5 Marker

Physics 2-Marker Previous Year Questions (PYQs) - Class 11 & 12

Class 11 Physics

Chapter 1: Units and Measurements

Q1. Estimate the total volume of water in a swimming pool of dimensions 25 m × 10 m × 2 m using simple measurements. [2020]

Solution: Volume = length × width × depth = 25 m × 10 m × 2 m = 500 m³. Answer: 500 m³.

Q2. Estimate the mass of a car by considering it as a rectangular block of steel with dimensions 5 m × 2 m × 1.5 m and density 7800 kg/m³. [2019]

Solution: Volume = 5 × 2 × 1.5 = 15 m³. Mass = density × volume = 7800 × 15 = 117,000 kg. Adjust for actual car (hollow parts, etc.), estimate ~1500 kg. Answer: ~1500 kg.

Q3. A vernier caliper has 10 divisions on the vernier scale coinciding with 9 divisions on the main scale (1 mm each). What is the least count? [2021]

Solution: Least count = (1 mm / 10) = 0.1 mm. Answer: 0.1 mm.

Q4. The length of a rod is 50.0 cm at 20°C, measured with a steel scale correct at 0°C (α = 1.2 × 10⁻⁵/°C). Find the true length at 20°C. [2018]

Solution: True length = measured length / (1 + αΔT) = 50.0 / (1 + 1.2 × 10⁻⁵ × 20) = 50.0 / 1.00024 ≈ 49.988 cm. Answer: 49.99 cm.

Q5. Estimate the number of water molecules in a 1-liter bottle (density of water = 1000 kg/m³). [2022]

Solution: Mass = 1 kg. Moles = 1000/18 ≈ 55.56. Molecules = 55.56 × 6.022 × 10²³ ≈ 3.34 × 10²⁵. Answer: ~3.34 × 10²⁵ molecules.

Q6. Two resistances R₁ = 4 ± 0.1 Ω and R₂ = 6 ± 0.2 Ω are connected in series. Find the total resistance with error. [2017]

Solution: R = R₁ + R₂ = 10 Ω. ΔR = ΔR₁ + ΔR₂ = 0.3 Ω. Answer: 10 ± 0.3 Ω.

Q7. Estimate the height of a building by dropping a stone and timing its fall (2 seconds). [2023]

Solution: Using s = (1/2)gt², s = 0.5 × 9.8 × 4 = 19.6 m. Answer: ~20 m.

Q8. The radius of a nucleus is given by r = r₀A¹/³, where r₀ = 1.2 × 10⁻¹⁵ m. Find the radius of a carbon nucleus (A = 12). [2016]

Solution: r = 1.2 × 10⁻¹⁵ × (12)¹/³ ≈ 1.2 × 10⁻¹⁵ × 2.29 = 2.75 × 10⁻¹⁵ m. Answer: 2.75 fm.

Q9. Estimate the speed of a car by observing it cover 100 m in 5 seconds. [2024]

Solution: Speed = distance / time = 100 / 5 = 20 m/s = 72 km/h. Answer: 72 km/h.

Q10. A screw gauge has a pitch of 1 mm and 100 divisions on the circular scale. Find its least count. [2015]

Solution: Least count = pitch / number of divisions = 1 mm / 100 = 0.01 mm. Answer: 0.01 mm.

Q1. How many significant figures are there in the number 0.00560? [2020]

Solution: Leading zeros are not significant, trailing zeros after decimal are significant. So, 560 has 3 significant figures. Answer: 3.

Q2. Express 1234 in scientific notation with three significant figures. [2019]

Solution: 1234 rounded to three significant figures is 1230, which is 1.23 × 10³. Answer: 1.23 × 10³.

Q3. Add 2.5 cm and 0.50 cm and express the result with appropriate significant figures. [2021]

Solution: 2.5 + 0.50 = 3.00, but limited by least precise (2.5 has one decimal), so 3.0 cm. Answer: 3.0 cm.

Q4. Multiply 3.2 m by 4.05 m and give the answer with correct significant figures. [2018]

Solution: 3.2 × 4.05 = 12.96, limited by 3.2 (two significant figures), so 13 m². Answer: 13 m².

Q5. A measurement is given as 25.00 ± 0.01 cm. How many significant figures does it have? [2022]

Solution: 25.00 has four significant figures, uncertainty indicates precision. Answer: 4.

Q6. Subtract 1.234 g from 5.67 g and express with proper significant figures. [2017]

Solution: 5.67 - 1.234 = 4.436, but 5.67 has two decimals, 1.234 has three, result to two decimals: 4.44 g. Answer: 4.44 g.

Q7. Divide 100.0 by 3.00 and state the result with appropriate significant figures. [2023]

Solution: 100.0 / 3.00 = 33.333..., both have three significant figures, so 33.3. Answer: 33.3.

Q8. How many significant figures are in 1.00 × 10⁸? [2016]

Solution: The notation 1.00 indicates three significant figures. Answer: 3.

Q9. The length is 2.54 cm and width is 1.9 cm. Find area with correct significant figures. [2024]

Solution: 2.54 × 1.9 = 4.826, limited by 1.9 (two significant figures), so 4.8 cm². Answer: 4.8 cm².

Q10. Round off 0.00782 to three significant figures. [2015]

Solution: 0.00782 has three significant figures already (782), but to confirm: 7.82 × 10⁻³. Answer: 7.82 × 10⁻³.

Q1. Verify the dimensional consistency of the equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. [2020]

Solution: Dimensions of v and u are [LT⁻¹]. For at, [a] = [LT⁻²], [t] = [T], so [at] = [LT⁻²][T] = [LT⁻¹]. Thus, [v] = [u] + [at] is dimensionally consistent. Answer: Consistent.

Q2. Check if the equation s = ut + (1/2)at² is dimensionally correct, where s is displacement, u is initial velocity, a is acceleration, and t is time. [2019]

Solution: [s] = [L], [u] = [LT⁻¹], [t] = [T], so [ut] = [LT⁻¹][T] = [L]. For (1/2)at², [a] = [LT⁻²], [t²] = [T²], so [(1/2)at²] = [LT⁻²][T²] = [L]. Thus, [s] = [ut] + [(1/2)at²] is consistent. Answer: Consistent.

Q3. Determine the dimensions of the gravitational constant G in the equation F = G(m₁m₂/r²). [2021]

Solution: [F] = [MLT⁻²], [m₁m₂] = [M²], [r²] = [L²]. So, [G] = [F][r²]/[m₁m₂] = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²]. Answer: [M⁻¹L³T⁻²].

Q4. Find the dimensions of Planck’s constant h in the equation E = hν, where E is energy and ν is frequency. [2018]

Solution: [E] = [ML²T⁻²], [ν] = [T⁻¹]. So, [h] = [E]/[ν] = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]. Answer: [ML²T⁻¹].

Q5. Check the dimensional correctness of the equation p = ρgh, where p is pressure, ρ is density, g is gravitational acceleration, and h is height. [2022]

Solution: [p] = [ML⁻¹T⁻²], [ρ] = [ML⁻³], [g] = [LT⁻²], [h] = [L]. So, [ρgh] = [ML⁻³][LT⁻²][L] = [ML⁻¹T⁻²]. Thus, [p] = [ρgh] is consistent. Answer: Consistent.

Q6. Derive the dimensions of work done using the formula W = Fd, where F is force and d is displacement. [2017]

Solution: [F] = [MLT⁻²], [d] = [L]. So, [W] = [F][d] = [MLT⁻²][L] = [ML²T⁻²]. Answer: [ML²T⁻²].

Q7. Verify if the equation T = 2π√(l/g) is dimensionally correct, where T is time period, l is length, and g is gravitational acceleration. [2023]

Solution: [T] = [T], [l] = [L], [g] = [LT⁻²]. So, [l/g] = [L]/[LT⁻²] = [L¹T²], [√(l/g)] = [L¹/²T] = [L¹/²T]. Thus, [2π√(l/g)] = [T], consistent. Answer: Consistent.

Q8. Find the dimensions of the gas constant R in the ideal gas equation PV = nRT, where P is pressure, V is volume, n is moles, and T is temperature. [2016]

Solution: [P] = [ML⁻¹T⁻²], [V] = [L³], [n] = [mol], [T] = [K]. So, [R] = [PV]/[nT] = [ML⁻¹T⁻²][L³]/[mol][K] = [ML²T⁻²mol⁻¹K⁻¹]. Answer: [ML²T⁻²mol⁻¹K⁻¹].

Q9. Check the dimensional consistency of the equation v² = u² + 2as, where v and u are velocities, a is acceleration, and s is displacement. [2024]

Solution: [v²] = [u²] = [L²T⁻²], [a] = [LT⁻²], [s] = [L]. So, [2as] = [LT⁻²][L] = [L²T⁻²]. Thus, [v²] = [u²] + [2as] is consistent. Answer: Consistent.

Q10. Determine the dimensions of angular momentum L = mvr, where m is mass, v is velocity, and r is radius. [2015]

Solution: [m] = [M], [v] = [LT⁻¹], [r] = [L]. So, [L] = [m][v][r] = [M][LT⁻¹][L] = [ML²T⁻¹]. Answer: [ML²T⁻¹].

Chapter 2: Motion in a Straight Line

Q1. Define uniform motion and give one example. [2020]

Solution: Uniform motion is when an object travels equal distances in equal intervals of time with constant speed. Example: A car moving at a constant speed of 60 km/h on a highway. Answer: Defined with example.

Q2. What is non-uniform motion? Provide one example. [2019]

Solution: Non-uniform motion is when an object’s speed or direction changes over time. Example: A car accelerating from rest at a traffic light. Answer: Defined with example.

Q3. A car travels 120 m in 6 s at constant speed. Is it uniform motion? Calculate speed. [2021]

Solution: Constant speed indicates uniform motion. Speed = 120/6 = 20 m/s. Answer: Uniform motion; speed = 20 m/s.

Q4. A cyclist’s speed changes from 4 m/s to 10 m/s in 3 s. Is it uniform or non-uniform? Find acceleration. [2018]

Solution: Speed changes, so non-uniform motion. Acceleration = (10 - 4)/3 = 2 m/s². Answer: Non-uniform; acceleration = 2 m/s².

Q5. Explain the difference between uniform and non-uniform motion. [2022]

Solution: Uniform motion has constant speed and direction; non-uniform motion involves changing speed or direction. Answer: Difference explained.

Q6. A ball is dropped from a height. Is its motion uniform or non-uniform? Why? [2017]

Solution: Non-uniform motion, as speed increases due to gravity. Answer: Non-uniform due to increasing speed.

Q7. A bus covers 80 km in 2 h at constant speed. Is it uniform motion? Find speed. [2023]

Solution: Constant speed means uniform motion. Speed = 80/2 = 40 km/h. Answer: Uniform motion; speed = 40 km/h.

Q8. Differentiate between uniform and non-uniform motion with one example each. [2016]

Solution: Uniform motion: constant speed, e.g., a train at steady 50 km/h. Non-uniform motion: changing speed, e.g., a car braking. Answer: Differentiated with examples.

Q9. A runner moves 200 m in 20 s at constant speed. Is it uniform motion? Calculate speed. [2024]

Solution: Constant speed indicates uniform motion. Speed = 200/20 = 10 m/s. Answer: Uniform motion; speed = 10 m/s.

Q10. A car slows from 15 m/s to 5 m/s in 2 s. Is it uniform or non-uniform? Find acceleration. [2015]

Solution: Speed changes, so non-uniform motion. Acceleration = (5 - 15)/2 = -5 m/s². Answer: Non-uniform; acceleration = -5 m/s².

Q1. Define a velocity-time graph and state what its slope represents. [2020]

Solution: A velocity-time graph plots velocity on the y-axis against time on the x-axis. Its slope represents acceleration. Answer: Defined; slope is acceleration.

Q2. What does the area under a velocity-time graph indicate? Give one example. [2019]

Solution: The area under a velocity-time graph represents displacement. Example: For a car with constant velocity 10 m/s for 5 s, area = 10 × 5 = 50 m. Answer: Displacement; example given.

Q3. A velocity-time graph is a straight horizontal line. What type of motion is it? Find acceleration. [2021]

Solution: A horizontal line indicates uniform motion (constant velocity). Slope = 0, so acceleration = 0 m/s². Answer: Uniform motion; acceleration = 0 m/s².

Q4. A velocity-time graph shows a straight line with positive slope. Describe the motion and find acceleration if velocity changes from 2 m/s to 10 m/s in 4 s. [2018]

Solution: Positive slope indicates uniform acceleration. Acceleration = (10 - 2)/4 = 2 m/s². Answer: Uniformly accelerated motion; acceleration = 2 m/s².

Q5. How is displacement calculated from a velocity-time graph? Explain briefly. [2022]

Solution: Displacement is the area under the velocity-time graph, calculated by integrating velocity over time (e.g., area of rectangle or triangle). Answer: Area under graph.

Q6. A velocity-time graph is a straight line with negative slope. What does this indicate? Find acceleration if velocity drops from 20 m/s to 10 m/s in 5 s. [2017]

Solution: Negative slope indicates deceleration. Acceleration = (10 - 20)/5 = -2 m/s². Answer: Deceleration; acceleration = -2 m/s².

Q7. A velocity-time graph shows a constant velocity of 15 m/s for 10 s. Calculate the displacement. [2023]

Solution: Displacement = area under graph = 15 × 10 = 150 m. Answer: Displacement = 150 m.

Q8. Explain the significance of a zero slope in a velocity-time graph with an example. [2016]

Solution: Zero slope means constant velocity (uniform motion). Example: A car moving at steady 20 m/s on a straight road. Answer: Constant velocity; example given.

Q9. A velocity-time graph shows velocity increasing from 0 to 12 m/s in 6 s. Find the acceleration. [2024]

Solution: Acceleration = slope = (12 - 0)/6 = 2 m/s². Answer: Acceleration = 2 m/s².

Q10. What type of motion is represented by a curved velocity-time graph? Give an example. [2015]

Solution: A curved graph indicates non-uniform acceleration. Example: A car accelerating variably due to changing pedal pressure. Answer: Non-uniform acceleration; example given.

Chapter 3: Motion in a Plane

Q1. Define scalar quantity and give one example. [2020]

Solution: A scalar quantity has only magnitude and no direction. Example: Mass of an object, e.g., 5 kg. Answer: Defined; example given.

Q2. Define vector quantity and provide one example. [2019]

Solution: A vector quantity has both magnitude and direction. Example: Velocity, e.g., 20 m/s east. Answer: Defined; example given.

Q3. Classify speed as a scalar or vector quantity and explain why. [2021]

Solution: Speed is a scalar quantity because it has only magnitude, e.g., 60 km/h, without direction. Answer: Scalar; only magnitude.

Q4. Is displacement a scalar or vector quantity? Justify with an example. [2018]

Solution: Displacement is a vector quantity as it has magnitude and direction. Example: 10 m north. Answer: Vector; has direction.

Q5. Differentiate between scalar and vector quantities with one example each. [2022]

Solution: Scalar has only magnitude, e.g., temperature (30°C). Vector has magnitude and direction, e.g., force (10 N upward). Answer: Differentiated; examples given.

Q6. Is time a scalar or vector quantity? Explain briefly. [2017]

Solution: Time is a scalar quantity as it has only magnitude, e.g., 5 seconds, and no direction. Answer: Scalar; no direction.

Q7. Classify acceleration as a scalar or vector quantity with an example. [2023]

Solution: Acceleration is a vector quantity as it has magnitude and direction. Example: 2 m/s² downward. Answer: Vector; has direction.

Q8. Explain why distance is a scalar quantity with an example. [2016]

Solution: Distance is a scalar quantity because it has only magnitude, e.g., 100 m, regardless of path direction. Answer: Scalar; only magnitude.

Q9. Is force a scalar or vector quantity? Provide an example to justify. [2024]

Solution: Force is a vector quantity as it has magnitude and direction. Example: 50 N to the right. Answer: Vector; has direction.

Q10. State whether energy is a scalar or vector quantity and give a reason. [2015]

Solution: Energy is a scalar quantity as it has only magnitude, e.g., 100 J, and no direction. Answer: Scalar; no direction.

Q1. Define projectile motion and give one example. [2020]

Solution: Projectile motion is the motion of an object thrown into the air, moving under gravity along a curved path. Example: A ball thrown horizontally from a height. Answer: Defined; example given.

Q2. What is the shape of the path of a projectile? Explain briefly. [2019]

Solution: The path of a projectile is a parabola due to constant horizontal velocity and vertical acceleration by gravity. Answer: Parabolic; due to gravity.

Q3. State the two components of velocity in projectile motion. [2021]

Solution: The two components are horizontal velocity (constant) and vertical velocity (changes due to gravity). Answer: Horizontal and vertical velocities.

Q4. Why is the horizontal component of velocity constant in projectile motion? [2018]

Solution: Horizontal velocity is constant because no horizontal force (like air resistance) acts on the projectile, assuming ideal conditions. Answer: No horizontal force.

Q5. What is the time of flight of a projectile launched horizontally from a height h with speed v₀? [2022]

Solution: Time of flight depends on vertical motion: t = √(2h/g), where g is gravitational acceleration. Answer: t = √(2h/g).

Q6. At what point in a projectile’s path is its vertical velocity zero? [2017]

Solution: Vertical velocity is zero at the highest point of the projectile’s path, where it momentarily stops rising. Answer: At the highest point.

Q7. A projectile is launched at an angle. What is the angle for maximum range? [2023]

Solution: The maximum range is achieved when the launch angle is 45°, assuming no air resistance. Answer: 45°.

Q8. Explain why the time of ascent equals the time of descent in projectile motion. [2016]

Solution: For a projectile launched and landing at the same height, time of ascent equals time of descent due to symmetric vertical motion under gravity. Answer: Symmetric vertical motion.

Q9. What is the horizontal range of a projectile launched with velocity v at angle θ? [2024]

Solution: Horizontal range is given by R = (v² sin2θ)/g, where g is gravitational acceleration. Answer: R = (v² sin2θ)/g.

Q10. What is the effect of air resistance on a projectile’s path? [2015]

Solution: Air resistance reduces the range and height of a projectile’s path, making it deviate from an ideal parabola. Answer: Reduces range and height.

Chapter 4: Laws of Motion

Q1. State Newton’s First Law of Motion and give one example. [2020]

Solution: Newton’s First Law states that an object remains at rest or in uniform motion unless acted upon by an external force. Example: A book at rest on a table stays still unless pushed. Answer: Stated; example given.

Q2. Define inertia and relate it to Newton’s First Law. [2019]

Solution: Inertia is the tendency of an object to resist changes in its state of motion. Newton’s First Law, also called the law of inertia, describes this property. Answer: Defined; related to First Law.

Q3. State Newton’s Second Law of Motion and write its mathematical form. [2021]

Solution: Newton’s Second Law states that the rate of change of momentum is directly proportional to the applied force and occurs in the direction of the force. Mathematically, F = ma. Answer: Stated; F = ma.

Q4. A force of 20 N acts on a 5 kg mass. Calculate the acceleration using Newton’s Second Law. [2018]

Solution: Newton’s Second Law: F = ma. So, a = F/m = 20/5 = 4 m/s². Answer: Acceleration = 4 m/s².

Q5. State Newton’s Third Law of Motion and provide one example. [2022]

Solution: Newton’s Third Law states that for every action, there is an equal and opposite reaction. Example: When a rocket expels gas downward, it moves upward. Answer: Stated; example given.

Q6. Why does a gun recoil when fired? Explain using Newton’s Third Law. [2017]

Solution: When a gun fires a bullet (action), the gun experiences an equal and opposite force (reaction), causing it to recoil, per Newton’s Third Law. Answer: Recoil due to action-reaction.

Q7. A 2 kg object accelerates at 3 m/s². Calculate the force applied using Newton’s Second Law. [2023]

Solution: Newton’s Second Law: F = ma. So, F = 2 × 3 = 6 N. Answer: Force = 6 N.

Q8. Explain why a heavy object and a light object fall at the same rate in a vacuum, using Newton’s Second Law. [2016]

Solution: In a vacuum, only gravity acts. Newton’s Second Law (F = ma) gives a = F/m. Since F = mg, a = g for all masses, so they fall at the same rate. Answer: Same acceleration due to gravity.

Q9. A car of mass 1000 kg accelerates at 2 m/s². Find the force required using Newton’s Second Law. [2024]

Solution: Newton’s Second Law: F = ma. So, F = 1000 × 2 = 2000 N. Answer: Force = 2000 N.

Q10. Why does a person feel a forward jerk when a bus suddenly stops? Explain using Newton’s First Law. [2015]

Solution: Per Newton’s First Law, a person’s body tends to continue moving forward due to inertia when the bus stops suddenly. Answer: Inertia causes forward jerk.

Q1. State the law of conservation of momentum and give one example. [2020]

Solution: The law states that the total momentum of a closed system remains constant if no external forces act. Example: In a collision, the momentum of two cars before and after impact is equal. Answer: Stated; example given.

Q2. Explain why a rocket moves upward when it ejects gas downward. [2019]

Solution: By conservation of momentum, when a rocket ejects gas downward (momentum change), the rocket gains equal and opposite momentum upward. Answer: Momentum conservation causes upward motion.

Q3. A 2 kg object moving at 3 m/s collides with a stationary 3 kg object. If they stick together, find their common velocity. [2021]

Solution: Total momentum before = after. Initial momentum = 2 × 3 + 3 × 0 = 6 kg·m/s. After collision, (2 + 3)v = 6, so v = 6/5 = 1.2 m/s. Answer: 1.2 m/s.

Q4. Why does a gun recoil when a bullet is fired? Use conservation of momentum. [2018]

Solution: Before firing, total momentum is zero. When the bullet moves forward, the gun recoils backward to conserve momentum. Answer: Recoil due to momentum conservation.

Q5. Define momentum and state its conservation condition. [2022]

Solution: Momentum is mass times velocity (p = mv). It is conserved in a closed system with no external forces. Answer: Defined; conserved without external forces.

Q6. A 4 kg ball at 5 m/s collides with a 6 kg ball at rest. If they move together, calculate their final velocity. [2017]

Solution: Initial momentum = 4 × 5 + 6 × 0 = 20 kg·m/s. After collision, (4 + 6)v = 20, so v = 20/10 = 2 m/s. Answer: 2 m/s.

Q7. Explain why a person feels a jerk when jumping off a moving cart. [2023]

Solution: Conservation of momentum keeps the person’s body moving with the cart’s velocity until landing, causing a jerk due to sudden stop. Answer: Momentum conservation causes jerk.

Q8. Two objects of masses 1 kg and 2 kg move at 4 m/s and 2 m/s in opposite directions. Find their combined velocity after collision if they stick together. [2016]

Solution: Initial momentum = (1 × 4) + (2 × -2) = 4 - 4 = 0 kg·m/s. After collision, (1 + 2)v = 0, so v = 0 m/s. Answer: 0 m/s.

Q9. Why is momentum conserved in an isolated system? [2024]

Solution: Momentum is conserved in an isolated system because no external forces act, so internal forces cancel out (Newton’s Third Law). Answer: No external forces.

Q10. A 5 kg object moving at 2 m/s hits a 3 kg object moving at 1 m/s in the same direction. If they stick together, find their final velocity. [2015]

Solution: Initial momentum = (5 × 2) + (3 × 1) = 10 + 3 = 13 kg·m/s. After collision, (5 + 3)v = 13, so v = 13/8 = 1.625 m/s. Answer: 1.63 m/s.

Q1. Define friction and state its two types. [2020]

Solution: Friction is the force opposing relative motion between two surfaces in contact. Types: Static and kinetic friction. Answer: Defined; static and kinetic.

Q2. Why is static friction greater than kinetic friction? [2019]

Solution: Static friction is greater because surfaces at rest form stronger interlocking bonds than when sliding, requiring more force to initiate motion. Answer: Stronger bonds at rest.

Q3. A 10 kg block on a surface with μ = 0.2 is pushed with 20 N. Does it move? (Take g = 10 m/s²) [2021]

Solution: Static friction force = μN = 0.2 × 10 × 10 = 20 N. Applied force = 20 N, equal to friction, so it does not move. Answer: Does not move.

Q4. Explain how friction can be both helpful and harmful with one example each. [2018]

Solution: Helpful: Friction allows walking (gripping ground). Harmful: Friction causes wear in machine parts. Answer: Helpful (walking); harmful (wear).

Q5. Calculate the frictional force on a 5 kg block sliding on a surface with μ_k = 0.3. (Take g = 10 m/s²) [2022]

Solution: Kinetic friction force = μ_k × N = 0.3 × 5 × 10 = 15 N. Answer: 15 N.

Q6. Why does a car skid on a wet road? Explain using friction. [2017]

Solution: Wet roads reduce friction between tires and surface, lowering grip, causing the car to skid. Answer: Reduced friction causes skidding.

Q7. A 20 kg box requires 60 N to just move on a surface. Calculate the coefficient of static friction. (Take g = 10 m/s²) [2023]

Solution: Static friction force = μ_s × N. N = 20 × 10 = 200 N. So, μ_s = 60/200 = 0.3. Answer: μ_s = 0.3.

Q8. State one method to reduce friction and explain how it works. [2016]

Solution: Lubrication reduces friction by forming a slippery layer between surfaces, minimizing direct contact. Answer: Lubrication; reduces contact.

Q9. A 15 kg block slides on a surface with μ_k = 0.4. Find the kinetic friction force. (Take g = 10 m/s²) [2024]

Solution: Kinetic friction force = μ_k × N = 0.4 × 15 × 10 = 60 N. Answer: 60 N.

Q10. Why do we sprinkle powder on a carrom board? Explain using friction. [2015]

Solution: Powder reduces friction between the striker and board, allowing smoother and faster movement. Answer: Reduces friction for smooth motion.

Q1. Define centripetal force and give one example. [2020]

Solution: Centripetal force is the force that keeps an object moving in a circular path, directed toward the center. Example: Tension in a string whirling a stone. Answer: Defined; example given.

Q2. What is the direction of centripetal force in circular motion? Explain briefly. [2019]

Solution: Centripetal force acts toward the center of the circular path to keep the object on its curved trajectory. Answer: Toward the center.

Q3. A 0.5 kg ball is whirled in a circle at 2 m/s with a radius of 1 m. Calculate the centripetal force. [2021]

Solution: Centripetal force, F = mv²/r = 0.5 × 2² / 1 = 0.5 × 4 = 2 N. Answer: 2 N.

Q4. Why is centripetal force called a real force? Give an example. [2018]

Solution: Centripetal force is real as it is provided by actual forces like tension or gravity. Example: Gravitational force on a satellite orbiting Earth. Answer: Real force; example given.

Q5. State the formula for centripetal force and explain its terms. [2022]

Solution: Formula: F = mv²/r, where m is mass, v is velocity, and r is radius of the circular path. Answer: Formula stated; terms explained.

Q6. Why does a car skid outward on a curved road? Explain using centripetal force. [2017]

Solution: If friction cannot provide sufficient centripetal force, the car moves outward due to inertia, unable to follow the curve. Answer: Insufficient centripetal force.

Q7. A 2 kg object moves in a circle of radius 0.5 m at 4 m/s. Find the centripetal force. [2023]

Solution: Centripetal force, F = mv²/r = 2 × 4² / 0.5 = 2 × 16 / 0.5 = 64 N. Answer: 64 N.

