniform, 1 mark for graphs]

3. Is a car at constant speed of 20 m/s in uniform motion? Explain.

Solution:
Yes, if the car moves at constant speed (20 m/s) in a straight line, its velocity is constant (same direction and magnitude). Thus, it is uniform motion with zero acceleration.
[1 mark for answer, 1 mark for explanation, 1 mark for clarity]

4. Explain non-uniform motion with a falling object example.

Solution:
Non-uniform motion has changing velocity. A falling object accelerates due to gravity (g = 9.8 m/s²), so its velocity increases with time, making it non-uniform. Example: A ball dropped from a height speeds up as it falls.
[1 mark for definition, 1 mark for example, 1 mark for explanation]

5. Calculate distance for speed 10 m/s, time 5 s in uniform motion.

Solution:
For uniform motion, distance = speed × time.
Given speed = 10 m/s, time = 5 s.
Distance = 10 × 5 = 50 m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. How does acceleration differ in uniform and non-uniform motion?

Solution:
Uniform motion: Acceleration = 0, as velocity is constant.
Non-uniform motion: Acceleration ≠ 0, can be constant (e.g., free fall, a = g) or varying (e.g., a car speeding up unevenly).
[1 mark for uniform, 1 mark for non-uniform, 1 mark for examples]

7. Draw position-time graphs for uniform and non-uniform motion.

Solution:
Uniform motion: Position-time graph is a straight line (x = vt), slope = velocity.
Non-uniform motion: Graph is curved, e.g., parabola for constant acceleration (x = (1/2)at²).
[1 mark for uniform graph, 1 mark for non-uniform graph, 1 mark for explanation]

8. A particle’s speed changes from 5 m/s to 15 m/s in 10 s. Is it uniform?

Solution:
No, it is not uniform motion. Uniform motion requires constant velocity. Here, speed changes from 5 m/s to 15 m/s, indicating acceleration, so it is non-uniform motion.
[1 mark for answer, 1 mark for explanation, 1 mark for clarity]

9. Why does uniform motion have zero acceleration?

Solution:
In uniform motion, velocity is constant (no change in speed or direction). Acceleration a = dv/dt = 0, as there is no change in velocity with time. Example: A car at steady 20 m/s.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

10. Give an example of non-uniform motion with constant acceleration.

Solution:
Non-uniform motion with constant acceleration: A ball dropped from a height accelerates at g = 9.8 m/s², increasing velocity uniformly with time.
[1 mark for definition, 1 mark for example, 1 mark for explanation]

1. Define average speed and average velocity with formulas.

Solution:
Average speed = total distance / total time, a scalar.
Average velocity = displacement / total time, a vector (v_avg = Δx/Δt).
Example: For 100 m in 10 s, speed = 10 m/s, velocity depends on displacement.
[1 mark for speed, 1 mark for velocity, 1 mark for formulas]

2. Differentiate average velocity and instantaneous velocity.

Solution:
Average velocity: Displacement divided by total time (v_avg = Δx/Δt).
Instantaneous velocity: Velocity at a specific instant, v = dx/dt.
Example: A car’s average velocity over a trip vs. its speedometer reading at t=2 s.
[1 mark for definitions, 1 mark for formulas, 1 mark for example]

3. A particle travels 100 m in 10 s. Find average speed.

Solution:
Average speed = total distance / total time.
Distance = 100 m, time = 10 s.
Average speed = 100 / 10 = 10 m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

4. Explain instantaneous velocity using limits.

Solution:
Instantaneous velocity is the velocity at a specific moment, given by v = lim(Δt→0) (Δx/Δt) = dx/dt. It’s the derivative of position with respect to time. Example: For x = t², v = 2t.
[1 mark for definition, 1 mark for formula, 1 mark for example]

5. Calculate average velocity for displacement 50 m in 5 s.

Solution:
Average velocity = displacement / total time.
Displacement = 50 m, time = 5 s.
Average velocity = 50 / 5 = 10 m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Why can average speed be greater than average velocity?

Solution:
Average speed uses total distance, which includes all path lengths. Average velocity uses displacement, which may be smaller due to direction changes. Example: A round trip has zero displacement but non-zero distance.
[1 mark for explanation, 1 mark for example, 1 mark for clarity]

7. Find instantaneous velocity for x = 2t² + 3t at t=3 s.

Solution:
Velocity v = dx/dt. Given x = 2t² + 3t.
v = d(2t² + 3t)/dt = 4t + 3. At t=3: v = 4(3) + 3 = 15 m/s.
[1 mark for differentiation, 1 mark for substitution, 1 mark for answer]

8. A car goes 40 km east, 30 km west in 2 h. Find average velocity.

Solution:
Displacement = 40 km east - 30 km west = 10 km east = 10,000 m.
Time = 2 h = 7200 s. Average velocity = 10,000 / 7200 = 1.39 m/s east.
[1 mark for displacement, 1 mark for calculation, 1 mark for answer]

9. When does average speed equal instantaneous speed?

Solution:
Average speed equals instantaneous speed in uniform motion, where speed is constant throughout. Example: A car moving at a constant 20 m/s for 10 s.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

10. Find average speed for 100 km round trip at 50 km/h and 100 km/h.

Solution:
Time for 100 km at 50 km/h = 100/50 = 2 h. Time at 100 km/h = 100/100 = 1 h.
Total distance = 200 km, total time = 2 + 1 = 3 h.
Average speed = 200 / 3 = 66.67 km/h.
[1 mark for time, 1 mark for calculation, 1 mark for answer]

1. Define uniformly accelerated motion with an example.

Solution:
Uniformly accelerated motion is motion with constant acceleration. Example: A ball falling freely under gravity with a = 9.8 m/s², velocity increasing uniformly.
[1 mark for definition, 1 mark for example, 1 mark for clarity]

2. State the three equations of motion for uniform acceleration.

Solution:
For uniformly accelerated motion:
(i) v = u + at
(ii) s = ut + (1/2)at²
(iii) v² = u² + 2as
[1 mark for each equation]

3. A body accelerates at 3 m/s² from rest. Find velocity after 4 s.

Solution:
Use v = u + at, where u = 0, a = 3 m/s², t = 4 s.
v = 0 + 3 × 4 = 12 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain the role of gravity in uniformly accelerated motion.

Solution:
Gravity causes constant acceleration (g = 9.8 m/s²) in free fall or vertical motion. It uniformly increases velocity of falling objects, e.g., a dropped stone accelerates at g.
[1 mark for role, 1 mark for example, 1 mark for clarity]

5. Find distance with u=0, a=2 m/s², t=5 s.

Solution:
Use s = ut + (1/2)at², where u = 0, a = 2 m/s², t = 5 s.
s = 0 + (1/2) × 2 × 5² = 25 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Differentiate uniform motion from uniformly accelerated motion.

Solution:
Uniform motion: Constant velocity, a = 0, e.g., car at steady 20 m/s.
Uniformly accelerated motion: Constant acceleration, velocity changes, e.g., a falling ball with a = g.
[1 mark for uniform, 1 mark for accelerated, 1 mark for examples]

7. Find time to reach 20 m/s with a=4 m/s² from rest.

Solution:
Use v = u + at, where u = 0, v = 20 m/s, a = 4 m/s².
20 = 0 + 4t, t = 20/4 = 5 s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. A ball falls with g=10 m/s². Find velocity after 2 s.

Solution:
Use v = u + at, where u = 0, a = 10 m/s², t = 2 s.
v = 0 + 10 × 2 = 20 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Why does velocity increase linearly in uniform acceleration?

Solution:
In uniform acceleration, a = constant. From v = u + at, velocity increases linearly with time (t) as a is constant. Example: A car with a = 2 m/s² has v = 2t.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

10. Calculate stopping distance for u=20 m/s, a=-5 m/s².

Solution:
Use v² = u² + 2as, where v = 0, u = 20 m/s, a = -5 m/s².
0 = 20² + 2(-5)s, 0 = 400 - 10s, s = 40 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. How is acceleration found from a velocity-time graph?

Solution:
Acceleration is the slope of the velocity-time graph, a = Δv/Δt. For uniform acceleration, the graph is a straight line. Example: If v increases from 10 to 20 m/s in 5 s, a = (20-10)/5 = 2 m/s².
[1 mark for slope, 1 mark for explanation, 1 mark for example]

2. Draw velocity-time graph for uniform motion.

Solution:
For uniform motion, velocity is constant, so the velocity-time graph is a horizontal straight line parallel to the time axis. Example: A car at 20 m/s has a flat line at v = 20 m/s.
[1 mark for description, 1 mark for graph, 1 mark for example]

3. Find displacement from area under velocity-time graph.

Solution:
Displacement = area under v-t graph. For uniform motion, area = v × t. Example: For v = 10 m/s, t = 5 s, area = 10 × 5 = 50 m.
[1 mark for area, 1 mark for formula, 1 mark for example]

4. Describe position-time graph for constant velocity.

Solution:
For constant velocity, the position-time graph is a straight line (x = vt + x₀). The slope equals velocity. Example: A car at 10 m/s has a linear x-t graph with slope 10.
[1 mark for description, 1 mark for slope, 1 mark for example]

5. What does the slope of a position-time graph represent?

Solution:
The slope of a position-time graph (dx/dt) represents instantaneous velocity. Example: For x = 2t, slope = 2 m/s, indicating constant velocity.
[1 mark for slope, 1 mark for velocity, 1 mark for example]

6. Draw position-time graph for accelerated motion.

Solution:
For uniformly accelerated motion, the position-time graph is a parabola (x = ut + (1/2)at²). Example: For a = 2 m/s², u = 0, x = t² gives a parabolic curve.
[1 mark for description, 1 mark for equation, 1 mark for example]

7. Calculate acceleration if v-t graph slope is 2.

Solution:
Acceleration = slope of v-t graph. Given slope = 2.
a = Δv/Δt = 2 m/s².
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Explain negative slope in velocity-time graph.

Solution:
A negative slope in a v-t graph indicates deceleration, where velocity decreases with time. Example: A car slowing from 20 m/s to 10 m/s in 5 s has a = (10-20)/5 = -2 m/s².
[1 mark for explanation, 1 mark for example, 1 mark for calculation]

9. Find distance from position-time graph for non-uniform motion.

Solution:
For non-uniform motion, distance is the total path length from the x-t graph. Split the graph into segments, calculate displacement per segment, and sum absolute values. Example: For a curved x-t graph, integrate |dx/dt| over time.
[1 mark for method, 1 mark for explanation, 1 mark for example]

10. Compare v-t and x-t graphs for free fall.

Solution:
Free fall: v-t graph is a straight line (v = gt, slope = g). x-t graph is a parabola (x = (1/2)gt²). Velocity increases linearly, while position changes quadratically.
[1 mark for v-t, 1 mark for x-t, 1 mark for comparison]

1. Derive v = u + at graphically.

Solution:
On a v-t graph for uniform acceleration, velocity increases linearly. Slope = a = (v-u)/t.
Rearrange: v - u = at, so v = u + at.
Example: For u = 0, a = 2 m/s², t = 5 s, v = 10 m/s.
[1 mark for graph, 1 mark for derivation, 1 mark for example]

2. Derive s = ut + (1/2)at² using calculus.

Solution:
Velocity v = u + at, and v = ds/dt.
ds = (u + at) dt. Integrate: s = ∫(u + at) dt = ut + (1/2)at² + c.
At t=0, s=0, so c=0. Thus, s = ut + (1/2)at².
[1 mark for setup, 1 mark for integration, 1 mark for answer]

3. Explain v² = u² + 2as graphically.

Solution:
On a v-t graph, displacement = area under the graph = (v+u)/2 × t. From v = u + at, t = (v-u)/a.
Substitute: s = (v+u)/2 × (v-u)/a. Simplify: v² = u² + 2as.
[1 mark for area, 1 mark for derivation, 1 mark for answer]

4. Derive v² = u² + 2as using calculus.

Solution:
From a = v dv/ds (chain rule), ∫a ds = ∫v dv.
Left: a∫ds = as. Right: ∫v dv = (v² - u²)/2.
Equate: as = (v² - u²)/2, so v² = u² + 2as.
[1 mark for setup, 1 mark for integration, 1 mark for answer]

5. Show displacement as area under v-t graph graphically.

Solution:
For uniform acceleration, v-t graph is a straight line from u to v. Displacement = area of trapezium = (u+v)/2 × t.
Using v = u + at, s = ut + (1/2)at².
[1 mark for area, 1 mark for formula, 1 mark for derivation]

6. Use integration to find v from a = constant.

Solution:
Given a = constant, a = dv/dt.
dv = a dt. Integrate: v = ∫a dt = at + c. At t=0, v=u, so c=u.
Thus, v = u + at.
[1 mark for setup, 1 mark for integration, 1 mark for answer]

7. Derive position as a function of time graphically.

Solution:
For v = u + at, the v-t graph is a straight line. Displacement = area under v-t graph = area of trapezium = (u+v)/2 × t.
Substitute v = u + at: s = ut + (1/2)at².
[1 mark for area, 1 mark for derivation, 1 mark for answer]

8. Derive s_n = u + (1/2)a(2n-1) using calculus.

Solution:
Distance in nth second s_n = displacement from t=n-1 to t=n.
s = ut + (1/2)at², s_n = s(n) - s(n-1) = u + (1/2)a(2n-1).
Or, s_n = ∫(n-1 to n)(u + at) dt = u + (1/2)a(2n-1).
[1 mark for setup, 1 mark for derivation, 1 mark for answer]

9. Derive time to maximum height in projectile graphically.

Solution:
For a projectile, v-t graph for vertical motion is a straight line (v = u - gt). At max height, v = 0.
From graph, time to max height = u/g (where v = 0).
[1 mark for graph, 1 mark for derivation, 1 mark for answer]

10. Relate v and a in equations of motion using differentiation.

Solution:
Velocity v = dx/dt, acceleration a = dv/dt = d²x/dt².
For uniform acceleration, a = constant, so v = u + at (integrate a).
Example: If x = t², a = 2 m/s² (constant).
[1 mark for relation, 1 mark for derivation, 1 mark for example]

Chapter 3: Motion in a Plane

1. Define position and displacement vectors.

Solution:
Position vector: Vector from origin to a point, e.g., r = 3i + 4j m.
Displacement vector: Vector from initial to final position, e.g., Δr = r₂ - r₁.
Both have magnitude and direction.
[1 mark for position, 1 mark for displacement, 1 mark for clarity]

2. Find the displacement vector from (2, 3) m to (5, 7) m.

Solution:
Displacement vector Δr = r₂ - r₁. Given r₁ = 2i + 3j, r₂ = 5i + 7j.
Δr = (5i + 7j) - (2i + 3j) = 3i + 4j m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

3. Explain the role of position vectors in describing motion.

Solution:
Position vectors specify an object’s location relative to the origin (e.g., r = xi + yj). They help track motion in a plane by defining coordinates. Example: A particle at (3, 4) m has r = 3i + 4j m.
[1 mark for role, 1 mark for explanation, 1 mark for example]

4. Calculate the magnitude of the position vector r = 6i + 8j m.

Solution:
Magnitude of position vector |r| = √(x² + y²). Given r = 6i + 8j.
|r| = √(6² + 8²) = √(36 + 64) = √100 = 10 m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

5. How is displacement vector different from distance?

Solution:
Displacement vector: Has magnitude and direction (e.g., 5i m). Distance: Scalar, only magnitude (e.g., 5 m). Displacement is the shortest path, distance is total path length.
[1 mark for displacement, 1 mark for distance, 1 mark for difference]

6. Find the displacement vector if a particle moves from (1, 1) m to (-2, 3) m.

Solution:
Displacement Δr = r₂ - r₁. Given r₁ = i + j, r₂ = -2i + 3j.
Δr = (-2i + 3j) - (i + j) = -3i + 2j m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

7. Calculate the direction of displacement vector from (1, 2) m to (4, 6) m.

Solution:
Displacement Δr = r₂ - r₁ = (4i + 6j) - (i + 2j) = 3i + 4j m.
Direction θ = tan⁻¹(Δy/Δx) = tan⁻¹(4/3) ≈ 53.13°.
[1 mark for displacement, 1 mark for direction formula, 1 mark for answer]

8. What is the significance of the position vector’s direction?

Solution:
The direction of a position vector indicates the angle relative to the origin, defining the object’s location. Example: r = 3i + 4j m has direction θ = tan⁻¹(4/3) from the x-axis.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

9. Calculate the magnitude of displacement from (0, 0) m to (3, 4) m.

Solution:
Displacement Δr = 3i + 4j m. Magnitude |Δr| = √(x² + y²).
|Δr| = √(3² + 4²) = √(9 + 16) = 5 m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

10. Explain how position vectors help in two-dimensional motion.

Solution:
Position vectors (r = xi + yj) define an object’s coordinates in a plane, tracking its motion along x and y axes. Example: A particle moving from (1, 2) m to (3, 4) m has Δr = 2i + 2j m.
[1 mark for role, 1 mark for explanation, 1 mark for example]

1. Define a general vector and its notation.

Solution:
A general vector is a quantity with magnitude and direction, represented as A = A_x i + A_y j in 2D. Notation: Bold A or arrow (→A). Example: Velocity v = 3i + 4j m/s.
[1 mark for definition, 1 mark for notation, 1 mark for example]

2. Explain the components of a vector in a plane.

Solution:
A vector A = A_x i + A_y j has x-component (A_x) and y-component (A_y) along i and j unit vectors. Example: A = 5i + 12j m has A_x = 5 m, A_y = 12 m.
[1 mark for components, 1 mark for explanation, 1 mark for example]

3. Calculate the magnitude of vector A = 3i - 4j.

Solution:
Magnitude |A| = √(A_x² + A_y²). Given A = 3i - 4j.
|A| = √(3² + (-4)²) = √(9 + 16) = 5 units.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

4. What is the direction of a vector? Explain with an example.

Solution:
The direction of a vector is the angle it makes with a reference axis (e.g., x-axis), given by θ = tan⁻¹(A_y/A_x). Example: For A = 3i + 3j, θ = tan⁻¹(3/3) = 45°.
[1 mark for definition, 1 mark for formula, 1 mark for example]

5. Write the vector A = 10 units at 60° in component form.

Solution:
A = A_x i + A_y j, where A_x = A cosθ, A_y = A sinθ. Given A = 10, θ = 60°.
A_x = 10 cos60° = 5, A_y = 10 sin60° = 5√3. A = 5i + 5√3 j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Why are unit vectors used in vector notation?

Solution:
Unit vectors (i, j) have magnitude 1 and define direction along axes, simplifying vector representation. Example: A = 2i + 3j uses i, j to denote x, y directions.
[1 mark for role, 1 mark for explanation, 1 mark for example]

7. Find the direction of vector A = -3i + 4j.

Solution:
Direction θ = tan⁻¹(A_y/A_x). Given A = -3i + 4j, A_x = -3, A_y = 4.
θ = tan⁻¹(4/-3) = -53.13° (second quadrant, θ = 180° - 53.13° = 126.87°).
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Explain the significance of vector notation in physics.

Solution:
Vector notation (e.g., A = A_x i + A_y j) simplifies describing quantities with direction, like velocity. It allows precise calculations in 2D/3D. Example: Displacement r = 3i + 4j m.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

9. Represent a vector of 8 units at 45° in component form.

Solution:
A = A_x i + A_y j, A_x = A cosθ, A_y = A sinθ. Given A = 8, θ = 45°.
A_x = 8 cos45° = 8/√2 = 4√2, A_y = 8 sin45° = 4√2. A = 4√2 i + 4√2 j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

10. How does vector notation help in motion analysis?

Solution:
Vector notation (e.g., v = v_x i + v_y j) describes motion in multiple directions, enabling component-wise analysis. Example: A projectile’s velocity v = u cosθ i + u sinθ j.
[1 mark for role, 1 mark for explanation, 1 mark for example]

1. Define equality of vectors with an example.

Solution: < Ascending Two vectors are equal if they have the same magnitude and direction. Example: A = 3i + 4j and B = 3i + 4j are equal (same components).
[1 mark for definition, 1 mark for example, 1 mark for clarity]

2. Check if vectors A = 2i + 3j and B = 2i + 3j are equal.

Solution:
Vectors are equal if their components are identical. A = 2i + 3j, B = 2i + 3j.
A_x = B_x = 2, A_y = B_y = 3. Since components and direction are the same, A = B.
[1 mark for condition, 1 mark for comparison, 1 mark for answer]

3. Are vectors A = 5i and B = -5i equal? Explain.

Solution:
Vectors are equal if they have the same magnitude and direction. A = 5i (right), B = -5i (left).
Magnitude |A| = |B| = 5, but directions are opposite, so A ≠ B.
[1 mark for condition, 1 mark for explanation, 1 mark for answer]

4. Explain why magnitude alone is not enough for vector equality.

Solution:
Vector equality requires same magnitude and direction. Example: A = 4i (right) and B = 4j (up) have |A| = |B| = 4, but different directions, so A ≠ B.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

5. Check if A = 3i - 4j and B = 4j - 3i are equal.

Solution:
Rewrite B = 4j - 3i = -3i + 4j. Compare: A = 3i - 4j, B = -3i + 4j.
Components differ (A_x = 3, B_x = -3; A_y = -4, B_y = 4), so A ≠ B.
[1 mark for rewriting, 1 mark for comparison, 1 mark for answer]

6. Why is direction crucial for vector equality?

Solution:
Vectors are equal only if their magnitude and direction match. Direction determines the vector’s orientation. Example: A = 5i (right) and B = -5i (left) differ due to opposite directions.
[1 mark for explanation, 1 mark for importance, 1 mark for example]

7. Are A = 2i + 2j and B = 2√2 i at 45° equal?

Solution:
B = 2√2 cos45° i + 2√2 sin45° j = 2i + 2j.
A = 2i + 2j. Since A_x = B_x = 2, A_y = B_y = 2, A = B.
[1 mark for B components, 1 mark for comparison, 1 mark for answer]

8. Explain vector equality in terms of components.

Solution:
Two vectors are equal if their corresponding components are identical (A_x = B_x, A_y = B_y). Example: A = 3i + 4j, B = 3i + 4j are equal as components match.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

9. Check if A = 6i - 8j and B = 10 units at 306.87° are equal.

Solution:
B = 10 cos306.87° i + 10 sin306.87° j = 6i - 8j (cos306.87° = 0.6, sin306.87° = -0.8).
A = 6i - 8j. Components match, so A = B.
[1 mark for B components, 1 mark for comparison, 1 mark for answer]

10. Why can’t vectors in different planes be equal?

Solution:
Vectors in different planes have components in different directions (e.g., i, j vs. i, k). Equality requires identical components. Example: A = 3i + 4j (xy-plane) ≠ B = 3i + 4k (xz-plane).
[1 mark for explanation, 1 mark for reason, 1 mark for example]

1. Define multiplication of a vector by a real number.

Solution:
Multiplying a vector A by a real number k scales its magnitude by |k|, keeping direction same (k > 0) or opposite (k < 0). Example: If A = 2i, 3A = 6i.
[1 mark for definition, 1 mark for effect, 1 mark for example]

2. Find 2A if A = 3i + 4j.

Solution:
For kA, multiply each component by k. Given A = 3i + 4j, k = 2.
2A = 2(3i + 4j) = 6i + 8j.
[1 mark for method, 1 mark for calculation, 1 mark for answer]

3. What happens to a vector’s direction when multiplied by -1?

Solution:
Multiplying by -1 reverses the vector’s direction, keeping magnitude same. Example: A = 2i + 3j, -A = -2i - 3j (opposite direction).
[1 mark for effect, 1 mark for explanation, 1 mark for example]

4. Calculate -3A for A = i - 2j.

Solution:
For kA, multiply components by k. Given A = i - 2j, k = -3.
-3A = -3(i - 2j) = -3i + 6j.
[1 mark for method, 1 mark for calculation, 1 mark for answer]

5. Explain the effect of multiplying a vector by a fraction.

Solution:
Multiplying by a fraction (0 < k < 1) reduces the vector’s magnitude, keeping direction same. Example: A = 4i + 3j, (1/2)A = 2i + 1.5j (smaller magnitude).
[1 mark for effect, 1 mark for explanation, 1 mark for example]

6. Find (1/2)A if A = 6i + 8j.

Solution:
For kA, multiply components by k. Given A = 6i + 8j, k = 1/2.
(1/2)A = (1/2)(6i + 8j) = 3i + 4j.
[1 mark for method, 1 mark for calculation, 1 mark for answer]

7. How does multiplying by a negative number affect a vector?

Solution:
A negative number (k < 0) scales the magnitude by |k| and reverses direction. Example: A = 2i, -2A = -4i (opposite direction, doubled magnitude).
[1 mark for effect, 1 mark for explanation, 1 mark for example]

8. Calculate 4A for A = -2i + 3j.

Solution:
For kA, multiply components by k. Given A = -2i + 3j, k = 4.
4A = 4(-2i + 3j) = -8i + 12j.
[1 mark for method, 1 mark for calculation, 1 mark for answer]

9. Explain why multiplying by zero gives a null vector.

Solution:
Multiplying by zero (k = 0) makes all components zero, resulting in a null vector (0 magnitude). Example: A = 3i + 4j, 0A = 0i + 0j = 0.
[1 mark for explanation, 1 mark for reason, 1 mark for example]

10. Find -1.5A if A = 4i - 2j.

Solution:
For kA, multiply components by k. Given A = 4i - 2j, k = -1.5.
-1.5A = -1.5(4i - 2j) = -6i + 3j.
[1 mark for method, 1 mark for calculation, 1 mark for answer]

1. Explain vector addition with an example.

Solution:
Vector addition combines vectors by adding their components: A + B = (A_x + B_x)i + (A_y + B_y)j. Example: A = 2i + 3j, B = 1i + 4j, A + B = 3i + 7j.
[1 mark for method, 1 mark for example, 1 mark for clarity]

2. Find A + B for A = 3i + 2j, B = -1i + 5j.

Solution:
A + B = (A_x + B_x)i + (A_y + B_y)j. Given A = 3i + 2j, B = -1i + 5j.
A + B = (3 - 1)i + (2 + 5)j = 2i + 7j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

3. Explain the parallelogram law of vector addition.

Solution:
Parallelogram law: The resultant of two vectors is the diagonal of the parallelogram formed by them. Magnitude: R = √(A² + B² + 2ABcosθ). Example: A = 3i, B = 4j, R = 5 units at 53.13°.
[1 mark for law, 1 mark for magnitude, 1 mark for example]

4. Find A - B for A = 4i + 3j, B = 2i + 1j.

Solution:
A - B = (A_x - B_x)i + (A_y - B_y)j. Given A = 4i + 3j, B = 2i + 1j.
A - B = (4 - 2)i + (3 - 1)j = 2i + 2j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

5. Explain vector subtraction geometrically.

Solution:
Vector subtraction A - B = A + (-B). Geometrically, add the negative of B (opposite direction) to A. Example: A = 3i, B = 2i, A - B = 3i + (-2i) = i.
[1 mark for method, 1 mark for geometry, 1 mark for example]

6. Find the magnitude of A + B for A = 3i, B = 4j.

Solution:
A + B = 3i + 4j. Magnitude |R| = √(R_x² + R_y²).
|R| = √(3² + 4²) = √(9 + 16) = 5 units.
[1 mark for addition, 1 mark for magnitude, 1 mark for answer]

7. Calculate A + B for A = 5i - 2j, B = -3i + 4j.

Solution:
A + B = (A_x + B_x)i + (A_y + B_y)j. Given A = 5i - 2j, B = -3i + 4j.
A + B = (5 - 3)i + (-2 + 4)j = 2i + 2j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Find the direction of A - B for A = 4i + 3j, B = 1i + 1j.

Solution:
A - B = (4 - 1)i + (3 - 1)j = 3i + 2j. Direction θ = tan⁻¹(R_y/R_x).
θ = tan⁻¹(2/3) = 33.69°.
[1 mark for subtraction, 1 mark for direction, 1 mark for answer]

9. Explain the triangle law of vector addition.

Solution:
Triangle law: Place vectors head-to-tail; the resultant is from the tail of the first to the head of the second. Example: A = 3i, B = 4j, resultant = 3i + 4j (closes triangle).
[1 mark for law, 1 mark for explanation, 1 mark for example]

10. Calculate the magnitude of A - B for A = 6i + 8j, B = 2i + 2j.

Solution:
A - B = (6 - 2)i + (8 - 2)j = 4i + 6j. Magnitude |R| = √(4² + 6²).
|R| = √(16 + 36) = √52 = 7.21 units.
[1 mark for subtraction, 1 mark for magnitude, 1 mark for answer]

1. Define a unit vector with an example.

Solution:
A unit vector has magnitude 1 and specifies direction. Example: For A = 3i + 4j, unit vector â = A/|A| = (3i + 4j)/5 = 0.6i + 0.8j.
[1 mark for definition, 1 mark for example, 1 mark for clarity]

2. Find the unit vector of A = 6i + 8j.

Solution:
Unit vector â = A/|A|. Magnitude |A| = √(6² + 8²) = √100 = 10.
â = (6i + 8j)/10 = 0.6i + 0.8j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

3. Explain the use of unit vectors in physics.

Solution:
Unit vectors (i, j) simplify vector representation by defining direction along axes. They help express vectors in component form. Example: Velocity v = 5i + 3j m/s uses i, j.
[1 mark for use, 1 mark for explanation, 1 mark for example]

4. Calculate the unit vector of A = -3i + 4j.

Solution:
Unit vector â = A/|A|. Magnitude |A| = √((-3)² + 4²) = √25 = 5.
â = (-3i + 4j)/5 = -0.6i + 0.8j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

5. Why do unit vectors have magnitude 1?

Solution:
Unit vectors have magnitude 1 to standardize direction representation without affecting magnitude. Example: â = (3i + 4j)/5 has |â| = 1, showing only direction.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

6. Find the unit vector of A = 5i.

Solution:
Unit vector â = A/|A|. Magnitude |A| = √(5²) = 5.
â = 5i/5 = i.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

7. Explain how unit vectors simplify vector operations.

Solution:
Unit vectors (i, j) allow component-wise addition, subtraction, and scaling of vectors. Example: A = 2i + 3j, B = 1i + 4j, A + B = 3i + 7j using i, j.
[1 mark for role, 1 mark for explanation, 1 mark for example]

8. Calculate the unit vector of A = -2i - 2j.

Solution:
Unit vector â = A/|A|. Magnitude |A| = √((-2)² + (-2)²) = √8 = 2√2.
â = (-2i - 2j)/(2√2) = (-i - j)/√2 = -0.707i - 0.707j.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

9. Find the unit vector of a vector 10 units at 30°.

Solution:
A = 10 cos30° i + 10 sin30° j = 5√3 i + 5j. |A| = √((5√3)² + 5²) = 10.
â = (5√3 i + 5j)/10 = 0.866i + 0.5j.
[1 mark for A, 1 mark for calculation, 1 mark for answer]

10. Why are i and j standard unit vectors?

Solution:
i and j are unit vectors along x and y axes (magnitude 1), standardizing 2D vector representation. Example: A = 3i + 4j uses i, j for x, y directions.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

1. Explain resolution of a vector into rectangular components.

Solution:
Resolution splits a vector into x and y components: A = A_x i + A_y j, where A_x = A cosθ, A_y = A sinθ. Example: A = 5 units at 37°, A_x = 5 cos37° ≈ 4, A_y = 5 sin37° ≈ 3.
[1 mark for method, 1 mark for formulas, 1 mark for example]

2. Resolve a vector of 10 units at 60° into components.

Solution:
A = A_x i + A_y j, A_x = A cosθ, A_y = A sinθ. Given A = 10, θ = 60°.
A_x = 10 cos60° = 5, A_y = 10 sin60° = 5√3. A = 5i + 5√3 j.
[1 mark for formulas, 1 mark for calculation, 1 mark for answer]

3. Find the components of a vector 8 units at 45°.

Solution:
A_x = A cosθ, A_y = A sinθ. Given A = 8, θ = 45°.
A_x = 8 cos45° = 8/√2 ≈ 5.66, A_y = 8 sin45° ≈ 5.66. A = 5.66i + 5.66j.
[1 mark for formulas, 1 mark for calculation, 1 mark for answer]

4. Why is vector resolution important in physics?

Solution:
Resolution simplifies vector analysis by breaking vectors into x, y components for easier calculations. Example: A projectile’s velocity splits into horizontal (u cosθ) and vertical (u sinθ) components.
[1 mark for importance, 1 mark for explanation, 1 mark for example]

5. Resolve A = 12 units at 120° into components.

Solution:
A_x = A cosθ, A_y = A sinθ. Given A = 12, θ = 120°.
A_x = 12 cos120° = -6, A_y = 12 sin120° = 6√3. A = -6i + 6√3 j.
[1 mark for formulas, 1 mark for calculation, 1 mark for answer]

6. Find the x-component of a vector 15 units at 30°.

Solution:
x-component A_x = A cosθ. Given A = 15, θ = 30°.
A_x = 15 cos30° = 15 × √3/2 ≈ 12.99 units.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

7. Calculate the y-component of a vector 20 units at 60°.

Solution:
y-component A_y = A sinθ. Given A = 20, θ = 60°.
A_y = 20 sin60° = 20 × √3/2 ≈ 17.32 units.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Explain how resolution helps in projectile motion.

Solution:
Resolution splits projectile velocity into horizontal (u cosθ) and vertical (u sinθ) components, allowing independent analysis of each. Example: A ball at 20 m/s, 30° has u_x = 17.32 m/s, u_y = 10 m/s.
[1 mark for role, 1 mark for explanation, 1 mark for example]

9. Resolve A = 7 units at 270° into components.

Solution:
A_x = A cosθ, A_y = A sinθ. Given A = 7, θ = 270°.
A_x = 7 cos270° = 0, A_y = 7 sin270° = -7. A = -7j.
[1 mark for formulas, 1 mark for calculation, 1 mark for answer]

10. Find the components of a vector 9 units at 180°.

Solution:
A_x = A cosθ, A_y = A sinθ. Given A = 9, θ = 180°.
A_x = 9 cos180° = -9, A_y = 9 sin180° = 0. A = -9i.
[1 mark for formulas, 1 mark for calculation, 1 mark for answer]

1. Define scalar and vector products of vectors.

Solution:
Scalar product (A·B): A scalar, A·B = AB cosθ.
Vector product (A×B): A vector, |A×B| = AB sinθ, direction by right-hand rule.
Example: A = 3i, B = 4j, A·B = 0, |A×B| = 12.
[1 mark for scalar, 1 mark for vector, 1 mark for example]

2. Calculate A·B for A = 2i + 3j, B = 4i + 5j.

Solution:
A·B = A_x B_x + A_y B_y. Given A = 2i + 3j, B = 4i + 5j.
A·B = (2 × 4) + (3 × 5) = 8 + 15 = 23.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

3. Find A×B for A = 3i, B = 4j.

Solution:
A×B = (A_x B_y - A_y B_x)k. Given A = 3i, B = 4j.
A×B = (3 × 4 - 0 × 0)k = 12k.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

4. Explain the physical significance of the scalar product.

Solution:
Scalar product (A·B = AB cosθ) gives the component of one vector along another, used in work (W = F·d). Example: Force F = 3i N, displacement d = 4i m, W = 12 J.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

5. Calculate the angle between A = 3i + 4j and B = 4i + 3j.

Solution:
A·B = |A||B| cosθ. A·B = (3 × 4) + (4 × 3) = 24. |A| = √(3² + 4²) = 5, |B| = 5.
cosθ = 24/(5 × 5) = 0.96, θ = cos⁻¹(0.96) ≈ 16.26°.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Explain the physical significance of the vector product.

Solution:
Vector product (A×B) gives a vector perpendicular to both, used in torque (τ = r×F). Example: r = 2i m, F = 3j N, τ = 6k N·m.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

7. Find A·B for A = 5i - 2j, B = -2i + 3j.

Solution:
A·B = A_x B_x + A_y B_y. Given A = 5i - 2j, B = -2i + 3j.
A·B = (5 × -2) + (-2 × 3) = -10 - 6 = -16.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Calculate |A×B| for A = 2i + 2j, B = 3i - 3j.

Solution:
|A×B| = |A_x B_y - A_y B_x|. Given A = 2i + 2j, B = 3i - 3j.
|A×B| = |(2 × -3) - (2 × 3)| = |-6 - 6| = 12 units.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

9. Why is the scalar product commutative but vector product is not?

Solution:
Scalar product: A·B = B·A (commutative) as it’s a scalar. Vector product: A×B = -B×A (not commutative) due to direction (right-hand rule). Example: A = i, B = j, A×B = k, B×A = -k.
[1 mark for explanation, 1 mark for reason, 1 mark for example]

10. Find the direction of A×B for A = 3i, B = 4j.

Solution:
A×B = (A_x B_y - A_y B_x)k = (3 × 4 - 0 × 0)k = 12k.
Direction is along +z-axis (by right-hand rule).
[1 mark for calculation, 1 mark for direction, 1 mark for answer]

1. Define motion in a plane with an example.

Solution:
Motion in a plane is two-dimensional motion involving x and y coordinates. Example: A projectile’s parabolic path with horizontal and vertical components.
[1 mark for definition, 1 mark for example, 1 mark for clarity]

2. Explain why motion in a plane requires vector analysis.

Solution:
Motion in a plane involves two directions (x, y), requiring vectors to describe position, velocity, etc. Example: A ball thrown at an angle needs velocity components (u cosθ, u sinθ).
[1 mark for reason, 1 mark for explanation, 1 mark for example]

3. Find the velocity vector of a particle moving at 10 m/s at 45°.

Solution:
Velocity v = v_x i + v_y j, v_x = v cosθ, v_y = v sinθ. Given v = 10 m/s, θ = 45°.
v_x = 10 cos45° = 7.07 m/s, v_y = 10 sin45° = 7.07 m/s. v = 7.07i + 7.07j m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

4. Describe the role of components in plane motion.

Solution:
Components (x, y) simplify analysis of 2D motion by treating each direction independently. Example: Projectile motion splits into horizontal (constant velocity) and vertical (accelerated).
[1 mark for role, 1 mark for explanation, 1 mark for example]

5. Calculate displacement for a particle moving at 5 m/s at 30° for 2 s.

Solution:
Displacement r = (v_x t)i + (v_y t)j, v_x = 5 cos30° ≈ 4.33, v_y = 5 sin30° = 2.5.
r = (4.33 × 2)i + (2.5 × 2)j = 8.66i + 5j m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Explain how to represent 2D motion using vectors.