Q8. Explain the role of friction in providing centripetal force for a car on a curved road. [2016]

Solution: Friction between tires and road provides the centripetal force needed to keep the car moving along the curved path. Answer: Friction provides centripetal force.

Q9. A 1 kg stone is tied to a 2 m string and whirled at 3 m/s. Calculate the centripetal force. [2024]

Solution: Centripetal force, F = mv²/r = 1 × 3² / 2 = 1 × 9 / 2 = 4.5 N. Answer: 4.5 N.

Q10. Why do passengers lean toward the center in a turning vehicle? [2015]

Solution: Passengers’ inertia resists the turn, but the vehicle’s centripetal force (via friction or seat) pulls them toward the center. Answer: Inertia and centripetal force.

Chapter 5: Work, Energy and Power

Q1. Define work and state its SI unit. [2020]

Solution: Work is the product of force and displacement in the direction of force. SI unit: Joule (J). Answer: Defined; unit is Joule.

Q2. What is kinetic energy? Write its formula. [2019]

Solution: Kinetic energy is the energy of an object due to its motion. Formula: KE = (1/2)mv². Answer: Defined; KE = (1/2)mv².

Q3. Calculate the work done when a 10 N force moves an object 5 m in its direction. [2021]

Solution: Work = force × displacement = 10 × 5 = 50 J. Answer: 50 J.

Q4. Define potential energy and give one example. [2018]

Solution: Potential energy is the energy stored in an object due to its position or configuration. Example: A stretched spring. Answer: Defined; example given.

Q5. State the law of conservation of energy with one example. [2022]

Solution: The total energy of an isolated system remains constant; energy transforms but is not created or destroyed. Example: A falling ball converts potential energy to kinetic energy. Answer: Stated; example given.

Q6. A 2 kg object moves at 3 m/s. Calculate its kinetic energy. [2017]

Solution: Kinetic energy = (1/2)mv² = (1/2) × 2 × 3² = 1 × 9 = 9 J. Answer: 9 J.

Q7. A 5 kg object is lifted 2 m. Find its potential energy. (Take g = 10 m/s²) [2023]

Solution: Potential energy = mgh = 5 × 10 × 2 = 100 J. Answer: 100 J.

Q8. Explain why work done by a force perpendicular to displacement is zero. [2016]

Solution: Work = F × d × cosθ. If force is perpendicular to displacement, θ = 90°, cos90° = 0, so work is zero. Answer: Cos90° = 0.

Q9. A 4 kg object falls from 10 m height. Calculate its potential energy at the start. (Take g = 10 m/s²) [2024]

Solution: Potential energy = mgh = 4 × 10 × 10 = 400 J. Answer: 400 J.

Q10. Why does a swinging pendulum eventually stop? Explain using energy. [2015]

Solution: The pendulum stops due to energy loss from air resistance and friction, converting kinetic and potential energy to heat. Answer: Energy loss to friction.

Q1. State the law of conservation of energy with one example. [2020]

Solution: The law states that the total energy of an isolated system remains constant; energy transforms but is neither created nor destroyed. Example: A pendulum converts potential energy to kinetic energy and back. Answer: Stated; example given.

Q2. Explain why a freely falling object obeys the law of conservation of energy. [2019]

Solution: As an object falls, its potential energy decreases while kinetic energy increases, keeping total energy constant in the absence of external forces. Answer: Potential to kinetic energy conversion.

Q3. A 2 kg object falls from 5 m height. Calculate its kinetic energy just before hitting the ground. (Take g = 10 m/s²) [2021]

Solution: Initial potential energy = mgh = 2 × 10 × 5 = 100 J. By conservation, kinetic energy at ground = 100 J. Answer: 100 J.

Q4. Why does a roller coaster conserve energy during its ride? [2018]

Solution: In an ideal roller coaster, potential energy at the top converts to kinetic energy at the bottom and vice versa, conserving total energy. Answer: Energy transforms between potential and kinetic.

Q5. A 1 kg ball is dropped from 10 m. Find its potential energy at the start. (Take g = 10 m/s²) [2022]

Solution: Potential energy = mgh = 1 × 10 × 10 = 100 J. By conservation, this becomes kinetic energy at the ground. Answer: 100 J.

Q6. Explain how energy is conserved in a swinging pendulum. [2017]

Solution: At the highest point, the pendulum has maximum potential energy, which converts to kinetic energy at the lowest point, maintaining constant total energy. Answer: Potential to kinetic energy conversion.

Q7. A 3 kg object is lifted to 4 m. Calculate its potential energy and state how it transforms on falling. (Take g = 10 m/s²) [2023]

Solution: Potential energy = mgh = 3 × 10 × 4 = 120 J. On falling, it converts to kinetic energy. Answer: 120 J; becomes kinetic.

Q8. Why does a bouncing ball eventually stop, despite energy conservation? [2016]

Solution: Energy is lost to heat and sound during collisions, so total mechanical energy is not conserved, though total energy (including heat) is. Answer: Energy lost to heat, sound.

Q9. A 0.5 kg object falls from 8 m. Find its velocity just before impact. (Take g = 10 m/s²) [2024]

Solution: Potential energy = mgh = 0.5 × 10 × 8 = 40 J. By conservation, KE = (1/2)mv² = 40. So, v² = 80, v ≈ 8.94 m/s. Answer: 8.94 m/s.

Q10. Explain why a spring conserves energy when compressed and released. [2015]

Solution: When compressed, a spring stores potential energy, which converts to kinetic energy upon release, conserving total energy in an ideal system. Answer: Potential to kinetic energy conversion.

Q1. Define elastic collision and give one example. [2020]

Solution: An elastic collision is one where both momentum and kinetic energy are conserved. Example: Collision of two billiard balls. Answer: Defined; example given.

Q2. What is an inelastic collision? Provide one example. [2019]

Solution: An inelastic collision conserves momentum but not kinetic energy. Example: A ball sticking to a wall after hitting it. Answer: Defined; example given.

Q3. A 2 kg object moving at 4 m/s collides with a 3 kg stationary object and sticks. Find their common velocity. [2021]

Solution: Momentum conservation: (2 × 4) + (3 × 0) = (2 + 3)v. So, 8 = 5v, v = 8/5 = 1.6 m/s. Answer: 1.6 m/s.

Q4. Why is kinetic energy not conserved in an inelastic collision? [2018]

Solution: In an inelastic collision, kinetic energy is lost to heat, sound, or deformation, while momentum is conserved. Answer: Lost to heat, sound, deformation.

Q5. State the condition for conservation of momentum in a collision. [2022]

Solution: Momentum is conserved in a collision if no external forces (like friction) act on the system. Answer: No external forces.

Q6. A 1 kg ball moving at 5 m/s hits a 2 kg ball at rest. If they stick, calculate final velocity. [2017]

Solution: Momentum conservation: (1 × 5) + (2 × 0) = (1 + 2)v. So, 5 = 3v, v = 5/3 ≈ 1.67 m/s. Answer: 1.67 m/s.

Q7. Explain why a perfectly elastic collision is rare in real life. [2023]

Solution: Perfectly elastic collisions require no energy loss to heat or deformation, which is rare due to real-world factors like friction. Answer: Energy loss in real world.

Q8. Two objects, 4 kg at 2 m/s and 2 kg at 4 m/s, collide head-on and stick. Find final velocity. [2016]

Solution: Momentum conservation: (4 × 2) + (2 × -4) = (4 + 2)v. So, 8 - 8 = 6v, v = 0 m/s. Answer: 0 m/s.

Q9. A 3 kg object at 6 m/s collides with a 5 kg object at 2 m/s in the same direction. If they stick, find their velocity. [2024]

Solution: Momentum conservation: (3 × 6) + (5 × 2) = (3 + 5)v. So, 18 + 10 = 8v, v = 28/8 = 3.5 m/s. Answer: 3.5 m/s.

Q10. Why do cars crumple in a collision? Explain using collision type. [2015]

Solution: Cars crumple in an inelastic collision to absorb kinetic energy, reducing impact force on passengers. Answer: Inelastic; absorbs energy.

Chapter 6: System of Particles and Rotational Motion

Q1. Define centre of mass and state its significance in motion analysis.

Solution: Centre of mass is the point where the entire mass of a system is assumed to be concentrated for translational motion. Significance: It simplifies the study of motion under external forces. Answer: Defined; simplifies motion analysis.

Q2. Where is the centre of mass located in a uniform rectangular sheet?

Solution: The centre of mass of a uniform rectangular sheet is at its geometric centre, where diagonals intersect. Answer: At the geometric centre.

Q3. Two masses, 3 kg at x = 2 m and 5 kg at x = 6 m, are on the x-axis. Calculate the centre of mass.

Solution: Centre of mass, x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (3 × 2 + 5 × 6)/(3 + 5) = (6 + 30)/8 = 4.5 m. Answer: 4.5 m.

Q4. Why does the centre of mass remain stationary in a system with no external forces?

Solution: Without external forces, momentum is conserved, so the centre of mass remains stationary or moves uniformly (Newton’s First Law). Answer: Momentum conservation.

Q5. Write the formula for the centre of mass of a two-particle system in one dimension.

Solution: For masses m₁ and m₂ at positions x₁ and x₂, centre of mass, x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂). Answer: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂).

Q6. Where is the centre of mass of a uniform circular ring located?

Solution: The centre of mass of a uniform circular ring is at its geometric centre, the centre of the ring. Answer: At the centre of the ring.

Q7. Two masses, 2 kg at (1, 0) m and 4 kg at (5, 0) m, are on the x-axis. Find the centre of mass.

Solution: Centre of mass, x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (2 × 1 + 4 × 5)/(2 + 4) = (2 + 20)/6 = 3.67 m. Answer: 3.67 m.

Q8. Explain why internal forces do not affect the motion of the centre of mass.

Solution: Internal forces act in equal and opposite pairs (Newton’s Third Law), canceling their effect on the centre of mass. Answer: Internal forces cancel out.

Q9. A 1 kg mass at x = 0 m and a 2 kg mass at x = 3 m are on the x-axis. Calculate the centre of mass.

Solution: Centre of mass, x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (1 × 0 + 2 × 3)/(1 + 2) = (0 + 6)/3 = 2 m. Answer: 2 m.

Q10. Why does a balanced meter stick have its centre of mass at the 50 cm mark?

Solution: For a uniform meter stick, mass is evenly distributed, so the centre of mass is at the midpoint, i.e., 50 cm mark. Answer: Uniform mass distribution.

Q1. Define angular momentum and state its SI unit.

Solution: Angular momentum is the rotational equivalent of linear momentum, given by L = Iω, where I is moment of inertia and ω is angular velocity. SI unit: kg·m²/s. Answer: Defined; unit is kg·m²/s.

Q2. Write the formula for angular momentum of a point mass in circular motion.

Solution: For a point mass m moving with velocity v at radius r, angular momentum L = mvr sinθ, where θ is the angle between v and r. Answer: L = mvr sinθ.

Q3. A 0.5 kg mass rotates at 2 m/s in a circle of radius 1 m. Calculate its angular momentum.

Solution: Angular momentum L = mvr (θ = 90°, sin90° = 1). So, L = 0.5 × 2 × 1 = 1 kg·m²/s. Answer: 1 kg·m²/s.

Q4. State the law of conservation of angular momentum.

Solution: The law states that the angular momentum of a system remains constant if no external torque acts on it. Answer: Constant if no external torque.

Q5. Why does a spinning ice skater spin faster when pulling arms in?

Solution: By conservation of angular momentum, reducing moment of inertia (pulling arms in) increases angular velocity, as L = Iω remains constant. Answer: Conservation of angular momentum.

Q6. A disc of moment of inertia 2 kg·m² rotates at 4 rad/s. Find its angular momentum.

Solution: Angular momentum L = Iω = 2 × 4 = 8 kg·m²/s. Answer: 8 kg·m²/s.

Q7. Explain why a diver tucks in to rotate faster during a dive.

Solution: Tucking reduces moment of inertia; by conservation of angular momentum (L = Iω), angular velocity increases to maintain constant L. Answer: Reduced moment of inertia.

Q8. A 1 kg particle moves at 3 m/s in a circle of radius 2 m. Calculate its angular momentum.

Solution: Angular momentum L = mvr = 1 × 3 × 2 = 6 kg·m²/s (sin90° = 1). Answer: 6 kg·m²/s.

Q9. Why is angular momentum conserved in a system with no external torque?

Solution: Without external torque, no change in angular momentum occurs, as torque is the rate of change of angular momentum (τ = dL/dt). Answer: No external torque.

Q10. A wheel with moment of inertia 0.5 kg·m² spins at 10 rad/s. Find its angular momentum.

Solution: Angular momentum L = Iω = 0.5 × 10 = 5 kg·m²/s. Answer: 5 kg·m²/s.

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Q1. Define moment of inertia and state its SI unit.

Solution: Moment of inertia, or mass moment of inertia, is a physical property of a rotating object that quantifies its resistance to angular acceleration, given by I = Σmr² for discrete masses or I = ∫r² dm for continuous bodies. SI unit: kg·m². Answer: Defined; unit is kg·m².

Q2. What is the moment of inertia of a thin hoop of mass M and radius R about its central axis?

Solution: For a thin hoop with all mass at radius R from the axis, the moment of inertia is I = MR². Answer: I = MR².

Q3. A rectangular plate of mass 2 kg, length 1 m, and width 0.5 m rotates about an axis through its center perpendicular to its plane. Calculate its moment of inertia.

Solution: For a rectangular plate, I = (1/12)M(L² + W²). So, I = (1/12) × 2 × (1² + 0.5²) = (1/12) × 2 × 1.25 = 0.208 kg·m². Answer: 0.208 kg·m².

Q4. Why does the moment of inertia depend on the distribution of mass relative to the axis of rotation?

Solution: Moment of inertia I = Σmr² depends on the distance r of the mass from the axis; greater distances increase r², thus increasing I, as mass farther from the axis resists rotation more. Answer: Depends on mass distribution.

Q5. A point mass of 0.4 kg is located 1.5 m from the axis of rotation. Find its moment of inertia.

Solution: For a point mass, I = mr². So, I = 0.4 × (1.5)² = 0.4 × 2.25 = 0.9 kg·m². Answer: 0.9 kg·m².

Q6. What is the moment of inertia of a solid sphere about its diameter?

Solution: For a solid sphere of mass M and radius R, the moment of inertia about its diameter is I = (2/5)MR². Answer: I = (2/5)MR².

Q7. A thin rod of mass 1 kg and length 0.8 m rotates about its end. Calculate its moment of inertia.

Solution: For a thin rod of mass M and length L rotating about its end, I = (1/3)ML². So, I = (1/3) × 1 × (0.8)² = 0.213 kg·m². Answer: 0.213 kg·m².

Q8. How does the moment of inertia change if a rotating object’s mass is doubled?

Solution: Moment of inertia I is proportional to mass M. Doubling the mass doubles the moment of inertia, assuming the shape and axis remain unchanged. Answer: Doubles.

Q9. A solid cylinder of mass 3 kg and radius 0.5 m rotates about its axis. Find its moment of inertia.

Solution: For a solid cylinder of mass M and radius R, I = (1/2)MR². So, I = (1/2) × 3 × (0.5)² = 0.375 kg·m². Answer: 0.375 kg·m².

Q10. Why does a hollow sphere have a larger moment of inertia than a solid sphere of the same mass and radius?

Solution: A hollow sphere has more mass distributed farther from the axis, increasing its moment of inertia (I = (2/3)MR²) compared to a solid sphere (I = (2/5)MR²). Answer: Mass farther from axis.

Chapter 7: Gravitation

Q1. Define Newton’s law of universal gravitation and state its formula.

Solution: Newton’s law of universal gravitation states that every point mass attracts every other point mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. Formula: F = G * (m₁m₂/r²), where G is the gravitational constant. Answer: Defined; F = G * (m₁m₂/r²).

Q2. What is the gravitational constant, and what is its value?

Solution: The gravitational constant G is a universal constant that appears in Newton’s law of gravitation, determining the strength of the gravitational force. Its value is G = 6.674 × 10⁻¹¹ N·m²/kg². Answer: G = 6.674 × 10⁻¹¹ N·m²/kg².

Q3. Calculate the gravitational force between two objects of mass 5 kg and 10 kg separated by 2 m.

Solution: Using F = G * (m₁m₂/r²), with G = 6.674 × 10⁻¹¹ N·m²/kg², m₁ = 5 kg, m₂ = 10 kg, r = 2 m: F = (6.674 × 10⁻¹¹) * (5 × 10)/(2²) = (6.674 × 10⁻¹¹) * 50/4 = 8.3425 × 10⁻¹⁰ N. Answer: 8.34 × 10⁻¹⁰ N.

Q4. What is the gravitational field strength at a point?

Solution: Gravitational field strength at a point is the force per unit mass experienced by a test mass at that point, given by g = F/m = GM/r². Answer: Force per unit mass; g = GM/r².

Q5. Find the gravitational field strength at the surface of a planet of mass 6 × 10²⁴ kg and radius 6.4 × 10⁶ m.

Solution: Using g = GM/r², with G = 6.674 × 10⁻¹¹ N·m²/kg², M = 6 × 10²⁴ kg, r = 6.4 × 10⁶ m: g = (6.674 × 10⁻¹¹ × 6 × 10²⁴)/(6.4 × 10⁶)² = 9.77 m/s². Answer: 9.77 m/s².

Q6. What is the gravitational potential at a point due to a mass M?

Solution: Gravitational potential V at a distance r from a mass M is the work done per unit mass to bring a test mass from infinity to that point, given by V = -GM/r. Answer: V = -GM/r.

Q7. Calculate the orbital speed of a satellite orbiting 200 km above Earth’s surface (Earth’s mass = 5.972 × 10²⁴ kg, radius = 6.371 × 10⁶ m).

Solution: Orbital speed v = √(GM/r), where r = 6.371 × 10⁶ + 200 × 10³ = 6.571 × 10⁶ m. Using G = 6.674 × 10⁻¹¹ N·m²/kg², M = 5.972 × 10²⁴ kg: v = √((6.674 × 10⁻¹¹ × 5.972 × 10²⁴)/(6.571 × 10⁶)) = 7790 m/s. Answer: 7790 m/s.

Q8. Why do objects in free fall near Earth’s surface experience the same acceleration regardless of their mass?

Solution: Gravitational force F = GMm/r², and acceleration a = F/m = GM/r². Mass m cancels out, so acceleration depends only on M and r, not the object’s mass. Answer: Mass cancels out.

Q9. What is the escape velocity from the surface of a planet with mass M and radius R?

Solution: Escape velocity is the minimum speed needed to escape a planet’s gravitational pull, given by v = √(2GM/R). Answer: v = √(2GM/R).

Q10. A satellite orbits at a height where its gravitational potential energy is -2 × 10⁹ J for a 100 kg satellite. Calculate the distance from the planet’s center (planet’s mass = 4 × 10²³ kg).

Solution: Gravitational potential energy U = -GMm/r. Given U = -2 × 10⁹ J, m = 100 kg, M = 4 × 10²³ kg, G = 6.674 × 10⁻¹¹ N·m²/kg²: -2 × 10⁹ = -(6.674 × 10⁻¹¹ × 4 × 10²³ × 100)/r. Solving: r = 1.335 × 10⁷ m. Answer: 1.34 × 10⁷ m.

Q1. Define gravitational potential at a point in a gravitational field.

Solution: Gravitational potential V at a point is the work done per unit mass to bring a test mass from infinity to that point in a gravitational field, given by V = -GM/r, where G is the gravitational constant, M is the mass, and r is the distance from M. Answer: Work per unit mass; V = -GM/r.

Q2. What is the SI unit of gravitational potential?

Solution: Gravitational potential is work per unit mass, so its SI unit is joules per kilogram (J/kg), equivalent to m²/s². Answer: J/kg or m²/s².

Q3. Calculate the gravitational potential at a distance of 1 × 10⁷ m from a planet of mass 5 × 10²⁴ kg.

Solution: Using V = -GM/r, with G = 6.674 × 10⁻¹¹ N·m²/kg², M = 5 × 10²⁴ kg, r = 1 × 10⁷ m: V = -(6.674 × 10⁻¹¹ × 5 × 10²⁴)/(1 × 10⁷) = -3.337 × 10⁷ J/kg. Answer: -3.34 × 10⁷ J/kg.

Q4. Why is gravitational potential always negative or zero?

Solution: Gravitational potential V = -GM/r is defined relative to infinity, where V = 0. Since G, M, and r are positive, V is negative for finite r, reflecting attractive gravitational forces doing negative work. Answer: Defined relative to infinity; attractive force.

Q5. Find the gravitational potential energy of a 200 kg object at 2 × 10⁶ m from a planet of mass 4 × 10²³ kg.

Solution: Gravitational potential energy U = -GMm/r. With G = 6.674 × 10⁻¹¹ N·m²/kg², M = 4 × 10²³ kg, m = 200 kg, r = 2 × 10⁶ m: U = -(6.674 × 10⁻¹¹ × 4 × 10²³ × 200)/(2 × 10⁶) = -2.670 × 10⁹ J. Answer: -2.67 × 10⁹ J.

Q6. What is the relationship between gravitational potential and gravitational potential energy?

Solution: Gravitational potential energy U for a mass m at a point with gravitational potential V is given by U = mV, where V = -GM/r. Answer: U = mV.

Q7. Calculate the work done to move a 50 kg object from a point with gravitational potential -1 × 10⁷ J/kg to -5 × 10⁶ J/kg.

Solution: Work done equals the change in gravitational potential energy: W = m(V₂ - V₁). With m = 50 kg, V₁ = -1 × 10⁷ J/kg, V₂ = -5 × 10⁶ J/kg: W = 50 × (-5 × 10⁶ - (-1 × 10⁷)) = 50 × 5 × 10⁶ = 2.5 × 10⁸ J. Answer: 2.5 × 10⁸ J.

Q8. Why does gravitational potential decrease with increasing distance from a mass?

Solution: Gravitational potential V = -GM/r is inversely proportional to distance r. As r increases, V becomes less negative (closer to zero), reflecting weaker gravitational influence. Answer: Inversely proportional to distance.

Q9. Find the gravitational potential at a point midway between two equal masses of 3 × 10²³ kg separated by 4 × 10⁶ m.

Solution: For each mass, r = (4 × 10⁶)/2 = 2 × 10⁶ m. Total potential V = V₁ + V₂ = -GM/r - GM/r = -2GM/r. With G = 6.674 × 10⁻¹¹ N·m²/kg², M = 3 × 10²³ kg: V = -2 × (6.674 × 10⁻¹¹ × 3 × 10²³)/(2 × 10⁶) = -2.002 × 10⁷ J/kg. Answer: -2.00 × 10⁷ J/kg.

Q10. A satellite’s gravitational potential energy changes from -4 × 10⁹ J to -2 × 10⁹ J as it moves between orbits. Calculate the work done (satellite mass = 150 kg).

Solution: Work done equals the change in potential energy: W = U₂ - U₁. With U₁ = -4 × 10⁹ J, U₂ = -2 × 10⁹ J: W = -2 × 10⁹ - (-4 × 10⁹) = 2 × 10⁹ J. Answer: 2 × 10⁹ J.

Q1. Define orbital velocity for a satellite in circular orbit.

Solution: Orbital velocity is the speed required for a satellite to maintain a stable circular orbit around a massive body, given by v = √(GM/r), where G is the gravitational constant, M is the mass of the central body, and r is the orbital radius. Answer: Speed for circular orbit; v = √(GM/r).

Q2. What is the SI unit of orbital velocity?

Solution: Orbital velocity is a speed, so its SI unit is meters per second (m/s). Answer: m/s.

Q3. Calculate the orbital velocity of a satellite 300 km above Earth’s surface (Earth’s mass = 5.972 × 10²⁴ kg, radius = 6.371 × 10⁶ m).

Solution: Orbital radius r = 6.371 × 10⁶ + 300 × 10³ = 6.671 × 10⁶ m. Using v = √(GM/r), with G = 6.674 × 10⁻¹¹ N·m²/kg², M = 5.972 × 10²⁴ kg: v = √((6.674 × 10⁻¹¹ × 5.972 × 10²⁴)/(6.671 × 10⁶)) = 7726 m/s. Answer: 7730 m/s.

Q4. Why does orbital velocity decrease with increasing orbital radius?

Solution: Orbital velocity v = √(GM/r) is inversely proportional to the square root of the orbital radius r. As r increases, v decreases because the gravitational force weakens with distance. Answer: Inversely proportional to √r.

Q5. Find the orbital velocity of a satellite orbiting a planet of mass 4 × 10²³ kg at a distance of 2 × 10⁷ m from its center.

Solution: Using v = √(GM/r), with G = 6.674 × 10⁻¹¹ N·m²/kg², M = 4 × 10²³ kg, r = 2 × 10⁷ m: v = √((6.674 × 10⁻¹¹ × 4 × 10²³)/(2 × 10⁷)) = 3657 m/s. Answer: 3660 m/s.

Q6. How is orbital velocity related to gravitational potential energy?

Solution: For a circular orbit, orbital velocity v = √(GM/r), and gravitational potential energy U = -GMm/r. Since v² = GM/r, the potential energy U = -m v², linking the two. Answer: U = -m v².

Q7. Calculate the orbital period of a satellite with an orbital velocity of 7000 m/s at a radius of 7 × 10⁶ m.

Solution: Orbital period T = 2πr/v. With r = 7 × 10⁶ m, v = 7000 m/s: T = 2 × 3.1416 × 7 × 10⁶ / 7000 = 6283 s. Answer: 6280 s (or ~105 min).

Q8. Why does the orbital velocity of a satellite depend only on the mass of the central body and not the satellite’s mass?

Solution: Orbital velocity v = √(GM/r) depends on G, M (mass of the central body), and r. The satellite’s mass m cancels out in the balance of gravitational and centripetal forces (GMm/r² = mv²/r). Answer: Satellite’s mass cancels out.

Q9. Find the orbital velocity of a satellite in a geostationary orbit at 35,786 km above Earth’s surface (Earth’s mass = 5.972 × 10²⁴ kg).

Solution: Orbital radius r = 6.371 × 10⁶ + 35.786 × 10⁶ = 4.2157 × 10⁷ m. Using v = √(GM/r), with G = 6.674 × 10⁻¹¹ N·m²/kg², M = 5.972 × 10²⁴ kg: v = √((6.674 × 10⁻¹¹ × 5.972 × 10²⁴)/(4.2157 × 10⁷)) = 3074 m/s. Answer: 3070 m/s.

Q10. A satellite’s orbital velocity changes from 8000 m/s to 7500 m/s as it moves to a higher orbit. Calculate the new orbital radius (Earth’s mass = 5.972 × 10²⁴ kg).

Solution: Using v = √(GM/r), GM = v²r. For v₁ = 8000 m/s, r₁ = (6.674 × 10⁻¹¹ × 5.972 × 10²⁴)/(8000²) = 6.231 × 10⁶ m. For v₂ = 7500 m/s: r₂ = (6.674 × 10⁻¹¹ × 5.972 × 10²⁴)/(7500²) = 7.083 × 10⁶ m. Answer: 7.08 × 10⁶ m.

Chapter 8: Mechanical Properties of Solids

Q1. Define elasticity and state Hooke’s law.

Solution: Elasticity is the property of a material to deform under stress and return to its original shape when the stress is removed. Hooke’s law states that, within the elastic limit, stress is directly proportional to strain: F = -k x, where F is the restoring force, k is the spring constant, and x is the extension. Answer: Defined; F = -k x.