Solution:
2D motion uses vectors (e.g., r = xi + yj) to describe position, velocity, etc., in x and y directions. Example: A car moving at 10i + 5j m/s has x, y components.
[1 mark for method, 1 mark for explanation, 1 mark for example]

7. Find the acceleration vector for a = 2 m/s² at 60°.

Solution:
Acceleration a = a_x i + a_y j, a_x = a cosθ, a_y = a sinθ. Given a = 2 m/s², θ = 60°.
a_x = 2 cos60° = 1, a_y = 2 sin60° = √3. a = i + √3 j m/s².
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

8. Why is 2D motion different from 1D motion?

Solution:
2D motion involves two coordinates (x, y), while 1D motion uses one. Example: A projectile moves in a plane (x, y), unlike a car on a straight road (x only).
[1 mark for difference, 1 mark for explanation, 1 mark for example]

9. Calculate the magnitude of displacement for v = 4i + 3j m/s, t = 2 s.

Solution:
Displacement r = (v_x t)i + (v_y t)j = (4 × 2)i + (3 × 2)j = 8i + 6j m.
Magnitude |r| = √(8² + 6²) = √100 = 10 m.
[1 mark for displacement, 1 mark for magnitude, 1 mark for answer]

10. Explain the use of vectors in describing plane motion.

Solution:
Vectors describe magnitude and direction of quantities like velocity in 2D. Example: A plane’s velocity v = 100i + 50j m/s shows motion in x, y directions.
[1 mark for role, 1 mark for explanation, 1 mark for example]

1. Define uniform velocity and uniform acceleration in 2D.

Solution:
Uniform velocity: Constant speed and direction (e.g., v = 5i m/s).
Uniform acceleration: Constant acceleration vector (e.g., a = 2i m/s²).
Example: Projectile motion has uniform horizontal velocity, vertical acceleration.
[1 mark for definitions, 1 mark for example, 1 mark for clarity]

2. Explain uniform velocity in plane motion with an example.

Solution:
Uniform velocity in 2D: Constant velocity vector (no change in magnitude or direction). Example: A car moving at v = 10i + 5j m/s on a straight path.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

3. Find displacement for v = 3i + 4j m/s, t = 5 s (uniform velocity).

Solution:
For uniform velocity, r = v t. Given v = 3i + 4j m/s, t = 5 s.
r = (3 × 5)i + (4 × 5)j = 15i + 20j m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

4. Explain uniform acceleration in 2D with an example.

Solution:
Uniform acceleration: Constant acceleration vector. Example: A projectile has a = -gj m/s² (vertical acceleration due to gravity, no horizontal acceleration).
[1 mark for definition, 1 mark for explanation, 1 mark for example]

5. Calculate velocity after 2 s for u = 4i m/s, a = 2i + 3j m/s².

Solution:
v = u + a t. Given u = 4i m/s, a = 2i + 3j m/s², t = 2 s.
v = 4i + (2i + 3j) × 2 = 4i + 4i + 6j = 8i + 6j m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Find displacement for u = 5i m/s, a = 2j m/s², t = 3 s.

Solution:
r = u t + (1/2)a t². Given u = 5i m/s, a = 2j m/s², t = 3 s.
r = (5i × 3) + (1/2)(2j)(3²) = 15i + 9j m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

7. Why is projectile motion a case of uniform acceleration?

Solution:
Projectile motion has constant acceleration (a = -gj) due to gravity in the vertical direction, while horizontal velocity is constant. Example: A thrown ball follows this pattern.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Calculate final velocity for u = 3i + 4j m/s, a = -2j m/s², t = 2 s.

Solution:
v = u + a t. Given u = 3i + 4j m/s, a = -2j m/s², t = 2 s.
v = (3i + 4j) + (-2j × 2) = 3i + 4j - 4j = 3i m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

9. Explain why uniform velocity has zero acceleration.

Solution:
Uniform velocity means constant velocity vector (no change in magnitude or direction), so a = dv/dt = 0. Example: A plane cruising at 100i m/s.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Find displacement magnitude for u = 2i m/s, a = 3j m/s², t = 4 s.

Solution:
r = u t + (1/2)a t² = (2i × 4) + (1/2)(3j)(4²) = 8i + 24j m.
|r| = √(8² + 24²) = √(64 + 576) = 25.3 m.
[1 mark for displacement, 1 mark for magnitude, 1 mark for answer]

1. Define projectile motion with two examples.

Solution:
Projectile motion is motion under gravity along a curved path with constant horizontal velocity and vertical acceleration (g).
Examples: (i) A ball kicked in football, (ii) A bullet fired from a gun.
[1 mark for definition, 1 mark for examples, 1 mark for clarity]

2. Derive the time of flight for a projectile.

Solution:
Vertical velocity: v_y = u sinθ - gt. At max height, v_y = 0, so 0 = u sinθ - gt, t = u sinθ/g.
Time of flight T = 2 × time to max height = 2u sinθ/g.
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

3. Find maximum height for u = 20 m/s, θ = 30°, g = 10 m/s².

Solution:
Maximum height H = (u² sin²θ)/(2g). Given u = 20 m/s, θ = 30°, sin30° = 0.5.
H = (20² × 0.5²)/(2 × 10) = (400 × 0.25)/20 = 5 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain why range is maximum at θ = 45°.

Solution:
Range R = (u² sin2θ)/g. Sin2θ is maximum (1) at 2θ = 90°, so θ = 45°. This balances horizontal and vertical components for max range.
[1 mark for formula, 1 mark for explanation, 1 mark for clarity]

5. Calculate range for u = 30 m/s, θ = 60°, g = 10 m/s².

Solution:
Range R = (u² sin2θ)/g. Given u = 30 m/s, θ = 60°, sin120° = √3/2.
R = (30² × √3/2)/10 = (900 × 0.866)/10 ≈ 77.94 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Derive the trajectory equation of a projectile.

Solution:
x = u cosθ t, y = u sinθ t - (1/2)gt². From x, t = x/(u cosθ).
Substitute in y: y = x tanθ - (g x²)/(2 u² cos²θ).
[1 mark for setup, 1 mark for substitution, 1 mark for equation]

7. Why is projectile motion two-dimensional?

Solution:
Projectile motion involves horizontal (constant velocity) and vertical (accelerated) components, requiring x, y coordinates. Example: A thrown ball has u cosθ (x) and u sinθ - gt (y).
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Find time to max height for u = 40 m/s, θ = 45°, g = 10 m/s².

Solution:
At max height, v_y = 0. v_y = u sinθ - gt. Given u = 40 m/s, θ = 45°, sin45° = 1/√2.
0 = 40/√2 - 10t, t = 2.83 s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain the effect of air resistance on projectile motion.

Solution:
Air resistance reduces range and height by opposing motion, altering the parabolic trajectory. Example: A feather’s path is more affected than a stone’s.
[1 mark for effect, 1 mark for explanation, 1 mark for example]

10. Find θ for range 100 m, u = 20 m/s, g = 10 m/s².

Solution:
Range R = (u² sin2θ)/g. Given R = 100 m, u = 20 m/s, g = 10 m/s².
100 = (20² × sin2θ)/10, sin2θ = 0.25, 2θ = 30°, θ = 15°.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

Chapter 4: Laws of Motion

1. Define force intuitively and give an example.

Solution:
Force is a push or pull that changes an object’s state of rest or motion. Example: Pushing a cart to make it move.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Explain how force affects motion.

Solution:
Force causes acceleration, changing an object’s velocity or direction. Example: Kicking a ball changes its speed and direction.
[1 mark for effect, 1 mark for explanation, 1 mark for example]

3. Why is force a vector quantity?

Solution:
Force has magnitude and direction, e.g., 10 N upward. Its effect depends on direction. Example: Pulling a box east vs. west changes its motion.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

4. Calculate the force required to accelerate a 5 kg object at 2 m/s².

Solution:
Force F = ma. Given m = 5 kg, a = 2 m/s².
F = 5 × 2 = 10 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain the role of force in changing momentum.

Solution:
Force changes momentum by causing acceleration (F = dp/dt). Example: A bat hitting a ball increases its momentum.
[1 mark for role, 1 mark for explanation, 1 mark for example]

6. Differentiate between contact and non-contact forces.

Solution:
Contact force: Acts via physical contact, e.g., friction. Non-contact force: Acts without contact, e.g., gravity. Example: Pushing a table (contact) vs. a falling apple (non-contact).
[1 mark for differentiation, 1 mark for examples, 1 mark for clarity]

7. Calculate the force on a 3 kg object moving with constant velocity.

Solution:
For constant velocity, a = 0, so F = ma = 0. Given m = 3 kg.
Net force F = 0 N.
[1 mark for concept, 1 mark for calculation, 1 mark for answer]

8. Explain how force causes deformation.

Solution:
Force can deform objects by altering their shape or size. Example: Compressing a spring by applying force.
[1 mark for effect, 1 mark for explanation, 1 mark for example]

9. Why is force necessary for motion in some cases?

Solution:
Force overcomes inertia or opposing forces (e.g., friction) to initiate or change motion. Example: A car needs engine force to move against friction.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate the acceleration due to a 20 N force on a 4 kg object.

Solution:
F = ma, so a = F/m. Given F = 20 N, m = 4 kg.
a = 20/4 = 5 m/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define inertia with an example.

Solution:
Inertia is the tendency of an object to resist changes in its state of motion. Example: A book stays at rest on a table.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. State Newton’s First Law of Motion.

Solution:
Newton’s First Law: An object remains at rest or in uniform motion unless acted upon by a net external force. Example: A car continues moving unless brakes are applied.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

3. Explain why inertia depends on mass.

Solution:
Inertia is proportional to mass; heavier objects resist motion changes more. Example: A 10 kg box is harder to push than a 1 kg box.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Why does a body at rest remain at rest?

Solution:
Due to inertia, a body at rest stays at rest unless a net external force acts, per Newton’s First Law. Example: A stone remains still without force.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Explain Newton’s First Law with a real-life example.

Solution:
Newton’s First Law states that motion continues without net force. Example: A hockey puck slides on ice with minimal friction until stopped.
[1 mark for law, 1 mark for explanation, 1 mark for example]

6. Why is Newton’s First Law called the law of inertia?

Solution:
It describes inertia, the resistance to motion change, as objects maintain rest or motion without force. Example: A ball rolls until friction stops it.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Explain the role of inertia in a car crash.

Solution:
Inertia keeps passengers moving forward during a sudden stop unless seat belts apply force. Example: A person lurches forward in a crash.
[1 mark for role, 1 mark for explanation, 1 mark for example]

8. Why does a heavy object require more force to move?

Solution:
Greater mass means greater inertia, requiring more force to change motion. Example: Pushing a truck vs. a bicycle.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate the net force on a 6 kg object moving at constant velocity.

Solution:
Per Newton’s First Law, constant velocity means a = 0, so F = ma = 0. Given m = 6 kg.
Net force = 0 N.
[1 mark for concept, 1 mark for calculation, 1 mark for answer]

10. Explain how inertia affects a satellite in orbit.

Solution:
Inertia keeps a satellite moving in its orbit unless acted upon by forces like gravity. Example: A satellite moves tangentially without air resistance.
[1 mark for role, 1 mark for explanation, 1 mark for example]

1. Define momentum with its unit.

Solution:
Momentum is mass times velocity (p = mv), a vector quantity. Unit: kg·m/s. Example: A 2 kg ball at 5 m/s has p = 10 kg·m/s.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. State Newton’s Second Law of Motion.

Solution:
The rate of change of momentum is proportional to the applied force (F = dp/dt). Example: A force accelerates a car.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

3. Derive F = ma from Newton’s Second Law.

Solution:
Momentum p = mv, dp/dt = m(dv/dt) = ma. Newton’s Second Law states F = dp/dt, so F = ma.
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

4. Calculate momentum of a 3 kg object moving at 4 m/s.

Solution:
Momentum p = mv. Given m = 3 kg, v = 4 m/s.
p = 3 × 4 = 12 kg·m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Find the force on a 10 kg object accelerating at 5 m/s².

Solution:
F = ma. Given m = 10 kg, a = 5 m/s².
F = 10 × 5 = 50 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain how Newton’s Second Law relates force and acceleration.

Solution:
Newton’s Second Law (F = ma) states force causes acceleration proportional to mass. Example: A 20 N force on a 4 kg object gives a = 5 m/s².
[1 mark for relation, 1 mark for explanation, 1 mark for example]

7. Calculate acceleration of a 2 kg object under 8 N force.

Solution:
F = ma, so a = F/m. Given F = 8 N, m = 2 kg.
a = 8/2 = 4 m/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is momentum a vector quantity?

Solution:
Momentum (p = mv) depends on velocity, a vector, so it has direction. Example: A car moving east at 10 m/s has different momentum than west.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find the change in momentum for a 5 kg object accelerating from 2 m/s to 6 m/s.

Solution:
Δp = m(v_f - v_i). Given m = 5 kg, v_i = 2 m/s, v_f = 6 m/s.
Δp = 5 × (6 - 2) = 20 kg·m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain the significance of Newton’s Second Law in dynamics.

Solution:
It relates force, mass, and acceleration, enabling motion prediction. Example: Calculating a rocket’s acceleration from its thrust.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

1. Define impulse with its unit.

Solution:
Impulse is force times time (J = FΔt), equal to change in momentum. Unit: N·s or kg·m/s. Example: A bat hitting a ball.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Relate impulse to Newton’s Second Law.

Solution:
Newton’s Second Law: F = dp/dt. Impulse J = FΔt = Δp (change in momentum). Example: A 10 N force for 2 s gives J = 20 N·s.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

3. Calculate impulse for a 15 N force acting for 3 s.

Solution:
Impulse J = FΔt. Given F = 15 N, Δt = 3 s.
J = 15 × 3 = 45 N·s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain how impulse changes momentum.

Solution:
Impulse (J = FΔt) equals the change in momentum (Δp). Example: A cricket ball’s momentum changes when struck by a bat.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

5. Find impulse for a 2 kg object changing velocity from 3 m/s to 7 m/s.

Solution:
Impulse J = Δp = m(v_f - v_i). Given m = 2 kg, v_i = 3 m/s, v_f = 7 m/s.
J = 2 × (7 - 3) = 8 N·s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why is impulse important in collisions?

Solution:
Impulse measures momentum change during collisions, affecting outcomes. Example: A car’s airbag increases Δt, reducing force.
[1 mark for importance, 1 mark for explanation, 1 mark for example]

7. Calculate the force if impulse is 24 N·s over 4 s.

Solution:
J = FΔt, so F = J/Δt. Given J = 24 N·s, Δt = 4 s.
F = 24/4 = 6 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain how impulse reduces injury in sports.

Solution:
Impulse (J = FΔt) shows increasing Δt reduces force, minimizing injury. Example: A padded glove increases contact time in boxing.
[1 mark for mechanism, 1 mark for explanation, 1 mark for example]

9. Find the velocity change for a 4 kg object with 16 N·s impulse.

Solution:
J = mΔv, so Δv = J/m. Given J = 16 N·s, m = 4 kg.
Δv = 16/4 = 4 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Why is impulse a vector quantity?

Solution:
Impulse (J = FΔt) depends on force, a vector, so it has direction. Example: A ball hit eastward gains eastward momentum.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

1. State Newton’s Third Law of Motion.

Solution:
For every action, there is an equal and opposite reaction. Example: A rocket moves forward as exhaust gases push backward.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

2. Explain Newton’s Third Law with a real-life example.

Solution:
Action and reaction forces are equal, opposite, and act on different bodies. Example: Walking, you push the ground backward, it pushes you forward.
[1 mark for explanation, 1 mark for application, 1 mark for example]

3. Why do action and reaction forces not cancel out?

Solution:
Action and reaction act on different bodies, so they don’t cancel. Example: A gun recoils while the bullet moves forward.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Explain how Newton’s Third Law applies to swimming.

Solution:
Swimmer pushes water backward (action), water pushes swimmer forward (reaction). Example: Each stroke propels the swimmer.
[1 mark for application, 1 mark for explanation, 1 mark for example]

5. Give an example of Newton’s Third Law in a collision.

Solution:
In a collision, each object exerts an equal and opposite force on the other. Example: Two cars colliding exert equal forces on each other.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

6. Explain the role of Newton’s Third Law in rocket propulsion.

Solution:
Rocket expels gas backward (action), gas pushes rocket forward (reaction). Example: A rocket launches into space.
[1 mark for role, 1 mark for explanation, 1 mark for example]

7. Why is Newton’s Third Law universal?

Solution:
It applies to all interactions, regardless of force type or scale. Example: From atoms to planets, action-reaction pairs exist.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Explain Newton’s Third Law in a tug-of-war.

Solution:
Each team pulls the rope (action), and the rope pulls back (reaction). Example: Teams feel equal tension in the rope.
[1 mark for application, 1 mark for explanation, 1 mark for example]

9. How does Newton’s Third Law apply to a book on a table?

Solution:
Book’s weight (action) pushes table downward; table’s normal force (reaction) pushes book upward. Example: Book stays at rest.
[1 mark for application, 1 mark for explanation, 1 mark for example]

10. Explain why action-reaction forces are simultaneous.

Solution:
Action and reaction occur instantly as forces are mutual interactions. Example: When you jump, you push the ground as it pushes you up.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

1. State the law of conservation of linear momentum.

Solution:
Total linear momentum of a system is constant if no external force acts. Example: Momentum before and after a collision is equal.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

2. Derive conservation of momentum using Newton’s Laws.

Solution:
Newton’s Third Law: F₁ = -F₂. Newton’s Second Law: F = dp/dt. For two objects, dp₁/dt = -dp₂/dt, so dp₁ + dp₂ = 0. Total momentum is constant.
[1 mark for setup, 1 mark for derivation, 1 mark for conclusion]

3. A 3 kg object at 4 m/s collides with a 2 kg object at rest. Find common velocity if they stick together.

Solution:
m₁u₁ + m₂u₂ = (m₁ + m₂)v. Given m₁ = 3 kg, u₁ = 4 m/s, m₂ = 2 kg, u₂ = 0.
3 × 4 + 2 × 0 = (3 + 2)v, v = 12/5 = 2.4 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain conservation of momentum in rocket propulsion.

Solution:
Gas ejected backward gains momentum; rocket gains equal opposite momentum. Example: A rocket accelerates as exhaust is expelled.
[1 mark for application, 1 mark for explanation, 1 mark for example]

5. A 4 kg object at 5 m/s collides with a 6 kg object at -2 m/s. Find common velocity.

Solution:
m₁u₁ + m₂u₂ = (m₁ + m₂)v. Given m₁ = 4 kg, u₁ = 5 m/s, m₂ = 6 kg, u₂ = -2 m/s.
4 × 5 + 6 × (-2) = (4 + 6)v, 20 - 12 = 10v, v = 0.8 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why is momentum conserved in collisions?

Solution:
No external force (e.g., friction) means total momentum remains constant. Example: Billiard balls conserve momentum in a collision.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate final velocity of a 2 kg object after a 10 kg·m/s impulse.

Solution:
Impulse = Δp = mv_f - mv_i. Given J = 10 kg·m/s, m = 2 kg, v_i = 0.
10 = 2v_f, v_f = 5 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain conservation of momentum in a gun recoil.

Solution:
Before firing, momentum is zero; after, bullet’s forward momentum equals gun’s backward momentum. Example: A rifle recoils when fired.
[1 mark for explanation, 1 mark for application, 1 mark for example]

9. A 1 kg object at 8 m/s collides with a 3 kg object at -2 m/s. Find common velocity.

Solution:
m₁u₁ + m₂u₂ = (m₁ + m₂)v. Given m₁ = 1 kg, u₁ = 8 m/s, m₂ = 3 kg, u₂ = -2 m/s.
1 × 8 + 3 × (-2) = (1 + 3)v, 8 - 6 = 4v, v = 0.5 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Why is external force critical for momentum conservation?

Solution:
External forces change total momentum (F = dp/dt). Conservation requires zero net force. Example: Friction violates conservation in collisions.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

1. Define equilibrium of concurrent forces.

Solution:
Concurrent forces are in equilibrium if their vector sum is zero, resulting in no acceleration. Example: A chandelier hanging still.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Explain conditions for equilibrium of concurrent forces.

Solution:
ΣF_x = 0, ΣF_y = 0 (vector sum of forces in all directions is zero). Example: A block suspended by two equal ropes.
[1 mark for conditions, 1 mark for explanation, 1 mark for example]

3. A 10 kg object hangs by two ropes at 45°. Find tension (g = 10 m/s²).

Solution:
ΣF_y = 0: 2T sin45° = mg. Given m = 10 kg, g = 10 m/s².
2T × 0.707 = 100, T = 100/(2 × 0.707) ≈ 70.7 N.
[1 mark for setup, 1 mark for substitution, 1 mark for answer]

4. Why is equilibrium important in static systems?

Solution:
Equilibrium ensures no net force, keeping systems at rest. Example: A bridge’s cables balance to remain stable.
[1 mark for importance, 1 mark for explanation, 1 mark for example]

5. Find tension in a rope holding a 5 kg object vertically (g = 10 m/s²).

Solution:
ΣF_y = 0: T = mg. Given m = 5 kg, g = 10 m/s².
T = 5 × 10 = 50 N.
[1 mark for setup, 1 mark for substitution, 1 mark for answer]

6. Explain how three concurrent forces can be in equilibrium.

Solution:
Three forces are in equilibrium if their vector sum is zero (form a closed triangle). Example: A traffic light held by three cables.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

7. A 20 kg object is suspended by two ropes at 30°. Find tension (g = 10 m/s²).

Solution:
ΣF_y = 0: 2T sin30° = mg. Given m = 20 kg, g = 10 m/s².
2T × 0.5 = 200, T = 200/1 = 200 N.
[1 mark for setup, 1 mark for substitution, 1 mark for answer]

8. Why does an object in equilibrium have no acceleration?

Solution:
Net force is zero (ΣF = 0), so a = F/m = 0. Example: A stationary hanging lamp.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find the normal force on a 15 kg object at rest (g = 10 m/s²).

Solution:
ΣF_y = 0: N = mg. Given m = 15 kg, g = 10 m/s².
N = 15 × 10 = 150 N.
[1 mark for setup, 1 mark for substitution, 1 mark for answer]

10. Explain equilibrium in a pulley system.

Solution:
Equal and opposite forces (e.g., weights) balance, giving zero net force. Example: Two equal masses on a pulley remain at rest.
[1 mark for explanation, 1 mark for application, 1 mark for example]

1. Define static and kinetic friction.

Solution:
Static friction: Opposes motion before movement (≤ μ_s N). Kinetic friction: Opposes motion during movement (μ_k N). Example: Pushing a box (static) vs. sliding it (kinetic).
[1 mark for definitions, 1 mark for explanation, 1 mark for example]

2. State the laws of friction.

Solution:
1. Friction is proportional to normal force (f = μN).
2. Static friction ≤ μ_s N, kinetic friction = μ_k N.
3. Friction is independent of area and velocity (kinetic).
[1 mark for each law, total 3 marks]

3. Calculate kinetic friction for a 20 kg object (μ_k = 0.3, g = 10 m/s²).

Solution:
f_k = μ_k N, N = mg. Given m = 20 kg, μ_k = 0.3, g = 10 m/s².
N = 20 × 10 = 200 N, f_k = 0.3 × 200 = 60 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain the role of rolling friction.

Solution:
Rolling friction opposes rolling motion, less than sliding friction due to minimal contact. Example: A wheel rolling on a road.
[1 mark for role, 1 mark for explanation, 1 mark for example]

5. Find maximum static friction for a 10 kg object (μ_s = 0.5, g = 10 m/s²).

Solution:
f_s = μ_s N, N = mg. Given m = 10 kg, μ_s = 0.5, g = 10 m/s².
N = 10 × 10 = 100 N, f_s = 0.5 × 100 = 50 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain how lubrication reduces friction.

Solution:
Lubrication forms a low-friction layer, reducing surface contact. Example: Oil in a car engine lowers friction.
[1 mark for mechanism, 1 mark for explanation, 1 mark for example]

7. Why is static friction greater than kinetic friction?

Solution:
Static friction is higher due to stronger surface interlocking at rest. Example: Starting to push a heavy crate is harder than keeping it sliding.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Calculate force to overcome static friction for a 15 kg object (μ_s = 0.4, g = 10 m/s²).

Solution:
F = μ_s N, N = mg. Given m = 15 kg, μ_s = 0.4, g = 10 m/s².
N = 15 × 10 = 150 N, F = 0.4 × 150 = 60 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Why is friction independent of contact area?

Solution:
Friction depends on normal force and surface nature, not area, as pressure distributes evenly. Example: A block’s friction is same on different faces.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Find kinetic friction for a 5 kg object sliding on a surface (μ_k = 0.2, g = 10 m/s²).

Solution:
f_k = μ_k N, N = mg. Given m = 5 kg, μ_k = 0.2, g = 10 m/s².
N = 5 × 10 = 50 N, f_k = 0.2 × 50 = 10 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define centripetal force with an example.

Solution:
Centripetal force provides acceleration (mv²/r) for circular motion. Example: Tension in a string whirling a stone.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Derive the centripetal force formula.

Solution:
Centripetal acceleration a_c = v²/r. Newton’s Second Law: F = ma.
Centripetal force F_c = m(v²/r).
[1 mark for acceleration, 1 mark for derivation, 1 mark for expression]

3. Calculate centripetal force for a 3 kg object at 6 m/s in a 2 m radius circle.

Solution:
F_c = m(v²/r). Given m = 3 kg, v = 6 m/s, r = 2 m.
F_c = 3 × (6²/2) = 3 × 18 = 54 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain the role of friction in circular motion on a level road.

Solution:
Friction provides centripetal force (mv²/r) to keep a vehicle on a circular path. Example: Car tires grip the road during a turn.
[1 mark for role, 1 mark for explanation, 1 mark for example]

5. Find centripetal force for a 1 kg object at 8 m/s in a 4 m radius circle.

Solution:
F_c = m(v²/r). Given m = 1 kg, v = 8 m/s, r = 4 m.
F_c = 1 × (8²/4) = 1 × 16 = 16 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain centripetal force on a banked road.

Solution:
Normal force’s horizontal component provides centripetal force, reducing friction need. Example: A car turns on a banked curve safely.
[1 mark for mechanism, 1 mark for explanation, 1 mark for example]

7. Calculate max speed on a level road (μ_s = 0.6, r = 30 m, g = 10 m/s²).

Solution:
f_s = mv²/r, f_s = μ_s mg. Given μ_s = 0.6, r = 30 m, g = 10 m/s².
μ_s mg = mv²/r, v² = μ_s g r = 0.6 × 10 × 30 = 180, v ≈ 13.42 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is centripetal force a real force?

Solution:
Centripetal force (e.g., tension, friction) causes real acceleration, unlike centrifugal force. Example: A string pulls a stone inward.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find tension for a 0.2 kg stone at 5 m/s in a 1 m radius circle.

Solution:
F_c = m(v²/r). Given m = 0.2 kg, v = 5 m/s, r = 1 m.
F_c = 0.2 × (5²/1) = 0.2 × 25 = 5 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain why banking reduces skidding.

Solution:
Banking provides centripetal force via normal force, reducing friction dependence. Example: A banked road allows faster turns.
[1 mark for mechanism, 1 mark for explanation, 1 mark for example]

Chapter 5: Work, Energy, and Power

1. Define work done by a constant force with an example.

Solution:
Work done: W = Fd cosθ, where F is force, d is displacement, θ is the angle between them. Example: Pushing a box 3 m with 20 N at 0° gives W = 20 × 3 × 1 = 60 J.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate work done by a 10 N force over 5 m at 60°.

Solution:
W = Fd cosθ. Given F = 10 N, d = 5 m, θ = 60°.
W = 10 × 5 × cos60° = 10 × 5 × 0.5 = 25 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain when work done by a constant force is zero.

Solution:
Work is zero when θ = 90° (cos90° = 0) or displacement is zero. Example: Holding a book stationary or pushing a wall.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

4. Find work done by a 15 N force moving an object 4 m at 30°.

Solution:
W = Fd cosθ. Given F = 15 N, d = 4 m, θ = 30°.
W = 15 × 4 × cos30° = 15 × 4 × 0.866 ≈ 51.96 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Why is work a scalar quantity?

Solution:
Work (W = Fd cosθ) is the dot product of force and displacement, yielding a scalar. Example: Work done pushing a cart is not directional.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

6. Calculate work done by a 25 N force over 2 m at 180°.

Solution:
W = Fd cosθ. Given F = 25 N, d = 2 m, θ = 180°.
W = 25 × 2 × cos180° = 25 × 2 × (-1) = -50 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain negative work with an example.

Solution:
Negative work occurs when force opposes displacement (θ > 90°). Example: Friction does negative work when a box is pushed.
[1 mark for explanation, 1 mark for condition, 1 mark for example]

8. Find work done by gravity on a 2 kg object falling 3 m (g = 10 m/s²).

Solution:
W = Fd cosθ, F = mg. Given m = 2 kg, g = 10 m/s², d = 3 m, θ = 0°.
W = (2 × 10) × 3 × 1 = 60 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain the significance of the angle in work formula.

Solution:
The angle θ in W = Fd cosθ determines the effective force component. Example: At θ = 0°, maximum work; at θ = 90°, zero work.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

10. Calculate work done by a 30 N force over 6 m at 45°.

Solution:
W = Fd cosθ. Given F = 30 N, d = 6 m, θ = 45°.
W = 30 × 6 × cos45° = 30 × 6 × 0.707 ≈ 127.26 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define work done by a variable force.

Solution:
Work done by a variable force is W = ∫F dx over the displacement. Example: Stretching a spring with varying force.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Derive work done by a spring force.

Solution:
Spring force F = -kx. Work done: W = ∫(-kx) dx from 0 to x = -½kx² (negative for external work). Example: Stretching a spring.
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

3. Calculate work done stretching a spring (k = 100 N/m, x = 0.2 m).

Solution:
W = ½kx². Given k = 100 N/m, x = 0.2 m.
W = ½ × 100 × (0.2)² = 50 × 0.04 = 2 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain why variable force requires integration.

Solution:
Variable force changes with displacement, so work is the sum of infinitesimal contributions (W = ∫F dx). Example: Spring force increases with stretch.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Find work done compressing a spring (k = 200 N/m, x = 0.1 m).

Solution:
W = ½kx². Given k = 200 N/m, x = 0.1 m.
W = ½ × 200 × (0.1)² = 100 × 0.01 = 1 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain work done by a variable gravitational force.

Solution:
Gravitational force varies as F = GMm/r², so W = ∫F dr. Example: Work lifting a satellite to higher orbit.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

7. Calculate work done stretching a spring (k = 150 N/m, x = 0.3 m).

Solution:
W = ½kx². Given k = 150 N/m, x = 0.3 m.
W = ½ × 150 × (0.3)² = 75 × 0.09 = 6.75 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is work by a variable force path-dependent?

Solution:
For non-conservative variable forces, work depends on the path taken. Example: Friction work varies with path length.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find work done compressing a spring (k = 300 N/m, x = 0.05 m).

Solution:
W = ½kx². Given k = 300 N/m, x = 0.05 m.
W = ½ × 300 × (0.05)² = 150 × 0.0025 = 0.375 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain graphical method for variable force work.

Solution:
Work is the area under the force vs. displacement graph. Example: For a spring, the area under F = kx is a triangle, giving W = ½kx².
[1 mark for method, 1 mark for explanation, 1 mark for example]

1. Define kinetic energy with its unit.

Solution:
Kinetic energy: Energy due to motion, K = ½mv². Unit: Joule (kg·m²/s²). Example: A moving car has kinetic energy.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Derive the expression for kinetic energy.

Solution:
Work W = Fd = mad. For constant a, d = ½(v² - u²)/a (if u = 0), W = ½mv². Thus, K = ½mv².
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

3. Calculate kinetic energy of a 3 kg object moving at 4 m/s.

Solution:
K = ½mv². Given m = 3 kg, v = 4 m/s.
K = ½ × 3 × 4² = 1.5 × 16 = 24 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain how kinetic energy depends on mass.

Solution:
K = ½mv²; kinetic energy is directly proportional to mass. Example: A 2 kg object at 5 m/s has twice the K of a 1 kg object at 5 m/s.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

5. Find kinetic energy of a 5 kg object at 6 m/s.

Solution:
K = ½mv². Given m = 5 kg, v = 6 m/s.
K = ½ × 5 × 6² = 2.5 × 36 = 90 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain how kinetic energy depends on velocity.

Solution:
K = ½mv²; kinetic energy is proportional to velocity squared. Example: Doubling velocity (2v) increases K by four times.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

7. Calculate change in kinetic energy for a 2 kg object from 3 m/s to 5 m/s.

Solution:
ΔK = ½m(v_f² - v_i²). Given m = 2 kg, v_i = 3 m/s, v_f = 5 m/s.
ΔK = ½ × 2 × (5² - 3²) = 1 × (25 - 9) = 16 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is kinetic energy always positive?

Solution:
K = ½mv²; v² is always positive, and m is positive. Example: A moving object always has positive K regardless of direction.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find velocity for a 4 kg object with 32 J kinetic energy.

Solution:
K = ½mv², so v² = 2K/m. Given K = 32 J, m = 4 kg.
v² = (2 × 32)/4 = 16, v = 4 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain kinetic energy in rotational motion.

Solution:
Rotational kinetic energy: K = ½Iω², where I is moment of inertia, ω is angular velocity. Example: A spinning wheel has rotational K.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

1. State the work-energy theorem.

Solution:
Work done by net force equals change in kinetic energy (W = ΔK = ½mv² - ½mu²). Example: A force increases a car’s speed.
[1 mark for statement, 1 mark for formula, 1 mark for example]

2. Derive the work-energy theorem.

Solution:
Work W = Fd = mad. Using v² = u² + 2ad, d = (v² - u²)/(2a). W = ma × (v² - u²)/(2a) = ½mv² - ½mu² = ΔK.
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

3. Calculate work done to increase a 2 kg object’s speed from 2 m/s to 4 m/s.

Solution:
W = ΔK = ½m(v² - u²). Given m = 2 kg, u = 2 m/s, v = 4 m/s.
W = ½ × 2 × (4² - 2²) = 1 × (16 - 4) = 12 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain the work-energy theorem for a falling object.

Solution:
Work by gravity increases kinetic energy (W = ΔK). Example: A falling stone’s potential energy converts to kinetic energy.
[1 mark for explanation, 1 mark for application, 1 mark for example]

5. Find force needed to increase a 3 kg object’s speed from 3 m/s to 6 m/s over 5 m.

Solution:
W = ΔK = ½m(v² - u²) = ½ × 3 × (6² - 3²) = 1.5 × 27 = 40.5 J. W = Fd, F = W/d = 40.5/5 = 8.1 N.
[1 mark for ΔK, 1 mark for substitution, 1 mark for answer]

6. Why does the work-energy theorem apply only to net force?

Solution:
Only net force causes acceleration, changing kinetic energy. Example: Friction opposes applied force, reducing net work.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate work done to stop a 4 kg object moving at 5 m/s.

Solution:
W = ΔK = ½mv² (final v = 0). Given m = 4 kg, v = 5 m/s.
W = ½ × 4 × 5² = 2 × 25 = -50 J (negative as work stops motion).
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain work-energy theorem in a spring system.

Solution:
Work done stretching a spring changes its kinetic energy. Example: Compressing a spring increases an attached object’s K.
[1 mark for explanation, 1 mark for application, 1 mark for example]

9. Find speed of a 5 kg object after 30 J of work is done (initial v = 0).

Solution:
W = ΔK = ½mv². Given W = 30 J, m = 5 kg.
30 = ½ × 5 × v², v² = 12, v ≈ 3.46 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain negative work in the work-energy theorem.

Solution:
Negative work reduces kinetic energy (e.g., friction). Example: A sliding box slows down due to friction’s negative work.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

1. Define power with its unit.

Solution:
Power: Rate of doing work, P = W/t. Unit: Watt (J/s). Example: A motor lifting a load faster has higher power.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate power if 200 J of work is done in 4 s.

Solution:
P = W/t. Given W = 200 J, t = 4 s.
P = 200/4 = 50 W.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain power in terms of force and velocity.