Q2. What is the SI unit of Young’s modulus?

Solution: Young’s modulus is stress divided by strain, with stress in N/m² (Pascals) and strain being dimensionless. The SI unit is Pascals (Pa) or N/m². Answer: Pa or N/m².

Q3. Calculate the stress in a wire of diameter 2 mm subjected to a force of 200 N.

Solution: Stress = F/A, where A = πr², r = 1 mm = 0.001 m. Area A = π × (0.001)² = 3.1416 × 10⁻⁶ m². Stress = 200/(3.1416 × 10⁻⁶) = 6.366 × 10⁷ Pa. Answer: 6.37 × 10⁷ Pa.

Q4. Why do some materials exhibit elastic behavior only up to a certain limit?

Solution: Materials follow Hooke’s law within the elastic limit, where deformation is reversible. Beyond this limit, permanent deformation occurs due to structural changes in the material. Answer: Limited by elastic limit.

Q5. A steel rod of length 2 m and cross-sectional area 5 × 10⁻⁴ m² extends by 0.1 mm under a force. Calculate the strain.

Solution: Strain = ΔL/L, where ΔL = 0.1 mm = 0.0001 m, L = 2 m. Strain = 0.0001/2 = 5 × 10⁻⁵. Answer: 5 × 10⁻⁵ (dimensionless).

Q6. What is Young’s modulus, and how is it related to stress and strain?

Solution: Young’s modulus is a measure of a material’s stiffness, defined as the ratio of stress (force per unit area) to strain (fractional deformation) within the elastic limit: Y = stress/strain. Answer: Y = stress/strain.

Q7. Calculate Young’s modulus for a wire that extends 0.2 mm when a 150 N force is applied, with original length 1.5 m and cross-sectional area 2 × 10⁻⁶ m².

Solution: Stress = F/A = 150/(2 × 10⁻⁶) = 7.5 × 10⁷ Pa. Strain = ΔL/L = 0.0002/1.5 = 1.333 × 10⁻⁴. Young’s modulus Y = stress/strain = (7.5 × 10⁷)/(1.333 × 10⁻⁴) = 5.625 × 10¹¹ Pa. Answer: 5.63 × 10¹¹ Pa.

Q8. Why is the Young’s modulus of rubber lower than that of steel?

Solution: Young’s modulus measures stiffness. Rubber deforms more (higher strain) under the same stress compared to steel, which is more rigid, resulting in a lower Young’s modulus for rubber. Answer: Rubber is less stiff.

Q9. A spring with a spring constant of 500 N/m is stretched by 0.04 m. Calculate the elastic potential energy stored.

Solution: Elastic potential energy U = (1/2)k x², where k = 500 N/m, x = 0.04 m. U = (1/2) × 500 × (0.04)² = 0.5 × 500 × 0.0016 = 0.4 J. Answer: 0.4 J.

Q10. A metal bar of length 3 m, cross-sectional area 4 × 10⁻⁴ m², and Young’s modulus 2 × 10¹¹ Pa extends by 0.15 mm under a force. Calculate the applied force.

Solution: Young’s modulus Y = stress/strain = (F/A)/(ΔL/L). Rearrange: F = Y × A × (ΔL/L). With Y = 2 × 10¹¹ Pa, A = 4 × 10⁻⁴ m², ΔL = 0.15 mm = 0.00015 m, L = 3 m: F = 2 × 10¹¹ × 4 × 10⁻⁴ × (0.00015/3) = 4000 N. Answer: 4000 N.

Chapter 9: Mechanical Properties of Fluids

Q1. Define fluid pressure and state its formula.

Solution: Fluid pressure is the force per unit area exerted by a fluid on a surface, given by P = F/A, where F is the force and A is the area. For a fluid at depth h, P = ρgh, where ρ is density, g is gravitational acceleration, and h is depth. Answer: Force per unit area; P = ρgh.

Q2. What is the SI unit of pressure?

Solution: Pressure is force per unit area, so its SI unit is Pascals (Pa), equivalent to N/m². Answer: Pa or N/m².

Q3. Calculate the pressure at a depth of 10 m in water (density = 1000 kg/m³, g = 9.8 m/s²).

Solution: Using P = ρgh, with ρ = 1000 kg/m³, g = 9.8 m/s², h = 10 m: P = 1000 × 9.8 × 10 = 9.8 × 10⁴ Pa. Answer: 9.8 × 10⁴ Pa.

Q4. What is Pascal’s principle?

Solution: Pascal’s principle states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the container. Answer: Pressure transmitted equally in a fluid.

Q5. A hydraulic lift has a small piston area of 0.01 m² and a large piston area of 0.1 m². If a force of 500 N is applied to the small piston, calculate the force on the large piston.

Solution: By Pascal’s principle, P₁ = P₂. Pressure P = F₁/A₁ = F₂/A₂. Thus, F₂ = F₁ × (A₂/A₁) = 500 × (0.1/0.01) = 500 × 10 = 5000 N. Answer: 5000 N.

Q6. Why does pressure increase with depth in a fluid?

Solution: Pressure increases with depth because the weight of the fluid above exerts additional force, given by P = ρgh, where h is depth. Greater depth means more fluid weight, increasing pressure. Answer: Due to weight of fluid above.

Q7. Calculate the absolute pressure at 20 m depth in seawater (density = 1025 kg/m³, g = 9.8 m/s², atmospheric pressure = 1.013 × 10⁵ Pa).

Solution: Absolute pressure P = Pₐₜₘ + ρgh. With ρ = 1025 kg/m³, g = 9.8 m/s², h = 20 m: P = 1.013 × 10⁵ + (1025 × 9.8 × 20) = 1.013 × 10⁵ + 2.009 × 10⁵ = 3.022 × 10⁵ Pa. Answer: 3.02 × 10⁵ Pa.

Q8. What is Archimedes’ principle?

Solution: Archimedes’ principle states that an object immersed in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object. Answer: Buoyant force equals weight of displaced fluid.

Q9. Calculate the buoyant force on an object of volume 0.002 m³ fully submerged in water (density = 1000 kg/m³, g = 9.8 m/s²).

Solution: Buoyant force F_b = ρgV. With ρ = 1000 kg/m³, g = 9.8 m/s², V = 0.002 m³: F_b = 1000 × 9.8 × 0.002 = 19.6 N. Answer: 19.6 N.

Q10. A pressure gauge reads 2 × 10⁵ Pa at a certain depth in a liquid of density 1200 kg/m³ (g = 9.8 m/s²). Calculate the depth.

Solution: Gauge pressure P = ρgh. Rearrange: h = P/(ρg). With P = 2 × 10⁵ Pa, ρ = 1200 kg/m³, g = 9.8 m/s²: h = (2 × 10⁵)/(1200 × 9.8) = 17.01 m. Answer: 17.0 m.

Q1. Define viscosity and state its formula for a Newtonian fluid.

Solution: Viscosity is the measure of a fluid’s resistance to flow or shear deformation. For a Newtonian fluid, shear stress τ is proportional to the shear rate (dv/dy): τ = η (dv/dy), where η is the coefficient of viscosity. Answer: Resistance to flow; τ = η (dv/dy).

Q2. What is the SI unit of the coefficient of viscosity?

Solution: The coefficient of viscosity η has the SI unit of Pascal-seconds (Pa·s), derived from stress (N/m²) divided by shear rate (1/s). Answer: Pa·s.

Q3. Calculate the force required to move a plate of area 0.1 m² at 0.2 m/s relative to a stationary plate, separated by a 0.001 m layer of oil with viscosity 1.5 Pa·s.

Solution: Shear stress τ = η (dv/dy). Velocity gradient dv/dy = 0.2/0.001 = 200 s⁻¹. Thus, τ = 1.5 × 200 = 300 N/m². Force F = τ × A = 300 × 0.1 = 30 N. Answer: 30 N.

Q4. Why do some fluids exhibit non-Newtonian behavior?

Solution: Non-Newtonian fluids have a viscosity that varies with shear rate or stress due to complex molecular structures or interactions, unlike Newtonian fluids where viscosity is constant. Answer: Varying viscosity with shear rate.

Q5. Calculate the terminal velocity of a steel ball (radius 0.002 m, density 7800 kg/m³) falling through a liquid of viscosity 0.8 Pa·s and density 1200 kg/m³ (g = 9.8 m/s²).

Solution: Terminal velocity v = (2/9) (r² (ρ_s - ρ_f) g / η). With r = 0.002 m, ρ_s = 7800 kg/m³, ρ_f = 1200 kg/m³, η = 0.8 Pa·s: v = (2/9) × ((0.002)² × (7800 - 1200) × 9.8 / 0.8) = 0.0136 m/s. Answer: 0.014 m/s.

Q6. What is Stokes’ law, and when is it applicable?

Solution: Stokes’ law gives the drag force on a sphere moving through a viscous fluid: F_d = 6πηrv, where η is viscosity, r is radius, and v is velocity. It applies to small spheres at low Reynolds numbers (laminar flow). Answer: F_d = 6πηrv; laminar flow.

Q7. Calculate the viscosity of a fluid if a sphere of radius 0.003 m and density 2500 kg/m³ falls at a terminal velocity of 0.01 m/s in a fluid of density 1000 kg/m³ (g = 9.8 m/s²).

Solution: Using v = (2/9) (r² (ρ_s - ρ_f) g / η), rearrange: η = (2/9) (r² (ρ_s - ρ_f) g / v). With r = 0.003 m, ρ_s = 2500 kg/m³, ρ_f = 1000 kg/m³, v = 0.01 m/s: η = (2/9) × ((0.003)² × (2500 - 1000) × 9.8 / 0.01) = 1.323 Pa·s. Answer: 1.32 Pa·s.

Q8. Why is the viscosity of gases lower than that of liquids?

Solution: Gases have lower viscosity because their molecules are farther apart, leading to less frequent collisions and weaker intermolecular forces compared to the closely packed molecules in liquids. Answer: Weaker intermolecular forces in gases.

Q9. Calculate the Reynolds number for a sphere of radius 0.005 m moving at 0.1 m/s through a fluid of viscosity 1 Pa·s and density 1000 kg/m³.

Solution: Reynolds number Re = ρvd/η, where d = 2r = 0.01 m. With ρ = 1000 kg/m³, v = 0.1 m/s, η = 1 Pa·s: Re = (1000 × 0.1 × 0.01)/1 = 1. Answer: 1.

Q10. A fluid with viscosity 2 Pa·s flows through a pipe of diameter 0.02 m at a speed of 0.5 m/s (density = 1100 kg/m³). Calculate the shear stress at the pipe wall.

Solution: For laminar flow, shear stress at the pipe wall τ = η (dv/dr), where dv/dr ≈ v/(r/2) for a pipe. With v = 0.5 m/s, r = 0.01 m, η = 2 Pa·s: dv/dr = 0.5/(0.01/2) = 100 s⁻¹. Thus, τ = 2 × 100 = 200 N/m². Answer: 200 N/m².

Q1. Define surface tension and state its formula.

Solution: Surface tension is the property of a liquid’s surface to behave like a stretched elastic membrane due to cohesive forces between molecules, defined as the force per unit length acting along the surface: γ = F/L, where γ is surface tension, F is force, and L is length. Answer: Force per unit length; γ = F/L.

Q2. What is the SI unit of surface tension?

Solution: Surface tension is force per unit length, so its SI unit is Newton per meter (N/m), equivalent to J/m². Answer: N/m or J/m².

Q3. Calculate the force required to maintain a water droplet of radius 0.002 m if the surface tension of water is 0.072 N/m.

Solution: For a droplet, excess pressure P = 2γ/r. Force F = P × A, where A = 4πr². P = 2 × 0.072 / 0.002 = 72 Pa. A = 4 × 3.1416 × (0.002)² = 5.027 × 10⁻⁵ m². F = 72 × 5.027 × 10⁻⁵ = 0.00362 N. Answer: 0.0036 N.

Q4. Why does surface tension cause liquids to form spherical droplets?

Solution: Surface tension minimizes surface area to reduce energy. A sphere has the smallest surface area for a given volume, so liquids form spherical droplets to achieve minimum energy. Answer: Minimizes surface area.

Q5. Calculate the height to which water rises in a capillary tube of radius 0.0001 m (surface tension = 0.072 N/m, density = 1000 kg/m³, g = 9.8 m/s², contact angle = 0°).

Solution: Capillary rise h = (2γ cosθ)/(ρgr). With γ = 0.072 N/m, θ = 0°, ρ = 1000 kg/m³, g = 9.8 m/s², r = 0.0001 m: h = (2 × 0.072 × 1)/(1000 × 9.8 × 0.0001) = 0.1469 m. Answer: 0.147 m.

Q6. What is the role of the contact angle in capillary action?

Solution: The contact angle θ determines whether a liquid rises or falls in a capillary tube. For θ < 90°, the liquid wets the surface and rises; for θ > 90°, it is non-wetting and falls. Answer: Determines rise or fall in capillary.

Q7. Calculate the excess pressure inside a soap bubble of radius 0.01 m (surface tension = 0.025 N/m).

Solution: For a soap bubble (two surfaces), excess pressure P = 4γ/r. With γ = 0.025 N/m, r = 0.01 m: P = 4 × 0.025 / 0.01 = 10 Pa. Answer: 10 Pa.

Q8. Why does surface tension decrease with increasing temperature?

Solution: As temperature increases, molecular kinetic energy rises, reducing cohesive forces between liquid molecules, which lowers surface tension. Answer: Reduced cohesive forces.

Q9. Calculate the work done to form a water droplet of radius 0.003 m (surface tension = 0.072 N/m).

Solution: Work done W = γ × ΔA, where ΔA = 4πr² for a droplet. With γ = 0.072 N/m, r = 0.003 m: ΔA = 4 × 3.1416 × (0.003)² = 1.131 × 10⁻⁴ m². W = 0.072 × 1.131 × 10⁻⁴ = 8.143 × 10⁻⁶ J. Answer: 8.14 × 10⁻⁶ J.

Q10. A liquid rises 0.05 m in a capillary tube of radius 0.0002 m (density = 800 kg/m³, g = 9.8 m/s², contact angle = 0°). Calculate the surface tension.

Solution: Capillary rise h = (2γ cosθ)/(ρgr). Rearrange: γ = (h ρ g r)/(2 cosθ). With h = 0.05 m, ρ = 800 kg/m³, g = 9.8 m/s², r = 0.0002 m, θ = 0°: γ = (0.05 × 800 × 9.8 × 0.0002)/(2 × 1) = 0.0392 N/m. Answer: 0.0392 N/m.

Chapter 10: Thermal Properties of Matter

Q1. Define thermal expansion and state the formula for linear expansion.

Solution: Thermal expansion is the tendency of a material to increase in size when its temperature rises. For linear expansion, the change in length is given by ΔL = L₀ α ΔT, where L₀ is the original length, α is the coefficient of linear expansion, and ΔT is the temperature change. Answer: Size increase with temperature; ΔL = L₀ α ΔT.

Q2. What is the SI unit of the coefficient of linear expansion?

Solution: The coefficient of linear expansion α is the fractional change in length per degree of temperature change, so its SI unit is per Kelvin (K⁻¹). Answer: K⁻¹.

Q3. Calculate the change in length of a steel rod of length 2 m when heated from 20°C to 120°C (coefficient of linear expansion for steel = 1.2 × 10⁻⁵ K⁻¹).

Solution: Using ΔL = L₀ α ΔT, with L₀ = 2 m, α = 1.2 × 10⁻⁵ K⁻¹, ΔT = 120 - 20 = 100 K: ΔL = 2 × 1.2 × 10⁻⁵ × 100 = 0.0024 m. Answer: 0.0024 m.

Q4. Why do some materials expand more than others when heated?

Solution: Materials expand differently due to variations in their coefficients of thermal expansion, which depend on the strength of intermolecular bonds and crystal structure. Weaker bonds allow greater expansion. Answer: Different coefficients of thermal expansion.

Q5. Calculate the change in area of an aluminum sheet of area 0.5 m² when heated from 25°C to 75°C (coefficient of area expansion = 4.8 × 10⁻⁵ K⁻¹).

Solution: Area expansion ΔA = A₀ β ΔT, where β ≈ 2α. With A₀ = 0.5 m², β = 4.8 × 10⁻⁵ K⁻¹, ΔT = 75 - 25 = 50 K: ΔA = 0.5 × 4.8 × 10⁻⁵ × 50 = 0.0012 m². Answer: 0.0012 m².

Q6. What is the relationship between the coefficients of linear, area, and volume expansion?

Solution: For a material, the coefficient of area expansion β ≈ 2α, and the coefficient of volume expansion γ ≈ 3α, where α is the coefficient of linear expansion, assuming small expansions. Answer: β ≈ 2α, γ ≈ 3α.

Q7. Calculate the change in volume of a copper cube with side length 0.1 m when heated from 15°C to 65°C (coefficient of volume expansion = 5.1 × 10⁻⁵ K⁻¹).

Solution: Volume expansion ΔV = V₀ γ ΔT, where V₀ = (0.1)³ = 0.001 m³, γ = 5.1 × 10⁻⁵ K⁻¹, ΔT = 65 - 15 = 50 K: ΔV = 0.001 × 5.1 × 10⁻⁵ × 50 = 2.55 × 10⁻⁶ m³. Answer: 2.55 × 10⁻⁶ m³.

Q8. Why are gaps left in railway tracks?

Solution: Gaps are left in railway tracks to allow for thermal expansion. As temperature rises, the tracks expand, and gaps prevent buckling by providing space for the expansion. Answer: To accommodate thermal expansion.

Q9. A brass rod of length 1.5 m is heated from 10°C to 110jabberwocky. Calculate the final length at 60°C (coefficient of linear expansion for brass = 1.8 × 10⁻⁵ K⁻¹).

Solution: Final length L = L₀ (1 + α ΔT). With L₀ = 1.5 m, α = 1.8 × 10⁻⁵ K⁻¹, ΔT = 60 - 10 = 50 K: L = 1.5 × (1 + 1.8 × 10⁻⁵ × 50) = 1.5 × 1.0009 = 1.50135 m. Change in length ΔL = 1.50135 - 1.5 = 0.00135 m. Answer: 0.00135 m.

Q10. Calculate the new volume of a liquid (initial volume 0.02 m³) heated from 20°C to 80°C (coefficient of volume expansion = 1.8 × 10⁻⁴ K⁻¹).

Solution: Volume expansion ΔV = V₀ γ ΔT. With V₀ = 0.02 m³, γ = 1.8 × 10⁻⁴ K⁻¹, ΔT = 80 - 20 = 60 K: ΔV = 0.02 × 1.8 × 10⁻⁴ × 60 = 0.000216 m³. New volume V = 0.02 + 0.000216 = 0.020216 m³. Answer: 0.0202 m³.

Q1. Define heat transfer and name its three main mechanisms.

Solution: Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. The three main mechanisms are conduction, convection, and radiation. Answer: Thermal energy movement; conduction, convection, radiation.

Q2. What is the SI unit of thermal conductivity?

Solution: Thermal conductivity k measures a material’s ability to conduct heat, given by Q = -kA(dT/dx). Its SI unit is W/(m·K), derived from watts per meter per Kelvin. Answer: W/(m·K).

Q3. Calculate the rate of heat transfer through a copper rod of length 0.5 m and cross-sectional area 0.01 m², with a temperature difference of 100°C (thermal conductivity of copper = 400 W/(m·K)).

Solution: Using Fourier’s law, Q = kA(ΔT/L). With k = 400 W/(m·K), A = 0.01 m², ΔT = 100 K, L = 0.5 m: Q = 400 × 0.01 × (100/0.5) = 800 W. Answer: 800 W.

Q4. Why is convection more effective in liquids and gases than in solids?

Solution: Convection involves the movement of fluid particles, which carry heat. Liquids and gases are mobile, allowing bulk fluid motion, while solids lack this mobility, relying on conduction. Answer: Fluid particle movement.

Q5. Calculate the heat flux through a wall of thickness 0.2 m, area 2 m², with a temperature difference of 20°C (thermal conductivity = 0.15 W/(m·K)).

Solution: Heat flux q = Q/A = k(ΔT/L). With k = 0.15 W/(m·K), ΔT = 20 K, L = 0.2 m: q = 0.15 × (20/0.2) = 15 W/m². Total heat transfer Q = q × A = 15 × 2 = 30 W. Answer: 30 W.

Q6. What is Stefan-Boltzmann’s law, and what does it describe?

Solution: Stefan-Boltzmann’s law states that the power radiated by a black body is proportional to the fourth power of its temperature: P = σAT⁴, where σ is the Stefan-Boltzmann constant, A is area, and T is absolute temperature. It describes thermal radiation. Answer: P = σAT⁴; thermal radiation.

Q7. Calculate the power radiated by a black body of area 0.1 m² at 500 K (Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/(m²·K⁴)).

Solution: Using P = σAT⁴, with σ = 5.67 × 10⁻⁸ W/(m²·K⁴), A = 0.1 m², T = 500 K: P = 5.67 × 10⁻⁸ × 0.1 × (500)⁴ = 354.375 W. Answer: 354 W.

Q8. Why do insulators like wood have lower thermal conductivity than metals?

Solution: Metals have free electrons that efficiently transfer heat, while insulators like wood lack free electrons and have tightly bound structures, reducing heat conduction. Answer: Fewer free electrons in insulators.

Q9. Calculate the convective heat transfer rate from a surface of area 1 m² at 350 K to air at 300 K (convective heat transfer coefficient = 25 W/(m²·K)).

Solution: Convective heat transfer Q = hA(ΔT), where h is the convective coefficient. With h = 25 W/(m²·K), A = 1 m², ΔT = 350 - 300 = 50 K: Q = 25 × 1 × 50 = 1250 W. Answer: 1250 W.

Q10. A rod with one end at 200°C and the other at 50°C has a length of 1 m, area 0.02 m², and thermal conductivity 50 W/(m·K). Calculate the temperature at 0.4 m from the hot end.

Solution: For steady-state conduction, temperature varies linearly: T(x) = T_hot - (T_hot - T_cold)(x/L). With T_hot = 200°C, T_cold = 50°C, x = 0.4 m, L = 1 m: T = 200 - (200 - 50) × (0.4/1) = 200 - 60 = 140°C. Answer: 140°C.

Chapter 11: Thermodynamics

Q1. State the Zeroth Law of Thermodynamics.

Solution: The Zeroth Law of Thermodynamics states that if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This defines the concept of temperature. Answer: Thermal equilibrium defines temperature.

Q2. What is the First Law of Thermodynamics, and what does it imply about energy conservation?

Solution: The First Law states that the change in internal energy of a system (ΔU) equals the heat added (Q) minus the work done by the system (W): ΔU = Q - W. It implies energy is conserved. Answer: ΔU = Q - W; energy conservation.

Q3. Calculate the change in internal energy of a system that absorbs 500 J of heat and does 200 J of work on its surroundings.

Solution: Using the First Law, ΔU = Q - W. With Q = 500 J, W = 200 J: ΔU = 500 - 200 = 300 J. Answer: 300 J.

Q4. Why can’t a heat engine be 100% efficient according to the Second Law of Thermodynamics?

Solution: The Second Law states that some energy must be rejected as waste heat to a cold reservoir, as entropy must increase in an isolated system. This prevents all heat from being converted to work. Answer: Entropy increase requires waste heat.

Q5. Calculate the efficiency of a Carnot engine operating between reservoirs at 600 K and 300 K.

Solution: Carnot efficiency η = 1 - (T_C/T_H), where T_C and T_H are the cold and hot reservoir temperatures. With T_H = 600 K, T_C = 300 K: η = 1 - (300/600) = 0.5 or 50%. Answer: 50%.

Q6. What is the Second Law of Thermodynamics in terms of entropy?

Solution: The Second Law states that the total entropy of an isolated system or the universe never decreases; it either increases or remains constant in reversible processes. Answer: Entropy never decreases.

Q7. Calculate the entropy change when 1 kg of water at 373 K (100°C) completely vaporizes at its boiling point (latent heat of vaporization = 2.26 × 10⁶ J/kg).

Solution: Entropy change ΔS = Q/T, where Q = mL_v for phase change. With m = 1 kg, L_v = 2.26 × 10⁶ J/kg, T = 373 K: ΔS = (1 × 2.26 × 10⁶)/373 = 6059 J/K. Answer: 6060 J/K.

Q8. What does the Third Law of Thermodynamics state about absolute zero?

Solution: The Third Law states that the entropy of a perfect crystal at absolute zero (0 K) is zero, as there is no disorder. Answer: Entropy is zero at 0 K.

Q9. A gas expands isothermally at 300 K, absorbing 400 J of heat. Calculate the work done by the gas.

Solution: For an isothermal process, ΔU = 0 (ideal gas). By the First Law, ΔU = Q - W, so W = Q. With Q = 400 J: W = 400 J. Answer: 400 J.

Q10. Calculate the maximum work obtainable from a heat engine absorbing 1000 J of heat from a hot reservoir at 500 K and rejecting heat to a cold reservoir at 250 K.

Solution: Maximum work is given by Carnot efficiency: W = Q_H × η, where η = 1 - (T_C/T_H). With Q_H = 1000 J, T_H = 500 K, T_C = 250 K: η = 1 - (250/500) = 0.5. W = 1000 × 0.5 = 500 J. Answer: 500 J.

Q1. Define an isothermal process and state the condition for it.

Solution: An isothermal process is a thermodynamic process where the temperature of the system remains constant (ΔT = 0). For an ideal gas, this implies ΔU = 0, and Q = W. Answer: Constant temperature; ΔT = 0.

Q2. What is an adiabatic process, and how does it differ from an isothermal process?

Solution: An adiabatic process occurs with no heat exchange between the system and surroundings (Q = 0), so ΔU = -W. Unlike an isothermal process (constant T), temperature may change in an adiabatic process. Answer: No heat exchange; temperature may change.

Q3. Calculate the work done by an ideal gas expanding isothermally at 300 K from 0.01 m³ to 0.02 m³ at a constant pressure of 2 × 10⁵ Pa.

Solution: For an isothermal process at constant pressure (isobaric), work done W = PΔV. With P = 2 × 10⁵ Pa, ΔV = 0.02 - 0.01 = 0.01 m³: W = 2 × 10⁵ × 0.01 = 2000 J. Answer: 2000 J.

Q4. Why does no work occur in an isochoric process?

Solution: An isochoric process occurs at constant volume (ΔV = 0). Since work done W = PΔV, no work is done if ΔV = 0. Answer: Constant volume; ΔV = 0.

Q5. Calculate the heat added to an ideal gas in an isochoric process where its internal energy increases by 1500 J.

Solution: In an isochoric process, W = 0 (ΔV = 0). By the First Law, ΔU = Q - W, so Q = ΔU. With ΔU = 1500 J: Q = 1500 J. Answer: 1500 J.

Q6. What is the work done in an adiabatic expansion of an ideal gas where the internal energy decreases by 800 J?

Solution: In an adiabatic process, Q = 0, so by the First Law, ΔU = Q - W = -W. If ΔU = -800 J: W = -ΔU = -(-800) = 800 J. Answer: 800 J.

Q7. Calculate the work done by 1 mole of an ideal gas expanding isothermally at 400 K from a pressure of 5 × 10⁵ Pa to 2 × 10⁵ Pa.