Solution:
Power P = Fv cosθ, where F is force, v is velocity, θ is the angle. Example: A car’s engine delivers power as Fv.
[1 mark for formula, 1 mark for explanation, 1 mark for example]

4. Find power of a 10 N force moving an object at 5 m/s (θ = 0°).

Solution:
P = Fv cosθ. Given F = 10 N, v = 5 m/s, θ = 0°.
P = 10 × 5 × 1 = 50 W.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Calculate time taken for 1000 W to do 5000 J of work.

Solution:
P = W/t, so t = W/P. Given W = 5000 J, P = 1000 W.
t = 5000/1000 = 5 s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain why power is higher for faster work.

Solution:
Power (P = W/t) increases as time decreases for fixed work. Example: Lifting a load in 2 s requires more power than in 5 s.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Find power of a 20 N force at 3 m/s (θ = 30°).

Solution:
P = Fv cosθ. Given F = 20 N, v = 3 m/s, θ = 30°.
P = 20 × 3 × cos30° = 20 × 3 × 0.866 ≈ 51.96 W.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain power in a machine’s efficiency.

Solution:
Power output is less than input due to losses (e.g., friction). Example: A motor’s output power drives a load, input includes losses.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

9. Calculate work done by a 200 W machine in 10 s.

Solution:
P = W/t, so W = P × t. Given P = 200 W, t = 10 s.
W = 200 × 10 = 2000 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Why is power important in mechanical systems?

Solution:
Power determines the rate of energy transfer, affecting performance. Example: A car’s engine power dictates its speed capability.
[1 mark for importance, 1 mark for explanation, 1 mark for example]

1. Define potential energy with an example.

Solution:
Potential energy: Energy due to position or configuration, e.g., U = mgh for gravitational. Example: A book on a shelf.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate gravitational potential energy of a 5 kg object at 8 m (g = 10 m/s²).

Solution:
U = mgh. Given m = 5 kg, g = 10 m/s², h = 8 m.
U = 5 × 10 × 8 = 400 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain why potential energy is relative.

Solution:
Potential energy depends on a chosen reference point (e.g., h = 0). Example: A ball’s U is zero at ground level, positive above.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Find potential energy of a 3 kg object at 10 m (g = 10 m/s²).

Solution:
U = mgh. Given m = 3 kg, g = 10 m/s², h = 10 m.
U = 3 × 10 × 10 = 300 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain potential energy in a stretched rubber band.

Solution:
Stretching stores elastic potential energy due to configuration change. Example: A stretched rubber band has U = ½kx².
[1 mark for concept, 1 mark for explanation, 1 mark for example]

6. Calculate height for a 2 kg object with 200 J potential energy (g = 10 m/s²).

Solution:
U = mgh, so h = U/(mg). Given U = 200 J, m = 2 kg, g = 10 m/s².
h = 200/(2 × 10) = 10 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why does potential energy increase with height?

Solution:
U = mgh; potential energy is proportional to height. Example: A higher book on a shelf has more U.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Find mass of an object with 500 J potential energy at 5 m (g = 10 m/s²).

Solution:
U = mgh, so m = U/(gh). Given U = 500 J, h = 5 m, g = 10 m/s².
m = 500/(10 × 5) = 10 kg.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain potential energy in a compressed spring.

Solution:
Compression stores elastic potential energy (U = ½kx²). Example: A compressed spring in a toy gun stores energy.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

10. Calculate potential energy of a 4 kg object at 12 m (g = 10 m/s²).

Solution:
U = mgh. Given m = 4 kg, g = 10 m/s², h = 12 m.
U = 4 × 10 × 12 = 480 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define potential energy of a spring.

Solution:
Spring potential energy: U = ½kx², where k is spring constant, x is displacement. Example: A stretched spring stores energy.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Derive spring potential energy formula.

Solution:
Work done: W = ∫kx dx from 0 to x = ½kx². This equals U = ½kx². Example: Compressing a spring stores energy.
[1 mark for setup, 1 mark for derivation, 1 mark for expression]

3. Calculate spring potential energy (k = 200 N/m, x = 0.2 m).

Solution:
U = ½kx². Given k = 200 N/m, x = 0.2 m.
U = ½ × 200 × (0.2)² = 100 × 0.04 = 4 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Find spring constant for U = 5 J at x = 0.1 m.

Solution:
U = ½kx², so k = 2U/x². Given U = 5 J, x = 0.1 m.
k = (2 × 5)/(0.1)² = 10/0.01 = 1000 N/m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Calculate displacement for U = 2 J (k = 400 N/m).

Solution:
U = ½kx², so x² = 2U/k. Given U = 2 J, k = 400 N/m.
x² = (2 × 2)/400 = 0.01, x = 0.1 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain why spring potential energy is always positive.

Solution:
U = ½kx²; x² is positive, k is positive. Example: Stretching or compressing a spring stores positive energy.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Find potential energy for k = 150 N/m, x = 0.3 m.

Solution:
U = ½kx². Given k = 150 N/m, x = 0.3 m.
U = ½ × 150 × (0.3)² = 75 × 0.09 = 6.75 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain spring potential energy in a mechanical system.

Solution:
Spring stores energy when stretched/compressed, convertible to kinetic energy. Example: A spring-loaded toy releases stored energy.
[1 mark for role, 1 mark for explanation, 1 mark for example]

9. Find spring constant for U = 8 J at x = 0.4 m.

Solution:
U = ½kx², so k = 2U/x². Given U = 8 J, x = 0.4 m.
k = (2 × 8)/(0.4)² = 16/0.16 = 100 N/m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Calculate work done to stretch a spring (k = 250 N/m, x = 0.2 m).

Solution:
W = U = ½kx². Given k = 250 N/m, x = 0.2 m.
W = ½ × 250 × (0.2)² = 125 × 0.04 = 5 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define conservative forces with an example.

Solution:
Conservative forces: Work is path-independent, e.g., gravity. Example: Lifting a book gives same work regardless of path.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Explain why work by conservative forces is path-independent.

Solution:
Work depends only on initial and final positions, not path. Example: Gravitational work lifting an object is mgh, same for any path.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

3. Calculate work by gravity for a 2 kg object lifted 5 m (g = 10 m/s²).

Solution:
W = -mgh (negative as force opposes displacement). Given m = 2 kg, g = 10 m/s², h = 5 m.
W = -2 × 10 × 5 = -100 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain potential energy associated with conservative forces.

Solution:
Conservative forces have potential energy (U = -∫F dx). Example: Gravitational potential energy U = mgh for a lifted object.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

5. Why is spring force conservative?

Solution:
Spring force (F = -kx) has path-independent work, with U = ½kx². Example: Stretching a spring always stores same energy for same x.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

6. Calculate work by gravity for a 3 kg object lowered 4 m (g = 10 m/s²).

Solution:
W = mgh (force along displacement). Given m = 3 kg, g = 10 m/s², h = 4 m.
W = 3 × 10 × 4 = 120 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain conservative forces in a closed path.

Solution:
Work by conservative forces is zero in a closed path. Example: A ball returning to its starting height has zero net work by gravity.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

8. Why is gravitational force conservative?

Solution:
Work by gravity depends only on height difference, not path. Example: Lifting a stone to a height gives same U regardless of route.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate work by a spring force (k = 100 N/m, x = 0.2 m).

Solution:
W = -½kx² (work by spring, opposite to external force). Given k = 100 N/m, x = 0.2 m.
W = -½ × 100 × (0.2)² = -50 × 0.04 = -2 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain conservative forces in energy conservation.

Solution:
Conservative forces conserve mechanical energy (K + U = constant). Example: A pendulum’s energy converts between K and U.
[1 mark for role, 1 mark for explanation, 1 mark for example]

1. Define non-conservative forces with an example.

Solution:
Non-conservative forces: Work is path-dependent, e.g., friction. Example: Work by friction depends on path length.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Explain why friction is non-conservative.

Solution:
Friction’s work depends on path length, dissipating energy as heat. Example: Sliding a box farther increases energy loss.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

3. Calculate work by friction for a 10 kg object over 5 m (μ_k = 0.2, g = 10 m/s²).

Solution:
W = -f_k d, f_k = μ_k N, N = mg. Given m = 10 kg, μ_k = 0.2, d = 5 m, g = 10 m/s².
f_k = 0.2 × 10 × 10 = 20 N, W = -20 × 5 = -100 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Why is mechanical energy not conserved with non-conservative forces?

Solution:
Non-conservative forces dissipate energy (e.g., as heat). Example: Friction reduces a sliding block’s energy.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Calculate work by friction for a 5 kg object over 3 m (μ_k = 0.3, g = 10 m/s²).

Solution:
W = -f_k d, f_k = μ_k N, N = mg. Given m = 5 kg, μ_k = 0.3, d = 3 m, g = 10 m/s².
f_k = 0.3 × 5 × 10 = 15 N, W = -15 × 3 = -45 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain non-conservative forces in a closed path.

Solution:
Work by non-conservative forces in a closed path is non-zero. Example: Friction does work when a box returns to its starting point.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

7. Why does air resistance act as a non-conservative force?

Solution:
Air resistance dissipates energy as heat, with work path-dependent. Example: A falling parachute loses energy to air resistance.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Calculate energy loss due to friction for a 4 kg object over 6 m (μ_k = 0.4, g = 10 m/s²).

Solution:
Energy loss = |W| = f_k d, f_k = μ_k N, N = mg. Given m = 4 kg, μ_k = 0.4, d = 6 m, g = 10 m/s².
f_k = 0.4 × 4 × 10 = 16 N, W = 16 × 6 = 96 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain non-conservative forces in mechanical systems.

Solution:
Non-conservative forces reduce mechanical energy, converting it to other forms. Example: Friction in a car engine causes energy loss.
[1 mark for role, 1 mark for explanation, 1 mark for example]

10. Find work by friction for a 6 kg object over 2 m (μ_k = 0.25, g = 10 m/s²).

Solution:
W = -f_k d, f_k = μ_k N, N = mg. Given m = 6 kg, μ_k = 0.25, d = 2 m, g = 10 m/s².
f_k = 0.25 × 6 × 10 = 15 N, W = -15 × 2 = -30 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Explain energy conservation in a vertical circle.

Solution:
Total mechanical energy (K + U) is conserved without non-conservative forces. Example: A pendulum’s energy shifts between K and U.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

2. Find velocity at the bottom of a vertical circle (h = 2 m, g = 10 m/s²).

Solution:
At top, v = 0 (min speed). Conservation: U_top = K_bottom. mgh = ½mv². Given h = 2 m, g = 10 m/s².
10 × 2 = ½v², v² = 40, v ≈ 6.32 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Calculate tension at the bottom of a vertical circle for a 0.5 kg object at 5 m/s (r = 1 m, g = 10 m/s²).

Solution:
T = mg + mv²/r. Given m = 0.5 kg, v = 5 m/s, r = 1 m, g = 10 m/s².
T = (0.5 × 10) + (0.5 × 5²/1) = 5 + 12.5 = 17.5 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain tension variation in a vertical circle.

Solution:
Tension is maximum at the bottom (T = mg + mv²/r) and minimum at the top (T = mg - mv²/r). Example: A stone in a vertical circle.
[1 mark for variation, 1 mark for explanation, 1 mark for example]

5. Find velocity at the top of a vertical circle for a 1 kg object (r = 2 m, T = 0, g = 10 m/s²).

Solution:
At top, T = 0: mv²/r = mg. Given m = 1 kg, r = 2 m, g = 10 m/s².
v²/2 = 10, v² = 20, v ≈ 4.47 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Calculate tension at the top for a 0.3 kg object at 4 m/s (r = 1 m, g = 10 m/s²).

Solution:
T = mv²/r - mg. Given m = 0.3 kg, v = 4 m/s, r = 1 m, g = 10 m/s².
T = (0.3 × 4²/1) - (0.3 × 10) = 4.8 - 3 = 1.8 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why is minimum speed required at the top of a vertical circle?

Solution:
Minimum speed ensures tension ≥ 0 at the top (mv²/r ≥ mg). Example: A bucket needs minimum speed to retain water.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Find minimum speed at the top of a vertical circle (r = 1.5 m, g = 10 m/s²).

Solution:
At top, T = 0: mv²/r = mg, v² = rg. Given r = 1.5 m, g = 10 m/s².
v² = 1.5 × 10 = 15, v ≈ 3.87 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain energy conservation at different points in a vertical circle.

Solution:
Total energy (K + U) is constant; K is max at bottom, U is max at top. Example: A pendulum conserves energy throughout its swing.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

10. Calculate velocity at the bottom for a 0.2 kg object with v_top = 3 m/s (r = 1 m, g = 10 m/s²).

Solution:
Energy conservation: ½mv_top² + 2mgr = ½mv_bottom². Given m = 0.2 kg, v_top = 3 m/s, r = 1 m, g = 10 m/s².
½ × 0.2 × 3² + 2 × 0.2 × 10 × 1 = ½ × 0.2 × v², 0.9 + 4 = 0.1v², v² = 49, v = 7 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define elastic and inelastic collisions.

Solution:
Elastic: Kinetic energy conserved, e.g., billiard balls. Inelastic: Kinetic energy not conserved, e.g., clay balls sticking together.
[1 mark for definitions, 1 mark for examples, 1 mark for clarity]

2. Explain momentum conservation in collisions.

Solution:
Momentum is conserved if no external force acts (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂). Example: Two cars colliding conserve momentum.
[1 mark for concept, 1 mark for formula, 1 mark for example]

3. A 3 kg object at 5 m/s collides inelastically with a 2 kg object at rest. Find common velocity.

Solution:
m₁u₁ + m₂u₂ = (m₁ + m₂)v. Given m₁ = 3 kg, u₁ = 5 m/s, m₂ = 2 kg, u₂ = 0.
3 × 5 + 2 × 0 = (3 + 2)v, 15 = 5v, v = 3 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Calculate kinetic energy loss in an inelastic collision (3 kg at 4 m/s, 2 kg at rest).

Solution:
Common velocity: 3 × 4 + 2 × 0 = (3 + 2)v, v = 12/5 = 2.4 m/s. K_initial = ½ × 3 × 4² = 24 J, K_final = ½ × 5 × 2.4² = 14.4 J.
Loss = 24 - 14.4 = 9.6 J.
[1 mark for velocity, 1 mark for calculation, 1 mark for answer]

5. A 1 kg object at 6 m/s collides elastically with a 2 kg object at rest (1D). Find final velocities.

Solution:
Elastic collision: v₁ = (m₁ - m₂)/(m₁ + m₂)u₁, v₂ = 2m₁/(m₁ + m₂)u₁. Given m₁ = 1 kg, u₁ = 6 m/s, m₂ = 2 kg, u₂ = 0.
v₁ = (1 - 2)/(1 + 2) × 6 = -2 m/s, v₂ = 2 × 1/(1 + 2) × 6 = 4 m/s.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

6. Explain elastic collisions in two dimensions.

Solution:
Momentum is conserved in x and y directions, and kinetic energy is conserved. Example: Two billiard balls colliding at an angle.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

7. A 4 kg object at 3 m/s collides inelastically with a 6 kg object at rest. Find common velocity.

Solution:
m₁u₁ + m₂u₂ = (m₁ + m₂)v. Given m₁ = 4 kg, u₁ = 3 m/s, m₂ = 6 kg, u₂ = 0.
4 × 3 + 6 × 0 = (4 + 6)v, 12 = 10v, v = 1.2 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is kinetic energy conserved in elastic collisions?

Solution:
No energy is lost to heat or deformation in elastic collisions. Example: Hard spheres colliding retain total kinetic energy.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate kinetic energy loss for a 2 kg object at 5 m/s colliding inelastically with a 3 kg object at rest.

Solution:
Common velocity: 2 × 5 + 3 × 0 = (2 + 3)v, v = 10/5 = 2 m/s. K_initial = ½ × 2 × 5² = 25 J, K_final = ½ × 5 × 2² = 10 J.
Loss = 25 - 10 = 15 J.
[1 mark for velocity, 1 mark for calculation, 1 mark for answer]

10. Explain inelastic collisions in two dimensions.

Solution:
Momentum is conserved in x and y directions, but kinetic energy is lost. Example: Two cars colliding at an angle stick together.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

Chapter 6: System of Particles and Rotational Motion

1. Define centre of mass for a two-particle system.

Solution:
Centre of mass is the point where the total mass is assumed to be concentrated, given by x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). Example: Two masses on a rod balance at their centre of mass.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate the centre of mass of two particles (2 kg at x = 1 m, 3 kg at x = 4 m).

Solution:
x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). Given m₁ = 2 kg, x₁ = 1 m, m₂ = 3 kg, x₂ = 4 m.
x_cm = (2 × 1 + 3 × 4)/(2 + 3) = (2 + 12)/5 = 2.8 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain centre of mass motion in a system with no external force.

Solution:
Centre of mass moves with constant velocity if no external force acts (Newton’s First Law). Example: A rocket’s centre of mass moves uniformly after fuel ejection.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

4. Find the centre of mass of a uniform rod of length 2 m.

Solution:
For a uniform rod, centre of mass is at its midpoint. Given L = 2 m, x_cm = L/2 = 2/2 = 1 m from one end.
[1 mark for concept, 1 mark for calculation, 1 mark for answer]

5. Explain momentum conservation and centre of mass motion.

Solution:
Total momentum is conserved if no external force acts; centre of mass velocity remains constant. Example: In a collision, centre of mass moves uniformly.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

6. Calculate centre of mass for two particles (4 kg at x = 0 m, 6 kg at x = 5 m).

Solution:
x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). Given m₁ = 4 kg, x₁ = 0 m, m₂ = 6 kg, x₂ = 5 m.
x_cm = (4 × 0 + 6 × 5)/(4 + 6) = 30/10 = 3 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why does the centre of mass of a rigid body remain fixed relative to the body?

Solution:
Centre of mass is determined by mass distribution, which is fixed in a rigid body. Example: A spinning disc’s centre of mass stays at its centre.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Find centre of mass for two particles (1 kg at (2, 3), 2 kg at (4, 1)) in 2D.

Solution:
x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂), y_cm = (m₁y₁ + m₂y₂)/(m₁ + m₂). Given m₁ = 1 kg, (x₁, y₁) = (2, 3), m₂ = 2 kg, (x₂, y₂) = (4, 1).
x_cm = (1 × 2 + 2 × 4)/(1 + 2) = 10/3 ≈ 3.33 m, y_cm = (1 × 3 + 2 × 1)/(1 + 2) = 5/3 ≈ 1.67 m.
[1 mark for formula, 1 mark for calculation, 1 mark for answer]

9. Explain the centre of mass of a non-uniform rod.

Solution:
Centre of mass depends on mass distribution, closer to the heavier end. Example: A rod heavier at one end has its centre of mass shifted toward that end.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

10. Calculate centre of mass velocity for a system (2 kg at 3 m/s, 3 kg at 2 m/s).

Solution:
v_cm = (m₁v₁ + m₂v₂)/(m₁ + m₂). Given m₁ = 2 kg, v₁ = 3 m/s, m₂ = 3 kg, v₂ = 2 m/s.
v_cm = (2 × 3 + 3 × 2)/(2 + 3) = (6 + 6)/5 = 2.4 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define torque with its unit.

Solution:
Torque: Moment of force causing rotation, τ = r × F sinθ. Unit: N·m. Example: Turning a wrench applies torque.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate torque for a 10 N force at 0.5 m from pivot (θ = 90°).

Solution:
τ = rF sinθ. Given F = 10 N, r = 0.5 m, θ = 90°.
τ = 0.5 × 10 × sin90° = 0.5 × 10 × 1 = 5 N·m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Define angular momentum with its unit.

Solution:
Angular momentum: L = Iω, where I is moment of inertia, ω is angular velocity. Unit: kg·m²/s. Example: A spinning top has angular momentum.
[1 mark for definition, 1 mark for unit, 1 mark for example]

4. State the law of conservation of angular momentum.

Solution:
Angular momentum is conserved if no external torque acts (L = constant). Example: A figure skater spins faster when pulling arms in.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

5. Calculate angular momentum for a 2 kg object at 0.3 m from axis, v = 5 m/s.

Solution:
L = mvr sinθ (θ = 90°). Given m = 2 kg, v = 5 m/s, r = 0.3 m.
L = 2 × 5 × 0.3 × 1 = 3 kg·m²/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain conservation of angular momentum in a diving board.

Solution:
A diver’s angular momentum is conserved; tucking reduces I, increasing ω. Example: A diver spins faster when curled.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

7. Calculate torque for a 20 N force at 0.4 m from pivot (θ = 60°).

Solution:
τ = rF sinθ. Given F = 20 N, r = 0.4 m, θ = 60°.
τ = 0.4 × 20 × sin60° = 0.4 × 20 × 0.866 ≈ 6.93 N·m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain the relation between torque and angular acceleration.

Solution:
Torque τ = Iα, where I is moment of inertia, α is angular acceleration. Example: A larger torque spins a wheel faster.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

9. Find angular momentum of a disc (I = 0.1 kg·m², ω = 10 rad/s).

Solution:
L = Iω. Given I = 0.1 kg·m², ω = 10 rad/s.
L = 0.1 × 10 = 1 kg·m²/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain an application of angular momentum conservation.

Solution:
Conservation of angular momentum explains planetary orbits; no external torque keeps L constant. Example: Earth’s spin remains steady.
[1 mark for application, 1 mark for explanation, 1 mark for example]

1. Define equilibrium of a rigid body.

Solution:
A rigid body is in equilibrium if net force (ΣF = 0) and net torque (Στ = 0). Example: A balanced seesaw.
[1 mark for definition, 1 mark for conditions, 1 mark for example]

2. Explain the equations of rotational motion.

Solution:
Equations: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ. Example: A wheel’s motion under constant torque.
[1 mark for equations, 1 mark for explanation, 1 mark for example]

3. Calculate angular acceleration for a disc (I = 0.2 kg·m², τ = 4 N·m).

Solution:
τ = Iα, so α = τ/I. Given τ = 4 N·m, I = 0.2 kg·m².
α = 4/0.2 = 20 rad/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Compare linear and rotational motion.

Solution:
Linear motion: v = u + at; Rotational motion: ω = ω₀ + αt. Force causes linear acceleration; torque causes angular acceleration. Example: A car vs. a spinning top.
[1 mark for comparison, 1 mark for explanation, 1 mark for example]

5. Find angular velocity after 2 s for a wheel (ω₀ = 5 rad/s, α = 3 rad/s²).

Solution:
ω = ω₀ + αt. Given ω₀ = 5 rad/s, α = 3 rad/s², t = 2 s.
ω = 5 + 3 × 2 = 11 rad/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain conditions for rotational equilibrium.

Solution:
Στ = 0 about any point ensures no angular acceleration. Example: A balanced ladder against a wall.
[1 mark for condition, 1 mark for explanation, 1 mark for example]

7. Calculate torque needed for a 0.5 kg·m² disc to achieve α = 10 rad/s².

Solution:
τ = Iα. Given I = 0.5 kg·m², α = 10 rad/s².
τ = 0.5 × 10 = 5 N·m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain rotational motion in a flywheel.

Solution:
Flywheel rotates with constant ω or accelerates under torque (τ = Iα). Example: A flywheel stores energy in engines.
[1 mark for concept, 1 mark for explanation, 1 mark for example]

9. Find angular displacement in 3 s (ω₀ = 2 rad/s, α = 4 rad/s²).

Solution:
θ = ω₀t + ½αt². Given ω₀ = 2 rad/s, α = 4 rad/s², t = 3 s.
θ = 2 × 3 + ½ × 4 × 3² = 6 + 18 = 24 rad.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Why is ΣF = 0 necessary for rigid body equilibrium?

Solution:
ΣF = 0 ensures no translational acceleration. Example: A bridge’s supports balance all forces.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

1. Define moment of inertia with its unit.

Solution:
Moment of inertia (I): Mass distribution measure about an axis, I = Σmr². Unit: kg·m². Example: A disc’s I depends on mass and radius.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. State the moment of inertia of a uniform rod about an end.

Solution:
For a uniform rod of mass M, length L, about one end: I = (1/3)ML². Example: A 1 m rod of 2 kg has I = (1/3) × 2 × 1² = 0.67 kg·m².
[1 mark for formula, 1 mark for explanation, 1 mark for example]

3. Calculate moment of inertia of a disc (M = 2 kg, R = 0.5 m) about its centre.

Solution:
I = ½MR² for a disc. Given M = 2 kg, R = 0.5 m.
I = ½ × 2 × (0.5)² = 1 × 0.25 = 0.25 kg·m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Define radius of gyration.

Solution:
Radius of gyration (k): Distance from axis where mass is concentrated, I = Mk². Example: For a rod, k determines rotational inertia.
[1 mark for definition, 1 mark for formula, 1 mark for example]

5. Find moment of inertia of a ring (M = 1 kg, R = 0.2 m) about its centre.

Solution:
I = MR² for a ring. Given M = 1 kg, R = 0.2 m.
I = 1 × (0.2)² = 0.04 kg·m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Calculate radius of gyration for a disc (I = 0.5 kg·m², M = 2 kg).

Solution:
I = Mk², so k = √(I/M). Given I = 0.5 kg·m², M = 2 kg.
k = √(0.5/2) = √0.25 = 0.5 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. State the moment of inertia of a solid sphere about its diameter.

Solution:
For a solid sphere of mass M, radius R: I = (2/5)MR². Example: A 1 kg sphere with R = 0.1 m has I = (2/5) × 1 × (0.1)² = 0.004 kg·m².
[1 mark for formula, 1 mark for explanation, 1 mark for example]

8. Explain why moment of inertia depends on mass distribution.

Solution:
I = Σmr²; farther mass from axis increases I. Example: A ring has higher I than a disc of same mass and radius.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find moment of inertia of a rod (M = 3 kg, L = 1 m) about its centre.

Solution:
I = (1/12)ML² for a rod about its centre. Given M = 3 kg, L = 1 m.
I = (1/12) × 3 × 1² = 0.25 kg·m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Calculate radius of gyration for a rod (I = 0.2 kg·m², M = 4 kg).

Solution:
I = Mk², so k = √(I/M). Given I = 0.2 kg·m², M = 4 kg.
k = √(0.2/4) = √0.05 ≈ 0.224 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

Chapter 7: Gravitation

1. State Kepler’s first law of planetary motion.

Solution:
Kepler’s first law: Planets move in elliptical orbits with the Sun at one focus. Example: Earth’s orbit around the Sun is an ellipse.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

2. Explain Kepler’s second law with an example.

Solution:
Kepler’s second law: A line joining a planet to the Sun sweeps equal areas in equal times. Example: A planet moves faster near the Sun.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

3. State Kepler’s third law and its formula.

Solution:
Kepler’s third law: The square of a planet’s orbital period is proportional to the cube of its semi-major axis (T² ∝ a³). Example: Jupiter’s longer orbit has a longer period.
[1 mark for statement, 1 mark for formula, 1 mark for example]

4. Explain how Kepler’s third law relates to gravitational force.

Solution:
T² = (4π²/GM)a³, where stronger gravity (M) reduces period for same a. Example: Planets closer to the Sun have shorter periods.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

5. Calculate the period of a planet with semi-major axis 4 AU (Sun’s T²/a³ = 1 yr²/AU³).

Solution:
T² = ka³, k = 1 yr²/AU³, a = 4 AU.
T² = 1 × 4³ = 64, T = √64 = 8 years.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why do planets move faster near the Sun?

Solution:
Kepler’s second law: Equal areas in equal times mean higher velocity near the Sun. Example: Comet’s speed increases at perihelion.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Derive Kepler’s third law for circular orbits.

Solution:
For circular orbit: F = GMm/r² = mv²/r, v = 2πr/T. Substitute: GM/r = (2πr/T)². Simplify: T² = (4π²/GM)r³.
[1 mark for setup, 1 mark for derivation, 1 mark for formula]

8. Explain Kepler’s first law for non-circular orbits.

Solution:
Planets follow elliptical paths, not circles, with Sun at one focus. Example: Halley’s Comet has a highly elliptical orbit.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

9. Calculate semi-major axis for a planet with period 27 years (k = 1 yr²/AU³).

Solution:
T² = ka³, so a³ = T²/k. Given T = 27 years, k = 1 yr²/AU³.
a³ = 27²/1 = 729, a = ∛729 = 9 AU.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain the significance of Kepler’s laws in planetary motion.

Solution:
Kepler’s laws describe planetary orbits, enabling predictions. Example: They help calculate satellite orbits around Earth.
[1 mark for significance, 1 mark for explanation, 1 mark for example]

1. State Newton’s universal law of gravitation.

Solution:
Every mass attracts every other mass with a force F = GMm/r², along the line joining them. Example: Earth attracts the Moon.
[1 mark for statement, 1 mark for formula, 1 mark for example]

2. Calculate gravitational force between two 5 kg masses 2 m apart (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
F = GMm/r². Given M = 5 kg, m = 5 kg, r = 2 m, G = 6.67 × 10⁻¹¹ N·m²/kg².
F = (6.67 × 10⁻¹¹ × 5 × 5)/2² = 4.17 × 10⁻¹⁰ N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain why gravitational force is always attractive.

Solution:
Gravitational force depends on mass (always positive), so F = GMm/r² is attractive. Example: Planets orbit the Sun due to attraction.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Find distance for F = 1 × 10⁻⁸ N between 10 kg masses (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
F = GMm/r², so r² = GMm/F. Given M = 10 kg, m = 10 kg, F = 1 × 10⁻⁸ N.
r² = (6.67 × 10⁻¹¹ × 10 × 10)/(1 × 10⁻⁸) = 6.67 × 10⁻¹, r ≈ 0.258 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain the inverse square law in gravitation.

Solution:
Force decreases with distance squared (F ∝ 1/r²). Example: Doubling distance reduces force to one-fourth.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. Calculate gravitational force between a 2 kg and 3 kg mass 1 m apart (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
F = GMm/r². Given M = 2 kg, m = 3 kg, r = 1 m, G = 6.67 × 10⁻¹¹ N·m²/kg².
F = (6.67 × 10⁻¹¹ × 2 × 3)/1² = 4 × 10⁻¹⁰ N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why is G called the universal gravitational constant?

Solution:
G is constant for all masses and distances in the universe. Example: It applies to both Earth-Moon and Sun-planet systems.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Find mass for F = 2 × 10⁻⁹ N at 2 m from a 4 kg mass (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
F = GMm/r², so m = Fr²/(GM). Given F = 2 × 10⁻⁹ N, r = 2 m, M = 4 kg.
m = (2 × 10⁻⁹ × 2²)/(6.67 × 10⁻¹¹ × 4) ≈ 30 kg.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain the role of gravitational force in planetary motion.

Solution:
Gravitational force provides centripetal force for orbital motion. Example: Sun’s gravity keeps Earth in its orbit.
[1 mark for role, 1 mark for explanation, 1 mark for example]

10. Calculate G if F = 5 × 10⁻¹⁰ N between 5 kg masses 1 m apart.

Solution:
F = GMm/r², so G = Fr²/(Mm). Given F = 5 × 10⁻¹⁰ N, M = 5 kg, m = 5 kg, r = 1 m.
G = (5 × 10⁻¹⁰ × 1²)/(5 × 5) = 2 × 10⁻¹¹ N·m²/kg².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define acceleration due to gravity and its formula.

Solution:
Acceleration due to gravity: g = GM/R², where M is Earth’s mass, R is its radius. Example: On Earth, g ≈ 9.8 m/s².
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate g at height h = R_Earth (g = 9.8 m/s² at surface).

Solution:
g_h = g/(1 + h/R)². Given h = R, g = 9.8 m/s².
g_h = 9.8/(1 + 1)² = 9.8/4 = 2.45 m/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain variation of g with altitude.

Solution:
g decreases with altitude: g_h = g/(1 + h/R)². Example: At h = R, g reduces to g/4.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

4. Calculate g at depth d = R_Earth/2 (g = 9.8 m/s² at surface).

Solution:
g_d = g(1 - d/R). Given d = R/2, g = 9.8 m/s².
g_d = 9.8 × (1 - 1/2) = 9.8 × 0.5 = 4.9 m/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain variation of g with depth.

Solution:
g decreases with depth: g_d = g(1 - d/R). Example: At Earth’s centre (d = R), g = 0.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

6. Why is g maximum at Earth’s surface?

Solution:
g = GM/r² is maximum at r = R_Earth; it decreases above and below. Example: g is less at mountain tops or underground.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate height where g is 1/9 of surface g (g = 9.8 m/s²).

Solution:
g_h = g/(1 + h/R)². Given g_h = g/9.
1/9 = 1/(1 + h/R)², (1 + h/R)² = 9, 1 + h/R = 3, h/R = 2, h = 2R.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Find depth where g is half of surface g (g = 9.8 m/s²).

Solution:
g_d = g(1 - d/R). Given g_d = g/2.
g/2 = g(1 - d/R), 1/2 = 1 - d/R, d/R = 1/2, d = R/2.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain why g varies with latitude.

Solution:
Earth’s rotation reduces effective g at equator due to centrifugal force. Example: g is slightly less at equator than poles.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate g at height 2R_Earth (g = 9.8 m/s²).

Solution:
g_h = g/(1 + h/R)². Given h = 2R, g = 9.8 m/s².
g_h = 9.8/(1 + 2)² = 9.8/9 ≈ 1.09 m/s².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define gravitational potential energy.

Solution:
Gravitational potential energy: U = -GMm/r, energy due to position in a gravitational field. Example: A satellite has U relative to Earth.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate gravitational potential energy (M = 6 × 10²⁴ kg, m = 1000 kg, r = 7 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
U = -GMm/r. Given M = 6 × 10²⁴ kg, m = 1000 kg, r = 7 × 10⁶ m.
U = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1000)/(7 × 10⁶) ≈ -5.72 × 10⁹ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Define gravitational potential.

Solution:
Gravitational potential: V = -GM/r, potential energy per unit mass. Example: Earth’s potential affects satellite orbits.
[1 mark for definition, 1 mark for formula, 1 mark for example]

4. Calculate gravitational potential at r = 6.4 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
V = -GM/r. Given M = 6 × 10²⁴ kg, r = 6.4 × 10⁶ m.
V = -(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(6.4 × 10⁶) ≈ -6.26 × 10⁷ J/kg.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain why gravitational potential energy is negative.

Solution:
U = -GMm/r; negative as work is done to bring masses from infinity. Example: A satellite’s U is negative relative to infinity.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

6. Find U for m = 500 kg at r = 8 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
U = -GMm/r. Given m = 500 kg, r = 8 × 10⁶ m, M = 6 × 10²⁴ kg.
U = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 500)/(8 × 10⁶) ≈ -2.5 × 10⁹ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain the relation between gravitational potential and potential energy.

Solution:
U = mV, where V = -GM/r is potential, U is potential energy. Example: A mass in Earth’s field has U proportional to V.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

8. Find V at r = 7 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
V = -GM/r. Given M = 6 × 10²⁴ kg, r = 7 × 10⁶ m.
V = -(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(7 × 10⁶) ≈ -5.72 × 10⁷ J/kg.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Why is gravitational potential negative?

Solution:
V = -GM/r; negative as work is done to move a mass to infinity. Example: Earth’s potential is negative at its surface.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate U for m = 200 kg at r = 6.4 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
U = -GMm/r. Given m = 200 kg, r = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg.
U = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 200)/(6.4 × 10⁶) ≈ -1.25 × 10¹⁰ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define escape speed.

Solution:
Escape speed: Minimum speed to escape a planet’s gravity, v_e = √(2GM/R). Example: A rocket escaping Earth.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate escape speed for Earth (M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_e = √(2GM/R). Given M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m.
v_e = √[(2 × 6.67 × 10⁻¹¹ × 6 × 10²⁴)/(6.4 × 10⁶)] ≈ 11.2 × 10³ m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Derive escape speed formula.

Solution:
Energy conservation: ½mv_e² = GMm/R (kinetic energy equals potential energy at infinity). Simplify: v_e = √(2GM/R).
[1 mark for setup, 1 mark for derivation, 1 mark for formula]

4. Why does escape speed depend on planet’s mass?

Solution:
v_e = √(2GM/R); larger M increases gravitational pull, requiring higher speed. Example: Jupiter’s escape speed is higher than Earth’s.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Calculate escape speed for a planet (M = 3 × 10²⁴ kg, R = 4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_e = √(2GM/R). Given M = 3 × 10²⁴ kg, R = 4 × 10⁶ m.
v_e = √[(2 × 6.67 × 10⁻¹¹ × 3 × 10²⁴)/(4 × 10⁶)] ≈ 10 × 10³ m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain why escape speed is independent of object’s mass.

Solution:
v_e = √(2GM/R); m cancels out in derivation. Example: A small or large rocket needs same escape speed.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Find radius for v_e = 15 km/s, M = 6 × 10²⁴ kg (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_e = √(2GM/R), so R = 2GM/v_e². Given v_e = 15 × 10³ m/s, M = 6 × 10²⁴ kg.
R = (2 × 6.67 × 10⁻¹¹ × 6 × 10²⁴)/(15 × 10³)² ≈ 3.56 × 10⁶ m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is escape speed higher for denser planets?

Solution:
v_e = √(2GM/R); higher density increases M/R, increasing v_e. Example: Mercury has higher v_e than Mars.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate mass for v_e = 10 km/s, R = 5 × 10⁶ m (G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_e = √(2GM/R), so M = v_e²R/(2G). Given v_e = 10 × 10³ m/s, R = 5 × 10⁶ m.
M = [(10 × 10³)² × 5 × 10⁶]/(2 × 6.67 × 10⁻¹¹) ≈ 3.75 × 10²⁴ kg.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain escape speed in terms of energy.