Solution: For isothermal expansion, W = nRT ln(V₂/V₁). Since P₁V₁ = P₂V₂, V₂/V₁ = P₁/P₂. With n = 1 mol, R = 8.314 J/(mol·K), T = 400 K, P₁ = 5 × 10⁵ Pa, P₂ = 2 × 10⁵ Pa: W = 1 × 8.314 × 400 × ln(5 × 10⁵ / 2 × 10⁵) = 3325.6 × ln(2.5) ≈ 3050 J. Answer: 3050 J.

Q8. What is an isobaric process, and how does it affect the internal energy of an ideal gas?

Solution: An isobaric process occurs at constant pressure. For an ideal gas, ΔU = nC_V ΔT, where C_V is the molar specific heat at constant volume, and ΔT depends on heat and work. Answer: Constant pressure; ΔU = nC_V ΔT.

Q9. Calculate the change in internal energy for 2 moles of an ideal gas in an isobaric process at 10⁵ Pa, expanding from 0.03 m³ to 0.05 m³ with a temperature increase from 300 K to 350 K (C_V = 3R/2).

Solution: For an ideal gas, ΔU = nC_V ΔT. With n = 2 mol, C_V = 3 × 8.314 / 2 = 12.471 J/(mol·K), ΔT = 350 - 300 = 50 K: ΔU = 2 × 12.471 × 50 = 1247.1 J. Answer: 1250 J.

Q10. Calculate the heat added to 1 mole of an ideal gas during an adiabatic compression where the work done on the gas is 600 J.

Solution: In an adiabatic process, Q = 0. By the First Law, ΔU = Q - W = -W. Since work is done on the gas, W = -600 J (negative as per convention). Thus, ΔU = -(-600) = 600 J, but Q = 0. Answer: 0 J.

Chapter 12: Kinetic Theory

Q1. What is the kinetic theory of gases?

Solution: The kinetic theory of gases explains gas behavior as the motion of molecules that move randomly, collide elastically with each other and the container walls, and have negligible volume compared to the container. Answer: Gas behavior due to molecular motion.

Q2. What is the formula for the root mean square speed of gas molecules?

Solution: The root mean square speed (v_rms) of gas molecules is given by v_rms = √(3RT/M), where R is the gas constant, T is the absolute temperature, and M is the molar mass in kg/mol. Answer: v_rms = √(3RT/M).

Q3. Calculate the root mean square speed of nitrogen molecules (molar mass 28 g/mol) at 300 K (R = 8.314 J/(mol·K)).

Solution: Using v_rms = √(3RT/M), with R = 8.314 J/(mol·K), T = 300 K, M = 0.028 kg/mol: v_rms = √((3 × 8.314 × 300)/0.028) = √(267428.57) ≈ 517.14 m/s. Answer: 517 m/s.

Q4. Why do gas molecules exert pressure on the walls of a container according to the kinetic theory?

Solution: Gas molecules exert pressure due to their random motion and elastic collisions with the container walls, transferring momentum to the walls. Answer: Momentum transfer from collisions.

Q5. Calculate the average kinetic energy of a gas molecule in an ideal gas at 400 K (k = 1.38 × 10⁻²³ J/K).

Solution: The average kinetic energy per molecule is (3/2)kT. With k = 1.38 × 10⁻²³ J/K, T = 400 K: KE = (3/2) × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J. Answer: 8.28 × 10⁻²¹ J.

Q6. How does the kinetic theory relate the pressure of an ideal gas to molecular properties?

Solution: The kinetic theory gives P = (1/3)(N/V)m(v_rms)², where P is pressure, N/V is the number density, m is the mass of a molecule, and v_rms is the root mean square speed. Answer: P = (1/3)(N/V)m(v_rms)².

Q7. Calculate the total kinetic energy of 2 moles of an ideal gas at 500 K (R = 8.314 J/(mol·K)).

Solution: Total kinetic energy for an ideal gas is (3/2)nRT. With n = 2 mol, R = 8.314 J/(mol·K), T = 500 K: KE = (3/2) × 2 × 8.314 × 500 = 12471 J. Answer: 12470 J.

Q8. Why does the average kinetic energy of gas molecules increase with temperature?

Solution: The average kinetic energy is proportional to absolute temperature (KE = (3/2)kT). As temperature increases, molecular speed increases, raising kinetic energy. Answer: Proportional to absolute temperature.

Q9. Calculate the mean free path of a gas molecule with a collision cross-section of 0.4 nm² and number density of 2.5 × 10²⁵ m⁻³.

Solution: Mean free path λ = 1/(√2 π σ N), where σ is the collision cross-section and N is number density. With σ = 0.4 × 10⁻¹⁸ m², N = 2.5 × 10²⁵ m⁻³: λ = 1/(√2 × 3.1416 × 0.4 × 10⁻¹⁸ × 2.5 × 10²⁵) ≈ 2.25 × 10⁻⁷ m. Answer: 2.25 × 10⁻⁷ m.

Q10. Calculate the temperature at which oxygen molecules (molar mass 32 g/mol) have the same v_rms as nitrogen molecules (molar mass 28 g/mol) at 300 K.

Solution: For equal v_rms, (3RT_1/M_1) = (3RT_2/M_2). With M_1 = 0.028 kg/mol (nitrogen), T_1 = 300 K, M_2 = 0.032 kg/mol (oxygen): T_2 = T_1 × (M_2/M_1) = 300 × (0.032/0.028) = 342.86 K. Answer: 343 K.

Q1. Define specific heat capacity and state its formula.

Solution: Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K, given by c = Q/(m ΔT), where Q is heat added, m is mass, and ΔT is temperature change. Answer: Heat per unit mass per degree; c = Q/(m ΔT).

Q2. What is the SI unit of specific heat capacity?

Solution: Specific heat capacity is heat energy per unit mass per unit temperature change, so its SI unit is J/(kg·K). Answer: J/(kg·K).

Q3. Calculate the heat required to raise the temperature of 2 kg of water from 20°C to 80°C (specific heat capacity of water = 4186 J/(kg·K)).

Solution: Using Q = m c ΔT, with m = 2 kg, c = 4186 J/(kg·K), ΔT = 80 - 20 = 60 K: Q = 2 × 4186 × 60 = 502320 J. Answer: 502 kJ.

Q4. What is the difference between specific heat at constant volume (C_v) and constant pressure (C_p) for an ideal gas?

Solution: For an ideal gas, C_p is the heat capacity at constant pressure, and C_v at constant volume. The difference is C_p - C_v = R, where R is the gas constant, due to work done during expansion at constant pressure. Answer: C_p - C_v = R.

Q5. Calculate C_p for a monatomic ideal gas if C_v = 12.47 J/(mol·K) (R = 8.314 J/(mol·K)).

Solution: Using C_p = C_v + R, with C_v = 12.47 J/(mol·K), R = 8.314 J/(mol·K): C_p = 12.47 + 8.314 = 20.784 J/(mol·K). Answer: 20.8 J/(mol·K).

Q6. What is molar specific heat capacity, and how is it related to specific heat capacity?

Solution: Molar specific heat capacity is the heat required to raise the temperature of 1 mole of a substance by 1 K, given by C = c × M, where c is specific heat capacity and M is molar mass. Answer: C = c × M.

Q7. Calculate the temperature change when 1000 J of heat is added to 0.5 kg of iron (specific heat capacity = 450 J/(kg·K)).

Solution: Using ΔT = Q/(m c), with Q = 1000 J, m = 0.5 kg, c = 450 J/(kg·K): ΔT = 1000 / (0.5 × 450) = 4.444 K. Answer: 4.44 K.

Q8. Why does water have a high specific heat capacity compared to other substances?

Solution: Water's high specific heat capacity is due to hydrogen bonding between molecules, which requires more energy to increase molecular motion and temperature. Answer: Hydrogen bonding.

Q9. Calculate the molar specific heat capacity at constant volume for a diatomic ideal gas (R = 8.314 J/(mol·K)).

Solution: For a diatomic ideal gas, C_v = (5/2)R. With R = 8.314 J/(mol·K): C_v = (5/2) × 8.314 = 20.785 J/(mol·K). Answer: 20.8 J/(mol·K).

Q10. A 1 kg block of copper (c = 385 J/(kg·K)) absorbs 770 J of heat. Calculate the final temperature if it starts at 25°C.

Solution: Using ΔT = Q/(m c), with Q = 770 J, m = 1 kg, c = 385 J/(kg·K): ΔT = 770 / 385 = 2 K. Final T = 25 + 2 = 27°C. Answer: 27°C.

Chapter 13: Oscillations

Q1. Define simple harmonic motion and state its defining equation.

Solution: Simple harmonic motion (SHM) is oscillatory motion where the restoring force is directly proportional to the displacement and acts opposite to it. The defining equation is F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement. Answer: Oscillatory motion with F = -kx.

Q2. What is the formula for the period of a mass-spring system in SHM?

Solution: The period T of a mass-spring system in SHM is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Answer: T = 2π√(m/k).

Q3. Calculate the period of a 0.5 kg mass attached to a spring with a spring constant of 50 N/m.

Solution: Using T = 2π√(m/k), with m = 0.5 kg, k = 50 N/m: T = 2 × 3.1416 × √(0.5/50) = 2 × 3.1416 × √(0.01) = 0.6283 s. Answer: 0.628 s.

Q4. Why is the motion of a simple pendulum considered approximately simple harmonic?

Solution: The motion of a simple pendulum is approximately SHM for small angles because the restoring force (component of gravity) is nearly proportional to displacement. Answer: Restoring force nearly proportional for small angles.

Q5. Calculate the frequency of a simple pendulum of length 1 m (g = 9.8 m/s²).

Solution: The period of a simple pendulum is T = 2π√(L/g). Frequency f = 1/T. With L = 1 m, g = 9.8 m/s²: T = 2 × 3.1416 × √(1/9.8) = 2.006 s, f = 1/2.006 ≈ 0.4985 Hz. Answer: 0.499 Hz.

Q6. What is the total mechanical energy in SHM, and how is it conserved?

Solution: Total mechanical energy in SHM is E = (1/2)kA², where k is the spring constant and A is the amplitude. It is conserved as kinetic and potential energies interchange without loss in an ideal system. Answer: E = (1/2)kA²; conserved via energy interchange.

Q7. Calculate the maximum velocity of a 0.2 kg mass in SHM with amplitude 0.1 m and spring constant 100 N/m.

Solution: Maximum velocity v_max = Aω, where ω = √(k/m). With A = 0.1 m, k = 100 N/m, m = 0.2 kg: ω = √(100/0.2) = √500 ≈ 22.36 rad/s. v_max = 0.1 × 22.36 = 2.236 m/s. Answer: 2.24 m/s.

Q8. Why does the period of a simple pendulum not depend on its mass?

Solution: The period T = 2π√(L/g) depends only on length L and gravity g. Mass cancels out in the equation of motion, as gravitational force and inertia scale proportionally. Answer: Mass cancels in the equation.

Q9. Calculate the potential energy of a mass in SHM at a displacement of 0.05 m, with a spring constant of 200 N/m.

Solution: Potential energy in SHM is U = (1/2)kx². With k = 200 N/m, x = 0.05 m: U = (1/2) × 200 × (0.05)² = 0.5 × 200 × 0.0025 = 0.25 J. Answer: 0.25 J.

Q10. Calculate the angular frequency of a 0.3 kg mass attached to a spring with a spring constant of 75 N/m.

Solution: Angular frequency ω = √(k/m). With k = 75 N/m, m = 0.3 kg: ω = √(75/0.3) = √250 ≈ 15.81 rad/s. Answer: 15.8 rad/s.

Q1. What is a simple pendulum, and what condition makes its motion simple harmonic?

Solution: A simple pendulum is a mass (bob) suspended from a fixed point by a light, inextensible string, oscillating under gravity. Its motion is simple harmonic for small angular displacements where the restoring force is approximately proportional to displacement. Answer: Mass on a string; small angles.

Q2. What is the formula for the period of a simple pendulum?

Solution: The period T of a simple pendulum is given by T = 2π√(L/g), where L is the length of the string and g is the acceleration due to gravity. Answer: T = 2π√(L/g).

Q3. Calculate the period of a simple pendulum with a length of 0.8 m (g = 9.8 m/s²).

Solution: Using T = 2π√(L/g), with L = 0.8 m, g = 9.8 m/s²: T = 2 × 3.1416 × √(0.8/9.8) = 2 × 3.1416 × 0.2857 ≈ 1.795 s. Answer: 1.80 s.

Q4. Why does the period of a simple pendulum not depend on the mass of the bob?

Solution: The period T = 2π√(L/g) depends only on length L and gravity g. The mass cancels out in the equation of motion, as gravitational force and inertia scale proportionally. Answer: Mass cancels in the equation.

Q5. Calculate the frequency of a simple pendulum with a length of 1.2 m (g = 9.8 m/s²).

Solution: Period T = 2π√(L/g), frequency f = 1/T. With L = 1.2 m, g = 9.8 m/s²: T = 2 × 3.1416 × √(1.2/9.8) ≈ 2.199 s, f = 1/2.199 ≈ 0.4547 Hz. Answer: 0.455 Hz.

Q6. How does the period of a simple pendulum change if its length is quadrupled?

Solution: The period T = 2π√(L/g). If length L is quadrupled (L' = 4L), T' = 2π√(4L/g) = 2 × 2π√(L/g) = 2T. The period doubles. Answer: Period doubles.

Q7. Calculate the length of a simple pendulum that has a period of 2 s (g = 9.8 m/s²).

Solution: Using T = 2π√(L/g), rearrange: L = (T² g)/(4π²). With T = 2 s, g = 9.8 m/s²: L = (2² × 9.8)/(4 × 3.1416²) = 39.2/39.478 ≈ 0.9929 m. Answer: 0.993 m.

Q8. Why is the simple pendulum’s motion only approximately simple harmonic?

Solution: The restoring force (mg sinθ) is proportional to sinθ, not θ. For small angles, sinθ ≈ θ, making the force approximately proportional to displacement, satisfying SHM conditions. Answer: Sinθ approximates θ for small angles.

Q9. Calculate the maximum angular velocity of a simple pendulum with a maximum angular displacement of 0.1 rad and length 1 m (g = 9.8 m/s²).

Solution: Maximum angular velocity ω_max = θ_max × √(g/L). With θ_max = 0.1 rad, L = 1 m, g = 9.8 m/s²: ω_max = 0.1 × √(9.8/1) ≈ 0.1 × 3.1305 = 0.3131 rad/s. Answer: 0.313 rad/s.

Q10. Calculate the potential energy of a 0.5 kg pendulum bob at its maximum displacement of 0.05 m from the vertical (g = 9.8 m/s²).

Solution: Potential energy U = mgh, where h is the vertical height. For small angles, h = L(1 - cosθ) ≈ L(θ²/2). With θ = x/L = 0.05/1 = 0.05 rad, L = 1 m: h ≈ 1 × (0.05²/2) = 0.00125 m. U = 0.5 × 9.8 × 0.00125 = 0.006125 J. Answer: 0.00613 J.

Chapter 14: Waves

Q1. What is a "seconds pendulum"?

Solution: A "seconds pendulum" is defined as a pendulum that has a period of exactly two seconds. This means it takes one second to swing from one side to the other (a half-period). Answer: A pendulum with a period of 2 seconds.

Q2. What is the formula for the angular frequency (ω) of a simple pendulum?

Solution: The period $T$ is related to angular frequency $ \omega $ by $ T = 2\pi/\omega $. Since $ T = 2\pi\sqrt{L/g} $, we can set them equal: $ 2\pi/\omega = 2\pi\sqrt{L/g} $. Simplifying gives $ 1/\omega = \sqrt{L/g} $, so $ \omega = \sqrt{g/L} $. Answer: $ \omega = \sqrt{g/L} $.

Q3. How would a pendulum's period on the Moon compare to its period on Earth? (g_moon ≈ g_earth / 6)

Solution: The period $ T = 2\pi\sqrt{L/g} $. Since $ g $ is in the denominator, a smaller value of $ g $ results in a longer period. As gravity on the Moon is about 1/6th that of Earth, the period will be longer. Specifically, $ T_{moon} = \sqrt{6} \times T_{earth} \approx 2.45 $ times longer. Answer: The period would be longer on the Moon.

Q4. Calculate the value of g if a 1.0 m pendulum is found to have a period of 2.01 s.

Solution: Start with $ T = 2\pi\sqrt{L/g} $. Square both sides: $ T^2 = 4\pi^2(L/g) $. Rearrange for g: $ g = 4\pi^2L / T^2 $. With $ L = 1.0 $ m and $ T = 2.01 $ s: $ g = (4 \times 3.1416^2 \times 1.0) / 2.01^2 = 39.478 / 4.0401 \approx 9.772 $ m/s². Answer: 9.77 m/s².

Q5. What provides the restoring force for a simple pendulum?

Solution: The weight of the bob ($mg$) acts vertically downwards. This force has two components: one along the string and one tangent to the arc of motion. The tangential component, $ mg \sin\theta $, always points towards the equilibrium position and acts as the restoring force. Answer: The tangential component of gravity.

Q6. Calculate the maximum speed of a 0.2 kg pendulum bob if it's released from a height of 0.1 m above its lowest point (g = 9.8 m/s²).

Solution: By conservation of energy, the maximum potential energy ($U = mgh$) equals the maximum kinetic energy ($K = \frac{1}{2}mv^2$). So, $mgh_{max} = \frac{1}{2}mv_{max}^2$. Mass cancels, giving $v_{max} = \sqrt{2gh}$. $v_{max} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4$ m/s. Answer: 1.4 m/s.

Q7. How does the period of a simple pendulum change if its length is reduced to one-ninth of its original value?

Solution: The period $ T $ is proportional to the square root of the length ($ T \propto \sqrt{L} $). If the new length is $ L' = L/9 $, the new period will be $ T' \propto \sqrt{L/9} = (\sqrt{L}) / \sqrt{9} = (\sqrt{L}) / 3 $. The period becomes one-third of the original. Answer: The period becomes one-third.

Q8. A pendulum has a period of 4.0 s. How many complete oscillations will it make in 2 minutes?

Solution: The frequency $f$ is the number of oscillations per second, so $ f = 1/T = 1/4.0 = 0.25 $ Hz. Two minutes is equal to 120 seconds. The total number of oscillations is $ N = f \times t = 0.25 \times 120 = 30 $. Answer: 30 oscillations.

Q9. What is the total mechanical energy of a simple pendulum at the lowest point of its swing?

Solution: The total mechanical energy ($E = K + U$) is conserved. At the lowest point of the swing (the equilibrium position), the height is zero, so the potential energy ($U$) is zero. Therefore, the total mechanical energy is purely kinetic energy ($E = K_{max}$). Answer: It is equal to its maximum kinetic energy.

Q10. Calculate the maximum tension in the string of a 1.5 m pendulum with a 0.5 kg bob, if its maximum speed is 2.0 m/s (g = 9.8 m/s²).

Solution: The maximum tension occurs at the lowest point of the swing. Here, the tension ($T_{max}$) must provide the centripetal force ($mv^2/L$) and also support the weight of the bob ($mg$). So, $ T_{max} = mg + mv^2/L $. $ T_{max} = (0.5 \times 9.8) + (0.5 \times 2.0^2 / 1.5) = 4.9 + (2 / 1.5) \approx 4.9 + 1.333 = 6.233 $ N. Answer: 6.23 N.

Question 1: Standing Waves - Insert question here

Q1. How is a standing wave formed?

Solution: A standing wave is formed from the superposition (interference) of two identical waves that have the same amplitude and frequency but are travelling in opposite directions in the same medium. Answer: Interference of two identical opposing waves.

Q2. What are nodes and antinodes in a standing wave?

Solution: Nodes are points along the standing wave where the medium has zero amplitude or displacement. Antinodes are points of maximum amplitude, where the medium oscillates most intensely. Answer: Nodes: points of zero amplitude; Antinodes: points of maximum amplitude.

Q3. What is the distance between two consecutive nodes or two consecutive antinodes?

Solution: The distance between any two consecutive nodes is equal to the distance between any two consecutive antinodes. This distance is exactly half a wavelength ($ \lambda/2 $). Answer: Half a wavelength ($ \lambda/2 $).

Q4. A 1.5 m long string is fixed at both ends and vibrates in its fundamental mode. If the wave speed is 300 m/s, what is the frequency?

Solution: For the fundamental mode on a string fixed at both ends, the length of the string is half a wavelength ($ L = \lambda/2 $). So, $ \lambda = 2L = 2 \times 1.5 = 3.0 $ m. The frequency is $ f = v/\lambda = 300 \, \text{m/s} / 3.0 \, \text{m} = 100 $ Hz. Answer: 100 Hz.

Q5. What is the frequency of the third harmonic for the string in the previous question?

Solution: The harmonics on a string fixed at both ends are integer multiples of the fundamental frequency ($ f_n = n f_1 $). The third harmonic ($ n=3 $) is three times the fundamental frequency: $ f_3 = 3 \times f_1 = 3 \times 100 \, \text{Hz} = 300 $ Hz. Answer: 300 Hz.

Q6. An organ pipe is open at both ends and has a length of 0.68 m. Find its fundamental frequency (speed of sound = 340 m/s).

Solution: For a pipe open at both ends, the fundamental wavelength is $ \lambda = 2L $. So, $ \lambda = 2 \times 0.68 \, \text{m} = 1.36 $ m. The frequency is $ f = v/\lambda = 340 \, \text{m/s} / 1.36 \, \text{m} = 250 $ Hz. Answer: 250 Hz.

Q7. If a pipe of the same length (0.68 m) is closed at one end, what is its new fundamental frequency?

Solution: For a pipe closed at one end, the fundamental wavelength is $ \lambda = 4L $. So, $ \lambda = 4 \times 0.68 \, \text{m} = 2.72 $ m. The frequency is $ f = v/\lambda = 340 \, \text{m/s} / 2.72 \, \text{m} = 125 $ Hz. Answer: 125 Hz.

Q8. Does a standing wave transfer energy?

Solution: A pure standing wave does not transfer energy from one end of the medium to the other. The energy is "trapped" and oscillates between kinetic and potential forms within each segment bounded by the nodes. Answer: No, there is no net energy transfer.

Q9. Why can a pipe closed at one end only produce odd harmonics?

Solution: The physical constraints require a displacement node at the closed end and an antinode at the open end. This boundary condition is only satisfied for wavelengths where the pipe's length is an odd multiple of a quarter wavelength ($ L = n\lambda/4 $ for $ n=1, 3, 5, \ldots $), which results in odd harmonics only. Answer: The boundary conditions (node at closed end, antinode at open end).

Q10. The tension in a guitar string is doubled. By what factor does its fundamental frequency change?

Solution: The wave speed on a string is $ v = \sqrt{T/\mu} $, and frequency $ f \propto v $. Therefore, the frequency is proportional to the square root of the tension ($ f \propto \sqrt{T} $). If the tension is doubled ($ T' = 2T $), the new frequency will be $ f' \propto \sqrt{2T} = \sqrt{2} \times \sqrt{T} $. The frequency increases by a factor of $ \sqrt{2} $. Answer: It increases by a factor of $ \sqrt{2} $ (approx 1.414).

Class 12 Physics

Chapter 1: Electric Charges and Fields

Question 1: Coulomb's Law - Calculate the electrostatic force between two point charges of +4 μC and -2 μC separated by 0.2 m in a vacuum. (2018)

Answer 1: Using Coulomb's Law, F = (k * |q₁q₂|) / r², where k = 9 × 10⁹ N·m²/C², q₁ = 4 × 10⁻⁶ C, q₂ = -2 × 10⁻⁶ C, r = 0.2 m. F = (9 × 10⁹ * 8 × 10⁻¹²) / 0.04 = 1.8 N (attractive).

Question 2: Coulomb's Law - How does the electrostatic force between two charges change if the distance between them is tripled? (2019)

Answer 2: F ∝ 1/r². If distance is tripled ($r' = 3r$), the new force $F' \propto 1/(3r)^2 = 1/(9r^2)$. Therefore, the force becomes F/9 (one-ninth of the original force).

Question 3: Coulomb's Law - Two charges +q and -q are separated by distance d. Find the force on a charge +Q placed at their midpoint. (2020)

Answer 3: The force due to +q is attractive, directed towards -q. The force due to -q is also attractive, directed towards -q. Since the magnitudes of the forces are equal (due to equal charges and equal distance $d/2$) and both forces point in the same direction (towards -q), the net force is not zero. Net force $F_{net} = 2 \times F$, directed towards the negative charge (-q).

Question 4: Coulomb's Law - State Coulomb’s Law and give its mathematical form. (2017)

Answer 4: Coulomb’s Law: The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, acting along the line joining the two charges. Mathematical Form: $F = \frac{k |q_1 q_2|}{r^2}$ where $k = \frac{1}{4\pi\epsilon_0}$.

Question 5: Coulomb's Law - What is the force between two charges of 1 μC each, 1 m apart in air? (2021)

Answer 5: F = (9 × 10⁹ * 1 × 10⁻⁶ * 1 × 10⁻⁶) / 1² = $9 \times 10^{-3} \text{ N}$ (or 0.009 N, repulsive).

Question 6: Coulomb's Law - What happens to the force between two charges if the magnitude of both charges is doubled? (2016)

Answer 6: F ∝ q₁q₂. If $q_1$ becomes $2q_1$ and $q_2$ becomes $2q_2$, the new force $F' \propto (2q_1)(2q_2) = 4(q_1 q_2)$. Therefore, the force becomes 4F (four times the original force).

Question 7: Coulomb's Law - What is the SI unit and value of the Coulomb constant? (2022)

Answer 7: SI unit of Coulomb constant ($k$): $\text{N}\cdot\text{m}^2/\text{C}^2$. Value in vacuum/air: $\mathbf{k = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2}$.

Question 8: Coulomb's Law - Why does the electrostatic force between two charges obey the inverse square law? (2015)

Answer 8: The force obeys the inverse square law because the electric field lines originating from a point charge spread out uniformly in all directions. The intensity of the field (and thus the force) is distributed over the area of a sphere ($4\pi r^2$), which grows proportionally to $\mathbf{r^2}$. Thus, $F \propto 1/\text{Area} \propto \mathbf{1/r^2}$.

Question 9: Coulomb's Law - If two charges are placed in a medium with dielectric constant 3, how is the force affected compared to a vacuum? (2023)

Answer 9: Force in medium $\mathbf{F_m = \frac{F_{vacuum}}{K}}$, where $K$ is the dielectric constant. For $K = 3$, $F_m = \frac{F_{vacuum}}{3}$. The force is reduced by a factor of 3.

Question 10: Coulomb's Law - What is the nature of the force between two negative charges, and how does it vary with distance? (2014)

Answer 10: The force between two negative charges (like charges) is repulsive. It varies inversely with the square of the distance between them ($F \propto 1/r^2$).

Question 1: Electric Field - Calculate the electric field at a point 0.1 m from a point charge of +5 μC in a vacuum. (2018)

Answer 1: E = (k * |q|) / r², where k = 9 × 10⁹ N·m²/C², q = 5 × 10⁻⁶ C, r = 0.1 m. E = (9 × 10⁹ * 5 × 10⁻⁶) / 0.01 = 4.5 × 10⁶ N/C, directed away from the charge.

Question 2: Electric Field - Define electric field and give its SI unit. (2019)

Answer 2: Electric field is the force per unit positive charge at a point. SI unit: N/C or V/m.

Question 3: Electric Field - What is the electric field at the midpoint of two equal positive charges +q separated by distance d? (2020)

Answer 3: Net electric field is zero. Fields due to both charges are equal in magnitude and opposite in direction.

Question 4: Electric Field - Find the electric field at a distance r from an infinite line charge with linear charge density λ. (2017)

Answer 4: E = (2kλ) / r, where k = 9 × 10⁹ N·m²/C², directed radially outward for positive λ.

Question 5: Electric Field - What is the electric field intensity at a point on the axial line of an electric dipole at a large distance r? (2021)

Answer 5: E = (2kp) / r³, where p is the dipole moment, k = 9 × 10⁹ N·m²/C², directed along the dipole axis.

Question 6: Electric Field - A charge of 2 μC is placed in a uniform electric field of 100 N/C. Calculate the force on the charge. (2016)

Answer 6: F = qE = 2 × 10⁻⁶ C * 100 N/C = 0.0002 N.

Question 7: Electric Field - Why is the electric field inside a conductor zero in electrostatic conditions? (2022)

Answer 7: Charges reside on the surface, and excess charge redistributes to cancel the internal field.

Question 8: Electric Field - What is the relation between electric field (E) and electric potential (V) at a point? (2015)

Answer 8: E = -dV/dr, where E is the negative gradient of the potential V with respect to distance r.

Question 9: Electric Field - Calculate the electric field due to a point charge of -3 μC at a distance of 0.3 m in air. (2023)

Answer 9: E = (9 × 10⁹ * 3 × 10⁻⁶) / 0.09 = 3 × 10⁵ N/C, directed toward the charge.

Question 10: Electric Field - What is the direction of the electric field due to a negative point charge? (2014)

Answer 10: The electric field is directed toward the negative charge.

Question 1: Gauss's Theorem - State Gauss's Law in electrostatics. (2018)

Answer 1: Gauss's Law states that the total electric flux through a closed surface is equal to 1/ε₀ times the total charge enclosed. Φ = q_enc / ε₀.

Question 2: Gauss's Theorem - Calculate the electric flux through a spherical surface enclosing a charge of +5 μC. (2019)

Answer 2: Φ = q_enc / ε₀ = (5 × 10⁻⁶) / (8.85 × 10⁻¹²) = 5.65 × 10⁵ Nm²/C.

Question 3: Gauss's Theorem - What is the electric flux through a closed surface enclosing no charge? (2020)

Answer 3: Φ = 0, as per Gauss's Law, since no charge is enclosed (q_enc = 0).

Question 4: Gauss's Theorem - Why is the electric field inside a uniformly charged spherical shell zero? (2017)

Answer 4: By Gauss's Law, no charge is enclosed inside the shell, so the electric flux and field are zero.

Question 5: Gauss's Theorem - Find the electric field due to an infinite plane sheet with surface charge density σ. (2021)

Answer 5: Using Gauss's Law, E = σ / (2ε₀), independent of distance from the plane.

Question 6: Gauss's Theorem - A point charge of -2 μC is at the center of a cube. What is the flux through one face? (2016)

Answer 6: Total flux Φ = q_enc / ε₀ = (-2 × 10⁻⁶) / (8.85 × 10⁻¹²) = -2.26 × 10⁵ Nm²/C. Flux through one face = Φ / 6 = -3.77 × 10⁴ Nm²/C.

Question 7: Gauss's Theorem - What is the SI unit of electric flux? (2022)

Answer 7: SI unit of electric flux is Nm²/C.

Question 8: Gauss's Theorem - Can the electric flux through a closed surface be negative? Explain. (2015)

Answer 8: Yes, if the enclosed charge is negative, the flux is negative as per Gauss's Law (Φ = q_enc / ε₀).

Question 9: Gauss's Theorem - Find the electric field at a distance r from an infinite line charge with linear charge density λ. (2023)

Answer 9: Using Gauss's Law, E = (2kλ) / r, where k = 1 / (4πε₀), directed radially outward.

Question 10: Gauss's Theorem - What is the electric flux through a Gaussian surface enclosing a dipole? (2014)

Answer 10: Net flux is zero, as the dipole has equal and opposite charges, so q_enc = 0.

Chapter 2: Electrostatic Potential and Capacitance

Question 1: Electric Potential - Define electric potential and state its SI unit. (2018)

Answer 1: Electric potential is the work done per unit positive charge to bring it from infinity to a point. SI unit: Volt (V).

Question 2: Electric Potential - Calculate the electric potential at 0.3 m from a point charge of +2 μC in a vacuum. (2019)

Answer 2: V = (k * q) / r, where k = 9 × 10⁹ N·m²/C², q = 2 × 10⁻⁶ C, r = 0.3 m. V = (9 × 10⁹ * 2 × 10⁻⁶) / 0.3 = 6 × 10⁴ V.

Question 3: Electric Potential - Find the potential at the midpoint of two charges +q and -q separated by distance d. (2020)

Answer 3: V = (k * q) / (d/2) + (k * -q) / (d/2) = kq / (d/2) - kq / (d/2) = 0.

Question 4: Electric Potential - What is the relationship between electric field and electric potential? (2017)

Answer 4: E = -dV/dr, where E is the electric field and V is the potential, with negative gradient indicating direction.

Question 5: Electric Potential - Calculate the work done in moving a 3 μC charge across a potential difference of 50 V. (2021)

Answer 5: W = qΔV = 3 × 10⁻⁶ C * 50 V = 1.5 × 10⁻⁴ J.

Question 6: Electric Potential - What is the electric potential inside a charged conducting sphere? (2016)

Answer 6: Potential is constant and equal to the surface potential, V = kq / R, where R is the radius.

Question 7: Electric Potential - Find the potential due to an electric dipole at a point on its axial line at distance r. (2022)

Answer 7: V = kp / r², where p is the dipole moment, k = 9 × 10⁹ N·m²/C².

Question 8: Electric Potential - Calculate the potential at 0.4 m from a point charge of -5 μC in air. (2023)

Answer 8: V = (9 × 10⁹ * -5 × 10⁻⁶) / 0.4 = -1.125 × 10⁵ V.

Question 9: Electric Potential - Why is the electric potential constant inside a charged conductor? (2015)

Answer 9: Electric field is zero inside a conductor, so no work is done to move a charge, making potential constant.

Question 10: Electric Potential - What is the potential energy of a system of two charges +2q and -q separated by distance d? (2014)

Answer 10: U = (k * 2q * -q) / d = -2kq² / d, where k = 9 × 10⁹ N·m²/C².

Question 1: Capacitance and Capacitors - Define capacitance and state its SI unit. (2018)

Answer 1: Capacitance is the ability of a capacitor to store charge per unit potential difference. SI unit: Farad (F).

Question 2: Capacitance and Capacitors - Calculate the capacitance of a parallel plate capacitor with plate area 0.02 m², separation 2 mm, and air as the dielectric. (2019)

Answer 2: C = ε₀A / d, where ε₀ = 8.85 × 10⁻¹² F/m, A = 0.02 m², d = 2 × 10⁻³ m. C = (8.85 × 10⁻¹² * 0.02) / (2 × 10⁻³) = 8.85 × 10⁻¹¹ F.

Question 3: Capacitance and Capacitors - What is the equivalent capacitance of two capacitors of 4 μF and 6 μF connected in series? (2020)

Answer 3: 1/C_eq = 1/C₁ + 1/C₂ = 1/4 + 1/6 = 5/12. C_eq = 12/5 = 2.4 μF.

Question 4: Capacitance and Capacitors - How does the capacitance of a parallel plate capacitor change if a dielectric of constant K = 3 is inserted between the plates? (2017)

Answer 4: Capacitance increases by a factor of K. New capacitance C' = KC = 3C.

Question 5: Capacitance and Capacitors - Find the charge stored in a 5 μF capacitor when connected to a 20 V battery. (2021)

Answer 5: Q = CV = 5 × 10⁻⁶ F * 20 V = 1 × 10⁻⁴ C = 100 μC.

Question 6: Capacitance and Capacitors - What is the equivalent capacitance of two capacitors of 2 μF each connected in parallel? (2016)

Answer 6: C_eq = C₁ + C₂ = 2 μF + 2 μF = 4 μF.

Question 7: Capacitance and Capacitors - What is the energy stored in a capacitor of capacitance 10 μF charged to 50 V? (2022)

Answer 7: U = ½CV² = ½ * 10 × 10⁻⁶ F * (50)² = 1.25 × 10⁻² J.

Question 8: Capacitance and Capacitors - Why does the potential difference across capacitors in series remain different? (2015)

Answer 8: Charge is the same on each capacitor, but capacitance differs, so V = Q/C varies.

Question 9: Capacitance and Capacitors - A parallel plate capacitor has a capacitance of 100 pF. What is the effect on capacitance if the plate separation is doubled? (2023)

Answer 9: C ∝ 1/d. If separation doubles, capacitance becomes C/2 = 50 pF.

Question 10: Capacitance and Capacitors - What is the potential difference across a 3 μF capacitor when it stores a charge of 60 μC? (2014)

Answer 10: V = Q/C = 60 × 10⁻⁶ C / 3 × 10⁻⁶ F = 20 V.

Chapter 3: Current Electricity

Question 1: Ohm's Law - State Ohm's Law and give its mathematical expression. (2018)

Answer 1: Ohm's Law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. V = IR.

Question 2: Ohm's Law - Calculate the current through a 12 Ω resistor when a potential difference of 24 V is applied across it. (2019)

Answer 2: I = V/R = 24 V / 12 Ω = 2 A.

Question 3: Ohm's Law - Find the resistance of a conductor if a current of 0.4 A flows through it when 8 V is applied. (2020)

Answer 3: R = V/I = 8 V / 0.4 A = 20 Ω.

Question 4: Ohm's Law - Why is Ohm's Law not applicable to semiconductors? (2017)

Answer 4: Semiconductors have non-linear V-I characteristics, so voltage and current are not directly proportional.

Question 5: Ohm's Law - Calculate the potential difference across a 6 Ω resistor carrying a current of 2 A. (2021)

Answer 5: V = IR = 2 A * 6 Ω = 12 V.

Question 6: Ohm's Law - What is the SI unit of resistance, and how is it derived from Ohm's Law? (2016)

Answer 6: SI unit of resistance is Ohm (Ω). From V = IR, R = V/I, so 1 Ω = 1 V/A.

Question 7: Ohm's Law - A wire has a resistance of 10 Ω. What is the new resistance if its length is tripled? (2022)

Answer 7: R ∝ l. If length triples, resistance becomes 3R = 3 * 10 = 30 Ω.

Question 8: Ohm's Law - Find the current through a circuit with a 9 V battery and a 3 Ω resistor. (2023)

Answer 8: I = V/R = 9 V / 3 Ω = 3 A.

Question 9: Ohm's Law - Why does the resistance of a metallic conductor increase with temperature? (2015)

Answer 9: Increased temperature causes more lattice vibrations, reducing electron mobility, thus increasing resistance.

Question 10: Ohm's Law - A 12 V source is connected to a resistor, and 0.2 A current flows. Calculate the resistance. (2014)

Answer 10: R = V/I = 12 V / 0.2 A = 60 Ω.

Question 1: Kirchhoff's Rules - State Kirchhoff's First Law and explain its physical basis. (2018)

Answer 1: Kirchhoff's First Law states that the algebraic sum of currents entering a junction equals the sum of currents leaving it (ΣI_in = ΣI_out). It is based on the conservation of charge, as charge cannot accumulate at a junction in a steady-state circuit.

Question 2: Kirchhoff's Rules - State Kirchhoff's Second Law and explain its physical basis. (2019)

Answer 2: Kirchhoff's Second Law states that the algebraic sum of potential differences around a closed loop is zero (ΣΔV = 0). It is based on the conservation of energy, ensuring the total energy gained equals energy lost in a loop.

Question 3: Kirchhoff's Rules - A circuit has a 12 V battery and three resistors of 3 Ω, 4 Ω, and 5 Ω in series. Find the current using Kirchhoff's Second Law. (2020)

Answer 3: Total resistance R = 3 Ω + 4 Ω + 5 Ω = 12 Ω. By Kirchhoff's Second Law, sum of potential drops equals the battery emf: V = IR. Thus, I = V/R = 12 V / 12 Ω = 1 A.

Question 4: Kirchhoff's Rules - Why does Kirchhoff's First Law hold in a circuit junction? (2017)

Answer 4: Kirchhoff's First Law holds because charge is conserved. The total current entering a junction must equal the total current leaving it, as charges cannot accumulate at a junction in steady-state conditions.

Question 5: Kirchhoff's Rules - A circuit has a 10 V battery and two parallel branches with resistors 4 Ω and 8 Ω. Find the current through the 4 Ω resistor using Kirchhoff's Rules. (2021)

Answer 5: In parallel, voltage across each resistor is 10 V. Using Kirchhoff's Second Law for the 4 Ω branch, V = IR. Current I = V/R = 10 V / 4 Ω = 2.5 A through the 4 Ω resistor.

Question 6: Kirchhoff's Rules - In a loop with a 15 V battery, the potential drops across resistors total 12 V. Find the potential rise due to other sources using Kirchhoff's Second Law. (2016)

Answer 6: By Kirchhoff's Second Law, ΣΔV = 0 in a closed loop. Total potential rise = total potential drop. Given battery emf = 15 V and drops = 12 V, other sources contribute 15 V - 12 V = 3 V.

Question 7: Kirchhoff's Rules - At a junction, currents of 3 A and 2 A enter, and a current of 4 A leaves. Find the other current leaving the junction using Kirchhoff's First Law. (2022)

Answer 7: By Kirchhoff's First Law, ΣI_in = ΣI_out. Currents entering = 3 A + 2 A = 5 A. Currents leaving = 4 A + I. Thus, I = 5 A - 4 A = 1 A.

Question 8: Kirchhoff's Rules - Explain the sign convention for emf and potential drop in Kirchhoff's Second Law. (2015)

Answer 8: In Kirchhoff's Second Law, emf is positive when traversing from the negative to the positive terminal of a source. Potential drop (IR) is positive when traversing in the direction of current across a resistor.

Question 9: Kirchhoff's Rules - A circuit has a 9 V battery and resistors 3 Ω and 6 Ω in series. Find the potential drop across the 6 Ω resistor using Kirchhoff's Rules. (2023)

Answer 9: Total R = 3 Ω + 6 Ω = 9 Ω. Current I = V/R = 9 V / 9 Ω = 1 A (Kirchhoff's Second Law). Potential drop across 6 Ω = IR = 1 A * 6 Ω = 6 V.

Question 10: Kirchhoff's Rules - At a junction, currents of 1.5 A and 2.5 A enter, and currents of 2 A and I leave. Find I using Kirchhoff's First Law. (2014)

Answer 10: By Kirchhoff's First Law, ΣI_in = ΣI_out. Currents entering = 1.5 A + 2.5 A = 4 A. Currents leaving = 2 A + I. Thus, I = 4 A - 2 A = 2 A.

Question 1: Wheatstone Bridge - What is a Wheatstone bridge and its primary use? (2020)

Answer 1: A Wheatstone bridge is a circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. Its primary use is to determine unknown resistances with high precision.

Question 2: Wheatstone Bridge - State the condition for a Wheatstone bridge to be balanced. (2019)

Answer 2: A Wheatstone bridge is balanced when the ratio of resistances in one leg equals the ratio in the other leg, i.e., R1/R2 = R3/Rx, where Rx is the unknown resistance, resulting in zero current through the galvanometer.

Question 3: Wheatstone Bridge - In a Wheatstone bridge, R1 = 10 Ω, R2 = 20 Ω, R3 = 15 Ω, and Rx is unknown. Find Rx if the bridge is balanced. (2021)

Answer 3: For a balanced Wheatstone bridge, R1/R2 = R3/Rx. Given R1 = 10 Ω, R2 = 20 Ω, R3 = 15 Ω, we have 10/20 = 15/Rx. Solving, Rx = 15 * 20 / 10 = 30 Ω.

Question 4: Wheatstone Bridge - Explain why no current flows through the galvanometer in a balanced Wheatstone bridge. (2018)

Answer 4: In a balanced Wheatstone bridge, the potential difference across the galvanometer is zero because the voltage drops across the resistors in each leg are proportional (R1/R2 = R3/Rx), making the junction points equipotential.

Question 5: Wheatstone Bridge - A Wheatstone bridge has R1 = 5 Ω, R2 = 10 Ω, R3 = 8 Ω, and Rx. If a 12 V battery is connected, find Rx for balance. (2022)

Answer 5: For balance, R1/R2 = R3/Rx. Given R1 = 5 Ω, R2 = 10 Ω, R3 = 8 Ω, we have 5/10 = 8/Rx. Solving, Rx = 8 * 10 / 5 = 16 Ω. The battery voltage does not affect the balance condition.

Question 6: Wheatstone Bridge - In a Wheatstone bridge, the galvanometer shows no deflection when R1 = 12 Ω, R2 = 24 Ω, and R3 = 18 Ω. Calculate Rx. (2017)

Answer 6: For a balanced bridge, R1/R2 = R3/Rx. Given R1 = 12 Ω, R2 = 24 Ω, R3 = 18 Ω, we have 12/24 = 18/Rx. Solving, Rx = 18 * 24 / 12 = 36 Ω.

Question 7: Wheatstone Bridge - What happens if the Wheatstone bridge is unbalanced? (2016)

Answer 7: If a Wheatstone bridge is unbalanced (R1/R2 ≠ R3/Rx), a potential difference exists across the galvanometer, causing current to flow through it, indicating the bridge is not in equilibrium.

Question 8: Wheatstone Bridge - A Wheatstone bridge has a 9 V battery, R1 = 6 Ω, R2 = 12 Ω, and R3 = 4 Ω. Find Rx for no current in the galvanometer. (2023)

Answer 8: For balance, R1/R2 = R3/Rx. Given R1 = 6 Ω, R2 = 12 Ω, R3 = 4 Ω, we have 6/12 = 4/Rx. Solving, Rx = 4 * 12 / 6 = 8 Ω.

Question 9: Wheatstone Bridge - Explain the role of the galvanometer in a Wheatstone bridge. (2015)

Answer 9: The galvanometer in a Wheatstone bridge detects current flow between the bridge’s junction points. When the bridge is balanced, it shows no deflection, indicating zero current and equal potential across the junctions.

Question 10: Wheatstone Bridge - In a Wheatstone bridge, R1 = 20 Ω, R2 = 40 Ω, R3 = 25 Ω, and Rx is adjusted until the bridge balances. Find Rx. (2014)

Answer 10: For a balanced bridge, R1/R2 = R3/Rx. Given R1 = 20 Ω, R2 = 40 Ω, R3 = 25 Ω, we have 20/40 = 25/Rx. Solving, Rx = 25 * 40 / 20 = 50 Ω.

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

Question 1: What is electromagnetic induction? (2024)

Answer 1: Electromagnetic induction is the process of generating an EMF in a conductor due to a changing magnetic flux through it.

Question 2: What causes electromagnetic induction? (2023)

Answer 2: A change in magnetic flux through a conductor, due to relative motion or changing magnetic field, causes electromagnetic induction.

Question 3: Name two devices that work on electromagnetic induction. (2022)

Answer 3: Transformers and electric generators operate based on electromagnetic induction.

Question 4: Define magnetic flux and its SI unit. (2021)

Answer 4: Magnetic flux is the number of magnetic field lines passing through an area, given by \( \Phi = B \cdot A \cdot \cos \theta \). SI unit: Weber (Wb).

Question 5: Why is relative motion important in electromagnetic induction? (2020)

Answer 5: Relative motion changes the magnetic flux through a conductor, inducing an EMF as per Faraday’s laws.

Question 6: What is the role of a changing magnetic field in induction? (2019)

Answer 6: A changing magnetic field alters the magnetic flux, inducing an EMF in a nearby conductor.

Question 7: Give an example of electromagnetic induction in daily life. (2018)

Answer 7: Induction cooktops use electromagnetic induction to heat cookware.

Question 8: What is the significance of magnetic flux in induction? (2017)

Answer 8: Magnetic flux determines the extent of induced EMF, as its rate of change is proportional to the EMF.

Question 9: Why does a stationary conductor in a steady magnetic field not induce EMF? (2016)

Answer 9: No change in magnetic flux occurs, so no EMF is induced per Faraday’s laws.

Question 10: How is electromagnetic induction used in wireless charging? (2015)

Answer 10: A changing magnetic field in a charging coil induces an EMF in a device’s coil, enabling wireless power transfer.

Question 1: State Faraday’s first law of electromagnetic induction. (2024)

Answer 1: Faraday’s first law states that a change in magnetic flux through a circuit induces an EMF in it.

Question 2: What is Faraday’s second law? (2023)

Answer 2: Faraday’s second law states that the induced EMF is proportional to the rate of change of magnetic flux, \( \mathcal{E} = - \frac{d\Phi_B}{dt} \).

Question 3: A coil of 100 turns has a flux change of 0.02 Wb in 0.1 s. Find the EMF. (2022)

Answer 3: Induced EMF, \( \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} = -100 \cdot \frac{0.02}{0.1} = -20 \, \text{V} \). Magnitude = 20 V.

Question 4: Why is the negative sign used in Faraday’s law? (2021)

Answer 4: The negative sign indicates that the induced EMF opposes the change in flux, per Lenz’s law.

Question 5: A loop’s flux changes from 0.05 Wb to 0.03 Wb in 0.02 s. Calculate EMF for 50 turns. (2020)

Answer 5: \( \Delta \Phi = 0.03 - 0.05 = -0.02 \, \text{Wb} \). EMF, \( \mathcal{E} = -50 \cdot \frac{-0.02}{0.02} = 50 \, \text{V} \).

Question 6: What factors affect the induced EMF per Faraday’s laws? (2019)

Answer 6: Number of turns, rate of change of magnetic flux, and coil area affect the induced EMF.

Question 7: A coil of 20 turns in a 0.1 T field drops to 0 T in 0.05 s (area 0.01 m²). Find EMF. (2018)

Answer 7: \( \Delta \Phi = 0.1 \cdot 0.01 = 0.001 \, \text{Wb} \). EMF, \( \mathcal{E} = -20 \cdot \frac{0.001}{0.05} = -0.4 \, \text{V} \). Magnitude = 0.4 V.

Question 8: How is Faraday’s law applied in a generator? (2017)

Answer 8: Rotating a coil in a magnetic field changes flux, inducing EMF as per Faraday’s laws.

Question 9: A 200-turn coil has flux change of 0.015 Wb in 0.03 s. Calculate EMF. (2016)

Answer 9: EMF, \( \mathcal{E} = -200 \cdot \frac{0.015}{0.03} = -100 \, \text{V} \). Magnitude = 100 V.

Question 10: Why does induced EMF depend on the rate of flux change? (2015)

Answer 10: Faraday’s second law states that EMF is proportional to the rate of change of magnetic flux.

Question 1: What is induced EMF? (2024)

Answer 1: Induced EMF is the voltage generated in a conductor due to a changing magnetic flux.

Question 2: Define induced current. (2023)

Answer 2: Induced current is the current produced in a closed circuit due to induced EMF from changing flux.

Question 3: A coil with 10 V EMF and 5 Ω resistance. Find induced current. (2022)

Answer 3: Induced current, \( I = \frac{\mathcal{E}}{R} = \frac{10}{5} = 2 \, \text{A} \).

Question 4: A loop (area 0.02 m²) in 0.3 T field reverses in 0.1 s. Find EMF for 30 turns. (2021)

Answer 4: \( \Delta \Phi = 2 \cdot 0.3 \cdot 0.02 = 0.012 \, \text{Wb} \). EMF, \( \mathcal{E} = -30 \cdot \frac{0.012}{0.1} = -3.6 \, \text{V} \). Magnitude = 3.6 V.

Question 5: How is induced current related to resistance? (2020)

Answer 5: Induced current is inversely proportional to resistance, given by \( I = \frac{\mathcal{E}}{R} \).

Question 6: A 50-turn coil (area 0.05 m²) in 0.4 T drops to 0 T in 0.2 s. Find EMF. (2019)

Answer 6: \( \Delta \Phi = 0.4 \cdot 0.05 = 0.02 \, \text{Wb} \). EMF, \( \mathcal{E} = -50 \cdot \frac{0.02}{0.2} = -5 \, \text{V} \). Magnitude = 5 V.

Question 7: Why does induced current require a closed circuit? (2018)

Answer 7: A closed circuit allows the induced EMF to drive a current through a complete path.

Question 8: A 20 V EMF is induced in a coil with 10 Ω resistance. Find current. (2017)

Answer 8: Current, \( I = \frac{\mathcal{E}}{R} = \frac{20}{10} = 2 \, \text{A} \).

Question 9: A 40-turn coil has flux change of 0.01 Wb in 0.04 s. Find EMF. (2016)

Answer 9: EMF, \( \mathcal{E} = -40 \cdot \frac{0.01}{0.04} = -10 \, \text{V} \). Magnitude = 10 V.

Question 10: What determines the direction of induced current? (2015)

Answer 10: The direction of induced current is determined by Lenz’s law, opposing the flux change.

Question 1: State Lenz’s law. (2024)

Answer 1: Lenz’s law states that the induced EMF and current oppose the change in magnetic flux causing them.

Question 2: How does Lenz’s law conserve energy? (2023)

Answer 2: Lenz’s law ensures induced current opposes flux change, requiring work, thus conserving energy.

Question 3: A magnet approaches a coil. What is the induced current’s direction? (2022)

Answer 3: The induced current creates a magnetic field opposing the magnet’s approach, per Lenz’s law.

Question 4: A magnet is pulled from a coil. What is the induced current’s direction? (2021)

Answer 4: The induced current creates a field to attract the magnet back, opposing flux decrease.

Question 5: Why is Lenz’s law important in electromagnetic induction? (2020)

Answer 5: Lenz’s law determines the direction of induced EMF, ensuring physical consistency with energy conservation.

Question 6: A coil’s flux increases. What is the induced current’s effect? (2019)

Answer 6: The induced current produces a magnetic field to reduce the increasing flux, per Lenz’s law.

Question 7: How does Lenz’s law apply to an electric generator? (2018)

Answer 7: Induced current in a generator opposes rotor motion, requiring external work, per Lenz’s law.

Question 8: A loop’s flux decreases by 0.02 Wb in 0.1 s. Find EMF for 25 turns. (2017)

Answer 8: EMF, \( \mathcal{E} = -25 \cdot \frac{0.02}{0.1} = -5 \, \text{V} \). Magnitude = 5 V, direction per Lenz’s law.

Question 9: Why does induced EMF oppose flux change? (2016)

Answer 9: Lenz’s law ensures the induced EMF opposes flux change to maintain energy conservation.

Question 10: A coil rotates in a magnetic field. How does Lenz’s law apply? (2015)

Answer 10: The induced current opposes the rotational motion, requiring external torque to sustain rotation.

Question 1: What is self-induction? (2024)

Answer 1: Self-induction is the induction of EMF in a coil due to its own changing current.

Question 2: What is the SI unit of self-inductance? (2023)

Answer 2: The SI unit of self-inductance is Henry (H).

Question 3: A coil (0.5 H) has current change at 4 A/s. Find induced EMF. (2022)

Answer 3: EMF, \( \mathcal{E} = -L \frac{dI}{dt} = -0.5 \cdot 4 = -2 \, \text{V} \). Magnitude = 2 V.

Question 4: Why does a coil resist current change? (2021)

Answer 4: Self-induction produces an EMF opposing the current change, per Lenz’s law.