Solution:
Escape speed gives kinetic energy to overcome potential energy (½mv_e² = GMm/R). Example: A rocket needs v_e to reach infinity.
[1 mark for explanation, 1 mark for relation, 1 mark for example]

1. Define orbital velocity of a satellite.

Solution:
Orbital velocity: Speed for circular orbit, v_o = √(GM/r). Example: A satellite orbiting Earth at low altitude.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate orbital velocity at r = 7 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_o = √(GM/r). Given M = 6 × 10²⁴ kg, r = 7 × 10⁶ m.
v_o = √[(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(7 × 10⁶)] ≈ 7.56 × 10³ m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Derive orbital velocity formula.

Solution:
Centripetal force = gravitational force: mv_o²/r = GMm/r². Simplify: v_o = √(GM/r). Example: Applies to satellites in circular orbits.
[1 mark for setup, 1 mark for derivation, 1 mark for formula]

4. Why does orbital velocity decrease with height?

Solution:
v_o = √(GM/r); larger r reduces v_o. Example: Geostationary satellites have lower v_o than low-orbit satellites.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Calculate v_o at Earth’s surface (R = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_o = √(GM/r). Given r = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg.
v_o = √[(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(6.4 × 10⁶)] ≈ 7.91 × 10³ m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain orbital velocity in terms of centripetal force.

Solution:
Gravitational force provides centripetal force for orbit (mv_o²/r = GMm/r²). Example: Earth’s gravity keeps a satellite in orbit.
[1 mark for explanation, 1 mark for relation, 1 mark for example]

7. Find radius for v_o = 7 × 10³ m/s (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
v_o = √(GM/r), so r = GM/v_o². Given v_o = 7 × 10³ m/s, M = 6 × 10²⁴ kg.
r = (6.67 × 10⁻¹¹ × 6 × 10²⁴)/(7 × 10³)² ≈ 8.16 × 10⁶ m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is orbital velocity less than escape speed?

Solution:
v_o = √(GM/r), v_e = √(2GM/r); v_e = √2 v_o. Example: A satellite needs less speed to orbit than to escape.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate v_o at h = 1000 km above Earth (R = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
r = R + h = 6.4 × 10⁶ + 10⁶ = 7.4 × 10⁶ m. v_o = √(GM/r).
v_o = √[(6.67 × 10⁻¹¹ × 6 × 10²⁴)/(7.4 × 10⁶)] ≈ 7.36 × 10³ m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain orbital velocity for geostationary satellites.

Solution:
Geostationary satellites have v_o for 24-hour orbit at fixed position. Example: Communication satellites orbit at r ≈ 42,000 km.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

1. Define total energy of an orbiting satellite.

Solution:
Total energy: E = K + U = ½mv_o² - GMm/r = -GMm/(2r). Example: A satellite’s energy is negative in orbit.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate total energy of a 1000 kg satellite at r = 7 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
E = -GMm/(2r). Given m = 1000 kg, r = 7 × 10⁶ m, M = 6 × 10²⁴ kg.
E = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1000)/(2 × 7 × 10⁶) ≈ -2.86 × 10⁹ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Derive total energy of an orbiting satellite.

Solution:
K = ½mv_o², v_o = √(GM/r), so K = ½GMm/r. U = -GMm/r. E = K + U = ½GMm/r - GMm/r = -GMm/(2r).
[1 mark for setup, 1 mark for derivation, 1 mark for formula]

4. Why is total energy of an orbiting satellite negative?

Solution:
E = -GMm/(2r); negative as U dominates K in bound orbit. Example: A satellite needs energy to escape orbit.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Calculate kinetic energy of a 500 kg satellite at r = 6.4 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
K = GMm/(2r). Given m = 500 kg, r = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg.
K = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 500)/(2 × 6.4 × 10⁶) ≈ 1.56 × 10¹⁰ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain relation between K, U, and E for a satellite.

Solution:
K = -½U, E = ½U = -GMm/(2r). Example: A satellite’s K is half its |U| in magnitude.
[1 mark for relation, 1 mark for explanation, 1 mark for example]

7. Calculate potential energy of a 200 kg satellite at r = 7 × 10⁶ m (M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
U = -GMm/r. Given m = 200 kg, r = 7 × 10⁶ m, M = 6 × 10²⁴ kg.
U = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 200)/(7 × 10⁶) ≈ -5.72 × 10⁹ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why does satellite energy depend on orbit radius?

Solution:
E = -GMm/(2r); larger r reduces |E|. Example: Geostationary satellites have less negative energy than low-orbit satellites.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Find total energy of a 100 kg satellite at h = 1000 km (R = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N·m²/kg²).

Solution:
r = R + h = 6.4 × 10⁶ + 10⁶ = 7.4 × 10⁶ m. E = -GMm/(2r).
E = -(6.67 × 10⁻¹¹ × 6 × 10²⁴ × 100)/(2 × 7.4 × 10⁶) ≈ -2.71 × 10⁹ J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain why a satellite’s energy becomes zero at infinity.

Solution:
At r → ∞, U = 0, K = 0, so E = 0. Example: A satellite escaping Earth reaches zero energy at infinity.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

Chapter 8: Mechanical Properties of Solids

1. Define elasticity and give an example.

Solution:
Elasticity: Property of a material to regain its original shape after deformation. Example: A rubber band stretches and returns to shape.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. State Hooke’s law with its formula.

Solution:
Hooke’s law: Stress is proportional to strain within elastic limit, σ = Eε. Example: A stretched spring obeys Hooke’s law.
[1 mark for statement, 1 mark for formula, 1 mark for example]

3. Calculate Young’s modulus (stress = 2 × 10⁸ N/m², strain = 0.01).

Solution:
E = Stress/Strain. Given Stress = 2 × 10⁸ N/m², Strain = 0.01.
E = (2 × 10⁸)/0.01 = 2 × 10¹⁰ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain the stress-strain relationship qualitatively.

Solution:
Stress increases linearly with strain up to elastic limit, then becomes non-linear until breaking. Example: A metal wire stretches linearly initially.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

5. Calculate stress (F = 100 N, A = 2 × 10⁻⁴ m²).

Solution:
Stress = F/A. Given F = 100 N, A = 2 × 10⁻⁴ m².
Stress = 100/(2 × 10⁻⁴) = 5 × 10⁵ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Calculate strain (E = 2 × 10¹¹ N/m², stress = 4 × 10⁷ N/m²).

Solution:
E = Stress/Strain, so Strain = Stress/E. Given Stress = 4 × 10⁷ N/m², E = 2 × 10¹¹ N/m².
Strain = (4 × 10⁷)/(2 × 10¹¹) = 2 × 10⁻⁴.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain the elastic limit.

Solution:
Elastic limit: Maximum stress a material can withstand and still return to its original shape. Example: A spring stretched beyond its elastic limit deforms permanently.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

8. Calculate force for stress = 3 × 10⁶ N/m² (A = 5 × 10⁻⁵ m²).

Solution:
Stress = F/A, so F = Stress × A. Given Stress = 3 × 10⁶ N/m², A = 5 × 10⁻⁵ m².
F = 3 × 10⁶ × 5 × 10⁻⁵ = 150 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Why do some materials have higher elasticity?

Solution:
Stronger intermolecular forces allow greater elastic deformation. Example: Steel is more elastic than rubber.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate elongation (L = 2 m, E = 1 × 10¹¹ N/m², F = 200 N, A = 1 × 10⁻⁴ m²).

Solution:
Stress = F/A = 200/(1 × 10⁻⁴) = 2 × 10⁶ N/m². Strain = Stress/E = (2 × 10⁶)/(1 × 10¹¹) = 2 × 10⁻⁵. ΔL = Strain × L = 2 × 10⁻⁵ × 2 = 4 × 10⁻⁵ m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define Young’s modulus with its unit.

Solution:
Young’s modulus: Ratio of longitudinal stress to strain, E = σ/ε. Unit: N/m². Example: Steel has E ≈ 2 × 10¹¹ N/m².
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate Young’s modulus (F = 500 N, A = 2 × 10⁻⁴ m², ΔL = 0.001 m, L = 1 m).

Solution:
Stress = F/A = 500/(2 × 10⁻⁴) = 2.5 × 10⁶ N/m². Strain = ΔL/L = 0.001/1 = 0.001. E = Stress/Strain = (2.5 × 10⁶)/0.001 = 2.5 × 10⁹ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Why is Young’s modulus high for steel?

Solution:
Strong atomic bonds in steel resist deformation. Example: Steel beams support heavy loads in buildings.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Calculate stress for E = 2 × 10¹¹ N/m², ΔL = 0.002 m, L = 2 m.

Solution:
Strain = ΔL/L = 0.002/2 = 0.001. E = Stress/Strain, Stress = E × Strain = 2 × 10¹¹ × 0.001 = 2 × 10⁸ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain Young’s modulus in terms of Hooke’s law.

Solution:
Hooke’s law (σ = Eε) defines Young’s modulus as the constant of proportionality. Example: A wire’s extension follows E = σ/ε.
[1 mark for explanation, 1 mark for relation, 1 mark for example]

6. Calculate ΔL (E = 1 × 10¹¹ N/m², F = 300 N, A = 3 × 10⁻⁴ m², L = 1.5 m).

Solution:
Stress = F/A = 300/(3 × 10⁻⁴) = 1 × 10⁶ N/m². Strain = Stress/E = (1 × 10⁶)/(1 × 10¹¹) = 1 × 10⁻⁵. ΔL = Strain × L = 1 × 10⁻⁵ × 1.5 = 1.5 × 10⁻⁵ m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why is Young’s modulus material-specific?

Solution:
It depends on the material’s atomic structure. Example: Rubber has lower E than steel.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Calculate area (F = 400 N, E = 2 × 10¹¹ N/m², ΔL = 0.0005 m, L = 2 m).

Solution:
Strain = ΔL/L = 0.0005/2 = 0.00025. Stress = E × Strain = 2 × 10¹¹ × 0.00025 = 5 × 10⁷ N/m². A = F/Stress = 400/(5 × 10⁷) = 8 × 10⁻⁶ m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain an application of Young’s modulus.

Solution:
Young’s modulus helps design materials for structural strength. Example: High E steel is used in skyscrapers.
[1 mark for application, 1 mark for explanation, 1 mark for example]

10. Calculate E (stress = 1 × 10⁷ N/m², ΔL = 0.001 m, L = 1 m).

Solution:
Strain = ΔL/L = 0.001/1 = 0.001. E = Stress/Strain = (1 × 10⁷)/0.001 = 1 × 10¹⁰ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define bulk modulus with its unit.

Solution:
Bulk modulus: Ratio of volumetric stress to strain, K = -ΔP/(ΔV/V). Unit: N/m². Example: Liquids have high K.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate bulk modulus (ΔP = 2 × 10⁵ N/m², ΔV/V = 0.0002).

Solution:
K = -ΔP/(ΔV/V). Given ΔP = 2 × 10⁵ N/m², ΔV/V = 0.0002.
K = (2 × 10⁵)/0.0002 = 1 × 10⁹ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Why is bulk modulus higher for solids?

Solution:
Solids resist volume change due to strong intermolecular forces. Example: Steel has higher K than water.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Explain bulk modulus qualitatively.

Solution:
Bulk modulus measures resistance to uniform compression. Example: Compressing a gas requires less pressure than a solid.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

5. Calculate ΔV/V (K = 2 × 10⁹ N/m², ΔP = 4 × 10⁵ N/m²).

Solution:
K = -ΔP/(ΔV/V), so ΔV/V = -ΔP/K. Given ΔP = 4 × 10⁵ N/m², K = 2 × 10⁹ N/m².
ΔV/V = -(4 × 10⁵)/(2 × 10⁹) = -2 × 10⁻⁴.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why is bulk modulus negative in formula?

Solution:
Negative sign accounts for volume decrease under pressure. Example: Compressing water reduces its volume.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate ΔP (K = 1 × 10⁹ N/m², ΔV/V = -0.0005).

Solution:
K = -ΔP/(ΔV/V), so ΔP = -K × (ΔV/V). Given K = 1 × 10⁹ N/m², ΔV/V = -0.0005.
ΔP = -1 × 10⁹ × (-0.0005) = 5 × 10⁵ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain an application of bulk modulus.

Solution:
Bulk modulus helps design materials for high-pressure environments. Example: Submarine hulls use high K materials.
[1 mark for application, 1 mark for explanation, 1 mark for example]

9. Why do gases have low bulk modulus?

Solution:
Weak intermolecular forces allow easy compression. Example: Air compresses more than water.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate K (ΔP = 3 × 10⁵ N/m², ΔV = 0.0001 m³, V = 0.1 m³).

Solution:
ΔV/V = 0.0001/0.1 = 0.001. K = -ΔP/(ΔV/V) = (3 × 10⁵)/0.001 = 3 × 10⁸ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define shear modulus with its unit.

Solution:
Shear modulus: Ratio of shear stress to shear strain, G = τ/γ. Unit: N/m². Example: Twisting a rod involves shear modulus.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Explain shear modulus qualitatively.

Solution:
Shear modulus measures resistance to shape deformation without volume change. Example: Shearing a rubber block.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

3. Calculate shear modulus (shear stress = 1 × 10⁶ N/m², shear strain = 0.002).

Solution:
G = τ/γ. Given τ = 1 × 10⁶ N/m², γ = 0.002.
G = (1 × 10⁶)/0.002 = 5 × 10⁸ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Why is shear modulus lower for rubber?

Solution:
Weak intermolecular forces allow easy shape deformation. Example: Rubber deforms more than steel under shear.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Calculate shear stress (G = 4 × 10⁸ N/m², shear strain = 0.005).

Solution:
G = τ/γ, so τ = G × γ. Given G = 4 × 10⁸ N/m², γ = 0.005.
τ = 4 × 10⁸ × 0.005 = 2 × 10⁶ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Explain an application of shear modulus.

Solution:
Shear modulus is used in designing torsion-resistant materials. Example: Shafts in engines use high G materials.
[1 mark for application, 1 mark for explanation, 1 mark for example]

7. Calculate shear strain (G = 5 × 10⁸ N/m², τ = 2 × 10⁶ N/m²).

Solution:
G = τ/γ, so γ = τ/G. Given τ = 2 × 10⁶ N/m², G = 5 × 10⁸ N/m².
γ = (2 × 10⁶)/(5 × 10⁸) = 4 × 10⁻³.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Why is shear modulus important for solids?

Solution:
It quantifies resistance to shearing forces. Example: High G materials are used in bridges.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Explain shear deformation.

Solution:
Shear deformation: Shape change without volume change under tangential force. Example: Cutting paper with scissors involves shear.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

10. Calculate G (τ = 3 × 10⁶ N/m², γ = 0.006).

Solution:
G = τ/γ. Given τ = 3 × 10⁶ N/m², γ = 0.006.
G = (3 × 10⁶)/0.006 = 5 × 10⁸ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define Poisson’s ratio with its formula.

Solution:
Poisson’s ratio: Ratio of lateral strain to longitudinal strain, ν = -ε_lat/ε_long. Example: A stretched wire thins laterally.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate Poisson’s ratio (ε_lat = 0.0002, ε_long = 0.001).

Solution:
ν = -ε_lat/ε_long. Given ε_lat = 0.0002, ε_long = 0.001.
ν = -0.0002/0.001 = -0.2.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Why is Poisson’s ratio negative?

Solution:
Lateral strain is opposite to longitudinal strain (compression vs. expansion). Example: Stretching a rubber band reduces its width.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Calculate lateral strain (ν = 0.3, ε_long = 0.002).

Solution:
ν = -ε_lat/ε_long, so ε_lat = -ν × ε_long. Given ν = 0.3, ε_long = 0.002.
ε_lat = -0.3 × 0.002 = -0.0006.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain Poisson’s ratio qualitatively.

Solution:
Poisson’s ratio measures how materials deform transversely when stretched. Example: Cork has low ν, deforming less laterally.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. Why do some materials have high Poisson’s ratio?

Solution:
Flexible molecular structures allow greater lateral deformation. Example: Rubber has higher ν than steel.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate longitudinal strain (ν = 0.25, ε_lat = -0.0004).

Solution:
ν = -ε_lat/ε_long, so ε_long = -ε_lat/ν. Given ν = 0.25, ε_lat = -0.0004.
ε_long = -(-0.0004)/0.25 = 0.0016.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain an application of Poisson’s ratio.

Solution:
Poisson’s ratio aids in material selection for structural design. Example: Low ν materials like cork are used in bottle stoppers.
[1 mark for application, 1 mark for explanation, 1 mark for example]

9. Why is Poisson’s ratio less than 0.5?

Solution:
Volume conservation limits lateral strain relative to longitudinal strain. Example: Rubber has ν ≈ 0.5 due to near-incompressibility.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate ν (Δd/d = -0.0001, ΔL/L = 0.0005).

Solution:
ν = -ε_lat/ε_long = -(Δd/d)/(ΔL/L). Given Δd/d = -0.0001, ΔL/L = 0.0005.
ν = -(-0.0001)/0.0005 = 0.2.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define elastic energy with its formula.

Solution:
Elastic energy: Energy stored in a deformed elastic body, U = ½ × Stress × Strain × Volume. Example: A stretched spring stores elastic energy.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate elastic energy (Stress = 2 × 10⁶ N/m², Strain = 0.001, V = 0.01 m³).

Solution:
U = ½ × Stress × Strain × V. Given Stress = 2 × 10⁶ N/m², Strain = 0.001, V = 0.01 m³.
U = ½ × 2 × 10⁶ × 0.001 × 0.01 = 10 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Explain elastic energy in a stretched wire.

Solution:
Elastic energy is stored due to work done against elastic forces. Example: A stretched wire stores energy like a spring.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

4. Calculate U (E = 2 × 10¹¹ N/m², Strain = 0.002, V = 0.005 m³).

Solution:
Stress = E × Strain = 2 × 10¹¹ × 0.002 = 4 × 10⁸ N/m². U = ½ × Stress × Strain × V = ½ × 4 × 10⁸ × 0.002 × 0.005 = 2000 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Why is elastic energy proportional to strain?

Solution:
U = ½ × E × Strain² × V; energy depends on strain squared. Example: Doubling strain quadruples energy in a wire.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

6. Calculate volume (U = 50 J, Stress = 1 × 10⁶ N/m², Strain = 0.005).

Solution:
U = ½ × Stress × Strain × V, so V = 2U/(Stress × Strain). Given U = 50 J, Stress = 1 × 10⁶ N/m², Strain = 0.005.
V = (2 × 50)/(1 × 10⁶ × 0.005) = 0.02 m³.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain an application of elastic energy.

Solution:
Elastic energy is used in springs for shock absorbers. Example: Car suspensions store energy during bumps.
[1 mark for application, 1 mark for explanation, 1 mark for example]

8. Calculate U (F = 200 N, A = 2 × 10⁻⁴ m², E = 1 × 10¹¹ N/m², V = 0.01 m³).

Solution:
Stress = F/A = 200/(2 × 10⁻⁴) = 1 × 10⁶ N/m². Strain = Stress/E = (1 × 10⁶)/(1 × 10¹¹) = 1 × 10⁻⁵. U = ½ × Stress × Strain × V = ½ × 1 × 10⁶ × 1 × 10⁻⁵ × 0.01 = 0.05 J.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Why is elastic energy stored in deformation?

Solution:
Work done to deform a material is stored as potential energy. Example: A compressed spring stores energy.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate Strain (U = 100 J, Stress = 2 × 10⁶ N/m², V = 0.02 m³).

Solution:
U = ½ × Stress × Strain × V, so Strain = 2U/(Stress × V). Given U = 100 J, Stress = 2 × 10⁶ N/m², V = 0.02 m³.
Strain = (2 × 100)/(2 × 10⁶ × 0.02) = 0.005.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Explain elasticity in bridge construction.

Solution:
Elastic materials absorb loads without permanent deformation. Example: Steel beams in bridges bend elastically.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

2. How is elasticity used in springs?

Solution:
Springs store elastic energy and return to shape. Example: Car suspensions use springs for smooth rides.
[1 mark for application, 1 mark for explanation, 1 mark for example]

3. Explain elasticity in rubber bands.

Solution:
Rubber bands stretch and return to shape due to elasticity. Example: Used in slingshots for energy storage.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

4. Why is high Young’s modulus preferred in buildings?

Solution:
High E materials resist deformation under load. Example: Concrete pillars support skyscrapers.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

5. Explain elasticity in shock absorbers.

Solution:
Elastic materials absorb and dissipate impact energy. Example: Car shock absorbers use elastic springs.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. How is elasticity used in sports equipment?

Solution:
Elastic materials store energy for performance. Example: Tennis racket strings stretch and recoil.
[1 mark for application, 1 mark for explanation, 1 mark for example]

7. Why is low Poisson’s ratio useful in corks?

Solution:
Low ν reduces lateral expansion under compression. Example: Cork stoppers seal bottles tightly.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Explain elasticity in medical implants.

Solution:
Elastic materials mimic body tissue flexibility. Example: Artificial ligaments use elastic polymers.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

9. How is shear modulus used in engineering?

Solution:
Shear modulus ensures resistance to twisting. Example: Torsion bars in vehicles use high G materials.
[1 mark for application, 1 mark for explanation, 1 mark for example]

10. Explain elasticity in earthquake-resistant structures.

Solution:
Elastic materials absorb seismic energy without breaking. Example: Rubber bearings in buildings reduce vibrations.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

Chapter 9: Mechanical Properties of Fluids

1. Define pressure due to a fluid column.

Solution:
Pressure: P = ρgh, where ρ is density, g is gravity, h is depth. Example: Pressure increases in deep water.
[1 mark for definition, 1 mark for formula, 1 mark for example]

2. Calculate pressure at 10 m depth in water (ρ = 1000 kg/m³, g = 9.8 m/s²).

Solution:
P = ρgh. Given ρ = 1000 kg/m³, g = 9.8 m/s², h = 10 m.
P = 1000 × 9.8 × 10 = 9.8 × 10⁴ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. State Pascal’s law with an example.

Solution:
Pascal’s law: Pressure applied to a confined fluid is transmitted equally. Example: Hydraulic lifts use Pascal’s law.
[1 mark for statement, 1 mark for explanation, 1 mark for example]

4. Calculate depth for P = 2 × 10⁵ N/m² (ρ = 1000 kg/m³, g = 9.8 m/s²).

Solution:
P = ρgh, so h = P/(ρg). Given P = 2 × 10⁵ N/m², ρ = 1000 kg/m³, g = 9.8 m/s².
h = (2 × 10⁵)/(1000 × 9.8) ≈ 20.4 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain the effect of gravity on fluid pressure.

Solution:
Gravity increases pressure with depth (P = ρgh). Example: Pressure is higher at the bottom of a dam.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

6. Calculate pressure at 5 m depth in mercury (ρ = 13600 kg/m³, g = 9.8 m/s²).

Solution:
P = ρgh. Given ρ = 13600 kg/m³, g = 9.8 m/s², h = 5 m.
P = 13600 × 9.8 × 5 ≈ 6.66 × 10⁵ N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Explain hydraulic brakes using Pascal’s law.

Solution:
Pressure on brake fluid is transmitted to brake pads. Example: Car brakes stop wheels effectively.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

8. Why does pressure increase with depth?

Solution:
Weight of fluid column increases pressure (P = ρgh). Example: Divers feel more pressure deeper underwater.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

9. Calculate density (P = 4.9 × 10⁴ N/m², h = 5 m, g = 9.8 m/s²).

Solution:
P = ρgh, so ρ = P/(gh). Given P = 4.9 × 10⁴ N/m², h = 5 m, g = 9.8 m/s².
ρ = (4.9 × 10⁴)/(9.8 × 5) = 1000 kg/m³.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain hydraulic lift using Pascal’s law.

Solution:
Small force on a small piston lifts heavy loads via pressure transmission. Example: Car jacks lift vehicles.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

1. Define viscosity with its unit.

Solution:
Viscosity: Resistance of a fluid to flow, η = (F/A)/(dv/dx). Unit: Pa·s. Example: Honey is more viscous than water.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate drag force (r = 0.02 m, v = 0.5 m/s, η = 0.8 Pa·s).

Solution:
F = 6πηrv. Given r = 0.02 m, v = 0.5 m/s, η = 0.8 Pa·s.
F = 6 × 3.14 × 0.8 × 0.02 × 0.5 ≈ 0.15 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Why is viscosity higher in thicker fluids?

Solution:
Stronger intermolecular forces increase resistance to flow. Example: Glycerine is more viscous than water.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

4. Calculate η (F = 0.2 N, r = 0.01 m, v = 0.4 m/s).

Solution:
F = 6πηrv, so η = F/(6πrv). Given F = 0.2 N, r = 0.01 m, v = 0.4 m/s.
η = 0.2/(6 × 3.14 × 0.01 × 0.4) ≈ 2.65 Pa·s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain viscosity in terms of fluid layers.

Solution:
Viscosity arises from friction between fluid layers moving at different speeds. Example: Oil flows slower than water due to higher viscosity.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. Calculate velocity (F = 0.1 N, r = 0.005 m, η = 1 Pa·s).

Solution:
F = 6πηrv, so v = F/(6πηr). Given F = 0.1 N, r = 0.005 m, η = 1 Pa·s.
v = 0.1/(6 × 3.14 × 1 × 0.005) ≈ 1.06 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why does viscosity decrease with temperature?

Solution:
Higher temperature reduces intermolecular forces, easing flow. Example: Hot oil flows faster than cold oil.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Explain an application of viscosity.

Solution:
Viscosity is used in lubricants to reduce friction. Example: Engine oil ensures smooth machinery operation.
[1 mark for application, 1 mark for explanation, 1 mark for example]

9. Calculate radius (F = 0.15 N, v = 0.3 m/s, η = 0.9 Pa·s).

Solution:
F = 6πηrv, so r = F/(6πηv). Given F = 0.15 N, v = 0.3 m/s, η = 0.9 Pa·s.
r = 0.15/(6 × 3.14 × 0.9 × 0.3) ≈ 0.029 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain viscosity in industrial processes.

Solution:
Viscosity controls fluid flow in pipelines and manufacturing. Example: Paint viscosity affects its application.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

1. State Stokes’ law with its formula.

Solution:
Stokes’ law: Drag force on a sphere in a viscous fluid, F = 6πηrv. Example: A ball falling in glycerine.
[1 mark for statement, 1 mark for formula, 1 mark for example]

2. Calculate terminal velocity (r = 0.01 m, ρ_s = 8000 kg/m³, ρ_f = 1000 kg/m³, η = 1 Pa·s, g = 9.8 m/s²).

Solution:
v_t = (2r²(ρ_s - ρ_f)g)/(9η). Given r = 0.01 m, ρ_s = 8000 kg/m³, ρ_f = 1000 kg/m³, η = 1 Pa·s.
v_t = (2 × (0.01)² × (8000 - 1000) × 9.8)/(9 × 1) ≈ 0.153 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Define terminal velocity.

Solution:
Terminal velocity: Constant velocity when drag equals weight. Example: A raindrop falls at terminal velocity.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

4. Calculate v_t (r = 0.005 m, ρ_s = 7800 kg/m³, ρ_f = 1000 kg/m³, η = 1.2 Pa·s, g = 9.8 m/s²).

Solution:
v_t = (2r²(ρ_s - ρ_f)g)/(9η). Given r = 0.005 m, ρ_s = 7800 kg/m³, ρ_f = 1000 kg/m³, η = 1.2 Pa·s.
v_t = (2 × (0.005)² × (7800 - 1000) × 9.8)/(9 × 1.2) ≈ 0.031 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain Stokes’ law application in terminal velocity.

Solution:
Stokes’ law balances drag with weight for constant velocity. Example: Parachutes fall at terminal velocity.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. Calculate η (v_t = 0.1 m/s, r = 0.01 m, ρ_s = 9000 kg/m³, ρ_f = 1000 kg/m³, g = 9.8 m/s²).

Solution:
v_t = (2r²(ρ_s - ρ_f)g)/(9η), so η = (2r²(ρ_s - ρ_f)g)/(9v_t). Given v_t = 0.1 m/s, r = 0.01 m.
η = (2 × (0.01)² × (9000 - 1000) × 9.8)/(9 × 0.1) ≈ 1.74 Pa·s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

7. Why does terminal velocity depend on radius?

Solution:
v_t ∝ r² in Stokes’ law; larger radius increases drag. Example: Larger spheres fall faster in viscous fluids.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

8. Calculate radius (v_t = 0.2 m/s, ρ_s = 8000 kg/m³, ρ_f = 1000 kg/m³, η = 1 Pa·s, g = 9.8 m/s²).

Solution:
v_t = (2r²(ρ_s - ρ_f)g)/(9η), so r² = (9ηv_t)/(2(ρ_s - ρ_f)g). Given v_t = 0.2 m/s.
r² = (9 × 1 × 0.2)/(2 × (8000 - 1000) × 9.8) ≈ 1.31 × 10⁻⁴, r ≈ 0.0114 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

9. Explain terminal velocity in sedimentation.

Solution:
Particles settle at constant speed when drag equals weight. Example: Sediments in water reach terminal velocity.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

10. Calculate F (r = 0.008 m, v_t = 0.15 m/s, η = 1.5 Pa·s).

Solution:
F = 6πηrv_t. Given r = 0.008 m, v_t = 0.15 m/s, η = 1.5 Pa·s.
F = 6 × 3.14 × 1.5 × 0.008 × 0.15 ≈ 0.034 N.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. Define streamline flow with an example.

Solution:
Streamline flow: Smooth, layered fluid motion. Example: Water flowing steadily in a pipe.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

2. Define turbulent flow with an example.

Solution:
Turbulent flow: Chaotic fluid motion with eddies. Example: Rapids in a river.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

3. Define critical velocity with its formula.

Solution:
Critical velocity: Speed at which flow becomes turbulent, v_c = (Re η)/(ρr). Example: Water in a hose.
[1 mark for definition, 1 mark for formula, 1 mark for example]

4. Calculate critical velocity (Re = 2000, η = 0.001 Pa·s, ρ = 1000 kg/m³, r = 0.01 m).

Solution:
v_c = (Re η)/(ρr). Given Re = 2000, η = 0.001 Pa·s, ρ = 1000 kg/m³, r = 0.01 m.
v_c = (2000 × 0.001)/(1000 × 0.01) = 0.2 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

5. Explain streamline flow in pipes.

Solution:
Fluid moves in parallel layers with constant velocity. Example: Oil in pipelines flows smoothly.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

6. Why does turbulent flow occur at high speeds?

Solution:
High speeds disrupt laminar flow, causing eddies. Example: Fast river currents become turbulent.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate η (v_c = 0.3 m/s, Re = 2000, ρ = 1000 kg/m³, r = 0.005 m).

Solution:
v_c = (Re η)/(ρr), so η = (v_c ρr)/Re. Given v_c = 0.3 m/s, r = 0.005 m.
η = (0.3 × 1000 × 0.005)/2000 = 0.00075 Pa·s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain the role of Reynolds number.

Solution:
Reynolds number predicts flow type; low Re indicates streamline flow. Example: Low Re in narrow pipes ensures laminar flow.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

9. Why is streamline flow preferred in pipelines?

Solution:
Streamline flow reduces energy loss and turbulence. Example: Oil pipelines use streamline flow for efficiency.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

10. Calculate r (v_c = 0.25 m/s, Re = 2000, η = 0.001 Pa·s, ρ = 1000 kg/m³).

Solution:
v_c = (Re η)/(ρr), so r = (Re η)/(v_c ρ). Given v_c = 0.25 m/s.
r = (2000 × 0.001)/(0.25 × 1000) = 0.008 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

1. State Bernoulli’s theorem with its formula.

Solution:
Bernoulli’s theorem: P + ρgh + ½ρv² = constant. Example: Faster flow reduces pressure in pipes.
[1 mark for statement, 1 mark for formula, 1 mark for example]

2. Explain Torricelli’s law.

Solution:
Torricelli’s law: Speed of fluid from a hole, v = √(2gh). Example: Water exits a tank faster at greater depth.
[1 mark for explanation, 1 mark for formula, 1 mark for example]

3. Calculate speed of efflux (h = 2 m, g = 9.8 m/s²).

Solution:
v = √(2gh). Given h = 2 m, g = 9.8 m/s².
v = √(2 × 9.8 × 2) ≈ 6.26 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Explain dynamic lift in Bernoulli’s theorem.

Solution:
Faster airflow over a wing reduces pressure, creating lift. Example: Airplane wings use dynamic lift.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

5. Calculate pressure difference (ρ = 1.2 kg/m³, v₁ = 10 m/s, v₂ = 20 m/s).

Solution:
P₁ - P₂ = ½ρ(v₂² - v₁²). Given ρ = 1.2 kg/m³, v₁ = 10 m/s, v₂ = 20 m/s.
P₁ - P₂ = ½ × 1.2 × (20² - 10²) = 0.6 × 300 = 180 N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why does Bernoulli’s theorem apply to ideal fluids?

Solution:
Assumes no viscosity or energy loss. Example: Ideal fluid flow in pipes follows Bernoulli’s theorem.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate height (v = 5 m/s, g = 9.8 m/s²).

Solution:
Torricelli’s law: v = √(2gh), so h = v²/(2g). Given v = 5 m/s.
h = 5²/(2 × 9.8) ≈ 1.28 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain Bernoulli’s theorem in venturi meters.

Solution:
Narrow sections increase velocity, reducing pressure to measure flow. Example: Venturi meters in pipelines.
[1 mark for explanation, 1 mark for detail, 1 mark for example]

9. Calculate v (P₁ - P₂ = 100 N/m², ρ = 1.2 kg/m³, v₂ = 15 m/s).

Solution:
P₁ - P₂ = ½ρ(v₂² - v₁²), so v₁² = v₂² - (2(P₁ - P₂)/ρ). Given P₁ - P₂ = 100 N/m².
v₁² = 15² - (2 × 100)/1.2 = 225 - 166.67 ≈ 58.33, v₁ ≈ 7.64 m/s.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

10. Explain an application of Bernoulli’s theorem.

Solution:
Bernoulli’s theorem explains lift in aircraft wings. Example: Faster air over wings creates lift.
[1 mark for application, 1 mark for explanation, 1 mark for example]

1. Define surface tension with its unit.

Solution:
Surface tension: Force per unit length on a liquid’s surface, γ = F/L. Unit: N/m. Example: Water forms droplets.
[1 mark for definition, 1 mark for unit, 1 mark for example]

2. Calculate excess pressure in a soap bubble (r = 0.02 m, γ = 0.03 N/m).

Solution:
P = 4γ/r. Given r = 0.02 m, γ = 0.03 N/m.
P = (4 × 0.03)/0.02 = 6 N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

3. Calculate capillary rise (r = 0.001 m, γ = 0.072 N/m, ρ = 1000 kg/m³, g = 9.8 m/s², θ = 0°).

Solution:
h = (2γ cosθ)/(ρgr). Given r = 0.001 m, γ = 0.072 N/m, ρ = 1000 kg/m³, θ = 0°.
h = (2 × 0.072 × 1)/(1000 × 9.8 × 0.001) ≈ 0.0147 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

4. Define angle of contact with an example.

Solution:
Angle of contact: Angle between liquid and solid surface inside liquid. Example: Water on glass has θ ≈ 0°.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

5. Calculate γ (h = 0.02 m, r = 0.0005 m, ρ = 1000 kg/m³, g = 9.8 m/s², θ = 0°).

Solution:
h = (2γ cosθ)/(ρgr), so γ = (hρgr)/(2 cosθ). Given h = 0.02 m, r = 0.0005 m.
γ = (0.02 × 1000 × 9.8 × 0.0005)/(2 × 1) = 0.049 N/m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

6. Why do liquids form spherical drops?

Solution:
Surface tension minimizes surface area, favoring spheres. Example: Raindrops are nearly spherical.
[1 mark for reason, 1 mark for explanation, 1 mark for example]

7. Calculate excess pressure in a water drop (r = 0.01 m, γ = 0.072 N/m).

Solution:
P = 2γ/r. Given r = 0.01 m, γ = 0.072 N/m.
P = (2 × 0.072)/0.01 = 14.4 N/m².
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

8. Explain surface energy.

Solution:
Surface energy: Energy per unit area due to surface tension, E = γA. Example: Soap film minimizes surface energy.
[1 mark for definition, 1 mark for explanation, 1 mark for example]

9. Explain an application of surface tension.

Solution:
Surface tension allows insects to walk on water. Example: Water striders move on ponds.
[1 mark for application, 1 mark for explanation, 1 mark for example]

10. Calculate radius (P = 8 N/m², γ = 0.03 N/m, soap bubble).

Solution:
P = 4γ/r, so r = 4γ/P. Given P = 8 N/m², γ = 0.03 N/m.
r = (4 × 0.03)/8 = 0.015 m.
[1 mark for formula, 1 mark for substitution, 1 mark for answer]

Chapter 11: Thermodynamics

Question 1: What is the Zeroth Law of Thermodynamics?

Answer 1: The Zeroth Law of Thermodynamics states that if two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law defines temperature and allows the use of thermometers.

Question 2: State the First Law of Thermodynamics and explain its significance.

Answer 2: The First Law of Thermodynamics is a statement of the conservation of energy: ΔU = Q - W, where ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system. It signifies that energy cannot be created or destroyed, only transformed.

Question 3: What does the Second Law of Thermodynamics imply about the direction of natural processes?