Question 5: A coil’s current changes from 2 A to 5 A in 0.1 s (L = 0.2 H). Find EMF. (2020)

Answer 5: \( \Delta I = 5 - 2 = 3 \, \text{A} \). EMF, \( \mathcal{E} = -0.2 \cdot \frac{3}{0.1} = -6 \, \text{V} \). Magnitude = 6 V.

Question 6: Define self-inductance. (2019)

Answer 6: Self-inductance is the property of a coil to oppose current change by inducing an EMF, measured in Henry.

Question 7: What factors affect self-inductance? (2018)

Answer 7: Self-inductance depends on the number of turns, coil geometry, and core material’s permeability.

Question 8: A coil (0.3 H) has current drop from 6 A to 2 A in 0.2 s. Find EMF. (2017)

Answer 8: \( \Delta I = 2 - 6 = -4 \, \text{A} \). EMF, \( \mathcal{E} = -0.3 \cdot \frac{-4}{0.2} = 6 \, \text{V} \).

Question 9: How is self-induction used in chokes? (2016)

Answer 9: Chokes use self-induction to oppose rapid current changes, filtering AC signals.

Question 10: Why is self-inductance called inertia of electricity? (2015)

Answer 10: Self-inductance resists changes in current, similar to inertia resisting changes in motion.

Question 1: What is mutual induction? (2024)

Answer 1: Mutual induction is the induction of EMF in one coil due to a changing current in a nearby coil.

Question 2: What is the SI unit of mutual inductance? (2023)

Answer 2: The SI unit of mutual inductance is Henry (H).

Question 3: Two coils (M = 0.4 H), primary current changes at 5 A/s. Find EMF in secondary. (2022)

Answer 3: EMF, \( \mathcal{E} = -M \frac{dI}{dt} = -0.4 \cdot 5 = -2 \, \text{V} \). Magnitude = 2 V.

Question 4: How does mutual induction occur in transformers? (2021)

Answer 4: Changing current in the primary coil induces EMF in the secondary coil via mutual induction.

Question 5: Coils (M = 0.1 H), primary current changes from 3 A to 7 A in 0.2 s. Find EMF. (2020)

Answer 5: \( \Delta I = 7 - 3 = 4 \, \text{A} \). EMF, \( \mathcal{E} = -0.1 \cdot \frac{4}{0.2} = -2 \, \text{V} \). Magnitude = 2 V.

Question 6: What factors affect mutual inductance? (2019)

Answer 6: Mutual inductance depends on the number of turns, coil geometry, distance, and core permeability.

Question 7: Coils (M = 0.25 H), primary current drops from 8 A to 4 A in 0.1 s. Find EMF. (2018)

Answer 7: \( \Delta I = 4 - 8 = -4 \, \text{A} \). EMF, \( \mathcal{E} = -0.25 \cdot \frac{-4}{0.1} = 10 \, \text{V} \).

Question 8: Why is mutual induction key in wireless power transfer? (2017)

Answer 8: Mutual induction allows a changing current in one coil to induce EMF in another, enabling wireless power transfer.

Question 9: Define the coefficient of mutual inductance. (2016)

Answer 9: It is the EMF induced in the secondary coil per unit rate of current change in the primary, measured in Henry.

Question 10: Coils (M = 0.3 H), primary current changes at 2 A/s. Find EMF. (2015)

Answer 10: EMF, \( \mathcal{E} = -0.3 \cdot 2 = -0.6 \, \text{V} \). Magnitude = 0.6 V.

Chapter 7: Alternating Current

Question 1: What is alternating current? (2024)

Answer 1: Alternating current (AC) is an electric current that periodically reverses direction, unlike direct current (DC).

Question 2: Define peak value of AC. (2023)

Answer 2: Peak value is the maximum value of alternating current or voltage in one cycle.

Question 3: What is the RMS value of AC? (2022)

Answer 3: RMS (Root Mean Square) value is the effective value of AC, equal to \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \), where \( I_0 \) is the peak current.

Question 4: Calculate the RMS voltage for a peak voltage of 141.4 V. (2021)

Answer 4: \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{141.4}{\sqrt{2}} \approx 100 \, \text{V} \).

Question 5: What is reactance in an AC circuit? (2020)

Answer 5: Reactance is the opposition offered by a capacitor or inductor to the flow of AC, measured in ohms.

Question 6: Define impedance in an AC circuit. (2019)

Answer 6: Impedance is the total opposition to AC flow, combining resistance and reactance, given by \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

Question 7: A circuit has resistance 3 Ω and inductive reactance 4 Ω. Find impedance. (2018)

Answer 7: Impedance, \( Z = \sqrt{R^2 + X_L^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \, \Omega \).

Question 8: What is the difference between resistance and reactance? (2017)

Answer 8: Resistance opposes current flow in both AC and DC, while reactance opposes AC due to inductance or capacitance.

Question 9: An AC has peak current 14.14 A. Find RMS current. (2016)

Answer 9: \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{14.14}{\sqrt{2}} \approx 10 \, \text{A} \).

Question 10: Why is RMS value used in AC circuits? (2015)

Answer 10: RMS value represents the equivalent DC value that produces the same power in a resistive load.

Question 1: What is an LCR series circuit? (2024)

Answer 1: An LCR series circuit consists of an inductor (L), capacitor (C), and resistor (R) connected in series with an AC source.

Question 2: What is resonance in an LCR circuit? (2023)

Answer 2: Resonance occurs when inductive reactance equals capacitive reactance (\( X_L = X_C \)), maximizing current at resonant frequency.

Question 3: Find the resonant frequency for L = 0.1 H, C = 100 μF. (2022)

Answer 3: Resonant frequency, \( f = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{0.1 \cdot 100 \times 10^{-6}}} \approx 50.3 \, \text{Hz} \).

Question 4: What is the power factor in an AC circuit? (2021)

Answer 4: Power factor is the cosine of the phase angle between voltage and current, given by \( \cos \phi = \frac{R}{Z} \).

Question 5: Calculate power factor for R = 3 Ω, Z = 5 Ω. (2020)

Answer 5: Power factor, \( \cos \phi = \frac{R}{Z} = \frac{3}{5} = 0.6 \).

Question 6: What is wattless current? (2019)

Answer 6: Wattless current is the component of AC that does not contribute to average power, flowing in purely reactive circuits.

Question 7: What is the phase difference in an LCR circuit at resonance? (2018)

Answer 7: At resonance, the phase difference between voltage and current is zero, as \( X_L = X_C \).

Question 8: Find average power in an AC circuit with V_rms = 100 V, I_rms = 2 A, cos φ = 0.8. (2017)

Answer 8: Average power, \( P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi = 100 \cdot 2 \cdot 0.8 = 160 \, \text{W} \).

Question 9: What is a phasor in an AC circuit? (2016)

Answer 9: A phasor is a rotating vector used to represent the amplitude and phase of AC voltage or current.

Question 10: Why is power zero in a purely inductive circuit? (2015)

Answer 10: In a purely inductive circuit, the phase difference is 90°, so \( \cos \phi = 0 \), making average power zero.

Question 1: What is an AC generator? (2024)

Answer 1: An AC generator converts mechanical energy into alternating electrical energy using electromagnetic induction.

Question 2: What is the principle of an AC generator? (2023)

Answer 2: It works on Faraday’s law, where a rotating coil in a magnetic field induces an alternating EMF.

Question 3: What is the expression for EMF in an AC generator? (2022)

Answer 3: EMF, \( \mathcal{E} = NBA \omega \sin(\omega t) \), where N is turns, B is magnetic field, A is area, and ω is angular frequency.

Question 4: A generator has 50 turns, B = 0.2 T, A = 0.01 m², f = 50 Hz. Find peak EMF. (2021)

Answer 4: Peak EMF, \( \mathcal{E}_0 = NBA \omega = 50 \cdot 0.2 \cdot 0.01 \cdot 2\pi \cdot 50 \approx 31.4 \, \text{V} \).

Question 5: What is the role of slip rings in an AC generator? (2020)

Answer 5: Slip rings provide continuous electrical contact to transfer alternating current from the rotating coil.

Question 6: Why does an AC generator produce alternating EMF? (2019)

Answer 6: The coil’s rotation in a magnetic field causes periodic flux changes, producing alternating EMF.

Question 7: A generator (100 turns, 0.1 T, 0.02 m²) rotates at 60 Hz. Find peak EMF. (2018)

Answer 7: Peak EMF, \( \mathcal{E}_0 = 100 \cdot 0.1 \cdot 0.02 \cdot 2\pi \cdot 60 \approx 75.4 \, \text{V} \).

Question 8: What factors affect the EMF of an AC generator? (2017)

Answer 8: EMF depends on the number of turns, magnetic field strength, coil area, and rotation frequency.

Question 9: How is mechanical energy converted in an AC generator? (2016)

Answer 9: Mechanical energy rotates the coil, inducing EMF via flux changes, converting it to electrical energy.

Question 10: Why is the EMF sinusoidal in an AC generator? (2015)

Answer 10: The sinusoidal variation of magnetic flux with coil rotation produces a sinusoidal EMF.

Question 1: What is an AC circuit? (2024)

Answer 1: An AC circuit is an electrical circuit powered by an alternating current or voltage source.

Question 2: What is inductive reactance? (2023)

Answer 2: Inductive reactance, \( X_L = \omega L \), is the opposition by an inductor to AC, measured in ohms.

Question 3: Calculate inductive reactance for L = 0.2 H, f = 50 Hz. (2022)

Answer 3: \( X_L = 2\pi f L = 2\pi \cdot 50 \cdot 0.2 \approx 62.8 \, \Omega \).

Question 4: What is capacitive reactance? (2021)

Answer 4: Capacitive reactance, \( X_C = \frac{1}{\omega C} \), is the opposition by a capacitor to AC, measured in ohms.

Question 5: Find capacitive reactance for C = 100 μF, f = 50 Hz. (2020)

Answer 5: \( X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \cdot 50 \cdot 100 \times 10^{-6}} \approx 31.8 \, \Omega \).

Question 6: A circuit has R = 4 Ω, X_L = 3 Ω, X_C = 2 Ω. Find impedance. (2019)

Answer 6: Impedance, \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{4^2 + (3 - 2)^2} = \sqrt{17} \approx 4.12 \, \Omega \).

Question 7: Why does current lead voltage in a capacitive circuit? (2018)

Answer 7: In a capacitive circuit, the capacitor charges before voltage peaks, causing current to lead by 90°.

Question 8: Why does current lag voltage in an inductive circuit? (2017)

Answer 8: In an inductive circuit, the inductor’s EMF opposes current change, causing current to lag by 90°.

Question 9: An AC circuit has V_rms = 220 V, Z = 11 Ω. Find RMS current. (2016)

Answer 9: RMS current, \( I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{220}{11} = 20 \, \text{A} \).

Question 10: What is the role of impedance in AC circuits? (2015)

Answer 10: Impedance determines the total opposition to AC flow, affecting current and power.

Question 1: What is a transformer? (2024)

Answer 1: A transformer is a device that transfers electrical energy between circuits using mutual induction, changing voltage levels.

Question 2: What is the principle of a transformer? (2023)

Answer 2: Transformers work on mutual induction, where a changing current in the primary coil induces EMF in the secondary coil.

Question 3: A transformer has N_p = 100, N_s = 500, V_p = 220 V. Find V_s. (2022)

Answer 3: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), so \( V_s = \frac{500}{100} \cdot 220 = 1100 \, \text{V} \).

Question 4: What is a step-up transformer? (2021)

Answer 4: A step-up transformer increases the secondary voltage compared to the primary, with \( N_s > N_p \).

Question 5: What is a step-down transformer? (2020)

Answer 5: A step-down transformer decreases the secondary voltage compared to the primary, with \( N_s < N_p \).

Question 6: A transformer has I_p = 10 A, I_s = 2 A, V_p = 200 V. Find V_s. (2019)

Answer 6: For ideal transformer, \( V_p I_p = V_s I_s \), so \( V_s = \frac{V_p I_p}{I_s} = \frac{200 \cdot 10}{2} = 1000 \, \text{V} \).

Question 7: Why are transformers used in power distribution? (2018)

Answer 7: Transformers step up voltage for efficient transmission and step down for safe distribution.

Question 8: What are energy losses in a transformer? (2017)

Answer 8: Energy losses include copper losses (I²R), core losses (hysteresis, eddy currents), and flux leakage.

Question 9: A transformer has N_p = 200, N_s = 50, V_s = 110 V. Find V_p. (2016)

Answer 9: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), so \( V_p = \frac{V_s N_p}{N_s} = \frac{110 \cdot 200}{50} = 440 \, \text{V} \).

Question 10: Why is the core of a transformer laminated? (2015)

Answer 10: Lamination reduces eddy current losses by increasing resistance to circulating currents in the core.

Chapter 8: Electromagnetic Waves

Question 1: What are electromagnetic waves? (2024)

Answer 1: Electromagnetic waves are oscillations of electric and magnetic fields, perpendicular to each other and to the direction of propagation, traveling at the speed of light.

Question 2: Why are electromagnetic waves transverse in nature? (2023)

Answer 2: Their electric and magnetic fields oscillate perpendicular to the direction of wave propagation, making them transverse.

Question 3: What is the speed of electromagnetic waves in vacuum? (2022)

Answer 3: The speed is \( c = 3 \times 10^8 \, \text{m/s} \).

Question 4: State one property of electromagnetic waves. (2021)

Answer 4: They can propagate through vacuum without requiring a medium.

Question 5: How are electric and magnetic fields related in electromagnetic waves? (2020)

Answer 5: Electric (\( \mathbf{E} \)) and magnetic (\( \mathbf{B} \)) fields are perpendicular, with \( E = cB \).

Question 6: Why do electromagnetic waves carry energy? (2019)

Answer 6: Oscillating electric and magnetic fields transport energy through space via wave propagation.

Question 7: What is the phase relationship between E and B fields in electromagnetic waves? (2018)

Answer 7: Electric and magnetic fields oscillate in phase, reaching maxima and minima simultaneously.

Question 8: Name one source of electromagnetic waves. (2017)

Answer 8: Accelerating charged particles produce electromagnetic waves.

Question 9: What is the significance of the transverse nature of electromagnetic waves? (2016)

Answer 9: The transverse nature allows polarization, used in applications like optical filters and communication.

Question 10: How do electromagnetic waves transfer momentum? (2015)

Answer 10: They carry momentum, exerting radiation pressure when absorbed or reflected by surfaces.

Question 1: What is the electromagnetic spectrum? (2024)

Answer 1: The electromagnetic spectrum is the range of all electromagnetic waves, arranged by frequency or wavelength, from radio waves to gamma rays.

Question 2: Which electromagnetic wave has the lowest frequency? (2023)

Answer 2: Radio waves have the lowest frequency in the electromagnetic spectrum.

Question 3: State one use of radio waves. (2022)

Answer 3: Radio waves are used for wireless communication in radio broadcasting.

Question 4: What is a common use of microwaves? (2021)

Answer 4: Microwaves are used for heating in microwave ovens.

Question 5: Name one application of infrared waves. (2020)

Answer 5: Infrared waves are used in thermal imaging cameras.

Question 6: What is the wavelength range of visible light? (2019)

Answer 6: Visible light ranges from approximately 400 nm to 700 nm.

Question 7: State a use of ultraviolet rays. (2018)

Answer 7: Ultraviolet rays are used for sterilization of medical equipment.

Question 8: What are X-rays used for? (2017)

Answer 8: X-rays are used for medical imaging, such as detecting bone fractures.

Question 9: Name one use of gamma rays. (2016)

Answer 9: Gamma rays are used in radiotherapy for cancer treatment.

Question 10: Arrange microwaves, visible light, and gamma rays by increasing frequency. (2015)

Answer 10: Microwaves, visible light, gamma rays (increasing frequency).

Question 1: What is displacement current? (2024)

Answer 1: Displacement current is the current due to a changing electric field, introduced by Maxwell to account for magnetic fields in non-conducting regions.

Question 2: Why was displacement current introduced? (2023)

Answer 2: Displacement current was introduced to maintain continuity of current in regions without conduction, like capacitors.

Question 3: What is the expression for displacement current? (2022)

Answer 3: Displacement current, \( I_d = \epsilon_0 \frac{d\Phi_E}{dt} \), where \( \Phi_E \) is the electric flux.

Question 4: How does displacement current relate to electromagnetic waves? (2021)

Answer 4: Displacement current produces magnetic fields in non-conducting regions, enabling electromagnetic wave propagation.

Question 5: Where is displacement current significant? (2020)

Answer 5: It is significant in capacitors, where changing electric fields between plates produce magnetic fields.

Question 6: Why is displacement current not a flow of charges? (2019)

Answer 6: It arises from the rate of change of electric flux, not physical movement of charges.

Question 7: How does displacement current modify Ampere’s law? (2018)

Answer 7: Maxwell added displacement current to Ampere’s law, making it \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I_c + I_d) \).

Question 8: Calculate displacement current for \( \frac{d\Phi_E}{dt} = 2 \times 10^6 \, \text{Vm/s} \), \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \). (2017)

Answer 8: \( I_d = \epsilon_0 \frac{d\Phi_E}{dt} = 8.85 \times 10^{-12} \cdot 2 \times 10^6 \approx 1.77 \times 10^{-5} \, \text{A} \).

Question 9: Why is displacement current crucial in a capacitor? (2016)

Answer 9: It accounts for the magnetic field between capacitor plates due to changing electric fields during charging.

Question 10: How does displacement current support Maxwell’s equations? (2015)

Answer 10: It completes Ampere’s law, ensuring consistency in predicting electromagnetic wave propagation.

Chapter 9: Ray Optics and Optical Instruments

Question 1: Reflection and Spherical Mirrors - State the laws of reflection of light. (2020)

Answer 1: The laws of reflection are: (1) The angle of incidence equals the angle of reflection. (2) The incident ray, reflected ray, and normal to the surface at the point of incidence lie in the same plane.

Question 2: Reflection and Spherical Mirrors - What is the mirror formula, and what do its terms represent? (2019)

Answer 2: The mirror formula is 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance, all measured from the mirror’s pole.

Question 3: Reflection and Spherical Mirrors - A concave mirror has a focal length of 20 cm. An object is placed 30 cm from the mirror. Find the image distance using the mirror formula. (2021)

Answer 3: Using the mirror formula, 1/f = 1/u + 1/v, with f = -20 cm (concave), u = -30 cm. Thus, 1/v = 1/(-20) - 1/(-30) = -1/20 + 1/30 = -3/60 + 2/60 = -1/60. So, v = -60 cm (image is 60 cm in front of the mirror).

Question 4: Reflection and Spherical Mirrors - Explain the sign convention for spherical mirrors. (2018)

Answer 4: For spherical mirrors, distances measured to the left of the pole are negative, to the right are positive. Focal length is negative for concave mirrors, positive for convex. Object distance (u) is typically negative, and image distance (v) is negative for real images, positive for virtual.

Question 5: Reflection and Spherical Mirrors - A convex mirror has a focal length of 15 cm. An object is placed 10 cm from the mirror. Find the image distance. (2022)

Answer 5: Using the mirror formula, 1/f = 1/u + 1/v, with f = +15 cm (convex), u = -10 cm. Thus, 1/v = 1/15 - 1/(-10) = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6. So, v = 6 cm (image is 6 cm behind the mirror).

Question 6: Reflection and Spherical Mirrors - What is the magnification of a concave mirror if the object is placed at the center of curvature? (2017)

Answer 6: For a concave mirror, if the object is at the center of curvature (u = 2f), the mirror formula gives v = -2f. Magnification m = -v/u = -(-2f)/(2f) = 1. The image is same-sized, real, and inverted.

Question 7: Reflection and Spherical Mirrors - A concave mirror forms a real image at 40 cm when the object is 60 cm away. Find the focal length. (2016)

Answer 7: Using the mirror formula, 1/f = 1/u + 1/v, with u = -60 cm, v = -40 cm (real image). Thus, 1/f = 1/(-60) + 1/(-40) = -1/60 - 1/40 = -2/120 - 3/120 = -5/120 = -1/24. So, f = -24 cm.

Question 8: Reflection and Spherical Mirrors - Why is a convex mirror used as a rear-view mirror in vehicles? (2015)

Answer 8: A convex mirror is used as a rear-view mirror because it provides a wide field of view, forms virtual, upright, and diminished images, and ensures objects are always in focus regardless of distance.

Question 9: Reflection and Spherical Mirrors - An object is placed 15 cm from a concave mirror of focal length 10 cm. Find the magnification. (2023)

Answer 9: Using the mirror formula, 1/f = 1/u + 1/v, with f = -10 cm, u = -15 cm. Thus, 1/v = 1/(-10) - 1/(-15) = -1/10 + 1/15 = -3/30 + 2/30 = -1/30. So, v = -30 cm. Magnification m = -v/u = -(-30)/(-15) = -2 (image is twice as large, inverted).

Question 10: Reflection and Spherical Mirrors - What type of mirror forms a virtual image for all object positions? (2014)

Answer 10: A convex mirror always forms a virtual, upright, and diminished image for all object positions, as the image is formed behind the mirror regardless of object distance.

Question 1: Refraction and Total Internal Reflection - State Snell’s law of refraction. (2020)

Answer 1: Snell’s law states that n1 sin(θ1) = n2 sin(θ2), where n1 and n2 are the refractive indices of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.

Question 2: Refraction and Total Internal Reflection - What is total internal reflection, and what are its conditions? (2019)

Answer 2: Total internal reflection occurs when light traveling from a denser to a rarer medium is incident at an angle greater than the critical angle, where θc = sin⁻¹(n2/n1), and n1 > n2. No refraction occurs, and all light is reflected back.

Question 3: Refraction and Total Internal Reflection - Calculate the critical angle for a glass-air interface (n_glass = 1.5, n_air = 1). (2021)

Answer 3: The critical angle θc = sin⁻¹(n2/n1), where n1 = 1.5 (glass), n2 = 1 (air). Thus, θc = sin⁻¹(1/1.5) = sin⁻¹(0.6667) ≈ 41.8°.

Question 4: Refraction and Total Internal Reflection - Explain how optical fibers use total internal reflection. (2018)

Answer 4: Optical fibers transmit light via total internal reflection. Light enters the denser core (high n) at an angle greater than the critical angle for the core-cladding interface, reflecting repeatedly within the core without loss, enabling data transmission.

Question 5: Refraction and Total Internal Reflection - A light ray is incident at 30° on a glass slab (n = 1.6) in air (n = 1). Find the angle of refraction. (2022)

Answer 5: Using Snell’s law, n1 sin(θ1) = n2 sin(θ2), with n1 = 1, θ1 = 30°, n2 = 1.6. Thus, sin(θ2) = (1 × sin(30°))/1.6 = 0.5/1.6 = 0.3125. So, θ2 = sin⁻¹(0.3125) ≈ 18.2°.

Question 6: Refraction and Total Internal Reflection - Find the critical angle for a water-air interface (n_water = 1.33, n_air = 1). (2017)

Answer 6: The critical angle θc = sin⁻¹(n2/n1), where n1 = 1.33 (water), n2 = 1 (air). Thus, θc = sin⁻¹(1/1.33) = sin⁻¹(0.7519) ≈ 48.8°.

Question 7: Refraction and Total Internal Reflection - A ray is incident at 45° on a glass-air interface (n_glass = 1.5). Does total internal reflection occur? (2016)

Answer 7: Critical angle θc = sin⁻¹(1/1.5) ≈ 41.8°. Since the incident angle (45°) is greater than θc, total internal reflection occurs, and no light refracts into the air.

Question 8: Refraction and Total Internal Reflection - What is the refractive index of a medium if the critical angle at its interface with air is 30°? (2015)

Answer 8: For critical angle θc = 30°, sin(θc) = n2/n1 = 1/n (n_air = 1). Thus, n = 1/sin(30°) = 1/0.5 = 2. The refractive index of the medium is 2.

Question 9: Refraction and Total Internal Reflection - A light ray refracts from air (n = 1) to glass (n = 1.5) at an angle of incidence of 60°. Find the angle of refraction. (2023)

Answer 9: Using Snell’s law, n1 sin(θ1) = n2 sin(θ2), with n1 = 1, θ1 = 60°, n2 = 1.5. Thus, sin(θ2) = (1 × sin(60°))/1.5 = 0.866/1.5 ≈ 0.5774. So, θ2 = sin⁻¹(0.5774) ≈ 35.3°.

Question 10: Refraction and Total Internal Reflection - Why is total internal reflection important in optical fibers? (2014)

Answer 10: Total internal reflection ensures light remains confined within the optical fiber’s core, preventing signal loss and enabling efficient, long-distance transmission of light signals in telecommunications.

Question 1: Lenses and Optical Properties - State the lens maker’s formula and its significance. (2020)

Answer 1: The lens maker’s formula is 1/f = (n - 1)(1/R1 - 1/R2), where f is the focal length, n is the refractive index, and R1, R2 are the radii of curvature. It relates lens properties to its focal length.

Question 2: Lenses and Optical Properties - What is the power of a lens, and how is it calculated? (2019)

Answer 2: The power of a lens is the reciprocal of its focal length in meters, P = 1/f, measured in diopters (D). It indicates the lens’s ability to converge or diverge light.

Question 3: Lenses and Optical Properties - A convex lens has a focal length of 25 cm. Find its power. (2021)

Answer 3: Power P = 1/f, where f = 25 cm = 0.25 m. Thus, P = 1/0.25 = 4 D. The power of the convex lens is 4 diopters.

Question 4: Lenses and Optical Properties - A convex lens forms a real image at 30 cm when the object is 20 cm away. Find the focal length using the thin lens formula. (2018)

Answer 4: Using the thin lens formula, 1/f = 1/u + 1/v, with u = 20 cm, v = 30 cm (real image, positive for convex lens). Thus, 1/f = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12. So, f = 12 cm.

Question 5: Lenses and Optical Properties - Two thin lenses of focal lengths 20 cm and 30 cm are in contact. Find the equivalent focal length. (2022)

Answer 5: For lenses in contact, 1/F = 1/f1 + 1/f2, with f1 = 20 cm, f2 = 30 cm. Thus, 1/F = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12. So, F = 12 cm.

Question 6: Lenses and Optical Properties - What is the magnification of a convex lens if the image is formed at 40 cm and the object is at 20 cm? (2017)

Answer 6: Magnification m = v/u, with v = 40 cm, u = 20 cm. Thus, m = 40/20 = 2. The image is twice as large as the object and real (positive m indicates same orientation for virtual images, but here it’s real).

Question 7: Lenses and Optical Properties - A concave lens has a focal length of 15 cm. An object is placed 10 cm away. Find the image distance. (2016)

Answer 7: Using the thin lens formula, 1/f = 1/u + 1/v, with f = -15 cm (concave), u = -10 cm. Thus, 1/v = 1/(-15) - 1/(-10) = -1/15 + 1/10 = -2/30 + 3/30 = 1/30. So, v = 30 cm (virtual image, behind lens).

Question 8: Lenses and Optical Properties - A lens has radii of curvature 20 cm and 30 cm, and n = 1.5. Find the focal length using the lens maker’s formula. (2015)

Answer 8: Using the lens maker’s formula, 1/f = (n - 1)(1/R1 - 1/R2), with n = 1.5, R1 = +20 cm (convex surface), R2 = -30 cm (concave surface). Thus, 1/f = (1.5 - 1)(1/20 - 1/(-30)) = 0.5(1/20 + 1/30) = 0.5(3/60 + 2/60) = 0.5 × 5/60 = 1/24. So, f = 24 cm.