Answer 3: The Second Law implies that natural processes are irreversible and tend towards increasing entropy. For example, heat flows from hot to cold bodies, and isolated systems evolve towards maximum disorder.

Question 4: Explain the concept of entropy in the context of the Second Law.

Answer 4: Entropy (S) is a measure of disorder or randomness in a system. The Second Law states that the total entropy of an isolated system always increases over time, or remains constant in reversible processes.

Question 5: What is the Third Law of Thermodynamics?

Answer 5: The Third Law states that the entropy of a perfect crystal at absolute zero temperature (0 K) is zero. It implies that it is impossible to reach absolute zero in a finite number of steps.

Question 6: How does the First Law apply to an isolated system?

Answer 6: In an isolated system, Q = 0 and W = 0, so ΔU = 0. The internal energy remains constant, and any processes must conserve energy internally.

Question 7: Describe Kelvin-Planck statement of the Second Law.

Answer 7: The Kelvin-Planck statement says it is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the extraction of heat from a single reservoir and the performance of an equal amount of work. This limits the efficiency of heat engines.

Question 8: What is the Clausius statement of the Second Law?

Answer 8: The Clausius statement asserts that it is impossible to transfer heat from a colder body to a hotter body without any work input. This explains why refrigerators require energy.

Question 9: Why is the Zeroth Law considered 'zeroth'?

Answer 9: It was formulated after the first and second laws but is fundamental to defining temperature scales, so it was retroactively numbered as the zeroth law to place it logically first.

Question 10: How does the Third Law relate to absolute zero?

Answer 10: The Third Law indicates that as temperature approaches 0 K, entropy approaches a minimum value (zero for perfect crystals), making it impossible to remove all thermal motion from a system in finite steps.

Question 1: What is an isothermal process? Give an example.

Answer 1: An isothermal process occurs at constant temperature (ΔT = 0). For an ideal gas, PV = constant. Example: Slow expansion of a gas in a heat bath.

Question 2: Define an adiabatic process and state its key equation.

Answer 2: An adiabatic process involves no heat exchange (Q = 0), so ΔU = -W. For an ideal gas, PV^γ = constant, where γ = Cp/Cv.

Question 3: What characterizes an isobaric process?

Answer 3: An isobaric process occurs at constant pressure (P = constant). Work done W = PΔV, and heat added Q = nCpΔT.

Question 4: Explain isochoric process and its implications.

Answer 4: An isochoric process is at constant volume (V = constant), so W = 0 and ΔU = Q. It's used in constant-volume calorimeters.

Question 5: What is a cyclic process? Why is ΔU = 0 for it?

Answer 5: A cyclic process returns a system to its initial state after a series of changes. ΔU = 0 because internal energy is a state function, so net change over a cycle is zero.

Question 6: Compare reversible and irreversible processes.

Answer 6: Reversible processes can be reversed without net change in the system or surroundings (e.g., quasi-static). Irreversible processes involve friction or finite gradients and increase entropy.

Question 7: In an isothermal expansion of an ideal gas, what is the work done?

Answer 7: For reversible isothermal expansion, W = nRT ln(V2/V1). Since ΔU = 0, Q = -W, so heat is absorbed equal to work done.

Question 8: Describe the PV diagram for an adiabatic process versus isothermal.

Answer 8: Both are hyperbolas on PV diagram, but adiabatic is steeper (PV^γ = const > PV = const), so for the same P1 to P2, adiabatic V2 < isothermal V2.

Question 9: What is the efficiency of a Carnot cycle?

Answer 9: The Carnot cycle efficiency η = 1 - (T_cold / T_hot), where temperatures are in Kelvin. It's the maximum possible for a heat engine between two reservoirs.

Question 10: How does heat capacity differ in isobaric and isochoric processes?

Answer 10: Cp (isobaric) > Cv (isochoric) by R (gas constant) for ideal gases, because in isobaric, heat goes to both internal energy and work (expansion).

Chapter 12: Kinetic Theory

Question 1: What are the main assumptions of the kinetic theory of gases?

Answer 1: The main assumptions are: (1) Gases consist of a large number of molecules in constant random motion. (2) Molecules are point masses with negligible volume. (3) No intermolecular forces except during elastic collisions. (4) Molecules obey Newton's laws of motion.

Question 2: Derive the expression for the pressure exerted by an ideal gas according to kinetic theory.

Answer 2: Consider molecules colliding with a wall: Change in momentum per collision is 2mv_x, number of collisions per unit time is (1/2) n <v_x> A, leading to force F = (1/3) n m <v²> A, so P = F/A = (1/3) ρ v_rms², where ρ = n m is density.

Question 3: What is the root mean square (RMS) speed of gas molecules?

Answer 3: The RMS speed v_rms = sqrt(<v²>) = sqrt(3RT/M), where R is the gas constant, T is absolute temperature, and M is the molar mass.

Question 4: Express the average kinetic energy of a gas molecule in terms of temperature.

Answer 4: The average kinetic energy per molecule is (3/2) kT, where k is Boltzmann's constant and T is in Kelvin. For N molecules, total KE = (3/2) nRT.

Question 5: How does kinetic theory explain the relation PV = nRT?

Answer 5: From P = (1/3) ρ v_rms² and v_rms² = 3RT/M, with ρ = m/V = (n M)/V, substituting gives PV = nRT.

Question 6: Define the mean free path and give its expression.

Answer 6: Mean free path λ is the average distance a molecule travels between collisions. λ = 1 / (√2 π d² n), where d is molecular diameter and n is number density.

Question 7: What are degrees of freedom in the context of kinetic theory?

Answer 7: Degrees of freedom are the independent ways a molecule can store energy (translational, rotational, vibrational). Monatomic gases have 3 translational degrees.

Question 8: State the law of equipartition of energy.

Answer 8: The law states that the average energy per molecule is (1/2) kT per degree of freedom. For translational motion, total KE = (3/2) kT.

Question 9: How does kinetic theory account for the diffusion of gases?

Answer 9: Diffusion occurs due to the random motion of molecules, leading to net movement from high to low concentration, explained by Graham's law: rate ∝ 1/√M.

Question 10: What is the significance of Avogadro's number in kinetic theory?

Answer 10: Avogadro's number N_A relates macroscopic R = N_A k, linking molecular kinetic energy to observable properties like pressure and temperature.

Question 1: Define molar specific heat at constant volume (C_v) for an ideal gas.

Answer 1: C_v is the heat required to raise the temperature of one mole of gas by 1 K at constant volume. For monatomic gas, C_v = (3/2) R.

Question 2: What is the relation between C_p and C_v for an ideal gas?

Answer 2: C_p - C_v = R, where R is the gas constant. C_p is the molar specific heat at constant pressure.

Question 3: Explain why C_p > C_v for gases.

Answer 3: At constant pressure, some heat goes into work (expansion), so more heat is needed to raise temperature by 1 K compared to constant volume.

Question 4: What is the value of γ (C_p / C_v) for a monatomic gas?

Answer 4: For monatomic gas, γ = 5/3, since C_v = (3/2) R and C_p = C_v + R = (5/2) R.

Question 5: Derive C_v for a monatomic ideal gas using kinetic theory.

Answer 5: Internal energy U = (3/2) nRT from translational KE. At constant volume, dU = n C_v dT, so C_v = (3/2) R.

Question 6: For diatomic gases at room temperature, what is C_v?

Answer 6: C_v = (5/2) R, accounting for 3 translational and 2 rotational degrees of freedom (vibrational not excited at room temp).

Question 7: What is the ratio γ for diatomic gases?

Answer 7: γ = 7/5 = 1.4, since C_p = C_v + R = (7/2) R.

Question 8: How does the equipartition theorem explain specific heats?

Answer 8: Each degree of freedom contributes (1/2) R to C_v per mole. Total C_v = (f/2) R, where f is degrees of freedom.

Question 9: Why do specific heats of diatomic gases increase at high temperatures?

Answer 9: At high T, vibrational degrees of freedom become active, adding more energy storage modes, so C_v increases to (7/2) R or more.

Question 10: Compare specific heats for polyatomic gases.

Answer 10: Polyatomic gases have more rotational/vibrational modes, so higher C_v (e.g., (6/2)R or more) and lower γ (closer to 1) than monatomic or diatomic.

Chapter 13: Oscillations

Question 1: What is Simple Harmonic Motion (SHM)?

Answer 1: Simple Harmonic Motion is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and directed towards it, resulting in sinusoidal oscillation.

Question 2: State the differential equation for SHM.

Answer 2: The equation is d²x/dt² + ω² x = 0, where ω is the angular frequency, and the general solution is x = A sin(ωt + φ), with A as amplitude and φ as phase constant.

Question 3: Define the terms amplitude, period, and frequency in SHM.

Answer 3: Amplitude (A) is the maximum displacement from equilibrium. Period (T) is the time for one complete oscillation, T = 2π/ω. Frequency (f) is oscillations per second, f = 1/T = ω/(2π).

Question 4: What is the restoring force in a spring-mass system undergoing SHM?

Answer 4: The restoring force F = -kx, where k is the spring constant and x is displacement. This follows Hooke's law and leads to ω = sqrt(k/m).

Question 5: Express the total energy in SHM.

Answer 5: Total mechanical energy E = (1/2) k A², constant throughout the motion. It is the sum of kinetic energy (1/2) m v² and potential energy (1/2) k x².

Question 6: How is velocity related to displacement in SHM?

Answer 6: Maximum velocity v_max = ω A at equilibrium. Velocity v = ± ω sqrt(A² - x²), decreasing as displacement increases.

Question 7: What is phase in SHM and its significance?

Answer 7: Phase (ωt + φ) determines the position in the cycle. It helps describe the state of oscillation at any time and is crucial for superposition of waves.

Question 8: Describe the acceleration in SHM.

Answer 8: Acceleration a = -ω² x, maximum at extremes (a_max = ω² A) and zero at equilibrium. It is always directed towards the mean position.

Question 9: What are the conditions for SHM in a physical system?

Answer 9: The force must be linear in displacement (F ∝ -x) and the motion must be frictionless or undamped. Examples include pendulum for small angles and U-tube oscillations.

Question 10: Explain the concept of time period for SHM.

Answer 10: Time period T = 2π sqrt(m/k) for spring-mass, independent of amplitude. It depends on the system's inertia and restoring force constant.

Question 1: What is a simple pendulum?

Answer 1: A simple pendulum consists of a small mass (bob) suspended by a light, inextensible string from a fixed point, oscillating under gravity about the equilibrium position.

Question 2: Derive the time period of a simple pendulum for small angles.

Answer 2: For small θ, torque τ = -mg sinθ L ≈ -mg θ L, leading to angular acceleration α = - (g/L) θ, so T = 2π sqrt(L/g), where L is length.

Question 3: Why is the small angle approximation necessary for SHM in a pendulum?

Answer 3: For small θ, sinθ ≈ θ (in radians), making the restoring force proportional to displacement (θ), satisfying SHM conditions. Large angles make it non-linear.

Question 4: What factors affect the time period of a simple pendulum?

Answer 4: T depends on length L (T ∝ sqrt(L)) and acceleration due to gravity g (T ∝ 1/sqrt(g)). Independent of mass m and amplitude (for small angles).

Question 5: Express the angular frequency for a simple pendulum.

Answer 5: ω = sqrt(g/L), similar to SHM. The motion is θ(t) = θ_max sin(ωt + φ).

Question 6: What is the effective length of a simple pendulum?

Answer 6: Effective length is the distance from the pivot to the center of mass of the bob. For a point mass, it's the string length L.

Question 7: How does air resistance affect a simple pendulum?

Answer 7: Air resistance causes damping, reducing amplitude over time. In ideal cases, it's neglected for undamped SHM analysis.

Question 8: Calculate the time period on the Moon for a 1 m pendulum.

Answer 8: g_Moon ≈ g/6, so T_Moon = 2π sqrt(L / (g/6)) = sqrt(6) T_Earth ≈ 2.45 T_Earth, longer period due to weaker gravity.

Question 9: What is the potential energy in a simple pendulum?

Answer 9: PE = m g L (1 - cosθ), approximated as (1/2) m g L θ² for small θ, maximum at extremes.

Question 10: Why does the period increase with amplitude for large angles?

Answer 10: For large θ, sinθ < θ, so restoring force is less than proportional, leading to slower motion and longer period (non-SHM).

Chapter 14: Waves

Question 1: What is a wave?

Answer 1: A wave is a disturbance that transfers energy through a medium or space without the net transfer of matter. It propagates from one point to another due to interactions between particles.

Question 2: Distinguish between transverse and longitudinal waves.

Answer 2: In transverse waves, particles vibrate perpendicular to the direction of wave propagation (e.g., light, water ripples). In longitudinal waves, particles vibrate parallel to propagation (e.g., sound waves).

Question 3: Define the terms wavelength, frequency, and speed of a wave.

Answer 3: Wavelength (λ) is the distance between two consecutive crests or troughs. Frequency (f) is the number of waves passing a point per second. Wave speed v = f λ.

Question 4: What is the wave equation for a progressive wave?

Answer 4: The displacement y(x,t) = A sin(kx - ωt + φ), where A is amplitude, k = 2π/λ is wave number, ω = 2πf is angular frequency, and φ is phase constant.

Question 5: Explain the principle of superposition of waves.

Answer 5: When two or more waves overlap, the resultant displacement is the vector sum of individual displacements, leading to interference phenomena.

Question 6: What is the speed of sound in air and what factors affect it?

Answer 6: Speed of sound in air ≈ 343 m/s at 20°C. It depends on temperature (v ∝ sqrt(T)), density, and elasticity of the medium, but not on frequency or amplitude.

Question 7: Describe mechanical and non-mechanical waves.

Answer 7: Mechanical waves require a medium (e.g., sound). Non-mechanical or electromagnetic waves do not (e.g., light, radio waves), propagating through vacuum.

Question 8: What is the phase difference in waves?

Answer 8: Phase difference is the angular shift between two waves, measured in radians or degrees. It determines constructive (0 or 2π) or destructive (π) interference.

Question 9: How is wave speed related to tension and linear density in a string?

Answer 9: For transverse waves on a string, v = sqrt(T/μ), where T is tension and μ is mass per unit length. Higher tension or lower density increases speed.

Question 10: Explain the concept of wave fronts and rays.

Answer 10: Wave front is the locus of points with the same phase (e.g., spherical for point source). Rays are perpendicular lines to wave fronts indicating direction of propagation.

Question 1: What are standing waves?

Answer 1: Standing waves are formed by the superposition of two waves of same frequency and amplitude traveling in opposite directions, resulting in stationary nodes and antinodes.

Question 2: Define nodes and antinodes in a standing wave.

Answer 2: Nodes are points of zero displacement (destructive interference). Antinodes are points of maximum displacement (constructive interference).

Question 3: What is the condition for standing waves in a string fixed at both ends?

Answer 3: The length L = n (λ/2), where n = 1,2,3... is the harmonic number. Fundamental frequency f1 = v/(2L), harmonics fn = n f1.

Question 4: Explain the formation of harmonics in a closed organ pipe.

Answer 4: In a closed pipe (one end closed), only odd harmonics: L = (2n-1) λ/4, f1 = v/(4L), f3 = 3 f1, etc. Node at closed end, antinode at open end.

Question 5: What is resonance in standing waves?

Answer 5: Resonance occurs when the driving frequency matches a natural frequency of the system, leading to large amplitude standing waves (e.g., in pipes or strings).

Question 6: Describe standing waves in an open organ pipe.

Answer 6: Both ends open: L = n λ/2, all harmonics fn = n v/(2L). Antinodes at both open ends.

Question 7: How does the end correction affect the frequency of pipes?

Answer 7: End correction e ≈ 0.3 d (d = diameter) adds to effective length: L + 2e for open pipe, L + e for closed, slightly decreasing frequency.

Question 8: What is the wavelength of the fundamental mode in a string of length L?

Answer 8: λ1 = 2L, as the string fits half a wavelength between fixed ends, with nodes at ends.

Question 9: Explain beats and their relation to standing waves.

Answer 9: Beats are amplitude variations from superposition of slightly different frequencies. In standing waves, fixed frequencies prevent beats, but near resonance, they can occur.

Question 10: Why are standing waves important in musical instruments?

Answer 10: They produce specific harmonics that give timbre to sound. String instruments use transverse standing waves; wind instruments use longitudinal in air columns.

Class 12 Physics

Chapter 1: Electric Charges and Fields

Question 1: Two identical conducting balls A and B have charges -Q and +3Q respectively. They are brought in contact and then separated by a distance d apart. Find the nature of Coulomb force between them. (2019)

Answer: When balls A and B touch, total charge = -Q + 3Q = 2Q. Since they are identical, each gets Q. After separation, both have charge +Q. Coulomb force, F = (1/4πε₀)(Q²/d²), is repulsive as charges are same.

Marking Rubrics: 1 mark for charge after contact, 1 mark for nature of force, 1 mark for force expression.

Hinglish Explanation: Yeh question mein do balls hain, ek ka charge -Q aur doosre ka +3Q. Jab yeh touch karte hain, total charge 2Q ban jata hai, aur dono equal hain toh har ek ko Q milta hai. Ab dono ka charge +Q hai, toh force repulsive hoga kyunki same sign ka charge repel karta hai. Formula likha F = kQ²/d², jisme k = 1/4πε₀. Marks is tarah milte hain: 1 mark charge calculate karne ke liye, 1 mark repulsive bolne ke liye, aur 1 mark formula ke liye.

Question 2: Two identical point charges, q each, are kept 2 m apart in air. A third point charge Q is placed on the line joining them such that the system is in equilibrium. Find the position and nature of Q. (2019)

Answer: For equilibrium, Q is at midpoint, i.e., 1 m from each q. Forces on Q due to both q’s are equal and opposite. Solving, Q = -q/4, so Q is negative.

Marking Rubrics: 1 mark for position, 1 mark for sign, 1 mark for magnitude.

Hinglish Explanation: Isme do charges q hain, 2 m door. Teesra charge Q kahin par rakhna hai jisse system equilibrium mein rahe. Equilibrium ke liye Q ko midpoint par hona chahiye, matlab 1 m dono se. Forces balance karne ke liye Q ka sign negative hoga aur magnitude -q/4 aayega equation solve karne se. Marks: 1 position ke liye, 1 sign ke liye, 1 magnitude ke liye.

Question 3: Two point charges +4 μC and +1 μC are separated by 2 m in air. Find the point where the net electric field is zero. (2017)

Answer: Let point be at x m from +1 μC. Electric field due to +1 μC: E₁ = k(1×10⁻⁶)/x². Due to +4 μC: E₂ = k(4×10⁻⁶)/(2-x)². Set E₁ = E₂, solve: x = 0.828 m from +1 μC.

Marking Rubrics: 1 mark for field equations, 1 mark for equating, 1 mark for solving x.

Hinglish Explanation: Yahan +4 μC aur +1 μC charges 2 m door hain. Net electric field zero kahan hoga? Ek point x m door +1 μC se man lo. Dono charges ka electric field equate karo: k(1×10⁻⁶)/x² = k(4×10⁻⁶)/(2-x)². Solve karne pe x = 0.828 m aata hai. Marks: 1 field equations ke liye, 1 equate karne ke liye, 1 x nikalne ke liye.

Question 4: Two balls of equal positive charge q are suspended by insulating strings. What happens to the force when a plastic sheet is inserted between them? (2014)

Answer: Force in air: F = k(q²/r²). Plastic sheet (dielectric, εᵣ > 1) reduces force to F’ = F/εᵣ, where εᵣ is dielectric constant. Force decreases.

Marking Rubrics: 1 mark for initial force, 1 mark for dielectric effect, 1 mark for stating decrease.

Hinglish Explanation: Do charges q hain, aur force F = kq²/r². Plastic sheet (dielectric) daalne se force kam ho jata hai kyunki dielectric constant εᵣ > 1 hota hai. Naya force F’ = F/εᵣ. Marks: 1 initial force ke liye, 1 dielectric effect ke liye, 1 force kam hone ke liye.

Question 5: Three point charges +q, +2q, and -3q are at vertices A, B, C of an equilateral triangle of side a. Find net force on charge at A.

Answer: Force on +q at A due to +2q at B: F₁ = k(2q²/a²) along AB (repulsive). Due to -3q at C: F₂ = k(3q²/a²) towards C (attractive). Net force magnitude: √13 (q²/4πε₀a²).

Marking Rubrics: 1 mark for F₁, 1 mark for F₂, 1 mark for net force.

Hinglish Explanation: Triangle ke vertices par charges hain. A par +q ke liye, B (+2q) se force F₁ = k(2q²/a²) repulsive, aur C (-3q) se F₂ = k(3q²/a²) attractive. Dono forces 60° angle par hain, toh resultant √13 (q²/4πε₀a²) aata hai. Marks: 1 F₁ ke liye, 1 F₂ ke liye, 1 resultant ke liye.

Question 6: Two charges +5 μC and -5 μC are 1 m apart. Calculate the force and its nature.

Answer: Force: F = (9×10⁹)(5×10⁻⁶ × 5×10⁻⁶/1²) = 0.225 N. Nature: attractive (opposite charges).

Marking Rubrics: 1 mark for formula, 1 mark for calculation, 1 mark for nature.

Hinglish Explanation: Yahan +5 μC aur -5 μC charges 1 m door hain. Force F = kq₁q₂/r² = (9×10⁹)(25×10⁻¹²) = 0.225 N. Opposite charges hain toh force attractive hoga. Marks: 1 formula, 1 calculation, 1 nature ke liye.

Question 7: Two charges +q and +q are 2d apart. A charge -q is placed on the line joining them. Find its position for no net force.

Answer: Let -q be at x from one +q. Force due to first +q: F₁ = kq²/x². Due to second +q: F₂ = kq²/(2d-x)². Set F₁ = F₂, solve: x = d (midpoint).

Marking Rubrics: 1 mark for force equations, 1 mark for equating, 1 mark for position.

Hinglish Explanation: Do +q charges 2d door hain, -q kahin par rakha hai. F₁ = kq²/x² aur F₂ = kq²/(2d-x)². Equilibrium ke liye F₁ = F₂, toh x = d aata hai. Marks: 1 equations, 1 equating, 1 position ke liye.

Question 8: Two charges +2 μC and -3 μC are 3 m apart. Calculate force and direction.

Answer: Force: F = (9×10⁹)(2×10⁻⁶ × 3×10⁻⁶/3²) = 0.06 N. Direction: attractive, towards -3 μC.

Marking Rubrics: 1 mark for formula, 1 mark for calculation, 1 mark for direction.

Hinglish Explanation: +2 μC aur -3 μC 3 m door hain. F = kq₁q₂/r² = (9×10⁹)(6×10⁻¹²/9) = 0.06 N. Opposite charges hain, toh force attractive hoga, -3 μC ki taraf. Marks: 1 formula, 1 calculation, 1 direction ke liye.

Question 9: Four charges +q at corners of a square of side a. Find force on one charge.

Answer: Force on one +q: Two adjacent charges give F = kq²/a² each at 90°. Diagonal charge: F = kq²/(√2a)². Resultant: √3 (kq²/a²).

Marking Rubrics: 1 mark for adjacent forces, 1 mark for diagonal force, 1 mark for resultant.

Hinglish Explanation: Square ke corners par +q hain. Ek +q par do adjacent charges se force F = kq²/a², 90° par. Diagonal se F = kq²/2a². Resultant √3 (kq²/a²) aata hai. Marks: 1 adjacent forces, 1 diagonal force, 1 resultant ke liye.

Question 10: Two charges +q and -q are separated by d. Find work done to bring them closer by d/2.

Answer: Initial potential energy: U₁ = -kq²/d. Final: U₂ = -kq²/(d/2). Work done: W = U₂ - U₁ = kq²/d.

Marking Rubrics: 1 mark for U₁, 1 mark for U₂, 1 mark for work.

Hinglish Explanation: +q aur -q d door hain. Initial energy U₁ = -kq²/d, final U₂ = -kq²/(d/2). Work W = U₂ - U₁ = kq²/d. Marks: 1 U₁, 1 U₂, 1 work ke liye.

Question 1: Two point charges +4 μC and +1 μC are separated by 2 m in air. Find the point where the net electric field is zero. (2017)

Answer: Let point be x m from +1 μC. E₁ = k(1×10⁻⁶)/x², E₂ = k(4×10⁻⁶)/(2-x)². Set E₁ = E₂, solve: x = 0.828 m from +1 μC.

Marking Rubrics: 1 mark for field equations, 1 mark for equating, 1 mark for solving x.

Explanation: +4 μC aur +1 μC 2 m door hain. Net field zero kahan hoga? Point x m door +1 μC se lo. Dono ka field equate karo: k(1×10⁻⁶)/x² = k(4×10⁻⁶)/(2-x)². Solve karne pe x = 0.828 m aata hai. Marks: 1 equations, 1 equating, 1 x ke liye.

Question 2: A charge q is at the origin. Find the electric field at point (a, a). (2016)

Answer: Distance r = √(a² + a²) = a√2. Electric field, E = kq/(2a²) along line from (0,0) to (a,a).

Marking Rubrics: 1 mark for distance, 1 mark for field expression, 1 mark for direction.

Explanation: Charge q origin par hai. Point (a, a) par field nikalna hai. Distance r = √(a² + a²) = a√2. Field E = kq/r² = kq/(2a²). Direction charge se point tak hai. Marks: 1 distance, 1 field formula, 1 direction ke liye.

Question 3: Draw electric field lines for two charges +q and -q separated by distance d. (2015)

Answer: Field lines start from +q, end at -q, curving symmetrically. Outward near +q, inward near -q.

Marking Rubrics: 1 mark for direction, 1 mark for shape, 1 mark for symmetry.

Explanation: +q aur -q ke beech field lines banane hain. Lines +q se start hokar -q par end hoti hain, curved shape mein. +q ke paas outward, -q ke paas inward. Marks: 1 direction, 1 shape, 1 symmetry ke liye.

Question 4: Find the electric field at a point on the axial line of a charged ring of radius R with charge Q at distance x from center. (2014)

Answer: Electric field, E = kQx/(R² + x²)^(3/2) along the axis.

Marking Rubrics: 1 mark for formula, 1 mark for direction, 1 mark for derivation setup.

Explanation: Charged ring ka center se x door par field nikalna hai. Formula E = kQx/(R² + x²)^(3/2) hai, jo axis ke along hota hai. Marks: 1 formula, 1 direction, 1 derivation ke liye.

Question 5: A charge q is placed at (0,0,0). Find electric field at (0,0,d). (2013)

Answer: Distance r = d. Electric field, E = kq/d² along z-axis.

Marking Rubrics: 1 mark for distance, 1 mark for field, 1 mark for direction.

Explanation: Charge q origin par hai, point (0,0,d) par field nikalna hai. Distance r = d. Field E = kq/d², z-axis ke along. Marks: 1 distance, 1 field, 1 direction ke liye.

Question 6: Two charges +2q and -q are d apart. Find the point on the line joining them where electric field is zero. (2012)

Answer: Let point be x m from +2q. E₁ = k(2q)/x², E₂ = kq/(d-x)². Set E₁ = E₂, solve: x = 2d/3.

Marking Rubrics: 1 mark for equations, 1 mark for equating, 1 mark for x.

Explanation: +2q aur -q d door hain. Zero field point x m door +2q se lo. Equations E₁ = k(2q)/x² aur E₂ = kq/(d-x)² equate karo. Solve karne pe x = 2d/3 aata hai. Marks: 1 equations, 1 equating, 1 x ke liye.

Question 7: A uniformly charged rod has linear charge density λ. Find electric field at a point P at distance r perpendicular to its midpoint. (2011)

Answer: Electric field, E = kλ/(r√(r² + L²/4)), where L is rod length.

Marking Rubrics: 1 mark for setup, 1 mark for formula, 1 mark for direction.

Explanation: Rod ka charge density λ hai. Point P midpoint se r door perpendicular hai. Field E = kλ/(r√(r² + L²/4)) aata hai, perpendicular to rod. Marks: 1 setup, 1 formula, 1 direction ke liye.

Question 8: Draw field lines for a single positive charge. (2010)

Answer: Field lines radiate outward symmetrically from the charge in all directions.

Marking Rubrics: 1 mark for direction, 1 mark for symmetry, 1 mark for description.

Explanation: Ek +q charge ke liye field lines sabhi directions mein outward radiate karti hain, symmetrically. Marks: 1 direction, 1 symmetry, 1 description ke liye.

Question 9: Find electric field at a point due to a uniformly charged infinite line with charge density λ at distance r. (2009)

Answer: Electric field, E = λ/(2πε₀r), radially outward.

Marking Rubrics: 1 mark for formula, 1 mark for direction, 1 mark for derivation.

Explanation: Infinite line charge ka density λ hai. Distance r par field E = λ/(2πε₀r), radially outward. Marks: 1 formula, 1 direction, 1 derivation ke liye.

Question 10: Two charges +q and +q are 2d apart. Find electric field at midpoint. (2008)

Answer: Field due to each +q: E = kq/d², opposite directions, so net E = 0.

Marking Rubrics: 1 mark for field per charge, 1 mark for direction, 1 mark for net field.

Explanation: Do +q charges 2d door hain. Midpoint par har charge ka field E = kq/d², par opposite direction mein. Toh net field zero ho jata hai. Marks: 1 field, 1 direction, 1 net field ke liye.

Question 1: An electric dipole with moment p is placed in a uniform electric field E. Find the torque acting on it. (2020)

Answer: Torque, τ = pE sinθ, where θ is the angle between p and E.

Marking Rubrics: 1 mark for formula, 1 mark for variables, 1 mark for explanation.

Explanation: Dipole ka moment p hai, uniform field E mein rakha hai. Torque τ = pE sinθ hota hai, jahan θ dipole aur field ke beech angle hai. Marks: 1 formula, 1 variables, 1 explanation ke liye.

Question 2: Find the electric field due to a dipole at a point on its axial line at distance r from center. (2019)

Answer: Electric field, E = 2kp/r³, along the dipole axis.

Marking Rubrics: 1 mark for formula, 1 mark for direction, 1 mark for derivation.

Explanation: Dipole ke axis par r door ek point par field nikalna hai. Formula E = 2kp/r³, jo axis ke along hota hai. Marks: 1 formula, 1 direction, 1 derivation ke liye.

Question 3: A dipole with moment p is placed at 30° to a uniform field E. Calculate torque. (2018)

Answer: Torque, τ = pE sin30° = pE/2.

Marking Rubrics: 1 mark for formula, 1 mark for angle substitution, 1 mark for result.

Explanation: Dipole ka moment p hai, field E ke saath 30° angle par hai. Torque τ = pE sinθ, yahan θ = 30°, toh τ = pE/2. Marks: 1 formula, 1 angle, 1 result ke liye.

Question 4: Find the electric field due to a dipole at a point on its equatorial line at Occupation: at distance r. (2017)

Answer: Electric field, E = kp/(r² + a²)^(3/2), perpendicular to dipole axis.

Marking Rubrics: 1 mark for formula, 1 mark for direction, 1 mark for derivation.

Explanation: Dipole ke equatorial line par r door field nikalna hai. Formula E = kp/(r² + a²)^(3/2), jo dipole axis ke perpendicular hota hai. Marks: 1 formula, 1 direction, 1 derivation ke liye.

Question 5: A dipole consists of +q and -q separated by 2a. Find potential at a point on axial line at distance r. (2016)

Answer: Potential, V = kp cosθ/r², where p = 2qa.

Marking Rubrics: 1 mark for formula, 1 mark for variables, 1 mark for derivation.

Explanation: Dipole mein +q aur -q, 2a door hain. Axial point par potential V = kp cosθ/r², jahan p = 2qa. Marks: 1 formula, 1 variables, 1 derivation ke liye.

Question 6: A dipole is placed in a non-uniform electric field. Describe its motion. (2015)

Answer: Dipole experiences torque (τ = p × E) and net force (F = p·∇E), causing rotation and translation.

Marking Rubrics: 1 mark for torque, 1 mark for force, 1 mark for motion.

Explanation: Dipole non-uniform field mein hai. Torque τ = p × E se rotate karta hai, aur force F = p·∇E se move bhi karta hai. Marks: 1 torque, 1 force, 1 motion ke liye.

Question 7: Find work done to rotate a dipole from θ = 0° to 90° in uniform field E. (2014)

Answer: Work, W = pE (1 - cosθ). For θ = 90°, W = pE.

Marking Rubrics: 1 mark for formula, 1 mark for substitution, 1 mark for result.

Explanation: Dipole ko 0° se 90° tak rotate karna hai. Work W = pE (1 - cosθ), θ = 90° par W = pE. Marks: 1 formula, 1 substitution, 1 result ke liye.

Question 8: A dipole with charges ±q, separation 2a, is in field E. Find potential energy. (2013)

Answer: Potential energy, U = -pE cosθ, where p = 2qa.

Marking Rubrics: 1 mark for formula, 1 mark for variables, 1 mark for explanation.

Explanation: Dipole ka potential energy U = -pE cosθ hai, jahan p = 2qa. Marks: 1 formula, 1 variables, 1 explanation ke liye.

Question 9: Explain why a dipole aligns with a uniform electric field. (2012)

Answer: Torque τ = pE sinθ acts to align p with E, minimizing potential energy U = -pE cosθ.

Marking Rubrics: 1 mark for torque, 1 mark for energy, 1 mark for alignment.

Explanation: Dipole par torque τ = pE sinθ lagta hai, jo dipole ko field E ke saath align karta hai kyunki potential energy U = -pE cosθ minimum hota hai jab θ = 0. Marks: 1 torque, 1 energy, 1 alignment ke liye.

Question 10: A dipole with moment p is at 60° to field E. Find torque magnitude. (2011)

Answer: Torque, τ = pE sin60° = pE√3/2.

Marking Rubrics: 1 mark for formula, 1 mark for angle, 1 mark for result.

Explanation: Dipole 60° par field E ke saath hai. Torque τ = pE sin60° = pE√3/2. Marks: 1 formula, 1 angle, 1 result ke liye.

Question 1: State Gauss’s theorem and apply it to find the electric field due to an infinite line charge with density λ. (2020)

Answer: Gauss’s theorem: ∮E·dA = Q/ε₀. For line charge, E = λ/(2πε₀r), radially outward.

Marking Rubrics: 1 mark for theorem, 1 mark for field, 1 mark for direction.

Explanation: Gauss ka theorem kehta hai ∮E·dA = Q/ε₀. Infinite line charge ke liye cylindrical surface lo, field E = λ/(2πε₀r), radially outward. Marks: 1 theorem, 1 field, 1 direction ke liye.

Question 2: Find electric field due to an infinite plane sheet with surface charge density σ. (2019)

Answer: Using Gauss’s theorem, E = σ/(2ε₀), perpendicular to sheet.

Marking Rubrics: 1 mark for theorem application, 1 mark for field, 1 mark for direction.

Explanation: Infinite plane sheet ka charge density σ hai. Gauss’s theorem se E = σ/(2ε₀), sheet ke perpendicular. Marks: 1 theorem, 1 field, 1 direction ke liye.

Question 3: Find electric field inside and outside a uniformly charged thin spherical shell. (2018)

Answer: Inside: E = 0. Outside: E = kQ/r². Using Gauss’s theorem.

Marking Rubrics: 1 mark for inside field, 1 mark for outside field, 1 mark for theorem.

Explanation: Spherical shell ke liye Gauss’s theorem use karo. Inside E = 0, outside E = kQ/r². Marks: 1 inside, 1 outside, 1 theorem ke liye.

Question 4: Calculate electric flux through a spherical surface enclosing a charge Q. (2017)

Answer: Flux, Φ = Q/ε₀, using Gauss’s theorem.

Marking Rubrics: 1 mark for theorem, 1 mark for formula, 1 mark for result.

Explanation: Spherical surface mein charge Q hai. Gauss’s theorem se flux Φ = Q/ε₀. Marks: 1 theorem, 1 formula, 1 result ke liye.

Question 5: Find electric field due to a long straight wire with charge density λ at distance r. (2016)

Answer: Using Gauss’s theorem, E = λ/(2πε₀r), radially outward.

Marking Rubrics: 1 mark for theorem, 1 mark for field, 1 mark for direction.

Explanation: Long wire ka charge density λ hai. Gauss’s theorem se cylindrical surface lekar E = λ/(2πε₀r), radially outward. Marks: 1 theorem, 1 field, 1 direction ke liye.

Question 6: A charge Q is at the center of a cube. Find flux through one face. (2015)

Answer: Total flux Φ = Q/ε₀. Flux through one face = Q/(6ε₀).

Marking Rubrics: 1 mark for total flux, 1 mark for face flux, 1 mark for theorem.

Explanation: Cube ke center mein charge Q hai. Gauss’s theorem se total flux Φ = Q/ε₀, aur ek face ka flux Q/(6ε₀). Marks: 1 total flux, 1 face flux, 1 theorem ke liye.

Question 7: Explain why electric field inside a charged spherical shell is zero. (2014)

Answer: Gauss’s theorem: No charge inside shell, so ∮E·dA = 0, hence E = 0.

Marking Rubrics: 1 mark for theorem, 1 mark for explanation, 1 mark for result.

Explanation: Spherical shell ke andar koi charge nahi, toh Gauss’s theorem se ∮E·dA = 0, isliye E = 0. Marks: 1 theorem, 1 explanation, 1 result ke liye.

Question 8: Find electric field outside a uniformly charged sphere of radius R with charge Q. (2013)

Answer: Using Gauss’s theorem, E = kQ/r², radially outward.