Question 9: Lenses and Optical Properties - A convex lens of power 5 D is combined with a concave lens of power -2 D. Find the power of the combination. (2023)

Answer 9: For lenses in contact, total power P = P1 + P2. Given P1 = 5 D (convex), P2 = -2 D (concave). Thus, P = 5 + (-2) = 3 D. The combination has a power of 3 diopters.

Question 10: Lenses and Optical Properties - Why do concave lenses always produce virtual images? (2014)

Answer 10: Concave lenses diverge light rays, causing them to appear to originate from a point on the same side as the object. The image is always virtual, upright, and diminished for all object positions.

Question 1: Prisms - What is the angle of deviation in a prism? (2020)

Answer 1: The angle of deviation in a prism is the angle between the incident ray and the emergent ray, caused by refraction at the two faces of the prism.

Question 2: Prisms - State the formula for the angle of minimum deviation in a prism. (2019)

Answer 2: For minimum deviation, the angle of deviation D = 2i - A, where i is the angle of incidence and A is the prism’s refracting angle. Also, n = sin((A + D)/2)/sin(A/2), where n is the refractive index.

Question 3: Prisms - A prism has a refracting angle of 60° and refractive index 1.5. Find the angle of minimum deviation. (2021)

Answer 3: Using n = sin((A + D)/2)/sin(A/2), with n = 1.5, A = 60°. Thus, 1.5 = sin((60 + D)/2)/sin(30°). Since sin(30°) = 0.5, sin((60 + D)/2) = 1.5 × 0.5 = 0.75. So, (60 + D)/2 = sin⁻¹(0.75) ≈ 48.6°. Thus, 60 + D = 97.2°, D ≈ 37.2°.

Question 4: Prisms - Explain why a prism disperses white light into colors. (2018)

Answer 4: A prism disperses white light because different wavelengths (colors) of light have different refractive indices in the prism material. Shorter wavelengths (e.g., blue) refract more than longer ones (e.g., red), causing separation into a spectrum.

Question 5: Prisms - A prism has an angle of 60°. If the angle of incidence is 45° and n = 1.6, find the angle of deviation. (2022)

Answer 5: For a prism, deviation D = i + e - A, where i = 45°, A = 60°. Using Snell’s law at the first face, sin(e) = n sin(r1), where r1 is found from n = sin(i)/sin(r1). Thus, 1.6 = sin(45°)/sin(r1) ≈ 0.7071/sin(r1), sin(r1) ≈ 0.442, r1 ≈ 26.2°. At the second face, r2 = A - r1 = 60 - 26.2 = 33.8°. Using Snell’s law, sin(e) = 1.6 sin(33.8°) ≈ 1.6 × 0.557 ≈ 0.891, e ≈ 62.9°. Thus, D = 45 + 62.9 - 60 ≈ 47.9°.

Question 6: Prisms - What is the condition for minimum deviation in a prism? (2017)

Answer 6: Minimum deviation occurs when the angle of incidence equals the angle of emergence (i = e), and the ray passes symmetrically through the prism, minimizing the deviation angle.

Question 7: Prisms - A prism with n = 1.7 has a minimum deviation of 40°. Find the refracting angle. (2016)

Answer 7: Using n = sin((A + D)/2)/sin(A/2), with n = 1.7, D = 40°. Let A/2 = x, so (A + 40)/2 = x + 20. Thus, 1.7 = sin(x + 20)/sin(x). Solving numerically, sin(x + 20) = 1.7 sin(x). Testing x ≈ 30° (A = 60°), sin(50°)/sin(30°) ≈ 0.766/0.5 ≈ 1.532 (close to 1.7, adjust for precision). A ≈ 60° satisfies approximately.

Question 8: Prisms - Why does a prism not produce dispersion for monochromatic light? (2015)

Answer 8: Monochromatic light has a single wavelength, so it refracts at a constant angle in a prism (same refractive index). Without variation in refractive indices for different wavelengths, no dispersion occurs.

Question 9: Prisms - A prism has a refracting angle of 45° and n = 1.5. Find the angle of minimum deviation. (2023)

Answer 9: Using n = sin((A + D)/2)/sin(A/2), with n = 1.5, A = 45°. Thus, 1.5 = sin((45 + D)/2)/sin(22.5°). Since sin(22.5°) ≈ 0.3827, sin((45 + D)/2) = 1.5 × 0.3827 ≈ 0.574. So, (45 + D)/2 ≈ sin⁻¹(0.574) ≈ 35°. Thus, 45 + D ≈ 70°, D ≈ 25°.

Question 10: Prisms - What is the significance of the angle of minimum deviation in a prism? (2014)

Answer 10: The angle of minimum deviation is significant because it allows the determination of the prism’s refractive index using n = sin((A + D)/2)/sin(A/2), providing a precise method to characterize the prism material.

Question 1: Optical Instruments - What is the principle of a compound microscope? (2020)

Answer 1: A compound microscope uses two convex lenses: an objective lens forms a real, magnified image, and an eyepiece further magnifies this image, producing a virtual, enlarged image for the observer.

Question 2: Optical Instruments - Derive the magnifying power of a refracting telescope in normal adjustment. (2019)

Answer 2: For a refracting telescope in normal adjustment, magnifying power M = -f_o/f_e, where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece. The negative sign indicates an inverted image.

Question 3: Optical Instruments - A compound microscope has an objective of focal length 2 cm and an eyepiece of 5 cm. If the image is formed at infinity, find the magnifying power. (2021)

Answer 3: For a microscope in normal adjustment, M = (L/f_o) × (D/f_e), where L ≈ 15 cm (tube length), f_o = 2 cm, f_e = 5 cm, D = 25 cm (near point). Thus, M = (15/2) × (25/5) = 7.5 × 5 = 37.5.

Question 4: Optical Instruments - Explain the difference between reflecting and refracting telescopes. (2018)

Answer 4: Reflecting telescopes use a concave mirror to collect and focus light, while refracting telescopes use a convex lens. Reflecting telescopes avoid chromatic aberration and are better for large apertures, while refracting telescopes are simpler but prone to lens aberrations.

Question 5: Optical Instruments - A refracting telescope has an objective of focal length 100 cm and an eyepiece of 5 cm. Find the magnifying power in normal adjustment. (2022)

Answer 5: For a refracting telescope, M = -f_o/f_e, with f_o = 100 cm, f_e = 04 cm. Thus, M = -100/5 = -20. The magnifying power is 20 (inverted image).

Question 6: Optical Instruments - Why is a reflecting telescope preferred for astronomical observations? (2017)

Answer 6: Reflecting telescopes are preferred because they eliminate chromatic aberration, allow larger apertures for better light collection, and are more cost-effective for large-scale observations compared to refracting telescopes.

Question 7: Optical Instruments - A microscope’s objective has a focal length of 1 cm, and the eyepiece has 4 cm. If L = 14 cm, find the magnifying power. (2016)

Answer 7: For a microscope, M = (L/f_o) × (D/f_e), with L = 14 cm, f_o = 1 cm, f_e = 4 cm, D = 25 cm. Thus, M = (14/1) × (25/4) = 14 × 6.25 = 87.5.

Question 8: Optical Instruments - What is the role of the eyepiece in a telescope? (2015)

Answer 8: The eyepiece in a telescope magnifies the real image formed by the objective lens or mirror, producing a virtual, enlarged image for the observer to view.

Question 9: Optical Instruments - A reflecting telescope has a mirror of focal length 120 cm and an eyepiece of 6 cm. Find the magnifying power. (2023)

Answer 9: For a reflecting telescope, M = -f_o/f_e, with f_o = 120 cm, f_e = 6 cm. Thus, M = -120/6 = -20. The magnifying power is 20 (inverted image).

Question 10: Optical Instruments - Why is the image in a telescope inverted in normal adjustment? (2014)

Answer 10: The image in a telescope is inverted because the objective lens or mirror forms a real, inverted image, and the eyepiece further magnifies this image without altering its orientation in normal adjustment.

Chapter 10: Wave Optics

Question 1: State Huygen’s principle. (CBSE PYQ 2020)

Answer 1: Huygen’s principle states that every point on a wavefront acts as a source of secondary spherical wavelets that spread out in the forward direction at the speed of light. The new wavefront is the tangential surface to all these secondary wavelets.

Question 2: How does Huygen’s principle explain the propagation of light waves? (CBSE PYQ 2019)

Answer 2: Huygen’s principle explains wave propagation by considering each point on the primary wavefront as a source of secondary wavelets. The envelope of these wavelets forms the new wavefront, allowing for straight-line propagation in homogeneous media.

Question 3: What is a wavefront? (CBSE PYQ 2021)

Answer 3: A wavefront is a surface over which an oscillation has the same phase at all points at a given instant. For plane waves, it is a flat surface; for spherical waves, it is a sphere.

Question 4: Explain the concept of secondary wavelets in Huygen’s principle. (CBSE PYQ 2018)

Answer 4: Secondary wavelets are hypothetical spherical waves emitted from every point on a primary wavefront. Their amplitude decreases with distance, and they interfere to form the new wavefront, but only in the forward direction due to the obliquity factor.

Question 5: Why is the backward propagation of light not observed in Huygen’s principle? (CBSE PYQ 2022)

Answer 5: Backward propagation is not observed because the amplitude of secondary wavelets is proportional to (1 + cosθ)/2, where θ is the angle from the normal. For backward directions (θ=180°), the amplitude is zero.

Question 6: Describe the shape of wavefronts for a point source in a homogeneous medium. (CBSE PYQ 2017)

Answer 6: For a point source in a homogeneous medium, the wavefronts are spherical, expanding outward at the speed of light, with the source at the center.

Question 7: How does Huygen’s principle apply to plane waves? (CBSE PYQ 2016)

Answer 7: For plane waves, each point on the flat wavefront emits secondary wavelets. The tangent to these wavelets forms another plane wavefront parallel to the original, propagating perpendicularly.

Question 8: What limitations does the original Huygen’s principle have? (CBSE PYQ 2015)

Answer 8: The original Huygen’s principle does not account for the wave nature fully, such as diffraction effects in detail, and assumes uniform amplitude for secondary wavelets, which was later modified by Fresnel.

Question 9: Explain Huygen-Fresnel principle briefly. (CBSE PYQ 2023)

Answer 9: The Huygen-Fresnel principle incorporates interference of secondary wavelets and an obliquity factor to explain diffraction and the forward propagation of light more accurately than the original Huygen’s principle.

Question 10: Why is Huygen’s principle important in wave optics? (CBSE PYQ 2014)

Answer 10: Huygen’s principle is important as it provides a geometric method to understand wave propagation, reflection, refraction, and forms the basis for explaining interference and diffraction phenomena.

Question 1: Using Huygen’s principle, prove the law of reflection. (CBSE PYQ 2020)

Answer 1: In Huygen’s proof for reflection, consider a plane wave incident on a mirror. Secondary wavelets from points on the wavefront reach the mirror at different times, but the reflected wavefront satisfies i = r, as the time for light to travel paths is equal.

Question 2: Using Huygen’s principle, prove Snell’s law of refraction. (CBSE PYQ 2019)

Answer 2: For refraction, Huygen’s principle shows that the time for secondary wavelets to form the refracted wavefront leads to n1 sin(i) = n2 sin(r), as the speed v = c/n differs in media, maintaining constant frequency.

Question 3: How does the wavefront change during reflection? (CBSE PYQ 2021)

Answer 3: During reflection, the incident plane wavefront reflects such that the reflected wavefront is also plane, with the direction reversed relative to the normal, maintaining the law i = r.

Question 4: Describe the change in wavefront during refraction. (CBSE PYQ 2018)

Answer 4: During refraction, the wavefront bends as parts of it travel at different speeds in the two media, resulting in a change in direction according to Snell’s law, with the refracted wavefront remaining plane if incident is plane.

Question 5: Why does light bend during refraction according to wave theory? (CBSE PYQ 2022)

Answer 5: Light bends during refraction because its speed changes in different media (v = c/n), causing the wavefront to tilt as portions in the denser medium slow down, altering the direction.

Question 6: Illustrate the proof of reflection law with a diagram (qualitative). (CBSE PYQ 2017)

Answer 6: In the diagram, incident wavefront AB hits mirror at A first. Secondary wavelet from A travels distance equal to the path from B to mirror and back, leading to reflected wavefront with i = r.

Question 7: In refraction proof, why is time constant important? (CBSE PYQ 2016)

Answer 7: Time constancy ensures phase coherence; the time for light to go from incident to refracted wavefront is the same for all paths, leading to sin(i)/v1 = sin(r)/v2, or n1 sin(i) = n2 sin(r).

Question 8: What happens to frequency during refraction? (CBSE PYQ 2015)

Answer 8: Frequency remains constant during refraction because the wave’s oscillation rate is determined by the source and doesn’t change at the interface, while wavelength adjusts with speed.

Question 9: Explain why wavelength changes in refraction. (CBSE PYQ 2023)

Answer 9: Wavelength changes in refraction because λ = v/f, and while f is constant, v changes with medium (v = c/n), so λ decreases in denser media where v is smaller.

Question 10: How does Huygen’s principle unify reflection and refraction? (CBSE PYQ 2014)

Answer 10: Huygen’s principle unifies them by treating both as wavefront reconstructions via secondary wavelets, with reflection involving same speed and refraction different speeds in media.

Question 1: What is interference of light? (CBSE PYQ 2020)

Answer 1: Interference is the phenomenon where two or more coherent light waves superimpose, resulting in a redistribution of intensity, forming bright (constructive) and dark (destructive) fringes.

Question 2: What are coherent sources? (CBSE PYQ 2019)

Answer 2: Coherent sources are light sources that maintain a constant phase difference over time, essential for sustained interference patterns, such as lasers or slits illuminated by the same source.

Question 3: Explain constructive and destructive interference. (CBSE PYQ 2021)

Answer 3: Constructive interference occurs when waves are in phase (path difference = mλ), maximizing intensity. Destructive interference occurs when out of phase (path difference = (m+1/2)λ), minimizing intensity.

Question 4: Why is sustained interference not observed with ordinary light sources? (CBSE PYQ 2018)

Answer 4: Ordinary light sources emit incoherent waves with random phase changes, so interference averages out over time, not producing sustained patterns.

Question 5: What is the condition for maximum intensity in interference? (CBSE PYQ 2022)

Answer 5: Maximum intensity occurs when the phase difference δ = 2π m (m integer), or path difference Δx = mλ, leading to I_max = (A1 + A2)^2.

Question 6: What is the condition for minimum intensity in interference? (CBSE PYQ 2017)

Answer 6: Minimum intensity occurs when δ = π (2m+1), or Δx = (m+1/2)λ, leading to I_min = (A1 - A2)^2.

Question 7: If two waves have amplitudes A and 2A, find the ratio of max to min intensity. (CBSE PYQ 2016)

Answer 7: I_max = (A + 2A)^2 = 9A^2, I_min = (2A - A)^2 = A^2. Ratio I_max/I_min = 9/1 = 9.

Question 8: Why must sources be coherent for interference? (CBSE PYQ 2015)

Answer 8: Coherence ensures a fixed phase relation, allowing stable constructive and destructive patterns; incoherent sources cause fluctuating phases, averaging to uniform intensity.

Question 9: What is temporal and spatial coherence? (CBSE PYQ 2023)

Answer 9: Temporal coherence relates to phase stability over time (monochromaticity), spatial coherence to phase correlation across the wavefront (point source-like).

Question 10: How does interference demonstrate the wave nature of light? (CBSE PYQ 2014)

Answer 10: Interference shows superposition, a wave property, where light adds or cancels, unlike particles, confirming light's wave nature.

Question 1: Describe Young's double slit experiment. (CBSE PYQ 2020)

Answer 1: In Young's experiment, light from a single source passes through two slits, acting as coherent sources, interfering on a screen to produce alternating bright and dark fringes.

Question 2: What is the expression for fringe width? (CBSE PYQ 2019)

Answer 2: The fringe width β = λ D / d, where λ is wavelength, D is distance from slits to screen, d is slit separation.

Question 3: If λ = 600 nm, d = 0.1 mm, D = 1 m, find β. (CBSE PYQ 2021)

Answer 3: β = (600 × 10^{-9} × 1) / (0.1 × 10^{-3}) = 6 × 10^{-3} m = 6 mm.

Question 4: How does fringe width change if d increases? (CBSE PYQ 2018)

Answer 4: Fringe width β decreases if d increases, as β ∝ 1/d; wider slit separation reduces the angular spread of fringes.

Question 5: What happens to fringes if white light is used? (CBSE PYQ 2022)

Answer 5: With white light, central fringe is white, but colored fringes appear due to different wavelengths having different β, leading to overlap and fading at edges.

Question 6: Find the position of the m-th bright fringe. (CBSE PYQ 2017)

Answer 6: The position y_m = m λ D / d from the central fringe, where m = 0, ±1, ±2, ...

Question 7: If D doubles, how does β change? (CBSE PYQ 2016)

Answer 7: β doubles, as β ∝ D; increasing screen distance spreads the pattern.

Question 8: Why are slits narrow and close? (CBSE PYQ 2015)

Answer 8: Narrow slits ensure diffraction spreads light for overlap; close spacing ensures sufficient path difference for observable fringes within the screen.

Question 9: If λ increases, how does β change? (CBSE PYQ 2023)

Answer 9: β increases, as β ∝ λ; longer wavelengths produce wider fringes.

Question 10: What is the significance of YDSE? (CBSE PYQ 2014)

Answer 10: YDSE provided first experimental evidence for the wave nature of light through observable interference patterns.

Question 1: What is diffraction of light? (CBSE PYQ 2020)

Answer 1: Diffraction is the bending of light waves around obstacles or through apertures, resulting in interference patterns, demonstrating wave nature.

Question 2: Describe single slit diffraction pattern. (CBSE PYQ 2019)

Answer 2: Single slit diffraction shows a central maximum flanked by secondary maxima and minima, with intensity decreasing away from center.

Question 3: What is the width of the central maximum in single slit diffraction? (CBSE PYQ 2021)

Answer 3: The angular width of central maximum is 2λ/a, where a is slit width; linear width on screen is 2λ D/a, with D screen distance.

Question 4: Why is the central maximum wider than secondary ones? (CBSE PYQ 2018)

Answer 4: The central maximum is wider because it results from constructive interference of all wavelets; secondary are narrower due to fewer in-phase contributions.

Question 5: If slit width a increases, how does central maximum change? (CBSE PYQ 2022)

Answer 5: Central maximum width decreases as it ∝ 1/a; wider slit approaches geometric shadow, reducing diffraction spread.

Question 6: Condition for first minimum in single slit diffraction. (CBSE PYQ 2017)

Answer 6: First minimum occurs at θ such that a sinθ = λ, where path difference across slit is λ, leading to destructive interference.

Question 7: If λ = 500 nm, a = 0.1 mm, D = 2 m, find central maximum width. (CBSE PYQ 2016)

Answer 7: Width = 2λ D / a = 2 × 500 × 10^{-9} × 2 / (0.1 × 10^{-3}) = 2 × 10^{-6} / 10^{-4} = 0.02 m = 2 cm.

Question 8: Why is diffraction more pronounced for smaller apertures? (CBSE PYQ 2015)

Answer 8: Smaller apertures cause greater bending relative to size, as diffraction angle θ ≈ λ/a increases with smaller a.

Question 9: Differentiate between interference and diffraction. (CBSE PYQ 2023)

Answer 9: Interference is superposition from distinct sources; diffraction is interference from wavelets within a single wavefront bent by an obstacle or aperture.

Question 10: What is qualitative treatment of single slit diffraction? (CBSE PYQ 2014)

Answer 10: Qualitatively, single slit diffraction arises from Huygen’s wavelets interfering; central path has zero difference (max), edges have increasing differences leading to minima at multiples of λ.

Chapter 11: Dual Nature of Radiation and Matter

Question 1: What is the photoelectric effect? (2024)

Answer 1: The photoelectric effect is the emission of electrons from a metal surface when exposed to light of sufficient frequency.

Question 2: What did Hertz observe in the photoelectric effect? (2023)

Answer 2: Hertz observed that sparks occurred more readily when electrodes were exposed to ultraviolet light, indicating electron emission.

Question 3: What was Lenard’s contribution to the photoelectric effect? (2022)

Answer 3: Lenard confirmed that electrons are emitted from a metal surface when light of certain frequencies strikes it.

Question 4: State Einstein’s photoelectric equation. (2021)

Answer 4: Einstein’s equation is \( h\nu = \phi + \frac{1}{2}mv^2_{\text{max}} \), where \( h\nu \) is photon energy, \( \phi \) is work function, and \( \frac{1}{2}mv^2_{\text{max}} \) is maximum kinetic energy.

Question 5: Calculate the maximum kinetic energy for \( h\nu = 6 \, \text{eV} \), \( \phi = 4 \, \text{eV} \). (2020)

Answer 5: \( K_{\text{max}} = h\nu - \phi = 6 - 4 = 2 \, \text{eV} \).

Question 6: What is the threshold frequency in the photoelectric effect? (2019)

Answer 6: Threshold frequency is the minimum frequency of light required to eject electrons, given by \( \nu_0 = \frac{\phi}{h} \).

Question 7: Why does the photoelectric effect support the particle nature of light? (2018)

Answer 7: The effect shows light delivers energy in discrete packets (photons), not as continuous waves.

Question 8: What is the stopping potential in the photoelectric effect? (2017)

Answer 8: Stopping potential is the minimum voltage needed to stop the most energetic photoelectrons, related by \( eV_s = \frac{1}{2}mv^2_{\text{max}} \).

Question 9: If \( \phi = 2 \, \text{eV} \), find threshold wavelength (\( h = 4.14 \times 10^{-15} \, \text{eV·s} \), \( c = 3 \times 10^8 \, \text{m/s} \)). (2016)

Answer 9: \( \lambda_0 = \frac{hc}{\phi} = \frac{4.14 \times 10^{-15} \cdot 3 \times 10^8}{2} \approx 621 \, \text{nm} \).

Question 10: Name one experimental observation of the photoelectric effect. (2015)

Answer 10: The number of photoelectrons increases with light intensity, but their energy depends on frequency.

Question 1: What is the de-Broglie hypothesis? (2024)

Answer 1: De-Broglie proposed that particles like electrons have a wave nature, with wavelength \( \lambda = \frac{h}{p} \), where \( p \) is momentum.

Question 2: State the de-Broglie relation. (2023)

Answer 2: The de-Broglie wavelength is \( \lambda = \frac{h}{p} \), where \( h \) is Planck’s constant and \( p \) is momentum.

Question 3: Calculate the de-Broglie wavelength of an electron moving at \( 3 \times 10^6 \, \text{m/s} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \), \( m = 9.1 \times 10^{-31} \, \text{kg} \)). (2022)

Answer 3: \( \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \cdot 3 \times 10^6} \approx 2.43 \times 10^{-10} \, \text{m} \).

Question 4: Why is the de-Broglie wavelength significant for microscopic particles? (2021)

Answer 4: For microscopic particles, the de-Broglie wavelength is comparable to their size, showing wave-like behavior.

Question 5: Find the de-Broglie wavelength for a particle with momentum \( 2 \times 10^{-24} \, \text{kg·m/s} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \)). (2020)

Answer 5: \( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{2 \times 10^{-24}} \approx 3.315 \times 10^{-10} \, \text{m} \).

Question 6: How does de-Broglie wavelength vary with particle mass? (2019)

Answer 6: De-Broglie wavelength is inversely proportional to mass, \( \lambda \propto \frac{1}{m} \), for constant velocity.

Question 7: Why don’t macroscopic objects show wave properties? (2018)

Answer 7: Their large mass results in a very small de-Broglie wavelength, making wave effects negligible.

Question 8: Calculate the de-Broglie wavelength of a 1 eV electron (\( h = 4.14 \times 10^{-15} \, \text{eV·s} \), \( m = 9.1 \times 10^{-31} \, \text{kg} \), \( c = 3 \times 10^8 \, \text{m/s} \)). (2017)

Answer 8: \( \lambda = \frac{h}{\sqrt{2mE}} \approx \frac{4.14 \times 10^{-15}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 9 \times 10^{16}}} \approx 1.23 \times 10^{-9} \, \text{m} \).

Question 9: What is the significance of the de-Broglie relation? (2016)

Answer 9: It establishes the wave-particle duality, linking particle momentum to wave-like properties.

Question 10: How does de-Broglie wavelength change with velocity? (2015)

Answer 10: De-Broglie wavelength is inversely proportional to velocity, \( \lambda \propto \frac{1}{v} \), for constant mass.

Question 1: What is meant by dual nature of radiation? (2024)

Answer 1: Radiation exhibits both wave-like (e.g., diffraction) and particle-like (e.g., photoelectric effect) properties.

Question 2: How does the photoelectric effect demonstrate particle nature? (2023)

Answer 2: It shows light delivers energy in discrete photons, ejecting electrons based on frequency, not intensity.

Question 3: Name one phenomenon showing wave nature of light. (2022)

Answer 3: Diffraction of light demonstrates its wave nature.

Question 4: What is a photon? (2021)

Answer 4: A photon is a quantum of light, behaving as a particle with energy \( E = h\nu \).

Question 5: Calculate the energy of a photon with frequency \( 5 \times 10^{14} \, \text{Hz} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \)). (2020)

Answer 5: \( E = h\nu = 6.63 \times 10^{-34} \cdot 5 \times 10^{14} \approx 3.315 \times 10^{-19} \, \text{J} \).

Question 6: Why does light show dual nature? (2019)

Answer 6: Light exhibits wave properties in propagation and particle properties in interactions with matter.

Question 7: What is the momentum of a photon with \( \lambda = 500 \, \text{nm} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \))? (2018)

Answer 7: Photon momentum, \( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{500 \times 10^{-9}} \approx 1.326 \times 10^{-27} \, \text{kg·m/s} \).

Question 8: Name one experiment supporting particle nature of light. (2017)

Answer 8: The photoelectric effect supports the particle nature of light.

Question 9: How does wave-particle duality apply to radiation? (2016)

Answer 9: Radiation shows wave-like behavior (e.g., interference) and particle-like behavior (e.g., photon interactions).

Question 10: Calculate photon energy for \( \lambda = 400 \, \text{nm} \) (\( h = 4.14 \times 10^{-15} \, \text{eV·s} \), \( c = 3 \times 10^8 \, \text{m/s} \)). (2015)

Answer 10: \( E = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \cdot 3 \times 10^8}{400 \times 10^{-9}} \approx 3.1 \, \text{eV} \).

Question 1: What are matter waves? (2024)

Answer 1: Matter waves are the wave-like behavior exhibited by particles, such as electrons, as proposed by de-Broglie, with wavelength \( \lambda = \frac{h}{p} \).

Question 2: What experiment demonstrates matter waves? (2023)

Answer 2: The Davisson-Germer experiment shows electron diffraction, confirming the wave nature of matter.

Question 3: Calculate the de-Broglie wavelength of a proton moving at \( 2 \times 10^6 \, \text{m/s} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \), \( m = 1.67 \times 10^{-27} \, \text{kg} \)). (2022)

Answer 3: \( \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \cdot 2 \times 10^6} \approx 1.985 \times 10^{-13} \, \text{m} \).

Question 4: How does the de-Broglie wavelength relate to quantum mechanics? (2021)

Answer 4: The de-Broglie wavelength forms the basis of wave-particle duality, a cornerstone of quantum mechanics.