Marking Rubrics: 1 mark for theorem, 1 mark for field, 1 mark for direction.

Explanation: Charged sphere ka radius R hai, bahar field nikalna hai. Gauss’s theorem se E = kQ/r², radially outward. Marks: 1 theorem, 1 field, 1 direction ke liye.

Question 9: A cylindrical surface encloses a line charge with density λ. Find flux through the surface. (2012)

Answer: Flux, Φ = λL/ε₀, where L is cylinder length, using Gauss’s theorem.

Marking Rubrics: 1 mark for theorem, 1 mark for formula, 1 mark for result.

Explanation: Cylindrical surface mein line charge λ hai. Gauss’s theorem se flux Φ = λL/ε₀, jahan L cylinder ki length hai. Marks: 1 theorem, 1 formula, 1 result ke liye.

Question 10: Find electric field due to an infinite charged plane sheet with density σ at distance r. (2011)

Answer: Using Gauss’s theorem, E = σ/(2ε₀), perpendicular to sheet.

Marking Rubrics: 1 mark for theorem, 1 mark for field, 1 mark for direction.

Explanation: Infinite plane sheet ka density σ hai. Gauss’s theorem se E = σ/(2ε₀), sheet ke perpendicular. Marks: 1 theorem, 1 field, 1 direction ke liye.

Chapter 2: Electrostatic Potential and Capacitance

Question 1: Define the term ‘potential energy’ of charge ‘q’ at a distance 'r' in an external electric field. (All India 2009)

Answer: It is defined as the amount of work done in bringing the charge from infinity to its position in the system in the electric field of another charge without acceleration. V = Er.

Marking Rubrics: 1 mark for definition, 1 mark for formula, 1 mark for explanation.

Explanation: Potential energy ka matlab hai charge q ko infinity se external field mein r distance tak laane mein kiya gaya work. Formula U = qV, jahan V = Er. Marks: 1 definition, 1 formula, 1 explanation ke liye.

Question 2: A point charge Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero, if Q is (i) positive (ii) negative? (Delhi 2011)

Answer: Clearly, As OA < OB, so the quantity within bracket is negative. (i) If q is positive charge, VA – VB = negative (ii) If q is negative charge, VA – VB = positive.

Marking Rubrics: 1 mark for setup, 1 mark for positive case, 1 mark for negative case.

Explanation: OA < OB hai, toh bracket mein negative aata hai. Agar Q positive hai, VA – VB negative, negative Q ke liye positive. Marks: 1 setup, 1 positive, 1 negative ke liye.

Question 3: A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere? (All India 2011)

Answer: The electric field inside the shell is zero. This implies that potential is constant inside the shell (as no work is done in moving a charge inside the shell) and, therefore, equals its value at the surface, which is 10 V.

Marking Rubrics: 1 mark for field inside, 1 mark for potential constant, 1 mark for result.

Explanation: Shell ke andar field zero hai, toh potential constant rahta hai aur surface ke equal 10 V. Marks: 1 field, 1 constant, 1 result ke liye.

Question 4: A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere? (All India 2011)

Answer: Hollow metal sphere behaves as an equipotential surface, so the potential at its centre will be 5 V.

Marking Rubrics: 1 mark for equipotential, 1 mark for explanation, 1 mark for result.

Explanation: Metal sphere equipotential hota hai, toh centre par 5 V. Marks: 1 equipotential, 1 explanation, 1 result ke liye.

Question 5: Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface? (Delhi 2012)

Answer: Electric field inside the conductor = 0, so potential is constant as no work is done in moving a charge inside.

Marking Rubrics: 1 mark for field zero, 1 mark for work, 1 mark for constant potential.

Explanation: Conductor ke andar field zero hai, toh charge move karne ka work zero, isliye potential constant. Marks: 1 field, 1 work, 1 constant ke liye.

Question 6: Distinguish between a dielectric and a conductor. (Comptt. Delhi 2012)

Answer: Dielectric: Insulating materials which transmit electric effects without conducting. Conductor: Substances which can carry or conduct electric charge from one place to the other.

Marking Rubrics: 1 mark for dielectric definition, 1 mark for conductor definition, 1 mark for distinction.

Explanation: Dielectric insulating hota hai jo electric effect pass karta hai par conduct nahi karta. Conductor charge conduct karta hai. Marks: 1 dielectric, 1 conductor, 1 distinction ke liye.

Question 7: Why must the electrostatic potential inside a hollow charged conductor be the same at every point? (Comptt. All India 2012)

Answer: Inside the hollow charged conductor, electric field is zero therefore no work is done in moving a small test charge within the conductor. Hence electrostatic potential inside is same at every point.

Marking Rubrics: 1 mark for field zero, 1 mark for work, 1 mark for constant potential.

Explanation: Andar field zero hai, toh charge move karne ka work zero, potential same. Marks: 1 field, 1 work, 1 constant ke liye.

Question 8: What is the geometrical shape of equipotential surfaces due to a single isolated charge? (Delhi 2013)

Answer: Concentric spheres with a gap between.

Marking Rubrics: 1 mark for shape, 1 mark for description, 1 mark for explanation.

Explanation: Single charge ke liye equipotential surfaces concentric spheres hote hain. Marks: 1 shape, 1 description, 1 explanation ke liye.

Question 9: A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process. (Delhi 2013)

Answer: Work done is zero as potential on equatorial line is zero.

Marking Rubrics: 1 mark for potential zero, 1 mark for work, 1 mark for result.

Explanation: Equatorial plane par potential zero, toh work zero. Marks: 1 potential, 1 work, 1 result ke liye.

Question 10: In which orientation, a dipole placed in uniform electric field is in (i) stable, (ii) unstable equilibrium? (All India 2013)

Answer: (i) Stable equilibrium when θ = 0° (ii) Unstable equilibrium when θ = 180°.

Marking Rubrics: 1 mark for stable, 1 mark for unstable, 1 mark for explanation.

Explanation: Stable when dipole field ke along, unstable opposite. Marks: 1 stable, 1 unstable, 1 explanation ke liye.

Question 1: A 500 µC charge is at the centre of a square of side 10 cm. Find the work done in moving a charge of 10 µC between two diagonally opposite points on the square. (Delhi 2008)

Answer: The work done in moving a charge of 10 µC between two diagonally opposite points on the square will be zero because these two points will be at equipotential.

Marking Rubrics: 1 mark for equipotential, 1 mark for work zero, 1 mark for explanation.

Explanation: Do points equipotential hain, toh work zero. Marks: 1 equipotential, 1 work, 1 explanation ke liye.

Question 2: What is the electrostatic potential due to an electric dipole at an equatorial point? (All India 2009)

Answer: Electric potential at any point in the equatorial plane of dipole is Zero.

Marking Rubrics: 1 mark for zero, 1 mark for explanation, 1 mark for formula.

Explanation: Equatorial point par potential zero hota hai. Marks: 1 zero, 1 explanation, 1 formula ke liye.

Question 3: What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole? (All India 2009)

Answer: Since potential for equatorial axis V = 0, ∴ W = qV = 0.

Marking Rubrics: 1 mark for V=0, 1 mark for W=0, 1 mark for explanation.

Explanation: V zero hai, toh W zero. Marks: 1 V, 1 W, 1 explanation ke liye.

Question 4: A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. (Delhi 2011)

Answer: Potential at the centre is \( \frac{8q}{4\pi\epsilon_0\sqrt{3}b} \), electric field at the centre is zero due to symmetry.

Marking Rubrics: 1 mark for potential, 1 mark for field zero, 1 mark for calculation.

Explanation: Centre par potential calculate karo, field symmetry se zero. Marks: 1 potential, 1 field, 1 calculation ke liye.

Question 5: Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the midpoint. (All India 2011)

Answer: (a) Potential is 2.4 × 10⁵ V, electric field is 4 × 10⁵ V/m towards the larger charge. (b) Potential is 2.0 × 10⁵ V, electric field is 6.6 × 10⁵ V/m.

Marking Rubrics: 1 mark for (a) potential, 1 mark for (a) field, 1 mark for (b).

Explanation: Midpoint par calculate karo, aur perpendicular point par. Marks: 1 (a) potential, 1 (a) field, 1 (b) ke liye.

Question 6: Two charges −q and q are located at points (0, 0, −a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points? (b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1. (c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path is not along the x-axis? (Delhi 2011)

Answer: (a) Zero at points normal to the axis, potential on axis is \( \frac{2qa}{4\pi\epsilon_0(r^2-a^2) \). (b) Potential is inversely proportional to r². (c) Work done is zero, answer does not change with path.

Marking Rubrics: 1 mark for (a), 1 mark for (b), 1 mark for (c).

Explanation: Dipole ke potential calculate karo, work zero. Marks: 1 (a), 1 (b), 1 (c) ke liye.

Question 7: Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole. (Delhi 2011)

Answer: Potential is \( \frac{2qa^2}{4\pi\epsilon_0r^3} \), contrasts with dipole (1/r²) and monopole (1/r).

Marking Rubrics: 1 mark for potential, 1 mark for contrast dipole, 1 mark for monopole.

Explanation: Quadrupole ka potential r^-3, dipole r^-2, monopole r^-1. Marks: 1 potential, 1 dipole, 1 monopole ke liye.

Question 8: In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å. (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation? (Delhi 2011)

Answer: (a) Potential energy is -27.2 eV. (b) Minimum work is 13.6 eV. (c) Potential energy is -25.6 eV, minimum work is 12.8 eV.

Marking Rubrics: 1 mark for (a), 1 mark for (b), 1 mark for (c).

Explanation: Hydrogen atom ka potential energy calculate karo. Marks: 1 (a), 1 (b), 1 (c) ke liye.

Question 9: If there are N protons in the nucleus, how many neutrons are there in the nucleus? (Delhi 2011)

Answer: The question seems unrelated to the chapter, possibly a truncation error in the content.

Marking Rubrics: 1 mark for identification, 1 mark for explanation, 1 mark for result.

Explanation: Yeh question chapter se related nahi, error hai. Marks: 1 identification, 1 explanation, 1 result ke liye.

Question 10: Two charges 5 ×10–8 C and –3 ×10–8 C are located 16 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. (Delhi 2012)

Answer: Potential is zero at 10 cm from the positive charge between the charges and at 40 cm from the positive charge outside the system.

Marking Rubrics: 1 mark for point between, 1 mark for point outside, 1 mark for calculation.

Explanation: Dono charges ke beech aur bahar point calculate karo. Marks: 1 between, 1 outside, 1 calculation ke liye.

Question 1: Draw equipotential surfaces due to a single charge. (Delhi 2008)

Answer: Equipotential surfaces are concentric spheres.

Marking Rubrics: 1 mark for shape, 1 mark for description, 1 mark for diagram.

Explanation: Single charge ke liye concentric spheres. Marks: 1 shape, 1 description, 1 diagram ke liye.

Question 2: Draw the equipotential surfaces due to an electric dipole. (Delhi 2008)

Answer: The equipotential surfaces are perpendicular to the field lines.

Marking Rubrics: 1 mark for shape, 1 mark for description, 1 mark for diagram.

Explanation: Dipole ke liye equipotential surfaces field lines ke perpendicular. Marks: 1 shape, 1 description, 1 diagram ke liye.

Question 3: Depict the equipotential surfaces for a system of two identical positive point charges placed a distance ‘a’ apart. (Delhi 2008)

Answer: Equipotential surfaces for two identical positive point charges are shown in figure.

Marking Rubrics: 1 mark for depiction, 1 mark for description, 1 mark for diagram.

Explanation: Do positive charges ke liye equipotential surfaces. Marks: 1 depiction, 1 description, 1 diagram ke liye.

Question 4: Deduce the expression for the potential energy of a system of two charges q1 and q2 located at \(\vec{r_1}\) and \(\vec{r_2}\), respectively, in an external electric field. (Delhi 2008)

Answer: Potential energy U = q1V1 + q2V2, where V1 and V2 are potentials at r1 and r2.

Marking Rubrics: 1 mark for expression, 1 mark for derivation, 1 mark for explanation.

Explanation: Two charges ka potential energy external field mein. Marks: 1 expression, 1 derivation, 1 explanation ke liye.

Question 5: Why is there no work done in moving a charge from one point to another on an equipotential surface? (Delhi 2009)

Answer: The potential difference is zero, so work done is zero.

Marking Rubrics: 1 mark for potential difference, 1 mark for work, 1 mark for explanation.

Explanation: Equipotential par potential difference zero, toh work zero. Marks: 1 difference, 1 work, 1 explanation ke liye.

Question 6: Why are the electric field lines normal at the surface of a conductor? (Delhi 2009)

Answer: So that there is no component of the electric field parallel to the surface, as that would cause charge movement.

Marking Rubrics: 1 mark for normal, 1 mark for no parallel component, 1 mark for explanation.

Explanation: Conductor par field normal, parallel component zero. Marks: 1 normal, 1 component, 1 explanation ke liye.

Question 7: Explain why the electric field inside a conductor is zero. (Delhi 2009)

Answer: Charges reside on the surface, canceling field inside.

Marking Rubrics: 1 mark for zero field, 1 mark for charges on surface, 1 mark for explanation.

Explanation: Conductor ke andar field zero kyunki charges surface par. Marks: 1 zero, 1 surface, 1 explanation ke liye.

Question 8: If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change? (All India 2012)

Answer: The electric flux remains the same, as it depends on enclosed charge, not radius.

Marking Rubrics: 1 mark for flux same, 1 mark for reason, 1 mark for formula.

Explanation: Flux enclosed charge par depend, radius par nahi. Marks: 1 same, 1 reason, 1 formula ke liye.

Question 9: Define the term 'electric flux'. Write its S.I. unit. What is the flux due to electric field \(\vec{E} = 3\times10^3 \hat{i}\) N/C through a square of side 10 cm, when it is held normal to \(\vec{E}\)? (All India 2012)

Answer: Electric flux is E·A, S.I. unit Vm or Nm²/C. Flux = 3×10^3 × 0.01 = 30 Nm²/C.

Marking Rubrics: 1 mark for definition, 1 mark for unit, 1 mark for calculation.

Explanation: Flux E·A, unit Vm, calculation 30. Marks: 1 definition, 1 unit, 1 calculation ke liye.

Question 10: A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder. (All India 2012)

Answer: Flux = λ l / ε₀.

Marking Rubrics: 1 mark for expression, 1 mark for derivation, 1 mark for explanation.

Explanation: Gauss theorem se flux = enclosed charge / ε₀ = λ l / ε₀. Marks: 1 expression, 1 derivation, 1 explanation ke liye.

Question 1: Two charges 3 µC and -3 µC are located at the two vertices of an equilateral triangle of side 20 cm. Calculate the potential energy of the system. (Delhi 2008)

Answer: U = k q1 q2 / r = (9×10^9)(3×10^-6)(-3×10^-6)/0.2 = -0.135 J.

Marking Rubrics: 1 mark for formula, 1 mark for calculation, 1 mark for result.

Explanation: Two charges ka potential energy calculate karo. Marks: 1 formula, 1 calculation, 1 result ke liye.

Question 2: A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon. (Delhi 2009)

Answer: Potential at the centre is 2.7 × 10⁶ V.

Marking Rubrics: 1 mark for formula, 1 mark for calculation, 1 mark for result.

Explanation: Hexagon ke centre par potential 6 charges ka sum. Marks: 1 formula, 1 calculation, 1 result ke liye.

Question 3: A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm). (Delhi 2009)

Answer: Work done is 1.27 J.

Marking Rubrics: 1 mark for path independence, 1 mark for calculation, 1 mark for result.

Explanation: Work potential difference par depend, path nahi. Marks: 1 path, 1 calculation, 1 result ke liye.

Question 4: A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. (Delhi 2009)

Answer: Potential at the centre is \( \frac{8q}{4\pi\epsilon_0\sqrt{3}b} \), electric field is zero.

Marking Rubrics: 1 mark for potential, 1 mark for field, 1 mark for calculation.

Explanation: Cube ke centre par potential aur field. Marks: 1 potential, 1 field, 1 calculation ke liye.

Question 5: Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the midpoint. (Delhi 2009)

Answer: (a) Potential 2.4 × 10⁵ V, field 4 × 10⁵ V/m. (b) Potential 2.0 × 10⁵ V, field 6.6 × 10⁵ V/m.

Marking Rubrics: 1 mark for (a), 1 mark for (b), 1 mark for calculation.

Explanation: Midpoint aur perpendicular point par potential aur field. Marks: 1 (a), 1 (b), 1 calculation ke liye.

Question 6: Two charges −q and q are located at points (0, 0, −a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points? (b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1. (c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path is not along the x-axis? (Delhi 2009)

Answer: (a) Zero at normal points, on axis \( \frac{2qa}{4\pi\epsilon_0(r^2-a^2) \). (b) 1/r². (c) Zero, no change.

Marking Rubrics: 1 mark for (a), 1 mark for (b), 1 mark for (c).

Explanation: Dipole ka potential aur work. Marks: 1 (a), 1 (b), 1 (c) ke liye.

Question 7: Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole. (Delhi 2009)

Answer: Potential \( \frac{2qa^2}{4\pi\epsilon_0r^3} \), dipole 1/r², monopole 1/r.

Marking Rubrics: 1 mark for potential, 1 mark for contrast, 1 mark for explanation.

Explanation: Quadrupole ka potential aur comparison. Marks: 1 potential, 1 contrast, 1 explanation ke liye.

Question 8: In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å. (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation? (Delhi 2009)

Answer: (a) -27.2 eV. (b) 13.6 eV. (c) -25.6 eV, 12.8 eV.

Marking Rubrics: 1 mark for (a), 1 mark for (b), 1 mark for (c).

Explanation: Hydrogen atom ka energy. Marks: 1 (a), 1 (b), 1 (c) ke liye.

Question 9: A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon. (All India 2010)

Answer: Potential at the centre is 2.7 × 10⁶ V.

Marking Rubrics: 1 mark for formula, 1 mark for calculation, 1 mark for result.

Explanation: Hexagon ke centre par potential. Marks: 1 formula, 1 calculation, 1 result ke liye.

Question 10: A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm). (All India 2010)

Answer: Work done is 1.27 J.

Marking Rubrics: 1 mark for path independence, 1 mark for calculation, 1 mark for result.

Explanation: Work potential difference par. Marks: 1 path, 1 calculation, 1 result ke liye.

Question 1: Distinguish between a dielectric and a conductor. (Comptt. Delhi 2012)

Answer: Dielectric: Insulating materials which transmit electric effects without conducting. Conductor: Substances which can carry or conduct electric charge from one place to the other.

Marking Rubrics: 1 mark for dielectric, 1 mark for conductor, 1 mark for distinction.

Explanation: Dielectric aur conductor ka difference. Marks: 1 dielectric, 1 conductor, 1 distinction ke liye.

Question 2: A parallel plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change: (i) electric field between the plates (ii) capacitance, and (iii) energy stored in the capacitor. (Delhi 2013)

Answer: (i) Electric field halves. (ii) Capacitance halves. (iii) Energy doubles.

Marking Rubrics: 1 mark for each part.

Explanation: Distance double karne ka effect. Marks: 1 (i), 1 (ii), 1 (iii) ke liye.

Question 3: Explain briefly how a capacitor stores energy. (Delhi 2013)

Answer: Energy is stored in the electric field between the plates.

Marking Rubrics: 1 mark for explanation, 1 mark for formula, 1 mark for description.

Explanation: Capacitor energy field mein store. Marks: 1 explanation, 1 formula, 1 description ke liye.

Question 4: Derive an expression for the energy stored in a capacitor. (Delhi 2013)

Answer: U = 1/2 CV^2.

Marking Rubrics: 1 mark for derivation, 1 mark for expression, 1 mark for explanation.

Explanation: Energy ka expression derive karo. Marks: 1 derivation, 1 expression, 1 explanation ke liye.

Question 5: A network of four capacitors each of 12 μF capacitance is connected to a 500 V supply as shown in the figure. Determine the equivalent capacitance of the network and the charge on each capacitor. (Delhi 2013)

Answer: Equivalent C = 9 μF, charge on each 3 μF is 1500 μC, on 12 μF is 6000 μC.

Marking Rubrics: 1 mark for equivalent, 1 mark for charge, 1 mark for calculation.

Explanation: Network ka equivalent aur charge. Marks: 1 equivalent, 1 charge, 1 calculation ke liye.

Question 6: A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? (Delhi 2013)

Answer: C' = K C / (K/4 + 3).

Marking Rubrics: 1 mark for expression, 1 mark for derivation, 1 mark for explanation.

Explanation: Dielectric slab ka effect on capacitance. Marks: 1 expression, 1 derivation, 1 explanation ke liye.

Question 7: A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change? (i) charge stored by the capacitor. (ii) Field strength between the plates. (iii) Energy stored by the capacitor. Justify your answer in each case. (All India 2013)

Answer: (i) Charge same. (ii) Field halves. (iii) Energy doubles.

Marking Rubrics: 1 mark for each part.

Explanation: Separation double karne ka effect. Marks: 1 (i), 1 (ii), 1 (iii) ke liye.

Question 8: Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of K = 4. (i) Calculate the potential difference between the plates of X and Y to store the same charge. (ii) What is the ratio of electrostatic energy stored in X and Y? (All India 2013)

Answer: (i) Vx = 4 Vy. (ii) Ratio 4:1.

Marking Rubrics: 1 mark for (i), 1 mark for (ii), 1 mark for calculation.

Explanation: Air aur dielectric ke capacitors compare. Marks: 1 (i), 1 (ii), 1 calculation ke liye.

Question 9: A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. A charge q is placed at the centre of the shell. (a) What is the charge on the inner and outer surface of the shell? (b) Derive the expression for the electric field at a point (i) inside the shell (ii) between the shell (iii) outside the shell. (All India 2013)

Answer: (a) Inner -q, outer Q + q. (b) (i) 0, (ii) q/4π ε0 r^2, (iii) (Q + q)/4π ε0 r^2.

Marking Rubrics: 1 mark for (a), 1 mark for (b i), 1 mark for (ii iii).

Explanation: Spherical shell ka charge and field. Marks: 1 (a), 1 (b), 1 (ii iii) ke liye.

Question 10: Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates. (All India 2013)

Answer: Capacitance C = Q/V, for parallel plate C = ε0 A / d.

Marking Rubrics: 1 mark for definition, 1 mark for expression, 1 mark for derivation.

Explanation: Capacitance ka definition aur expression. Marks: 1 definition, 1 expression, 1 derivation ke liye.

Chapter 3: Current Electricity

Question 1: State Ohm's Law and write its mathematical expression.

Answer 1: Ohm's Law states that the current (I) through a conductor is directly proportional to the voltage (V) across it, provided temperature remains constant. Mathematically, V = IR, where R is resistance.
Explanation (Hinglish): Ohm's Law bolta hai ki agar temperature same rahe, toh current aur voltage ek doosre ke proportional hote hain. Formula V = IR se hum resistance, voltage ya current nikal sakte hain.

Question 2: A wire has a resistance of 10 Ω. If a potential difference of 20 V is applied, what is the current?

Answer 2: 2 A
Explanation (Hinglish): V = IR use karo. V = 20 V, R = 10 Ω, toh I = V/R = 20/10 = 2 A. Yeh simple Ohm's Law ka application hai.

Question 3: Why does a non-ohmic conductor not follow Ohm's Law?

Answer 3: Non-ohmic conductors have non-linear V-I characteristics because their resistance changes with voltage or current.
Explanation (Hinglish): Non-ohmic conductors mein V aur I ka graph straight line nahi hota, kyunki resistance fixed nahi rahta, jaise filament lamp mein heat se resistance badhta hai.

Question 4: What is the slope of the V-I graph for an ohmic conductor?

Answer 4: The slope is resistance (R).
Explanation (Hinglish): V-I graph mein slope = ΔV/ΔI = R hota hai, kyunki V = IR se yeh direct proportion dikhta hai.

Question 5: A 5 Ω resistor is connected to a 10 V supply. Find the power dissipated.

Answer 5: 20 W
Explanation (Hinglish): Power P = V²/R. V = 10 V, R = 5 Ω, toh P = 10²/5 = 100/5 = 20 W. Yeh electrical power ka formula hai.

Question 6: How does temperature affect the resistance of a conductor?

Answer 6: Resistance increases with temperature for conductors.
Explanation (Hinglish): Temperature badhne se metal ke atoms vibrate zyada karte hain, jo electrons ke flow ko resist karta hai, toh resistance badhta hai.

Question 7: What is the unit of resistivity?

Answer 7: Ohm-meter (Ω·m)
Explanation (Hinglish): Resistivity ka formula ρ = RA/l hai, toh unit = (Ω·m²)/m = Ω·m. Yeh material ki property hai jo resistance ko define karta hai.

Question 8: A conductor’s resistance is 2 Ω. If its length is doubled and area halved, find new resistance.

Answer 8: 8 Ω
Explanation (Hinglish): R = ρl/A. Agar l double ho (2l) aur A half ho (A/2), toh new R = ρ(2l)/(A/2) = 4ρl/A = 4R = 4×2 = 8 Ω.

Question 9: Define electrical conductivity.

Answer 9: It is the reciprocal of resistivity, measuring how well a material conducts electricity. Unit: Siemens per meter (S/m).
Explanation (Hinglish): Conductivity σ = 1/ρ hota hai. Yeh batata hai ki material kitna achha current conduct karta hai.

Question 10: Why do we use alloys like manganin in resistors?

Answer 10: Alloys like manganin have low temperature coefficient of resistance.
Explanation (Hinglish): Manganin ka resistance temperature ke saath kam change hota hai, isliye standard resistors mein use hota hai taaki accurate measurements mile.

Question 1: State Kirchhoff’s first rule.

Answer 1: The algebraic sum of currents at a junction is zero.
Explanation (Hinglish): Kirchhoff’s first rule (Junction Rule) kehta hai ki junction par jitna current enter karta hai, utna hi exit karta hai. Yeh charge conservation ka result hai.

Question 2: State Kirchhoff’s second rule.

Answer 2: The algebraic sum of potential differences in a closed loop is zero.
Explanation (Hinglish): Loop Rule kehta hai ki kisi closed loop mein emf aur voltage drops ka sum zero hota hai. Yeh energy conservation se aata hai.

Question 3: In a circuit, a 6 V battery and a 12 V battery are connected in series with a 3 Ω resistor. Find the current.

Answer 3: 6 A
Explanation (Hinglish): Series mein net emf = 6 + 12 = 18 V. Total resistance = 3 Ω. Toh I = V/R = 18/3 = 6 A. Kirchhoff’s loop rule se confirm hota hai.

Question 4: Why is Kirchhoff’s first rule called the junction rule?

Answer 4: It applies at a junction where currents meet.
Explanation (Hinglish): Junction rule isliye kehte hain kyunki yeh junction par currents ke balance ko define karta hai, jaise pani ka flow pipe ke junction par.

Question 5: A 10 V battery with 2 Ω internal resistance is connected to a 8 Ω resistor. Find the terminal voltage using Kirchhoff’s rules.

Answer 5: 8 V
Explanation (Hinglish): Total R = 2 + 8 = 10 Ω. Current I = 10/10 = 1 A. Terminal voltage V = emf - Ir = 10 - 1×2 = 8 V. Loop rule se yeh confirm hota hai.

Question 6: In a parallel circuit with two branches (3 Ω and 6 Ω), find the current ratio using Kirchhoff’s rules.

Answer 6: 2:1
Explanation (Hinglish): Parallel mein voltage same hota hai. Current I ∝ 1/R. Toh I_3Ω / I_6Ω = 6/3 = 2:1. Junction rule se total current divide hota hai.

Question 7: Why do we need Kirchhoff’s rules in complex circuits?

Answer 7: To solve for unknown currents and voltages in multi-loop circuits.
Explanation (Hinglish): Complex circuits mein multiple loops aur junctions hote hain, jahan Ohm’s law akela kaam nahi karta. Kirchhoff’s rules se equations banakar currents aur voltages nikalte hain.

Question 8: A circuit has two batteries (5 V and 10 V) opposing each other with a 5 Ω resistor. Find the current.

Answer 8: 1 A
Explanation (Hinglish): Net emf = 10 - 5 = 5 V. Total R = 5 Ω. Current I = 5/5 = 1 A. Direction higher emf waali battery se hoga, loop rule apply karke.

Question 9: What is conserved in Kirchhoff’s first rule?

Answer 9: Electric charge
Explanation (Hinglish): Junction rule charge conservation par based hai. Jitna charge (current) aata hai, utna hi jaata hai, koi charge loss nahi hota.

Question 10: In a circuit with a junction, 3 A and 2 A enter, what is the outgoing current?

Answer 10: 5 A
Explanation (Hinglish): Junction rule se, incoming currents ka sum = outgoing current. Toh 3 A + 2 A = 5 A outgoing hoga.

Question 1: What is the condition for a Wheatstone bridge to be balanced?

Answer 1: P/Q = R/S
Explanation (Hinglish): Wheatstone bridge balanced hota hai jab ratio of resistances in two arms equal ho, i.e., P/Q = R/S. Isse galvanometer mein no current flow hota hai.

Question 2: In a Wheatstone bridge, P = 2 Ω, Q = 3 Ω, R = 4 Ω. Find S for balance.

Answer 2: 6 Ω
Explanation (Hinglish): Balance condition P/Q = R/S. Toh 2/3 = 4/S. Solve karo, S = 4×3/2 = 6 Ω.

Question 3: What is the use of a Wheatstone bridge?

Answer 3: To measure unknown resistance accurately.
Explanation (Hinglish): Wheatstone bridge unknown resistance ko measure karne ke liye use hota hai jab bridge balanced hoti hai, kyunki yeh precise measurement deta hai.

Question 4: Why is there no current through the galvanometer in a balanced Wheatstone bridge?

Answer 4: Because the potential difference across the galvanometer is zero.
Explanation (Hinglish): Jab P/Q = R/S, dono arms ke potentials equal ho jaate hain, toh galvanometer ke across koi voltage nahi, isliye no current.

Question 5: A Wheatstone bridge has resistances 10 Ω, 15 Ω, 20 Ω, and unknown S. Find S for balance.

Answer 5: 30 Ω
Explanation (Hinglish): P/Q = R/S. Agar P = 10 Ω, Q = 15 Ω, R = 20 Ω, toh S = R×Q/P = 20×15/10 = 30 Ω.

Question 6: What happens if a Wheatstone bridge is not balanced?

Answer 6: Current flows through the galvanometer.
Explanation (Hinglish): Agar bridge balanced nahi hai, toh arms ke potentials equal nahi hote, aur galvanometer mein current flow hota hai.

Question 7: Can a Wheatstone bridge be used with AC supply?

Answer 7: Yes, if resistances are replaced by impedances.
Explanation (Hinglish): AC ke liye resistances ki jagah impedances use karte hain, aur balance condition same rehti hai, lekin AC galvanometer chahiye.

Question 8: In a Wheatstone bridge, if P = Q = R = S, is it balanced?

Answer 8: Yes
Explanation (Hinglish): Jab P = Q = R = S, toh P/Q = R/S (1 = 1). Yeh balance condition satisfy karta hai, toh bridge balanced hai.

Question 9: How is a meter bridge related to a Wheatstone bridge?

Answer 9: A meter bridge is a practical application of the Wheatstone bridge.
Explanation (Hinglish): Meter bridge ek simplified Wheatstone bridge hai, jisme wire ke length ratios se resistance measure karte hain using same balance condition.

Question 10: In a balanced Wheatstone bridge, what is the potential difference across the bridge?

Answer 10: Zero
Explanation (Hinglish): Balanced bridge mein galvanometer ke across potential difference zero hota hai, kyunki dono arms ke potentials equal hote hain.

Chapter 4: Moving Charges and Magnetism

Question 1: What is Oersted's experiment? What does it demonstrate? (CBSE 2018)

Answer 1: Oersted's experiment showed that a current-carrying wire deflects a magnetic needle placed nearby, demonstrating that a current produces a magnetic field.

Question 2: Describe Oersted’s experiment to show that electric current produces a magnetic field. (CBSE 2017)

Answer 2: A wire carrying current placed near a compass causes the needle to deflect, indicating a magnetic field around the wire due to the current.

Question 3: What is the significance of Oersted’s experiment in electromagnetism? (CBSE 2016)

Answer 3: It established that electric currents produce magnetic fields, laying the foundation for electromagnetism.

Question 4: Explain how Oersted’s experiment demonstrates the magnetic effect of current. (CBSE 2015)

Answer 4: A compass needle deflects when placed near a current-carrying conductor, showing that current generates a circular magnetic field.

Question 5: State Oersted’s observation in his experiment with a current-carrying conductor. (CBSE 2014)

Answer 5: Oersted observed that a magnetic needle deflects when placed near a wire carrying current, indicating a magnetic field.

Question 6: What is meant by the magnetic effect of current? (CBSE 2013)

Answer 6: The magnetic effect of current is the production of a magnetic field around a conductor carrying electric current, as shown by Oersted.

Question 7: Describe the principle behind Oersted’s experiment. (CBSE 2012)

Answer 7: A current-carrying wire creates a magnetic field, causing a nearby magnetic needle to deflect, proving current’s magnetic effect.

Question 8: What did Oersted conclude from his experiment? (CBSE 2011)

Answer 8: Oersted concluded that a current-carrying conductor produces a magnetic field perpendicular to the direction of current.

Question 9: How does Oersted’s experiment relate electric current to magnetism? (CBSE 2010)

Answer 9: It shows that an electric current in a wire generates a magnetic field, linking electricity and magnetism.

Question 10: Explain the observation in Oersted’s experiment when current direction is reversed. (CBSE 2019)

Answer 10: Reversing the current reverses the magnetic field direction, causing the compass needle to deflect in the opposite direction.

Question 1: State Biot-Savart law and express it in vector form. (CBSE 2019)

Answer 1: Biot-Savart law: \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \), where \( d\vec{B} \) is the magnetic field due to current element \( I d\vec{l} \), \( \mu_0 \) is permeability, \( r \) is distance, and \( \hat{r} \) is the unit vector.

Question 2: Derive the magnetic field at the centre of a circular coil using Biot-Savart law. (CBSE 2018)

Answer 2: For a coil of radius \( R \), current \( I \), \( dB = \frac{\mu_0 I dl}{4\pi R^2} \), \( \sin\theta = 1 \). Integrating: \( B = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R = \frac{\mu_0 I}{2 R} \).

Question 3: State Biot-Savart law and find the magnetic field at the centre of a circular coil. (CBSE 2017)

Answer 3: Biot-Savart law as above. At centre: \( B = \frac{\mu_0 I}{2 R} \).

Question 4: Derive the magnetic field on the axis of a current-carrying circular loop using Biot-Savart law. (CBSE 2016)

Answer 4: For a point at distance \( x \), \( dB_{\text{axial}} = \frac{\mu_0 I R dl}{4\pi (R^2 + x^2)^{3/2}} \). Integrating: \( B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \).

Question 5: State Biot-Savart law and derive the magnetic field at the centre and on the axis of a circular loop. (CBSE 2015)

Answer 5: Biot-Savart law as above. Centre: \( B = \frac{\mu_0 I}{2 R} \). Axis: \( B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \).

Question 6: Write the Biot-Savart law and the expression for the magnetic field at the centre of a circular loop. (CBSE 2014)

Answer 6: Biot-Savart law: \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \). Centre: \( B = \frac{\mu_0 I}{2 R} \).

Question 7: Why is Biot-Savart law analogous to Coulomb’s law? (CBSE 2013)

Answer 7: Both are inverse-square laws, relating field (magnetic for Biot-Savart, electric for Coulomb) to source (current element or charge).

Question 8: Write the expression for the magnetic field on the axis of a circular loop. (CBSE 2012)

Answer 8: \( B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \).

Question 9: State Biot-Savart law in vector form. (CBSE 2011)

Answer 9: \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \).

Question 10: Derive the magnetic field at the centre of a circular loop using Biot-Savart law. (CBSE 2010)

Answer 10: \( B = \frac{\mu_0 I}{2 R} \) (as derived in Q2).

Question 1: State Ampere’s circuital law and use it to find the magnetic field inside a long solenoid. (CBSE 2019)

Answer 1: Ampere’s law: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \). Inside solenoid: \( B = \mu_0 n I \) (qualitatively, field is uniform).

Question 2: State Ampere’s circuital law and explain the magnetic field inside and outside a solenoid. (CBSE 2018)

Answer 2: Ampere’s law as above. Inside: \( B = \mu_0 n I \). Outside: \( B \approx 0 \) (negligible field).

Question 3: Use Ampere’s circuital law to find the magnetic field due to a straight conductor. (CBSE 2017)

Answer 3: Ampere’s law: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I \). For a wire: \( B (2\pi r) = \mu_0 I \), so \( B = \frac{\mu_0 I}{2\pi r} \).

Question 4: State Ampere’s circuital law. (CBSE 2016)

Answer 4: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \).

Question 5: Explain the magnetic field inside a solenoid using Ampere’s circuital law. (CBSE 2015)

Answer 5: Inside solenoid, field is uniform: \( B = \mu_0 n I \). Outside, field is nearly zero.

Question 6: State Ampere’s circuital law and its significance. (CBSE 2014)

Answer 6: Ampere’s law relates magnetic field to enclosed current. It’s used to find fields in symmetric configurations.

Question 7: Use Ampere’s circuital law to explain the magnetic field due to a straight wire. (CBSE 2013)

Answer 7: \( B = \frac{\mu_0 I}{2\pi r} \) (as derived in Q3).

Question 8: Explain qualitatively the magnetic field inside a solenoid using Ampere’s law. (CBSE 2012)

Answer 8: Field inside solenoid is uniform and strong, \( B = \mu_0 n I \), due to enclosed current in Amperian loop.

Question 9: State Ampere’s circuital law and apply it to a straight wire. (CBSE 2011)

Answer 9: Ampere’s law as above. For wire: \( B = \frac{\mu_0 I}{2\pi r} \).

Question 10: Explain the magnetic field outside a solenoid using Ampere’s law. (CBSE 2010)

Answer 10: Outside, \( I_{\text{enc}} = 0 \), so \( \oint \vec{B} \cdot d\vec{l} = 0 \), implying \( B \approx 0 \).

Question 1: Write the expression for the force on a charge moving in a magnetic field. (CBSE 2019)

Answer 1: Force: \( \vec{F} = q (\vec{v} \times \vec{B}) \), where \( q \) is charge, \( \vec{v} \) is velocity, \( \vec{B} \) is magnetic field.

Question 2: State the expression for the force on a charged particle in combined electric and magnetic fields. (CBSE 2018)

Answer 2: Lorentz force: \( \vec{F} = q \vec{E} + q (\vec{v} \times \vec{B}) \), where \( \vec{E} \) is electric field.

Question 3: What is the force on a charged particle moving perpendicular to a magnetic field? (CBSE 2017)

Answer 3: \( F = q v B \), as \( \sin\theta = 1 \) when velocity is perpendicular to magnetic field.

Question 4: Write the expression for the force on a charge in a magnetic field and state its direction. (CBSE 2016)

Answer 4: \( \vec{F} = q (\vec{v} \times \vec{B}) \). Direction is given by right-hand rule.

Question 5: Why does a charged particle move in a circular path in a magnetic field? (CBSE 2015)

Answer 5: The force \( \vec{F} = q (\vec{v} \times \vec{B}) \) is perpendicular to velocity, providing centripetal force for circular motion.

Question 6: What is the Lorentz force on a charged particle? (CBSE 2014)

Answer 6: \( \vec{F} = q \vec{E} + q (\vec{v} \times \vec{B}) \).

Question 7: Why is no work done by the magnetic force on a moving charge? (CBSE 2013)

Answer 7: Magnetic force is perpendicular to velocity, so \( \vec{F} \cdot \vec{v} = 0 \), hence no work is done.

Question 8: Write the expression for the force on a charged particle in a uniform magnetic field. (CBSE 2012)

Answer 8: \( \vec{F} = q (\vec{v} \times \vec{B}) \).

Question 9: Explain the motion of a charged particle in a magnetic field. (CBSE 2011)

Answer 9: If velocity is perpendicular to \( \vec{B} \), the particle follows a circular path due to centripetal force \( q v B \).

Question 10: What is the effect of an electric field on a moving charge? (CBSE 2010)

Answer 10: Electric field exerts force \( \vec{F} = q \vec{E} \), causing acceleration along the field direction.

Question 1: Write the expression for the force on a current-carrying conductor in a magnetic field. (CBSE 2019)

Answer 1: \( \vec{F} = I (\vec{l} \times \vec{B}) \), where \( I \) is current, \( \vec{l} \) is length vector, \( \vec{B} \) is magnetic field.

Question 2: Derive the force on a current-carrying conductor in a uniform magnetic field. (CBSE 2018)

Answer 2: For a conductor with current \( I \), force on charge \( q \): \( \vec{F} = q (\vec{v} \times \vec{B}) \). Since \( I = n q v A \), total force: \( \vec{F} = I (\vec{l} \times \vec{B}) \).

Question 3: What is the force on a straight conductor carrying current in a magnetic field? (CBSE 2017)

Answer 3: \( F = I l B \sin\theta \), where \( \theta \) is the angle between conductor and field.

Question 4: Explain the direction of force on a current-carrying conductor in a magnetic field. (CBSE 2016)

Answer 4: Direction is given by right-hand rule: fingers along \( \vec{l} \), palm towards \( \vec{B} \), thumb gives force direction.

Question 5: Write the expression for force on a current-carrying conductor and its direction. (CBSE 2015)

Answer 5: \( \vec{F} = I (\vec{l} \times \vec{B}) \). Direction by right-hand rule.

Question 6: Why does a current-carrying conductor experience a force in a magnetic field? (CBSE 2014)

Answer 6: Moving charges in the conductor experience \( \vec{F} = q (\vec{v} \times \vec{B}) \), resulting in net force on the conductor.

Question 7: What is the magnitude of force on a conductor in a magnetic field? (CBSE 2013)

Answer 7: \( F = I l B \sin\theta \).

Question 8: State the principle behind the force on a current-carrying conductor. (CBSE 2012)

Answer 8: Force arises from the interaction of moving charges in the conductor with the magnetic field, given by \( \vec{F} = I (\vec{l} \times \vec{B}) \).

Question 9: Explain why a current-carrying conductor experiences a force in a magnetic field. (CBSE 2011)

Answer 9: Charges moving with drift velocity experience a magnetic force, resulting in net force on the conductor.

Question 10: Write the formula for force on a current-carrying conductor. (CBSE 2010)

Answer 10: \( \vec{F} = I (\vec{l} \times \vec{B}) \).

Question 1: Define one ampere using the force between parallel current-carrying conductors. (CBSE 2019)

Answer 1: One ampere is the current in two parallel conductors 1 m apart in vacuum, producing a force of \( 2 \times 10^{-7} \, \text{N/m} \).

Question 2: Derive the force per unit length between two parallel current-carrying conductors. (CBSE 2018)

Answer 2: Force per unit length: \( \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} \), due to magnetic field of one conductor acting on the other.

Question 3: What is the force per unit length between two parallel conductors? (CBSE 2017)

Answer 3: \( \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} \).

Question 4: How is the ampere defined using parallel conductors? (CBSE 2016)

Answer 4: Ampere is defined as the current causing a force of \( 2 \times 10^{-7} \, \text{N/m} \) between two parallel conductors 1 m apart.

Question 5: Write the expression for force between two parallel current-carrying conductors. (CBSE 2015)

Answer 5: \( \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} \).

Question 6: Explain why parallel current-carrying conductors attract or repel. (CBSE 2014)

Answer 6: Like currents produce opposite fields, causing attraction; unlike currents produce same-direction fields, causing repulsion.

Question 7: What is the nature of force between parallel conductors with currents in the same direction? (CBSE 2013)

Answer 7: Attractive, as magnetic fields create forces pulling conductors together.

Question 8: Derive the expression for force between parallel conductors. (CBSE 2012)

Answer 8: \( \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} \) (as in Q2).

Question 9: Define ampere based on the force between current-carrying conductors. (CBSE 2011)

Answer 9: Same as Q1: \( 2 \times 10^{-7} \, \text{N/m} \) for 1 m apart conductors.

Question 10: Why do parallel conductors with opposite currents repel? (CBSE 2010)

Answer 10: Opposite currents produce magnetic fields in the same direction between conductors, causing repulsion.

Question 1: Derive the expression for torque on a current loop in a magnetic field. (CBSE 2019)

Answer 1: Torque: \( \vec{\tau} = \vec{m} \times \vec{B} \), where \( \vec{m} = I \vec{A} \). For a loop, forces on opposite sides create a couple.

Question 2: Write the expression for torque on a current loop in a uniform magnetic field. (CBSE 2018)

Answer 2: \( \vec{\tau} = \vec{m} \times \vec{B} \), or \( \tau = m B \sin\theta \).

Question 3: Explain why a current loop experiences torque in a magnetic field. (CBSE 2017)

Answer 3: Forces on opposite sides of the loop are equal and opposite but non-collinear, producing a torque.

Question 4: What is the torque on a rectangular current loop in a magnetic field? (CBSE 2016)

Answer 4: \( \tau = I A B \sin\theta \), where \( A \) is area, \( \theta \) is angle between normal to loop and \( \vec{B} \).

Question 5: Derive the torque on a current-carrying loop in a uniform magnetic field. (CBSE 2015)

Answer 5: \( \vec{\tau} = \vec{m} \times \vec{B} \), where \( \vec{m} = I \vec{A} \) (as in Q1).

Question 6: Write the expression for torque on a current loop and its direction. (CBSE 2014)

Answer 6: \( \vec{\tau} = \vec{m} \times \vec{B} \). Direction by right-hand rule.

Question 7: Why does a current loop behave like a magnetic dipole? (CBSE 2013)

Answer 7: It produces a magnetic field similar to a dipole and experiences torque in a magnetic field.

Question 8: What is the maximum torque on a current loop in a magnetic field? (CBSE 2012)

Answer 8: Maximum torque: \( \tau = I A B \), when \( \theta = 90^\circ \).

Question 9: Explain the torque on a current loop in a magnetic field. (CBSE 2011)

Answer 9: Torque arises due to couple formed by forces on opposite sides of the loop.

Question 10: Write the formula for torque on a current loop. (CBSE 2010)

Answer 10: \( \vec{\tau} = \vec{m} \times \vec{B} \).

Question 1: What is the magnetic dipole moment of a current loop? (CBSE 2019)

Answer 1: Magnetic dipole moment: \( \vec{m} = I \vec{A} \), where \( \vec{A} \) is the area vector of the loop.

Question 2: Why is a current loop considered a magnetic dipole? (CBSE 2018)

Answer 2: It produces a magnetic field similar to a bar magnet and experiences torque in a magnetic field.

Question 3: Write the expression for the magnetic moment of a current loop. (CBSE 2017)

Answer 3: \( \vec{m} = I \vec{A} \).

Question 4: Explain how a current loop behaves as a magnetic dipole. (CBSE 2016)

Answer 4: It has a magnetic moment \( \vec{m} = I \vec{A} \) and produces a field like a dipole, with north and south poles.

Question 5: What is the magnetic moment of a rectangular current loop? (CBSE 2015)

Answer 5: \( \vec{m} = I \vec{A} \), where \( \vec{A} \) is area vector (length × breadth).

Question 6: Why does a current loop act like a magnetic dipole? (CBSE 2014)

Answer 6: It generates a magnetic field pattern similar to a bar magnet and responds to external fields like a dipole.

Question 7: Write the formula for the magnetic moment of a current loop. (CBSE 2013)

Answer 7: \( \vec{m} = I \vec{A} \).

Question 8: How is the magnetic moment of a current loop defined? (CBSE 2012)

Answer 8: It is the product of current and area vector: \( \vec{m} = I \vec{A} \).

Question 9: Explain the magnetic dipole behaviour of a current loop. (CBSE 2011)

Answer 9: Current loop produces a magnetic field like a dipole and aligns in an external field due to torque.

Question 10: What is the direction of the magnetic moment of a current loop? (CBSE 2010)

Answer 10: Direction of \( \vec{m} \) is perpendicular to the loop, given by right-hand rule.

Question 1: What is the principle of a moving coil galvanometer? (CBSE 2019)

Answer 1: It works on the principle that a current-carrying coil in a magnetic field experiences torque, causing deflection.

Question 2: Define current sensitivity of a galvanometer. (CBSE 2018)

Answer 2: Current sensitivity: Deflection per unit current, \( \theta/I = \frac{N A B}{k} \), where \( k \) is torsion constant.

Question 3: How is a galvanometer converted into an ammeter? (CBSE 2017)

Answer 3: Connect a low resistance (shunt) in parallel to allow excess current to bypass the coil.

Question 4: How is a galvanometer converted into a voltmeter? (CBSE 2016)

Answer 4: Connect a high resistance in series to limit current and measure voltage across it.

Question 5: What is the current sensitivity of a moving coil galvanometer? (CBSE 2015)

Answer 5: \( \theta/I = \frac{N A B}{k} \), where \( N \) is number of turns, \( A \) is area, \( B \) is field, \( k \) is torsion constant.

Question 6: Explain the working of a moving coil galvanometer. (CBSE 2014)

Answer 6: Current in coil produces torque \( \tau = N I A B \), balanced by spring torque \( k \theta \), causing deflection proportional to current.

Question 7: How does a galvanometer measure current? (CBSE 2013)

Answer 7: Deflection of coil in magnetic field is proportional to current, \( \theta \propto I \).

Question 8: Why is a shunt used in converting a galvanometer to an ammeter? (CBSE 2012)

Answer 8: Shunt reduces current through the galvanometer, allowing it to measure large currents safely.

Question 9: Why is a high resistance used in converting a galvanometer to a voltmeter? (CBSE 2011)

Answer 9: High resistance limits current, enabling the galvanometer to measure voltage.

Question 10: What is the role of the radial magnetic field in a galvanometer? (CBSE 2010)

Answer 10: Radial field ensures torque \( \tau = N I A B \) is constant, making deflection proportional to current.

Chapter 5: Magnetism and Matter

Question 1: What is a bar magnet and its basic properties? (CBSE 2019)

Answer 1: A bar magnet is a ferromagnetic material with permanent north and south poles, producing a magnetic field strongest at poles.

Question 2: Define the magnetic poles of a bar magnet. (CBSE 2018)

Answer 2: Poles are regions at the ends where the magnetic field is strongest; like poles repel, unlike poles attract.

Question 3: What is the magnetic moment of a bar magnet? (CBSE 2017)

Answer 3: Magnetic moment \( \vec{m} = m \cdot l \), where \( m \) is pole strength, \( l \) is length between poles, directed south to north.

Question 4: Why does a bar magnet have a magnetic moment? (CBSE 2016)

Answer 4: Aligned magnetic domains create a net dipole moment, behaving like a magnetic dipole.

Question 5: How does a bar magnet behave in a uniform magnetic field? (CBSE 2015)

Answer 5: Experiences torque to align with the field, but no net force in a uniform field.

Question 6: Why does a bar magnet always have two poles? (CBSE 2014)

Answer 6: Magnetic field lines form closed loops; cutting a magnet creates new poles, ensuring north and south poles.

Question 7: What is pole strength in a bar magnet? (CBSE 2013)

Answer 7: Pole strength is the magnetic force exerted by a pole, proportional to magnetic moment divided by length.

Question 8: Describe the magnetic field of a bar magnet. (CBSE 2012)

Answer 8: Field lines emerge from the north pole, curve, and enter the south pole, strongest near poles.

Question 9: Why does a bar magnet align in the earth’s magnetic field? (CBSE 2011)

Answer 9: Torque \( \vec{\tau} = \vec{m} \times \vec{B} \) aligns the magnetic moment with the earth’s field.

Question 10: What are the characteristics of a bar magnet’s magnetic field? (CBSE 2010)

Answer 10: Strongest at poles, directed from north to south externally, forms closed loops.

Question 1: Why is a bar magnet equivalent to a solenoid? (CBSE 2019)

Answer 1: Both produce similar magnetic fields, with poles at ends, due to aligned magnetic moments.

Question 2: Explain qualitatively the equivalence of a bar magnet to a solenoid. (CBSE 2018)

Answer 2: A solenoid’s current loops create a magnetic field like a bar magnet, with north and south poles at ends.

Question 3: How does a solenoid resemble a bar magnet? (CBSE 2017)

Answer 3: A solenoid’s field lines exit one end (north) and enter the other (south), mimicking a bar magnet.

Question 4: Compare the magnetic field of a bar magnet and a solenoid. (CBSE 2016)

Answer 4: Both have similar field patterns: lines exit north pole, enter south pole, strongest at ends.

Question 5: Why does a solenoid behave like a bar magnet? (CBSE 2015)

Answer 5: Current loops in a solenoid produce a magnetic moment, creating a dipole field like a bar magnet.

Question 6: Describe the magnetic field similarity between a solenoid and a bar magnet. (CBSE 2014)

Answer 6: Both have north and south poles, with field lines exiting one end and entering the other.

Question 7: What is the magnetic moment of a solenoid compared to a bar magnet? (CBSE 2013)

Answer 7: Solenoid’s moment \( \vec{m} = N I A \), similar to bar magnet’s \( \vec{m} = m \cdot l \).

Question 8: Why does a solenoid’s field resemble a bar magnet’s? (CBSE 2012)

Answer 8: Current loops act as magnetic dipoles, producing a field pattern like a bar magnet’s.

Question 9: Explain the equivalence of a solenoid’s magnetic field to a bar magnet. (CBSE 2011)

Answer 9: Solenoid’s field mimics a bar magnet due to cumulative effect of current loops forming a dipole.

Question 10: How is a solenoid’s magnetic behaviour similar to a bar magnet? (CBSE 2010)

Answer 10: Both act as magnetic dipoles with similar field lines and pole-like ends.

Question 1: Describe qualitatively the magnetic field on the axis of a bar magnet. (CBSE 2019)

Answer 1: Field is along the axis, strongest near poles, decreases with distance from the magnet.

Question 2: Explain the magnetic field perpendicular to a bar magnet’s axis. (CBSE 2018)

Answer 2: Field is weaker, opposite to axial field, and decreases rapidly with distance.

Question 3: What is the direction of the magnetic field on a bar magnet’s axis? (CBSE 2017)

Answer 3: Parallel to the axis, directed from north to south outside the magnet.

Question 4: Describe the magnetic field intensity along a magnetic dipole’s axis. (CBSE 2016)

Answer 4: Strongest at poles, decreases with distance, directed along the axis.

Question 5: How does the magnetic field vary perpendicular to a bar magnet’s axis? (CBSE 2015)

Answer 5: Weaker, opposite to axial field, decreases as the cube of distance.

Question 6: Explain the nature of the magnetic field on a bar magnet’s axis. (CBSE 2014)

Answer 6: Field is strong near poles, along the axis, weakens with increasing distance.

Question 7: What is the magnetic field perpendicular to a bar magnet’s axis? (CBSE 2013)

Answer 7: Weaker, directed opposite to axial field, follows dipole pattern.

Question 8: Describe the field intensity of a magnetic dipole on its axis. (CBSE 2012)

Answer 8: Strong near poles, decreases with distance, parallel to the axis.

Question 9: How does the magnetic field behave perpendicular to a dipole’s axis? (CBSE 2011)

Answer 9: Opposite to axial field, weaker, decreases rapidly with distance.

Question 10: What is the qualitative nature of a bar magnet’s axial field? (CBSE 2010)

Answer 10: Strong near poles, along the axis, weakens with distance.

Question 1: Explain qualitatively the torque on a bar magnet in a uniform magnetic field. (CBSE 2019)

Answer 1: Torque \( \vec{\tau} = \vec{m} \times \vec{B} \) aligns the magnet with the field, maximum when perpendicular.

Question 2: Why does a bar magnet experience torque in a magnetic field? (CBSE 2018)

Answer 2: Magnetic moment interacts with the field, creating a couple to align the magnet.

Question 3: Write the expression for torque on a magnetic dipole. (CBSE 2017)

Answer 3: \( \vec{\tau} = \vec{m} \times \vec{B} \), or \( \tau = m B \sin\theta \).

Question 4: Describe the torque on a bar magnet in a uniform field. (CBSE 2016)

Answer 4: Torque aligns the magnet with the field, maximum at \( \theta = 90^\circ \).

Question 5: Why is there no net force on a bar magnet in a uniform field? (CBSE 2015)

Answer 5: Equal and opposite forces on poles cancel, leaving only torque.

Question 6: What is the maximum torque on a magnetic dipole? (CBSE 2014)

Answer 6: Maximum torque: \( \tau = m B \), when \( \theta = 90^\circ \).

Question 7: Explain the direction of torque on a bar magnet. (CBSE 2013)

Answer 7: Direction of \( \vec{\tau} = \vec{m} \times \vec{B} \) is given by right-hand rule.

Question 8: Why does a magnetic dipole experience torque? (CBSE 2012)

Answer 8: Forces on poles form a couple, producing torque to align the dipole.

Question 9: Write the formula for torque on a bar magnet. (CBSE 2011)

Answer 9: \( \vec{\tau} = \vec{m} \times \vec{B} \).

Question 10: Describe the effect of torque on a bar magnet. (CBSE 2010)

Answer 10: Torque rotates the magnet to align with the external field.

Question 1: What are magnetic field lines and their properties? (CBSE 2019)

Answer 1: Imaginary lines showing magnetic field direction; form closed loops, never intersect, denser where field is stronger.

Question 2: Draw the magnetic field lines of a bar magnet. (CBSE 2018)

Answer 2: Lines emerge from north pole, curve, and enter south pole, forming closed loops inside.

Question 3: Why do magnetic field lines form closed loops? (CBSE 2017)

Answer 3: Absence of magnetic monopoles ensures continuous loops starting and ending at poles.

Question 4: What is the direction of magnetic field lines of a bar magnet? (CBSE 2016)

Answer 4: Exit north pole, enter south pole externally, form closed loops.

Question 5: Why don’t magnetic field lines intersect? (CBSE 2015)

Answer 5: Intersection implies two field directions at a point, which is impossible.

Question 6: Describe the properties of magnetic field lines. (CBSE 2014)

Answer 6: Form closed loops, denser at stronger fields, never cross.

Question 7: How do magnetic field lines show field strength? (CBSE 2013)

Answer 7: Closer lines indicate stronger field; wider spacing indicates weaker field.

Question 8: What is the significance of magnetic field lines? (CBSE 2012)

Answer 8: Indicate direction and relative strength of the magnetic field.

Question 9: Draw field lines for a bar magnet and indicate poles. (CBSE 2011)

Answer 9: Lines exit north pole, enter south pole, form closed loops.

Question 10: Why are magnetic field lines continuous? (CBSE 2010)

Answer 10: No magnetic monopoles; field lines form continuous closed loops.

Question 1: Distinguish between paramagnetic and diamagnetic substances with examples. (CBSE 2019)

Answer 1: Paramagnetic: Weakly attracted, \( \chi > 0 \) (e.g., aluminium). Diamagnetic: Weakly repelled, \( \chi < 0 \) (e.g., bismuth).

Question 2: Differentiate between ferromagnetic and paramagnetic substances. (CBSE 2018)

Answer 2: Ferromagnetic: Strong magnetism, permanent (e.g., iron). Paramagnetic: Weak attraction, no permanent magnetism (e.g., oxygen).

Question 3: What are diamagnetic substances? Give an example. (CBSE 2017)

Answer 3: Weakly repelled by magnetic fields, \( \chi < 0 \) (e.g., copper).

Question 4: Define ferromagnetic substances with an example. (CBSE 2016)

Answer 4: Strongly attracted, form permanent magnets (e.g., iron).

Question 5: Name one example each of para-, dia-, and ferromagnetic substances. (CBSE 2015)

Answer 5: Paramagnetic: Aluminium. Diamagnetic: Bismuth. Ferromagnetic: Cobalt.

Question 6: Why are paramagnetic substances attracted to magnetic fields? (CBSE 2014)

Answer 6: Unpaired electrons align with the field, causing weak attraction.

Question 7: What are ferromagnetic materials? Give an example. (CBSE 2013)

Answer 7: Materials with strong magnetism due to domains (e.g., nickel).

Question 8: Explain the behaviour of diamagnetic materials in a magnetic field. (CBSE 2012)

Answer 8: Weakly repelled due to induced moments opposing the external field.

Question 9: Compare paramagnetic and diamagnetic substances. (CBSE 2011)

Answer 9: Paramagnetic: Attracted, \( \chi > 0 \). Diamagnetic: Repelled, \( \chi < 0 \).

Question 10: Why do ferromagnetic materials show permanent magnetism? (CBSE 2010)

Answer 10: Aligned magnetic domains retain magnetism after external field removal.

Question 1: Define magnetization of a material. (CBSE 2019)

Answer 1: Magnetization \( \vec{M} \) is the magnetic moment per unit volume of a material.

Question 2: What is magnetic susceptibility? (CBSE 2018)

Answer 2: Magnetic susceptibility \( \chi = M/H \), ratio of magnetization to magnetic field intensity.

Question 3: Define magnetic susceptibility and its unit. (CBSE 2017)

Answer 3: \( \chi = M/H \), unitless as \( M \) and \( H \) have same units (A/m).

Question 4: Explain magnetization in magnetic materials. (CBSE 2016)

Answer 4: Magnetization is the alignment of atomic magnetic moments in an external field, measured as \( \vec{M} \).

Question 5: What is the relation between magnetization and magnetic field? (CBSE 2015)

Answer 5: \( \vec{M} = \chi \vec{H} \), where \( \chi \) is susceptibility.

Question 6: What is the significance of magnetic susceptibility? (CBSE 2014)

Answer 6: \( \chi = M/H \) indicates how a material responds to an external magnetic field.

Question 7: What is the unit of magnetization? (CBSE 2013)

Answer 7: Magnetization \( \vec{M} \) has unit \( \text{A/m} \) (ampere per meter).

Question 8: How does magnetization occur in materials? (CBSE 2012)

Answer 8: Atomic magnetic moments align with an external magnetic field, inducing magnetization.

Question 9: What is the susceptibility of ferromagnetic materials? (CBSE 2011)

Answer 9: \( \chi \gg 0 \), indicating strong magnetization in the field direction.

Question 10: Define magnetization and its significance. (CBSE 2010)

Answer 10: Magnetization is magnetic moment per unit volume, shows extent of material’s magnetic response.

Question 1: How does temperature affect ferromagnetic materials? (CBSE 2019)

Answer 1: Above Curie temperature, ferromagnetic materials become paramagnetic due to domain disruption.

Question 2: Explain the effect of temperature on paramagnetic materials. (CBSE 2018)

Answer 2: Susceptibility decreases with temperature, \( \chi \propto 1/T \), as per Curie’s law.

Question 3: What is the effect of temperature on diamagnetic materials? (CBSE 2017)

Answer 3: Diamagnetic susceptibility is nearly independent of temperature.

Question 4: How does temperature affect magnetism in ferromagnetic substances? (CBSE 2016)

Answer 4: Magnetism decreases; above Curie point, they become paramagnetic.

Question 5: State Curie’s law for paramagnetic materials. (CBSE 2015)

Answer 5: \( \chi = C/T \), susceptibility inversely proportional to temperature.

Question 6: Why do ferromagnetic materials lose magnetism at high temperatures? (CBSE 2014)

Answer 6: Thermal energy disrupts magnetic domain alignment above Curie temperature.

Question 7: How does temperature affect paramagnetic susceptibility? (CBSE 2013)

Answer 7: Susceptibility decreases inversely with temperature (\( \chi \propto 1/T \)).

Question 8: What happens to ferromagnetic materials above Curie temperature? (CBSE 2012)

Answer 8: They become paramagnetic, losing permanent magnetism.

Question 9: Explain the temperature dependence of diamagnetic materials. (CBSE 2011)

Answer 9: Diamagnetic susceptibility remains nearly constant with temperature.

Question 10: How does temperature affect magnetic properties? (CBSE 2010)

Answer 10: Ferromagnetic: Lose magnetism above Curie point. Paramagnetic: \( \chi \propto 1/T \). Diamagnetic: Unaffected.

Chapter 6: Electromagnetic Induction

Q1: State Faraday’s first and second laws of electromagnetic induction.

Faraday’s First Law: Whenever magnetic flux linked with a circuit changes, an emf is induced in it.
Faraday’s Second Law: The magnitude of induced emf is equal to the rate of change of magnetic flux linked with the circuit.
Mathematically: ε = −dΦ/dt

Q2: Define magnetic flux. Write its SI unit and dimensional formula.

Magnetic flux (Φ) through a surface is defined as Φ = B⋅A = BA cosθ
SI unit: weber (Wb)
1 Wb = 1 T·m² = 1 V·s
Dimensional formula: [ML²T⁻²A⁻¹]

Q3: A coil of area 0.1 m² is placed perpendicular to a uniform magnetic field of 0.2 T. If the field is reduced to zero in 0.1 s, find the induced emf.

Initial flux Φ₁ = BA = 0.2 × 0.1 = 0.02 Wb
Final flux Φ₂ = 0
ΔΦ = 0.02 Wb, Δt = 0.1 s
Induced emf ε = ΔΦ/Δt = 0.02/0.1 = 0.2 V

Q4: Explain Lenz’s law. How does it follow the law of conservation of energy?

Lenz’s law: The induced current flows in such a direction that it opposes the very cause which produces it.
It is based on conservation of energy. If induced current helped the change, energy would be created — which violates conservation of energy.

Q5: A square coil of side 10 cm has 50 turns. It is placed in a magnetic field of 0.5 T perpendicular to its plane. The coil is rotated by 180° in 0.1 s. Calculate the average induced emf.

Initial flux = NBA = 50 × 0.5 × 0.01 = 0.25 Wb
Final flux = −0.25 Wb (after 180° rotation)
Total change in flux = 0.25 − (−0.25) = 0.5 Wb
ε_avg = ΔΦ/Δt = 0.5 / 0.1 = 5 V

Q6: Distinguish between self-inductance and mutual inductance.

Self-inductance: Emf induced in a coil due to change in its own current. ε = −L dI/dt
Mutual inductance: Emf induced in one coil due to change in current in another nearby coil. ε₂ = −M dI₁/dt
Unit of both: henry (H)

Q7: A rod of length 0.5 m moves perpendicular to a magnetic field of 0.8 T with velocity 4 m/s. Calculate the induced emf across its ends.

Induced emf ε = Bℓv (when velocity ⊥ B and rod ⊥ B)
ε = 0.8 × 0.5 × 4 = 1.6 V

Q8: Why is the negative sign included in Faraday’s law (ε = −dΦ/dt)?

The negative sign indicates the direction of induced emf and is according to Lenz’s law. It shows that induced emf opposes the change in magnetic flux.

Q9: A circular coil of radius 5 cm and 100 turns is rotated at 300 rpm in a uniform magnetic field of 0.02 T perpendicular to its axis. Find maximum induced emf.

ω = 300 rpm = 300 × 2π / 60 = 10π rad/s
ε_max = NBAω
= 100 × 0.02 × π(0.05)² × 10π
= 100 × 0.02 × 0.0025π × 10π
= 0.5π² V ≈ 4.93 V

Q10: What is eddy current? Give two applications and two methods to reduce it.

Eddy currents are circulating currents induced in a conductor when it experiences changing magnetic field.
Applications: Induction furnace, electromagnetic damping
Methods to reduce: Lamination of core, using high resistivity material

Q1: State Lenz’s law. How does it follow the law of conservation of energy?

Lenz’s Law: The direction of induced emf (or induced current) is such that it always opposes the change in magnetic flux that causes it.
It is based on conservation of energy. If induced current helped the change, mechanical energy would be created from nothing — which is impossible. Hence it opposes to conserve energy.

Q2: A bar magnet is moved towards a coil with its north pole facing the coil. Predict the direction of induced current and deflection in galvanometer.

Flux is increasing into the coil → induced current will produce magnetic field out of the coil (to oppose increase).
By right-hand thumb rule: Induced current will be anticlockwise (if seen from magnet side).
Galvanometer deflects to the left.

Q3: A bar magnet is pulled away from a coil. What will be the direction of induced current?

Flux is decreasing → induced current will try to increase the flux (oppose decrease).
If north pole was facing coil, induced current will produce north pole towards magnet → clockwise current.

Q4: A metallic ring is placed in a changing magnetic field. Explain the direction of induced current using Lenz’s law.

If B is increasing into the page (×××), ring will produce B out of page (•••) → current will be anticlockwise.
If B is decreasing, current will be clockwise (to maintain original flux).

Q5: Why does a metal sheet experience a braking force when it falls between the poles of a magnet?

As sheet falls, flux changes → eddy currents are induced → these currents oppose the motion (Lenz’s law) → braking/retarding force is produced.

Q6: A coil is connected to a galvanometer. When the north pole of a magnet is pushed into the coil, the galvanometer deflects to the left. What will happen when the magnet is pulled out?

When pushed in (N-pole): deflection left (say anticlockwise current).
When pulled out: flux decreases → induced current will be in opposite direction → galvanometer deflects to the right.

Q7: Justify that Lenz’s law is a consequence of the law of conservation of energy with an example of a magnet falling through a copper tube.

Magnet falling through copper tube induces current → current produces opposing magnetic field → magnet falls slowly.
If induced current helped the motion, magnet would accelerate without external work → energy would be created → violates conservation of energy. Hence Lenz’s law is necessary.

Q8: Why is the negative sign introduced in Faraday’s law (ε = –dΦ/dt)?

The negative sign indicates the direction of induced emf and represents Lenz’s law mathematically. It ensures that induced emf opposes the change in flux.

Q9: A circular loop is placed in a uniform magnetic field. Why is no emf induced when the loop is rotated about its diameter?

When rotated about diameter, magnetic flux through the loop does not change (area vector direction changes symmetrically).
ΔΦ = 0 → induced emf = 0 (even though Lenz’s law applies, there is no change to oppose).

Q10: Give two practical applications of eddy currents and two methods to reduce them.

Applications:
1. Electromagnetic braking (in trains)
2. Induction furnace & metal detectors
Methods to reduce:
1. Laminating the core (thin insulated sheets)
2. Using high-resistivity material

Chapter 7: Alternating Current

Q1: Define the terms: (i) rms value (ii) peak value of alternating current/voltage.

(i) RMS value: I₍rms₎ = I₀/√2 = 0.707 I₀
(ii) Peak value: Maximum value attained by alternating quantity in one cycle (I₀ or V₀).
RMS value is the effective value that produces same heating effect as DC.

Q2: In a pure resistive AC circuit, draw the phasor diagram and waveforms of voltage and current. Write the phase difference.

In pure resistive circuit:
→ Voltage and current are in phase
→ Phase difference φ = 0°
→ Power factor = cosφ = 1 (maximum power dissipation)

Q3: In a purely inductive AC circuit, what is the phase difference between voltage and current? Draw phasor diagram.

Voltage leads the current by 90° (or current lags voltage by 90°)
V = V₀ sinωt, I = I₀ sin(ωt – 90°)
Average power consumed = 0 (energy stored and returned every cycle)

Q4: In a purely capacitive circuit, what is the phase relationship between voltage and current?

Current leads the voltage by 90° (or voltage lags current by 90°)
I = I₀ sinωt, V = V₀ sin(ωt – 90°)
Average power = 0

Q5: An AC voltage V = V₀ sinωt is applied to a pure inductor L. Obtain expression for current and show that current lags voltage by π/2.

XL = ωL
I₀ = V₀ / XL = V₀ / ωL
I = I₀ sin(ωt – π/2)
→ Current lags voltage by 90° (proven using phasor or derivative method)

Q6: Define impedance (Z) and write its formula in series LCR circuit.

Impedance (Z) is the effective opposition offered by LCR circuit to the flow of AC.
Z = √[R² + (Xₗ – Xc)²]
Xₗ = ωL, Xc = 1/ωC

Q7: What is resonance in series LCR circuit? Write the condition and expression for resonant frequency.

Resonance occurs when Xₗ = Xc → impedance is minimum (= R) → current is maximum.
Condition: ωL = 1/ωC
Resonant frequency: f₀ = 1/(2π√(LC))
At resonance, circuit behaves as purely resistive.

Q8: Define power factor. Write its value for (i) pure R (ii) pure L (iii) pure C circuit.

Power factor = cosφ = Real power / Apparent power = R/Z
(i) Pure resistor → cosφ = 1
(ii) Pure inductor → cosφ = 0
(iii) Pure capacitor → cosφ = 0

Q9: An AC source of 200 V, 50 Hz is connected to a series combination of R = 100 Ω, L = 0.5 H, C = 20 μF. Will the current lead or lag the voltage? (Calculate phase angle)

ω = 2π×50 = 314 rad/s
Xₗ = ωL = 314×0.5 = 157 Ω
Xc = 1/ωC = 1/(314×20×10⁻⁶) ≈ 159.2 Ω
Xc > Xₗ → capacitive circuit → current leads voltage
tanφ = (Xc – Xₗ)/R ≈ 2.2/100 → φ ≈ 1.26° (lead)

Q10: Draw the graph of variation of (i) inductive reactance (ii) capacitive reactance with frequency.

Xₗ = ωL → straight line passing through origin (increases with f)
Xc = 1/ωC → rectangular hyperbola (decreases with f)
They intersect at resonance frequency f₀.

Q1: State the working principle of a transformer. Define transformation ratio.

Principle: Transformer works on the principle of mutual induction. When alternating current flows in primary coil, changing magnetic flux is produced which links with secondary coil and induces emf in it.
Transformation ratio (K): K = Nₛ/Nₚ = Vₛ/Vₚ = Iₚ/Iₛ

Q2: Why is the core of a transformer laminated?

Core is laminated to reduce eddy current losses. Thin insulated laminations increase resistance in the path of eddy currents → eddy current becomes very small → heat loss (I²R) is minimised.

Q3: A transformer has 100 turns in primary and 400 turns in secondary. Input voltage is 220 V. Find output voltage if it is (i) step-up (ii) step-down transformer.

K = Nₛ/Nₚ = 400/100 = 4
(i) Step-up → Vₛ = K Vₚ = 4 × 220 = 880 V
(ii) Step-down → Vₛ = (100/400) × 220 = 55 V

Q4: Write any three energy losses in a transformer and how they are minimised.

1. Copper loss (I²R) → minimised by using thick copper wires
2. Iron/Hysteresis loss → minimised by using soft iron core with narrow hysteresis loop
3. Eddy current loss → minimised by laminated core

Q5: The efficiency of a transformer is 96%. Input power is 5000 W. Find the output power.

Efficiency η = (Output power / Input power) × 100
96 = (Pₒᵤₜ / 5000) × 100
Pₒᵤₜ = 5000 × 0.96 = 4800 W

Q6: Why can a transformer not be used with direct current (DC)?

Transformer works on mutual induction which requires changing magnetic flux. DC produces constant flux → no induced emf in secondary → transformer does not work with DC.

Q7: A transformer has 500 turns in primary and 100 turns in secondary. If primary is connected to 220 V AC, find secondary voltage and type of transformer.

K = Nₛ/Nₚ = 100/500 = 1/5
Vₛ = K × Vₚ = (1/5) × 220 = 44 V
Since Vₛ < Vₚ → it is a step-down transformer

Q8: An ideal transformer has voltage ratio 11,000 V / 220 V. If primary current is 2 A, find secondary current.

For ideal transformer: Vₚ Iₚ = Vₛ Iₛ
11,000 × 2 = 220 × Iₛ
Iₛ = (11,000 × 2) / 220 = 100 A

Q9: Why is soft iron used as core material in a transformer?

Soft iron has:
1. High permeability → strong magnetic field
2. Low coercivity & narrow hysteresis loop → low hysteresis loss
3. Low eddy current loss when laminated

Q10: A transformer is rated as 11 kV/220 V, 50 Hz, 5 kVA. What does this rating signify?

→ Primary voltage = 11 kV, Secondary voltage = 220 V
→ Frequency = 50 Hz
→ Maximum apparent power it can handle safely = 5 kVA
It is a step-down transformer used in power distribution.

Chapter 8: Electromagnetic Waves

Q1: State two main properties of electromagnetic waves.

1. They are transverse in nature.
2. They do not require any material medium for propagation.
3. They travel with speed c = 3×10⁸ m/s in vacuum.
4. They follow the relation: c = fλ

Q2: Arrange the following in increasing order of frequency: Radio waves, Microwaves, Infrared, Visible light, UV rays, X-rays, Gamma rays.

Radio waves < Microwaves < Infrared < Visible light < UV rays < X-rays < Gamma rays

Q3: The speed of EM waves in vacuum is 3×10⁸ m/s. Calculate the wavelength of a wave having frequency 6×10¹⁴ Hz.

c = fλ
λ = c/f = 3×10⁸ / 6×10¹⁴ = 5×10⁻⁷ m = 500 nm
→ This lies in visible region (violet)

Q4: Write Maxwell’s four equations that led to prediction of electromagnetic waves (in words or symbols).

1. ∮E·dl = Q/ε₀ → Gauss’s law for electricity
2. ∮B·dl = μ₀I + μ₀ε₀ dΦ_E/dt → Ampere-Maxwell law (with displacement current)
3. ∮E·dA = 0 → Gauss’s law for magnetism
4. ∮B·dA = –dΦ_B/dt → Faraday’s law

Q5: What is displacement current? Why was it introduced by Maxwell?

Displacement current: I_d = ε₀ dΦ_E/dt
Maxwell introduced it to make Ampere’s law consistent between the plates of a charging capacitor where conduction current is zero, but changing electric field produces magnetic field → necessary for prediction of EM waves.

Q6: In EM waves, what is the phase difference between electric field (E) and magnetic field (B)?

Electric field (E) and magnetic field (B) are in same phase.
Both reach maximum and zero at the same time and same place.

Q7: The amplitude of electric field in an EM wave is 100 V/m. Find the amplitude of magnetic field.

E₀/B₀ = c
B₀ = E₀/c = 100 / (3×10⁸) = 3.33×10⁻⁷ T = 0.333 μT

Q8: Name the part of EM spectrum used in: (i) radar (ii) treating muscular pain (iii) eye surgery (iv) preserving food.

(i) Radar → Microwaves
(ii) Muscular pain → Infrared
(iii) Eye surgery → UV rays
(iv) Preserving food → Gamma rays / UV

Q9: Why are microwaves used in radar systems?

1. They travel in straight line (line of sight)
2. They can be focused into narrow beams using small antennas
3. They are reflected by metallic surfaces
4. Short wavelength → high resolution

Q10: An EM wave is travelling in vacuum. Write the relation between its wave number (k), angular frequency (ω) and speed (c).

Speed of EM wave: c = ω/k
Also, c = fλ → ω = ck
And c = 1/√(μ₀ε₀) → proved by Maxwell

Question 1: Electromagnetic Spectrum - Insert question here

Q1: Write the increasing order of wavelength in electromagnetic spectrum.

Gamma rays < X-rays < Ultraviolet < Visible < Infrared < Microwaves < Radio waves

Q2: Name the electromagnetic waves used in: (i) Remote sensing (ii) Photography in fog (iii) Destroying cancer cells (iv) Greenhouse effect

(i) Remote sensing → Microwaves
(ii) Photography in fog → Infrared
(iii) Destroying cancer cells → Gamma rays
(iv) Greenhouse effect → Infrared

Q3: Which part of the electromagnetic spectrum is absorbed by ozone layer? Write its wavelength range.

Ultraviolet (UV) rays are absorbed by ozone layer.
Wavelength range: 200 nm – 400 nm (especially UV-B: 280–315 nm)

Q4: Name the EM waves having wavelength range: (i) 400 nm – 700 nm (ii) 10⁻³ m to 1 m (iii) 10⁻¹² m to 10⁻⁸ m

(i) 400 nm – 700 nm → Visible light
(ii) 1 mm to 1 m → Microwaves
(iii) 10⁻¹² m to 10⁻⁸ m → X-rays & Gamma rays

Q5: Why is the ionosphere important for long-distance radio communication?

Ionosphere reflects radio waves (frequency < 30 MHz) back to Earth → enables long-distance communication by sky wave propagation.

Q6: Give one use each of: (i) Infrared rays (ii) UV rays (iii) Microwaves (iv) Gamma rays

(i) Infrared → TV remote, thermal imaging
(ii) UV → Sterilisation, vitamin D formation
(iii) Microwaves → Radar, microwave oven
(iv) Gamma rays → Cancer treatment, food preservation

Q7: Which part of EM spectrum has highest frequency and highest penetrating power?

Highest frequency → Gamma rays
Highest penetrating power → Gamma rays (can pass through thick concrete and lead)

Q8: Why can radio waves be transmitted to long distances but not ultraviolet rays?

Radio waves have long wavelength → low frequency → diffract around obstacles and reflect from ionosphere.
UV rays have short wavelength → absorbed by atmosphere (ozone layer) → cannot travel far.

Q9: Arrange the following in decreasing order of wavelength: Visible light, X-rays, Microwaves, Infrared

Microwaves > Infrared > Visible light > X-rays

Q10: Match the following: A. 10⁻¹⁰ m → 1. Gamma rays B. 500 nm → 2. Visible light C. 1 cm → 3. Microwaves D. 300 nm → 4. Ultraviolet

A → 1 (Gamma rays)
B → 2 (Visible light)
C → 3 (Microwaves)
D → 4 (Ultraviolet)

Chapter 9: Ray Optics and Optical Instruments

Question 1: Reflection and Refraction - Insert question here

Q1: State the two laws of reflection of light.

1. Angle of incidence = Angle of reflection (i = r)
2. Incident ray, reflected ray and normal lie in the same plane.

Q2: Define refractive index. Write the relation between refractive index and speed of light.

Refractive index (μ) of a medium = Speed of light in vacuum (c) / Speed of light in medium (v)
μ = c/v
Also, μ = sin i / sin r (Snell’s law)

Q3: State Snell’s law of refraction. A ray of light passes from air to glass (μ = 1.5) at an angle of incidence 30°. Find the angle of refraction.

Snell’s law: μ₁ sin i = μ₂ sin r
1 × sin 30° = 1.5 × sin r
sin r = 0.5 / 1.5 = 1/3 ≈ 0.333
r = sin⁻¹(0.333) ≈ 19.5°

Q4: What is total internal reflection? Write its two essential conditions.

When light goes from denser to rarer medium and angle of incidence > critical angle, light is completely reflected back into the denser medium.
Conditions:
1. Light must travel from denser to rarer medium
2. Angle of incidence > critical angle (i > i_c)

Q5: Define critical angle. Derive the relation between critical angle and refractive index.

Critical angle (i_c) is the angle of incidence in denser medium for which angle of refraction in rarer medium is 90°.
At i = i_c, r = 90°
μ = sin i_c / sin 90° ⇒ sin i_c = 1/μ

Q6: A glass slab (μ = 1.5) is placed in air. Find the critical angle for glass-air interface.

sin i_c = 1/μ = 1/1.5 = 2/3 ≈ 0.666
i_c = sin⁻¹(0.666) ≈ 41.8°

Q7: Give any three applications of total internal reflection.

1. Optical fibres (telecommunication, endoscopy)
2. Mirage formation
3. Brilliance of diamond (multiple TIR inside)

Q8: A ray of light travelling in water (μ = 4/3) is incident on water-air interface at 45°. Will it undergo total internal reflection? Justify.

Critical angle for water-air: sin i_c = 1/(4/3) = 3/4 = 0.75
i_c ≈ 48.8°
Given i = 45° < 48.8° → No total internal reflection, light will be refracted into air.

Q9: Why does a swimmer inside water appear nearer than actual depth?

Due to refraction at water-air interface, rays bend away from normal → apparent depth < real depth.
Apparent depth = Real depth / μ
For water μ = 4/3 → apparent depth = 3/4 of real depth.

Q10: Draw ray diagram to show the refraction of light through a glass slab and prove that lateral shift occurs.

Lateral shift (d) = t × sin(i − r) / cos r
Where t = thickness of slab
Emergent ray is parallel to incident ray but laterally shifted.

Q1: Define angular magnification. Write its expression for a simple microscope in normal adjustment.

Angular magnification (m) = Angle subtended by image at eye / Angle subtended by object at eye
For simple microscope (normal adjustment):
m = D/fₑ + 1 (D = least distance of distinct vision = 25 cm)

Q2: A compound microscope has objective and eyepiece of focal lengths 1 cm and 5 cm respectively. Distance between them is 20 cm. Find magnifying power when final image is at 25 cm.

vₒ = ? uₒ = ? fₒ = 1 cm, fₑ = 5 cm, D = 25 cm
For objective: 1/vₒ – 1/uₒ = 1/1 → vₒ – uₒ = 1
vₒ + 5 = 20 → vₒ = 15 cm → uₒ = 14 cm
m = (vₒ/uₒ) × (D/fₑ + 1) = (15/14) × (25/5 + 1) = (15/14) × 6 ≈ 6.43

Q3: Write any three differences between compound microscope and astronomical telescope (in normal adjustment).

Compound Microscope	Astronomical Telescope
Object beyond fₒ	Object at infinity
Final image inverted, magnified	Final image inverted, at infinity
m = (vₒ/uₒ)(D/fₑ + 1)	m = fₒ/fₑ

Q4: Define resolving power of a microscope and telescope. On what factors does it depend?

Microscope: Ability to show two close objects as separate.
R.P. = 2μ sinβ / λ ∝ 1/λ
Telescope: Ability to show two close stars as separate.
R.P. = D / (1.22 λ) ∝ D (diameter of objective)

Q5: The focal lengths of objective and eyepiece of an astronomical telescope are 100 cm and 5 cm. Find magnifying power and length of tube in normal adjustment.

Normal adjustment: m = fₒ/fₑ = 100/5 = 20
Length of tube L = fₒ + fₑ = 100 + 5 = 105 cm

Q6: Why is the objective of a telescope of large focal length and large aperture?

Large focal length → high magnifying power (m = fₒ/fₑ)
Large aperture (diameter) → more light gathering power → brighter image + higher resolving power

Q7: A person with normal near point (25 cm) uses a compound microscope with objective of fₒ = 0.8 cm and eyepiece of fₑ = 2.5 cm. Find maximum magnifying power.

For maximum m (image at D):
m = (L/fₒ) × (D/fₑ + 1)
Usually L ≈ 16–18 cm (take 17 cm)
m ≈ (17/0.8) × (25/2.5 + 1) = 21.25 × 11 = approximately 234

Q8: What is the advantage of reflecting telescope over refracting telescope?

1. No chromatic aberration (mirrors reflect all colours equally)
2. No spherical aberration if parabolic mirror is used
3. Large aperture possible → higher resolving power and brighter image

Q9: Why is eyepiece of a telescope made of short focal length?

Magnifying power of telescope m = fₒ/fₑ
To get large magnification, fₑ should be small.

Q10: Name the optical instrument used by doctors to see inside the stomach. How does it work?

Endoscope (or Fibroscope)
Works on principle of total internal reflection in optical fibres.
One bundle carries light inside, another bundle brings reflected image out.

Chapter 10: Wave Optics

Q1: State the two conditions for sustained interference of light.

1. Two sources must be coherent (constant phase difference).
2. Sources should be monochromatic or of same wavelength.

Q2: What is meant by constructive and destructive interference? Give the condition of path difference in both cases.

Constructive: Bright fringe → path difference = nλ (n = 0,1,2…)
Destructive: Dark fringe → path difference = (2n–1)λ/2

Q3: In Young’s double-slit experiment, fringes are obtained using light of wavelength 600 nm. If the slit separation is 0.3 mm and screen is 1.5 m away, find fringe width (β).

β = λD / d
β = (600×10⁻⁹ × 1.5) / (0.3×10⁻³) = 3×10⁻³ m = 3 mm

Q4: In YDSE, the distance between two slits is 1 mm and distance of screen is 2 m. If wavelength is 600 nm, find the distance of 3rd bright fringe from central fringe.

For nth bright fringe: xₙ = nλD / d
n = 3 → x₃ = 3 × 600×10⁻⁹ × 2 / (1×10⁻³) = 3.6×10⁻³ m = 3.6 mm

Q5: In YDSE, what happens to fringe width if: (i) wavelength is doubled (ii) entire setup is immersed in water (μ=4/3)

β ∝ λ and β ∝ 1/μ
(i) λ becomes 2λ → β becomes double
(ii) In water → β becomes β/(4/3) = (3/4) times

Q6: In Young’s experiment, monochromatic light is replaced by white light. What will be observed on the screen?

Central fringe will be white (path difference zero for all λ).
Few coloured fringes on both sides, then uniform illumination (because different wavelengths produce overlapping fringes).

Q7: In YDSE, one slit is covered with a thin transparent sheet of refractive index 1.5 and thickness 3 μm. If λ = 600 nm, find the number of fringes that will shift.

Shift = (μ–1)t D / d
Number of fringes shifted = shift / β = (μ–1)t / λ
= (1.5–1)×3×10⁻⁶ / 600×10⁻⁹ = 0.5×10⁻⁶ / 600×10⁻⁹ = 5 fringes

Q8: Why are coherent sources necessary to produce sustained interference pattern?

Coherent sources have constant phase difference → intensity remains maximum or minimum at fixed points → sustained (stable) interference pattern is obtained.
Incoherent sources → phase difference changes randomly → fringes disappear.

Q9: In YDSE, the screen is moved away from the slits by 50 cm. What happens to fringe width?

β = λD / d
D increases → fringe width increases (directly proportional to D)

Q10: What is the shape of interference fringes in Young’s double-slit experiment using a point source and cylindrical wavefront?

When slits are illuminated by a point source → wavefront is cylindrical → interference fringes are hyperbolic.
But if source is far or lens is used → plane wavefront → fringes are straight.

Q1: Define diffraction of light. Give one example from daily life.

Bending of light around the corners of an obstacle or aperture is called diffraction.
Example: We can hear sound even when the speaker is behind a wall (sound diffracts), but cannot see it clearly (light diffracts very little due to small λ).

Q2: State the condition for central maximum in single-slit diffraction pattern.

Path difference between rays from ends of slit = 0 → central maximum on the axis.
Occurs when θ = 0 → sinθ = 0 → central bright fringe

Q3: In single-slit diffraction, derive the condition for first minimum (dark fringe).

For first minimum: path difference = λ
a sinθ = λ (a = slit width)
→ sinθ = λ/a (first dark fringe on either side)

Q4: In Fraunhofer diffraction using single slit of width 0.2 mm, wavelength 600 nm, find angular position of first minimum if screen is at large distance.

a sinθ = λ
sinθ ≈ θ (small angle)
θ = λ/a = 600×10⁻⁹ / 0.2×10⁻³ = 3×10⁻³ rad
θ ≈ 0.172°

Q5: Write the expression for angular width of central maximum in single-slit diffraction. On what factors does it depend?

Angular width of central maximum = 2θ = 2λ/a
Depends on:
1. Wavelength λ (directly proportional)
2. Slit width a (inversely proportional)

Q6: In single-slit diffraction experiment, if slit width is doubled, what happens to the width of central maximum?

Width of central maximum ∝ 1/a
If a becomes 2a → width becomes half

Q7: Why is diffraction not observed with a wide slit or door, but bending of sound is easily observed?

Diffraction is prominent when slit size ≈ wavelength.
λ of light ≈ 500 nm → very small → wide slit (mm) → no noticeable diffraction.
λ of sound ≈ metres → comparable to door size → diffraction easily observed.

Q8: In single-slit diffraction, the intensity of central maximum is I₀. Find the intensity at a point where path difference is λ.

Path difference = λ → first minimum → intensity = zero
(Complete destructive interference between half portions of the slit)

Q9: Distinguish between Fresnel and Fraunhofer diffraction (any three points).

Fresnel	Fraunhofer
Source & screen at finite distance	Source & screen at infinity
Curved wavefronts	Plane wavefronts
No lens required	Lenses used

Q10: In single-slit diffraction, red light is replaced by violet light. What will be the effect on the diffraction pattern?

λ_violet < λ_red
Width of central maximum ∝ λ
→ Central maximum becomes narrower
→ Fringes become closer

Chapter 11: Dual Nature of Radiation and Matter

Q1: Define the terms: (i) Threshold frequency (ii) Stopping potential (iii) Work function.

(i) Threshold frequency (ν₀): Minimum frequency of incident radiation required to eject electrons.
(ii) Stopping potential (V₀): Minimum negative potential given to collector to stop fastest photoelectron.
(iii) Work function (φ): Minimum energy required to eject an electron from metal surface. φ = hν₀

Q2: Write Einstein’s photoelectric equation and explain each term.

hν = φ + Kₘₐₓ
hν = Incident photon energy
φ = Work function
Kₘₐₓ = ½ mvₘₐₓ² = Maximum kinetic energy of photoelectron
Also, eV₀ = hν – φ → eV₀ = h(ν – ν₀)

Q3: The work function of caesium is 2.14 eV. Find (i) threshold frequency (ii) threshold wavelength.

(i) φ = hν₀ → ν₀ = φ/h = (2.14 × 1.6×10⁻¹⁹)/(6.626×10⁻³⁴) ≈ 5.16×10¹⁴ Hz
(ii) λ₀ = c/ν₀ = 3×10⁸ / 5.16×10¹⁴ ≈ 581 nm

Q4: State any three experimental observations of photoelectric effect that cannot be explained by wave theory.

1. Existence of threshold frequency
2. Photoelectric emission is instantaneous
3. Maximum KE of photoelectrons depends only on frequency, not intensity
4. Intensity affects only number of electrons, not their KE

Q5: Light of frequency 1.5 times threshold frequency falls on a metal surface. If stopping potential is 2 V, find work function in eV.

Given: ν = 1.5 ν₀
eV₀ = h(ν – ν₀) → 2 eV = h(1.5ν₀ – ν₀) = h(0.5ν₀)
hν₀ = φ → φ = 2 / 0.5 = 4 eV

Q6: Draw graphs showing variation of (i) stopping potential with frequency (ii) photoelectric current with intensity.

(i) V₀ vs ν → straight line with slope h/e, intercepts ν₀ on x-axis
(ii) I vs Intensity → straight line passing through origin (saturation current ∝ intensity)

Q7: The stopping potential for a metal is 3 V when light of frequency 8×10¹⁴ Hz is incident. Find the maximum kinetic energy of photoelectrons in eV.

Kₘₐₓ = eV₀ = 1 × 3 = 3 eV
(1 eV = 1.6×10⁻¹⁹ J, but directly in eV: Kₘₐₓ = V₀ in volts)

Q8: Why is photoelectric effect not observed below threshold frequency even with very high intensity?

Below ν₀, energy of each photon hν < φ → no single photon has enough energy to eject an electron, no matter how many photons (intensity) strike the surface.

Q9: The threshold wavelength for a metal is 500 nm. Find the maximum kinetic energy of photoelectrons when light of wavelength 400 nm is incident.

φ = hc/λ₀ = hc/500 nm
Kₘₐₓ = hc/λ – hc/λ₀ = hc (1/400 – 1/500) × 10⁹
= 1240 eV·nm (1/400 – 1/500) = 1240 (0.0025 – 0.002) = 1240 × 0.0005 = 0.62 eV

Q10: Name the scientist who explained photoelectric effect using quantum theory. Write the name of the particle nature of light he proposed.

Albert Einstein explained photoelectric effect in 1905 using quantum theory.
He proposed that light consists of discrete packets of energy called photons.

Q1: State de-Broglie hypothesis. Write the expression for de-Broglie wavelength.

de-Broglie hypothesis: Every moving particle has wave nature associated with it.
de-Broglie wavelength: λ = h/p = h/(mv)
h = Planck’s constant = 6.626 × 10⁻³⁴ J s

Q2: Calculate de-Broglie wavelength of an electron accelerated through 100 V.

KE = eV = 100 eV = 100 × 1.6×10⁻¹⁹ J
p = √(2mE) → λ = h/√(2m eV)
λ = 1.227 / √V nm = 1.227 / √100 = 0.1227 nm (or 1.227 Å)

Q3: An electron and a proton have same kinetic energy. Which has greater de-Broglie wavelength? Why?

λ = h/√(2m KE)
KE same → λ ∝ 1/√m
mₑ < mₚ → electron has greater de-Broglie wavelength

Q4: An electron and a photon have same de-Broglie wavelength. Compare their kinetic energies.

For electron: λ = h/(mv) → p = h/λ → KEₑ = p²/2mₑ
For photon: λ = h/p → p = h/λ (same p)
KEₚ = pc = (h/λ)c
KEₑ = (h/λ)² / (2mₑ) → KEₚ >> KEₑ

Q5: Name the experiment which confirmed wave nature of electron. Who performed it?

Davisson-Germer experiment (1927)
Performed by Clinton Davisson and Lester Germer
They observed diffraction pattern when electrons were scattered from nickel crystal → proved wave nature of electrons.

Q6: Find de-Broglie wavelength of a neutron having kinetic energy 0.025 eV (thermal neutron).

Use λ (in Å) = 0.286 / √E (E in eV)
λ = 0.286 / √0.025 = 0.286 / 0.158 ≈ 1.81 Å
(Comparable to interatomic spacing → used in neutron diffraction)

Q7: Why is de-Broglie wavelength of a cricket ball negligible?

Mass of cricket ball ≈ 160 g = 0.16 kg
Even at 100 km/h (≈28 m/s), p = mv ≈ 4.48 kg m/s
λ = h/p ≈ 6.6×10⁻³⁴ / 4.48 ≈ 10⁻³⁴ m → negligible (macroscopic objects show no wave nature)

Q8: In Davisson-Germer experiment, maximum electron diffraction intensity was observed at 50 V. Calculate the wavelength of electrons.

λ = 1.227 / √V nm
λ = 1.227 / √50 = 1.227 / 7.07 ≈ 0.1735 nm = 1.735 Å
(Matched with theoretical value → confirmed de-Broglie hypothesis)

Q9: An α-particle and a proton are accelerated through same potential difference. Find the ratio of their de-Broglie wavelengths.

KE same → p = √(2m KE) → λ = h/p ∝ 1/√m
mₐ = 4 mₚ → λₐ/λₚ = √(mₚ/mₐ) = √(1/4) = 1:2

Q10: Write any three applications of electron microscope over optical microscope.

1. Resolving power ≈ 1000 times better (λₑ ≈ 10⁻¹² m)
2. Magnification up to 10⁶ times
3. Can resolve atomic structure (viruses, DNA, etc.)

Chapter 12: Atoms

Q1: State any three postulates of Bohr’s atomic model.

1. Electrons revolve only in certain discrete orbits called stationary orbits without radiating energy.
2. Angular momentum is quantised: mvr = n h/(2π) (n = 1,2,3…)
3. When electron jumps from higher to lower orbit, it emits a photon: hν = E₂ – E₁

Q2: Write the expressions for (i) radius of nth orbit (ii) energy of electron in nth orbit of hydrogen atom.

(i) rₙ = 0.529 × n² Å (or rₙ = n² a₀, a₀ = 0.529 Å)
(ii) Eₙ = –13.6 / n² eV
For ground state (n=1): r = 0.529 Å, E = –13.6 eV

Q3: Find the ratio of radius and velocity of electron in first two orbits of hydrogen atom.

r ∝ n² → r₂/r₁ = 4/1 = 4 : 1
v ∝ 1/n → v₂/v₁ = 1/2 = 1 : 2

Q4: The ground state energy of hydrogen atom is –13.6 eV. Find the energy of electron in second excited state.

Second excited state → n = 3
E₃ = –13.6 / 3² = –13.6 / 9 = –1.51 eV

Q5: Calculate the wavelength of radiation emitted when electron jumps from n=4 to n=2 in hydrogen atom.

1/λ = R (1/2² – 1/4²) = R (1/4 – 1/16) = R (4/16) = R/4
λ = 4/R = 4 / (1.097×10⁷) ≈ 486 nm (Balmer series, blue line)

Q6: Using Bohr’s model, derive the expression for radius of nth orbit.

Balancing forces: mv²/r = kZe²/r²
mv² = kZe²/r
Angular momentum: mvr = nh/(2π)
→ r = n² h²/(4π² m k Ze²) → rₙ ∝ n² → rₙ = n² × 0.529 Å

Q7: Find the ratio of kinetic energy, potential energy and total energy of electron in any orbit.

KE = +13.6/n² eV
PE = –27.2/n² eV
Total E = –13.6/n² eV
Ratio → KE : |PE| : |E| = 1 : 2 : 1

Q8: An electron in hydrogen atom jumps from n=3 to n=1. Find the energy released and series name.

E₃ = –1.51 eV, E₁ = –13.6 eV
Energy released = 13.6 – 1.51 = 12.09 eV
Transition to n=1 → Lyman series (UV region)

Q9: Using Bohr’s postulate of quantization, show that radius is proportional to n².

From mvr = nh/(2π) → v = nh/(2πmr)
Centripetal force = kZe²/r² = mv²/r
→ kZe²/r² = m [nh/(2πmr)]² / r
Solving → r ∝ n²

Q10: Name the series of hydrogen spectrum lying in visible region. Write its wavelength limits.

Balmer series (n₂ = 2, n₁ = 3,4,5…)
Wavelength range: 400 nm to 700 nm
First line (Hα) = 656.3 nm, Last line = 364.6 nm

Chapter 13: Nuclei

Q1: Define atomic mass unit (amu). Find its energy equivalent in MeV.

1 amu = 1/12th mass of carbon-12 atom
= 1.66054 × 10⁻²⁷ kg
Energy equivalent: 1 amu = 931.5 MeV (or ≈ 931 MeV)

Q2: Write the relation between mass number (A), atomic number (Z) and neutron number (N).

Mass number A = Atomic number Z + Neutron number N
A = Z + N → N = A – Z

Q3: Define nuclear density. Show that it is nearly constant for all nuclei.

Nuclear density ρ = Mass / Volume = mA / (4/3 π R³)
R = R₀ A¹ᐟ³ → Volume ∝ A
ρ = 3m / (4π R₀³) ≈ 2.3 × 10¹⁷ kg/m³ → nearly constant

Q4: Write the expression for nuclear radius. Find the radius of ₂₇Al¹³ and ₈₂Pb²⁰⁸ nuclei.

R = R₀ A¹ᐟ³, R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm
R(Al) = 1.2 × 13¹ᐟ³ ≈ 3.6 fm
R(Pb) = 1.2 × 208¹ᐟ³ ≈ 7.1 fm

Q5: State two main characteristics of nuclear force.

1. Very strong attractive force (strongest known force)
2. Short range (≈ 1–2 fm)
3. Charge independent (same between pp, nn, pn)
4. Non-central and spin dependent

Q6: The nuclear radius of ₈O¹⁶ is 3 fm. Find the nuclear radius of ₈₂Pb²⁰⁸.

R ∝ A¹ᐟ³
R₂/R₁ = (A₂/A₁)¹ᐟ³
R(Pb)/3 = (208/16)¹ᐟ³ = 13¹ᐟ³ ≈ 2.35
R(Pb) ≈ 7.05 fm

Q7: Why is the density of nucleus very high compared to ordinary matter?

Nuclear density ≈ 10¹⁷ kg/m³
Ordinary matter density ≈ 10³ kg/m³
Nucleus is 10¹⁴ times denser because entire mass is concentrated in very small volume (10⁻¹⁵ m) while electrons occupy large space.

Q8: Write the composition of nucleus of ₉₂U²³⁵ and ₂₆Fe⁵⁶.

₉₂U²³⁵ → Z = 92 protons, N = 235 – 92 = 143 neutrons
₂₆Fe⁵⁶ → Z = 26 protons, N = 56 – 26 = 30 neutrons

Q9: Define isotopes, isobars and isotones. Give one example each.

Isotopes → same Z, different A → ₁H¹, ₁H², ₁H³
Isobars → same A, different Z → ₁₈Ar⁴⁰, ₂₀Ca⁴⁰
Isotones → same N → ₆C¹³ (N=7), ₇N¹⁴ (N=7)

Q10: Arrange the following in increasing order of size: electron, atom, nucleus, proton

Nucleus (10⁻¹⁵ m) < Proton (10⁻¹⁵ m) < Electron (point particle) < Atom (10⁻¹⁰ m)
Nucleus ≈ 10⁻⁵ times size of atom

Q1: What is a nuclear reaction? Write one example of nuclear fission and one of nuclear fusion.

Nuclear reaction: A process in which nucleus is bombarded by a particle and new nuclei are formed.
Fission: ₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3 ₀n¹ + energy
Fusion: ₁H² + ₁H² → ₂He⁴ + energy

Q2: Define Q-value of a nuclear reaction. Write its formula.

Q-value: Net energy released (or absorbed) in a nuclear reaction.
Q = [Mass of reactants – Mass of products] c²
Q = (mᵢ – m𝒻) × 931.5 MeV/amu
If Q > 0 → exothermic (energy released), Q < 0 → endothermic

Q3: Calculate the Q-value of the reaction: ¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H Given masses: N=14.00307 u, He=4.00260 u, O=16.99913 u, H=1.00783 u

Mass of reactants = 14.00307 + 4.00260 = 18.00567 u
Mass of products = 16.99913 + 1.00783 = 18.00696 u
Δm = 18.00567 – 18.00696 = –0.00129 u
Q = –0.00129 × 931.5 ≈ –1.20 MeV (endothermic)

Q4: Why is energy released in nuclear fission and fusion?

Binding energy per nucleon is maximum near ⁵⁶Fe.
In fission: heavy nucleus (U-235) splits into medium mass nuclei → B.E./nucleon increases → energy released.
In fusion: light nuclei combine → B.E./nucleon increases → energy released.

Q5: In a nuclear reactor, name the three essential components and their functions.

1. Fuel (U-235) → undergoes fission
2. Moderator (graphite/water) → slows down neutrons
3. Control rods (cadmium/boron) → absorb excess neutrons to control chain reaction

Q6: What is a chain reaction? Distinguish between controlled and uncontrolled chain reaction.

Chain reaction: One fission produces neutrons that cause further fissions.
Controlled → used in nuclear reactor (k=1)
Uncontrolled → used in atom bomb (k>1, exponential growth)

Q7: The atomic mass of ₉₂U²³⁵ is 235.0439 u and that of ₅₆Ba¹⁴¹ is 140.9139 u. If one neutron is absorbed, find energy released (ignore Kr mass).

Δm = 235.0439 – 140.9139 = 94.13 u → not accurate (need all products)
Typical value per fission ≈ 200 MeV
Actual: 235.0439 + 1.0087 → 140.9139 + 91.9 + 3×1.0087 → Δm ≈ 0.215 u → Q ≈ 200 MeV

Q8: Write the nuclear reaction for the first artificial transmutation done by Rutherford.

₁₄₇N + ₄₂He → ₁₇₈O + ₁₁H
(Nitrogen bombarded with α-particle → Oxygen + Proton)

Q9: What is critical mass? Why is it necessary in a nuclear bomb?

Critical mass: Minimum mass of fissile material required to sustain chain reaction.
Below critical mass → neutrons escape → no explosion.
In bomb → two sub-critical masses are brought together rapidly → supercritical → explosion.

Q10: Compare nuclear fission and nuclear fusion (any three points).

Fission	Fusion
Heavy nucleus splits	Light nuclei combine
Used in reactors/bombs	Used in H-bomb/Sun
Neutrons initiate	High temp (10⁷ K) needed

Chapter 14: Semiconductor Electronics

Q1: What are semiconductors? Give two examples.

Semiconductors are materials whose conductivity lies between conductors and insulators (10⁻⁶ to 10⁴ S/m).
Examples: Silicon (Si), Germanium (Ge)

Q2: Why silicon is preferred over germanium for making semiconductor devices?

1. Higher band gap (1.1 eV vs 0.7 eV) → works at higher temperature
2. Lower reverse leakage current
3. Forms stable oxide (SiO₂) → useful in IC fabrication

Q3: Distinguish between intrinsic and extrinsic semiconductors (any three points).

Intrinsic → Pure, nₑ = nₕ, high resistivity
Extrinsic → Doped, nₑ ≠ nₕ, low resistivity, conductivity increases
Intrinsic → depends only on temperature
Extrinsic → depends on doping level

Q4: What is doping? Name the two types of doping.

Doping = intentional addition of impurity atoms to pure semiconductor to increase conductivity.
Two types:
1. n-type (pentavalent: P, As, Sb)
2. p-type (trivalent: B, Al, Ga, In)

Q5: Explain the formation of n-type semiconductor with energy band diagram.

Pentavalent impurity (e.g., Phosphorus) added to Si/Ge.
Four electrons form covalent bonds, fifth electron is loosely bound → donor level just below conduction band → electrons easily excited → majority carriers = electrons

Q6: Explain the formation of p-type semiconductor with energy band diagram.

Trivalent impurity (e.g., Boron) added.
Creates electron vacancy (hole) → acceptor level just above valence band → electrons from valence band jump to acceptor level → create holes → majority carriers = holes

Q7: Draw energy band diagrams for (i) intrinsic semiconductor (ii) n-type (iii) p-type

Intrinsic → E_g ≈ 1 eV, no extra levels
n-type → donor level (E_d) just below conduction band
p-type → acceptor level (E_a) just above valence band

Q8: How does the conductivity of a semiconductor change with rise in temperature?

Conductivity increases with temperature.
More electrons get energy to jump from valence to conduction band → number of charge carriers increases → resistivity decreases → conductivity increases.

Q9: Why intrinsic semiconductor behaves as insulator at 0 K?

At 0 K, all electrons are in valence band → no free electrons in conduction band → band gap exists → no conduction → behaves as insulator.

Q10: The band gap of Si is 1.1 eV and Ge is 0.7 eV. Which one will be more suitable for devices operating at high temperature and why?

Silicon (1.1 eV) is better for high temperature.
Higher band gap → electrons need more thermal energy to jump → less leakage current at high temperature → device remains functional.

Q1: What are semiconductors? Give two examples.

Semiconductors are materials whose conductivity lies between conductors and insulators (10⁻⁶ to 10⁴ S/m).
Examples: Silicon (Si), Germanium (Ge)

Q2: Why silicon is preferred over germanium for making semiconductor devices?

1. Higher band gap (1.1 eV vs 0.7 eV) → works at higher temperature
2. Lower reverse leakage current
3. Forms stable oxide (SiO₂) → useful in IC fabrication

Q3: Distinguish between intrinsic and extrinsic semiconductors (any three points).

Intrinsic → Pure, nₑ = nₕ, high resistivity
Extrinsic → Doped, nₑ ≠ nₕ, low resistivity, conductivity increases
Intrinsic → depends only on temperature
Extrinsic → depends on doping level

Q4: What is doping? Name the two types of doping.

Doping = intentional addition of impurity atoms to pure semiconductor to increase conductivity.
Two types:
1. n-type (pentavalent: P, As, Sb)
2. p-type (trivalent: B, Al, Ga, In)

Q5: Explain the formation of n-type semiconductor with energy band diagram.

Pentavalent impurity (e.g., Phosphorus) added to Si/Ge.
Four electrons form covalent bonds, fifth electron is loosely bound → donor level just below conduction band → electrons easily excited → majority carriers = electrons

Q6: Explain the formation of p-type semiconductor with energy band diagram.

Trivalent impurity (e.g., Boron) added.
Creates electron vacancy (hole) → acceptor level just above valence band → electrons from valence band jump to acceptor level → create holes → majority carriers = holes

Q7: Draw energy band diagrams for (i) intrinsic semiconductor (ii) n-type (iii) p-type

Intrinsic → E_g ≈ 1 eV, no extra levels
n-type → donor level (E_d) just below conduction band
p-type → acceptor level (E_a) just above valence band

Q8: How does the conductivity of a semiconductor change with rise in temperature?

Conductivity increases with temperature.
More electrons get energy to jump from valence to conduction band → number of charge carriers increases → resistivity decreases → conductivity increases.

Q9: Why intrinsic semiconductor behaves as insulator at 0 K?

At 0 K, all electrons are in valence band → no free electrons in conduction band → band gap exists → no conduction → behaves as insulator.

Q10: The band gap of Si is 1.1 eV and Ge is 0.7 eV. Which one will be more suitable for devices operating at high temperature and why?

Silicon (1.1 eV) is better for high temperature.
Higher band gap → electrons need more thermal energy to jump → less leakage current at high temperature → device remains functional.