Question 5: Find the de-Broglie wavelength of a particle with kinetic energy \( 2 \, \text{eV} \) (\( m = 9.1 \times 10^{-31} \, \text{kg} \), \( h = 4.14 \times 10^{-15} \, \text{eV·s} \)). (2020)

Answer 5: \( \lambda = \frac{h}{\sqrt{2mE}} \approx \frac{4.14 \times 10^{-15}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 2 \cdot 1.6 \times 10^{-19}}} \approx 8.68 \times 10^{-10} \, \text{m} \).

Question 6: Why are matter waves not observable for large objects? (2019)

Answer 6: Large objects have large mass, resulting in extremely small de-Broglie wavelengths, making wave effects undetectable.

Question 7: What is the role of matter waves in electron microscopes? (2018)

Answer 7: Electron microscopes use the short de-Broglie wavelength of electrons to achieve high-resolution imaging.

Question 8: Calculate the de-Broglie wavelength of a neutron with momentum \( 1 \times 10^{-23} \, \text{kg·m/s} \) (\( h = 6.63 \times 10^{-34} \, \text{J·s} \)). (2017)

Answer 8: \( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1 \times 10^{-23}} \approx 6.63 \times 10^{-11} \, \text{m} \).

Question 9: How do matter waves differ from electromagnetic waves? (2016)

Answer 9: Matter waves are associated with particles having mass, while electromagnetic waves are massless photon-based waves.

Question 10: What is the significance of matter waves in quantum theory? (2015)

Answer 10: Matter waves underpin the wave-particle duality, explaining phenomena like electron diffraction and quantum behavior.

Chapter 12: Atoms

Question 1: Alpha-particle Scattering Experiment (2015) - What was the key observation in the alpha-particle scattering experiment?

Answer 1: Most alpha particles passed through the gold foil undeflected, but a few were scattered at large angles, indicating a small, dense, positively charged nucleus.

Question 2: Alpha-particle Scattering Experiment (2016) - Why did Rutherford use gold foil in the experiment?

Answer 2: Gold foil was used because it could be made very thin, allowing alpha particles to pass through and reveal the atomic structure via scattering patterns.

Question 3: Alpha-particle Scattering Experiment (2017) - What does the large-angle scattering of alpha particles suggest?

Answer 3: Large-angle scattering suggests the presence of a small, massive, positively charged nucleus that repels the positively charged alpha particles.

Question 4: Alpha-particle Scattering Experiment (2018) - How did the experiment disprove Thomson’s model?

Answer 4: Thomson’s model predicted uniform charge distribution, but large-angle scattering showed a concentrated positive charge in a small nucleus, contradicting it.

Question 5: Alpha-particle Scattering Experiment (2019) - What is the impact parameter in the context of the experiment?

Answer 5: The impact parameter is the perpendicular distance between the alpha particle’s path and the nucleus’s center, affecting the scattering angle.

Question 6: Alpha-particle Scattering Experiment (2020) - Why were alpha particles chosen for the experiment?

Answer 6: Alpha particles, being massive and positively charged, could penetrate the foil and interact strongly with the nucleus, revealing its properties.

Question 7: Alpha-particle Scattering Experiment (2021) - What role did the fluorescent screen play?

Answer 7: The fluorescent screen detected scattered alpha particles by producing light flashes, allowing observation of their scattering angles.

Question 8: Alpha-particle Scattering Experiment (2022) - How did Rutherford estimate nuclear size?

Answer 8: Rutherford estimated nuclear size by calculating the distance of closest approach, where the alpha particle’s kinetic energy equals the electrostatic potential energy, yielding ~10⁻¹⁵ m.

Question 9: Alpha-particle Scattering Experiment (2023) - Why did most alpha particles pass undeflected?

Answer 9: Most alpha particles passed undeflected because the atom is mostly empty space, with the nucleus occupying a tiny volume.

Question 10: Alpha-particle Scattering Experiment (2024) - What evidence supported the nuclear model?

Answer 10: The large-angle scattering of a few alpha particles provided evidence for a small, dense, positively charged nucleus in the atom.

Question 1: Rutherford’s Model of Atom (2015) - What are the main postulates of Rutherford’s atomic model?

Answer 1: The atom has a small, dense, positively charged nucleus containing most of its mass, with electrons orbiting it, and the atom is mostly empty space.

Question 2: Rutherford’s Model of Atom (2016) - Why is it called the planetary model?

Answer 2: It’s called the planetary model because electrons orbit the nucleus like planets orbit the sun, held by electrostatic forces.

Question 3: Rutherford’s Model of Atom (2017) - What is a major drawback of Rutherford’s model?

Answer 3: It cannot explain atomic stability, as classical physics predicts orbiting electrons would radiate energy and spiral into the nucleus.

Question 4: Rutherford’s Model of Atom (2018) - How does the model explain alpha-particle scattering?

Answer 4: The model explains scattering by a small, positively charged nucleus that repels alpha particles, causing large-angle deflections for close encounters.

Question 5: Rutherford’s Model of Atom (2019) - Why is the atom electrically neutral in Rutherford’s model?

Answer 5: The positive charge of the nucleus is balanced by the negative charge of the orbiting electrons, ensuring overall neutrality.

Question 6: Rutherford’s Model of Atom (2020) - What is the approximate size of the nucleus in the model?

Answer 6: The nucleus is about 10⁻¹⁵ m, much smaller than the atom’s size of ~10⁻¹⁰ m.

Question 7: Rutherford’s Model of Atom (2021) - Why can’t the model explain atomic spectra?

Answer 7: It lacks quantized energy levels, so it cannot account for the discrete spectral lines observed in atomic spectra.

Question 8: Rutherford’s Model of Atom (2022) - What force governs electron motion in the model?

Answer 8: The electrostatic force of attraction between the nucleus and electrons governs their orbital motion.

Question 9: Rutherford’s Model of Atom (2023) - How did Rutherford’s model improve on Thomson’s?

Answer 9: It introduced a dense nucleus, explaining large-angle scattering, unlike Thomson’s uniform charge distribution model.

Question 10: Rutherford’s Model of Atom (2024) - Why was Rutherford’s model a significant advancement?

Answer 10: It provided evidence for a nuclear structure, correcting earlier models and laying the foundation for modern atomic theory.

Question 1: Bohr Model of Hydrogen Atom (2015) - What is the Bohr radius and its significance?

Answer 1: The Bohr radius (~0.529 × 10⁻¹⁰ m) is the most probable electron-proton distance in the ground state (n=1), setting the scale for atomic dimensions.

Question 2: Bohr Model of Hydrogen Atom (2016) - How does the Bohr model ensure atomic stability?

Answer 2: It uses quantization of angular momentum (L = nħ), allowing electrons to orbit in stable, non-radiating energy levels.

Question 3: Bohr Model of Hydrogen Atom (2017) - What is the quantization condition in the Bohr model?

Answer 3: The angular momentum of the electron is quantized as L = nħ, where n is the principal quantum number and ħ is the reduced Planck constant.

Question 4: Bohr Model of Hydrogen Atom (2018) - Why is the Bohr model limited to hydrogen-like atoms?

Answer 4: It assumes a single electron orbiting a nucleus, ignoring electron-electron interactions, so it applies only to hydrogen and ions like He⁺.

Question 5: Bohr Model of Hydrogen Atom (2019) - What is the energy of an electron in the nth orbit?

Answer 5: The energy is Eₙ = -13.6/n² eV, derived from the balance of kinetic and potential energies in the quantized orbits.

Question 6: Bohr Model of Hydrogen Atom (2020) - How does the Bohr model explain spectral lines?

Answer 6: Spectral lines arise from electrons transitioning between quantized energy levels, emitting or absorbing photons with energy ΔE = hν.

Question 7: Bohr Model of Hydrogen Atom (2021) - What is the ionization energy of hydrogen?

Answer 7: The ionization energy, from n=1 to n=∞, is E = 13.6 eV, as Eₙ = -13.6/n² eV.

Question 8: Bohr Model of Hydrogen Atom (2022) - What role does the principal quantum number play?

Answer 8: The principal quantum number (n) determines the energy, radius, and velocity of the electron’s orbit, with higher n indicating higher energy.

Question 9: Bohr Model of Hydrogen Atom (2023) - Why does the model fail for multi-electron atoms?

Answer 9: It ignores electron-electron interactions and quantum effects like spin, making it unsuitable for multi-electron atoms.

Question 10: Bohr Model of Hydrogen Atom (2024) - How does the Bohr model improve on Rutherford’s model?

Answer 10: It introduces quantized energy levels, explaining atomic stability and the hydrogen spectrum, unlike Rutherford’s unstable orbits.

Question 1: Expression for Radius of nth Possible Orbit (2015) - Derive the radius of the nth Bohr orbit.

Answer 1: Using mvr = nħ and mv²/r = ke²/r², the radius is rₙ = (n²ħ²ε₀)/(πme²), where n is the principal quantum number, ħ is the reduced Planck constant, ε₀ is permittivity, m is electron mass, and e is electron charge.

Question 2: Expression for Radius of nth Possible Orbit (2016) - What is the radius of the first Bohr orbit?

Answer 2: For n=1, r₁ = (ħ²ε₀)/(πme²) ≈ 0.529 × 10⁻¹⁰ m, known as the Bohr radius.

Question 3: Expression for Radius of nth Possible Orbit (2017) - How does the radius scale with n?

Answer 3: The radius scales as rₙ ∝ n², increasing quadratically with the principal quantum number.

Question 4: Expression for Radius of nth Possible Orbit (2018) - Find the radius ratio for n=2 and n=3.

Answer 4: Since rₙ ∝ n², the ratio r₂:r₃ = 2²:3² = 4:9.

Question 5: Expression for Radius of nth Possible Orbit (2019) - What constants are involved in the radius expression?

Answer 5: The expression rₙ = (n²ħ²ε₀)/(πme²) includes ħ, ε₀, m, and e.

Question 6: Expression for Radius of nth Possible Orbit (2020) - Calculate the radius for n=2 in hydrogen.

Answer 6: For n=2, r₂ = 4 × r₁ = 4 × 0.529 × 10⁻¹⁰ m ≈ 2.116 × 10⁻¹⁰ m.

Question 7: Expression for Radius of nth Possible Orbit (2021) - How does the radius change for a hydrogen-like atom with atomic number Z?

Answer 7: For a hydrogen-like atom, rₙ = (n²ħ²ε₀)/(πmZe²), so radius is inversely proportional to Z.

Question 8: Expression for Radius of nth Possible Orbit (2022) - Why is the Bohr radius significant?

Answer 8: It represents the typical size of the hydrogen atom in its ground state, serving as a fundamental atomic scale.

Question 9: Expression for Radius of nth Possible Orbit (2023) - What is the radius for He⁺ in the ground state?

Answer 9: For He⁺ (Z=2, n=1), r₁ = (ħ²ε₀)/(πm·2e²) = 0.529 × 10⁻¹⁰ m / 2 ≈ 0.265 × 10⁻¹⁰ m.

Question 10: Expression for Radius of nth Possible Orbit (2024) - How does the radius formula incorporate quantization?

Answer 10: The radius formula uses the quantization condition mvr = nħ, leading to discrete orbit radii proportional to n².

Question 1: Velocity and Energy of Electron in nth Orbit (2015) - Derive the velocity of an electron in the nth orbit.

Answer 1: Using mvr = nħ and mv²/r = ke²/r², the velocity is vₙ = (e²)/(2ε₀nh), where e is the electron charge, ε₀ is permittivity, n is the principal quantum number, and h is Planck’s constant.

Question 2: Velocity and Energy of Electron in nth Orbit (2016) - What is the velocity in the first Bohr orbit?

Answer 2: For n=1, v₁ = (e²)/(2ε₀h) ≈ 2.19 × 10⁶ m/s.

Question 3: Velocity and Energy of Electron in nth Orbit (2017) - Derive the energy of an electron in the nth orbit.

Answer 3: Total energy Eₙ = KE + PE = (1/2)mv² - ke²/r. Using vₙ and rₙ, Eₙ = -13.6/n² eV.

Question 4: Velocity and Energy of Electron in nth Orbit (2018) - How does velocity scale with n?

Answer 4: Velocity scales as vₙ ∝ 1/n, decreasing inversely with the principal quantum number.

Question 5: Velocity and Energy of Electron in nth Orbit (2019) - What is the energy of the ground state?

Answer 5: For n=1, E₁ = -13.6/1² = -13.6 eV.

Question 6: Velocity and Energy of Electron in nth Orbit (2020) - Find the velocity ratio for n=1 and n=2.

Answer 6: Since vₙ ∝ 1/n, the ratio v₁:v₂ = 1:1/2 = 2:1.

Question 7: Velocity and Energy of Electron in nth Orbit (2021) - What is the energy difference between n=1 and n=2?

Answer 7: E₁ = -13.6 eV, E₂ = -3.4 eV. The difference is E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV.

Question 8: Velocity and Energy of Electron in nth Orbit (2022) - How does energy scale with n?

Answer 8: Energy scales as Eₙ ∝ -1/n², becoming less negative as n increases.

Question 9: Velocity and Energy of Electron in nth Orbit (2023) - What is the velocity for He⁺ in the ground state?

Answer 9: For He⁺ (Z=2, n=1), v₁ = Z × (e²)/(2ε₀h) = 2 × 2.19 × 10⁶ ≈ 4.38 × 10⁶ m/s.

Question 10: Velocity and Energy of Electron in nth Orbit (2024) - Why is the energy negative in the Bohr model?

Answer 10: The energy is negative because the electron is bound in the atom, with potential energy dominating over kinetic energy, referenced to zero at infinity.

Question 1: Hydrogen Line Spectra (2015) - What causes the line spectra in hydrogen?

Answer 1: Line spectra result from electrons transitioning between quantized energy levels, emitting or absorbing photons with specific energies.

Question 2: Hydrogen Line Spectra (2016) - Why are the spectral lines discrete?

Answer 2: Discrete lines occur because electron energy levels are quantized, so only specific energy differences (ΔE = hν) produce photons of distinct wavelengths.

Question 3: Hydrogen Line Spectra (2017) - What is the Lyman series?

Answer 3: The Lyman series corresponds to electron transitions from higher levels (n>1) to n=1, producing ultraviolet spectral lines.

Question 4: Hydrogen Line Spectra (2018) - What is the Balmer series?

Answer 4: The Balmer series involves transitions from higher levels (n>2) to n=2, producing visible light spectral lines.

Question 5: Hydrogen Line Spectra (2019) - Why do different series appear in different spectral regions?

Answer 5: Different series correspond to transitions to different final energy levels, with energy differences determining the photon wavelengths (e.g., UV for Lyman, visible for Balmer).

Question 6: Hydrogen Line Spectra (2020) - What is the significance of the Rydberg formula?

Answer 6: The Rydberg formula, 1/λ = R(1/n₁² - 1/n₂²), predicts the wavelengths of hydrogen spectral lines based on energy level transitions.

Question 7: Hydrogen Line Spectra (2021) - Why are emission and absorption spectra related?

Answer 7: Emission occurs when electrons drop to lower levels, releasing photons, while absorption occurs when electrons jump to higher levels, absorbing the same photon energies.

Question 8: Hydrogen Line Spectra (2022) - What is the Paschen series?

Answer 8: The Paschen series involves transitions from higher levels (n>3) to n=3, producing infrared spectral lines.

Question 9: Hydrogen Line Spectra (2023) - How does the Bohr model explain the hydrogen spectrum?

Answer 9: The Bohr model explains the spectrum through quantized energy levels, where electron transitions produce photons with energies corresponding to spectral lines.

Question 10: Hydrogen Line Spectra (2024) - Why do spectral lines have specific wavelengths?

Answer 10: Specific wavelengths arise because the energy difference between quantized levels is fixed, and photon energy (E = hc/λ) determines the wavelength.

Chapter 13: Nuclei

Question 1: Composition and Size of Nucleus (2015) - What particles make up the atomic nucleus?

Answer 1: The nucleus is composed of protons (positively charged) and neutrons (neutral), collectively called nucleons, bound by the strong nuclear force.

Question 2: Composition and Size of Nucleus (2016) - How is the size of a nucleus related to its mass number?

Answer 2: The nuclear radius is given by R = R₀A^(1/3), where R₀ ≈ 1.2 fm and A is the mass number, indicating size increases with A.

Question 3: Composition and Size of Nucleus (2017) - Why is the nucleus much smaller than the atom?

Answer 3: The nucleus is ~10⁻¹⁵ m (1 fm) in size, while the atom is ~10⁻¹⁰ m, as electrons orbit far from the dense nuclear core.

Question 4: Composition and Size of Nucleus (2018) - Estimate the radius of a nucleus with mass number A = 64.

Answer 4: Using R = 1.2 A^(1/3) fm, for A = 64, R = 1.2 × (64)^(1/3) = 1.2 × 4 ≈ 4.8 fm.

Question 5: Composition and Size of Nucleus (2019) - Why is nuclear density nearly constant across elements?

Answer 5: Nuclear volume ∝ A (since R ∝ A^(1/3)), and mass ∝ A, so density (mass/volume) is roughly constant, ~2.3 × 10¹⁷ kg/m³.

Question 6: Composition and Size of Nucleus (2020) - What determines the number of protons in a nucleus?

Answer 6: The number of protons (atomic number Z) determines the element’s identity and its positive charge, balancing electron charge in a neutral atom.

Question 7: Composition and Size of Nucleus (2021) - How does the neutron-to-proton ratio affect nuclear size?

Answer 7: The neutron-to-proton ratio affects mass number (A = Z + N), increasing nuclear size via R ∝ A^(1/3), but not density.

Question 8: Composition and Size of Nucleus (2022) - Why is the nucleus spherical for most elements?

Answer 8: The strong nuclear force is isotropic and short-range, distributing nucleons symmetrically, leading to a roughly spherical shape for most nuclei.

Question 9: Composition and Size of Nucleus (2023) - How was nuclear size experimentally determined?

Answer 9: Nuclear size was determined via scattering experiments (e.g., alpha-particle or electron scattering), analyzing deflection angles to estimate radius.

Question 10: Composition and Size of Nucleus (2024) - What is the significance of the constant R₀ in the nuclear radius formula?

Answer 10: R₀ ≈ 1.2 fm is an empirical constant reflecting the effective range of the nuclear force, scaling nuclear size with A^(1/3).

Question 1: Nuclear Force (2015) - What are the key characteristics of the nuclear force?

Answer 1: The nuclear force is strong, short-range (~1–2 fm), charge-independent, attractive at intermediate distances, and repulsive at very short distances.

Question 2: Nuclear Force (2016) - Why is the nuclear force charge-independent?

Answer 2: The nuclear force acts equally between proton-proton, neutron-neutron, and proton-neutron pairs, unlike the electromagnetic force, due to its origin in quark interactions.

Question 3: Nuclear Force (2017) - Why is the nuclear force short-range?

Answer 3: The nuclear force is mediated by massive mesons, limiting its range to ~1–2 fm, unlike the infinite-range electromagnetic force.

Question 4: Nuclear Force (2018) - How does the nuclear force overcome Coulomb repulsion in the nucleus?

Answer 4: The nuclear force is ~100 times stronger than the electromagnetic force at short distances, binding protons and neutrons despite Coulomb repulsion.

Question 5: Nuclear Force (2019) - What is the role of the repulsive core in the nuclear force?

Answer 5: The repulsive core at very short distances (<0.5 fm) prevents nucleons from collapsing into each other, maintaining nuclear stability.

Question 6: Nuclear Force (2020) - Why doesn’t the nuclear force affect electrons?

Answer 6: The nuclear force acts only between nucleons (protons and neutrons) via the strong interaction, while electrons, governed by electromagnetic forces, are unaffected.

Question 7: Nuclear Force (2021) - How does the nuclear force contribute to nuclear stability?

Answer 7: The nuclear force binds nucleons together, counteracting the repulsive Coulomb force between protons, stabilizing the nucleus.

Question 8: Nuclear Force (2022) - Why is the nuclear force non-central?

Answer 8: The nuclear force depends on the spin and relative orientation of nucleons, making it non-central, unlike the radial Coulomb force.

Question 9: Nuclear Force (2023) - How was the nuclear force’s range estimated?

Answer 9: The range (~1–2 fm) was estimated from scattering experiments and the uncertainty principle, relating to the mass of exchanged mesons.

Question 10: Nuclear Force (2024) - Why is the nuclear force stronger than gravity in the nucleus?

Answer 10: The nuclear force is ~10³⁸ times stronger than gravity at nuclear scales, dominating nucleon interactions and binding the nucleus.

Question 1: Mass-Energy Relation and Mass Defect (2015) - What is the mass defect of a nucleus?

Answer 1: Mass defect is the difference between the sum of individual nucleon masses and the mass of the nucleus, converted to binding energy via E = mc².

Question 2: Mass-Energy Relation and Mass Defect (2016) - How is the mass-energy relation applied to nuclei?

Answer 2: Einstein’s E = mc² converts the mass defect into binding energy, which holds the nucleus together, or energy released in nuclear reactions.

Question 3: Mass-Energy Relation and Mass Defect (2017) - Calculate the energy equivalent of a mass defect of 0.1 u.

Answer 3: Energy = Δm × 931.5 MeV/u = 0.1 × 931.5 = 93.15 MeV.

Question 4: Mass-Energy Relation and Mass Defect (2018) - Why is the mass of a nucleus less than its nucleons’ masses?

Answer 4: The mass defect arises because some mass is converted to binding energy, stabilizing the nucleus via the strong nuclear force.

Question 5: Mass-Energy Relation and Mass Defect (2019) - What is the mass defect for a nucleus with 2 protons and 2 neutrons if its mass is 4.0015 u?

Answer 5: Proton mass ≈ 1.0078 u, neutron mass ≈ 1.0087 u. Total mass = 2 × 1.0078 + 2 × 1.0087 = 4.0330 u. Mass defect = 4.0330 - 4.0015 = 0.0315 u.

Question 6: Mass-Energy Relation and Mass Defect (2020) - How does mass defect relate to nuclear stability?

Answer 6: A larger mass defect corresponds to greater binding energy, making the nucleus more stable against disintegration.

Question 7: Mass-Energy Relation and Mass Defect (2021) - What is the energy equivalent of 1 atomic mass unit?

Answer 7: 1 u = 931.5 MeV/c², so the energy equivalent is E = 1 × 931.5 = 931.5 MeV.

Question 8: Mass-Energy Relation and Mass Defect (2022) - Why is mass defect significant in nuclear reactions?

Answer 8: Mass defect determines the energy released or absorbed in nuclear reactions, as per E = mc², driving processes like fission and fusion.

Question 9: Mass-Energy Relation and Mass Defect (2023) - Calculate the binding energy for a mass defect of 0.05 u.

Answer 9: Binding energy = 0.05 × 931.5 = 46.575 MeV.

Question 10: Mass-Energy Relation and Mass Defect (2024) - How is the mass defect measured experimentally?

Answer 10: Mass defect is measured using mass spectrometry to determine the precise mass of the nucleus and comparing it to the sum of individual nucleon masses.

Question 1: Binding Energy per Nucleon (2015) - What is binding energy per nucleon?

Answer 1: Binding energy per nucleon is the total binding energy of a nucleus divided by its mass number (A), indicating nuclear stability.

Question 2: Binding Energy per Nucleon (2016) - How does binding energy per nucleon vary with mass number?

Answer 2: It increases for light nuclei, peaks at ~8.8 MeV for A ≈ 56 (e.g., iron), and slightly decreases for heavier nuclei due to Coulomb repulsion.

Question 3: Binding Energy per Nucleon (2017) - Why is iron-56 the most stable nucleus?

Answer 3: Iron-56 has the highest binding energy per nucleon (~8.8 MeV), minimizing energy, making it the most stable against nuclear reactions.

Question 4: Binding Energy per Nucleon (2018) - Calculate the binding energy per nucleon for a nucleus with 40 nucleons and binding energy 320 MeV.

Answer 4: Binding energy per nucleon = 320 / 40 = 8 MeV.

Question 5: Binding Energy per Nucleon (2019) - Why do heavy nuclei have lower binding energy per nucleon?

Answer 5: Heavy nuclei experience increased Coulomb repulsion between protons, reducing the net binding energy per nucleon compared to mid-sized nuclei.

Question 6: Binding Energy per Nucleon (2020) - How does binding energy per nucleon affect nuclear reactions?

Answer 6: Reactions like fission and fusion increase binding energy per nucleon, releasing energy as nuclei move toward the stability peak (A ≈ 56).

Question 7: Binding Energy per Nucleon (2021) - What is the binding energy per nucleon for helium-4 (mass defect 0.0304 u, A=4)?

Answer 7: Binding energy = 0.0304 × 931.5 = 28.3176 MeV. Per nucleon = 28.3176 / 4 ≈ 7.08 MeV.

Question 8: Binding Energy per Nucleon (2022) - Why do light nuclei have lower binding energy per nucleon?

Answer 8: Light nuclei have fewer nucleons, so surface effects and fewer nuclear force interactions reduce binding energy per nucleon compared to mid-sized nuclei.

Question 9: Binding Energy per Nucleon (2023) - How is binding energy per nucleon related to nuclear stability?

Answer 9: Higher binding energy per nucleon indicates greater stability, as more energy is required to break the nucleus into its constituent nucleons.

Question 10: Binding Energy per Nucleon (2024) - Why does the binding energy curve peak at A ≈ 56?

Answer 10: The peak at A ≈ 56 balances the strong nuclear force’s binding effect with Coulomb repulsion, maximizing stability for mid-sized nuclei like iron.

Question 1: Nuclear Fission (2015) - What is nuclear fission?

Answer 1: Nuclear fission is the splitting of a heavy nucleus (e.g., U-235) into two lighter nuclei, releasing energy, neutrons, and gamma rays, often induced by neutron absorption.

Question 2: Nuclear Fission (2016) - Why does fission release energy?

Answer 2: Fission products have higher binding energy per nucleon (~8 MeV) than the parent nucleus (~7.6 MeV), releasing energy via E = mc².

Question 3: Nuclear Fission (2017) - What is a chain reaction in fission?

Answer 3: A chain reaction occurs when neutrons from one fission event trigger additional fissions, sustaining the process, as in reactors or bombs.

Question 4: Nuclear Fission (2018) - Calculate the energy released in a fission reaction with a mass defect of 0.18 u.

Answer 4: Energy = 0.18 × 931.5 = 167.67 MeV.

Question 5: Nuclear Fission (2019) - Why is U-235 commonly used in fission reactors?

Answer 5: U-235 is fissile, readily undergoing fission with thermal neutrons, producing neutrons to sustain a chain reaction, unlike U-238.

Question 6: Nuclear Fission (2020) - What role do moderators play in fission reactors?

Answer 6: Moderators (e.g., water, graphite) slow down fast neutrons to thermal energies, increasing the probability of fission in U-235.

Question 7: Nuclear Fission (2021) - Why is a critical mass needed for a fission chain reaction?

Answer 7: Critical mass ensures enough fissile material is present so that neutrons trigger sufficient fissions to sustain the chain reaction without excessive loss.

Question 8: Nuclear Fission (2022) - What are fission products typically like?

Answer 8: Fission products are two medium-mass nuclei (A ≈ 90–140), neutrons, gamma rays, and are often radioactive, with higher binding energy per nucleon.

Question 9: Nuclear Fission (2023) - How does fission differ from radioactive decay?

Answer 9: Fission is the induced or spontaneous splitting of a heavy nucleus into two fragments, releasing energy, while radioactive decay involves emission of particles with no splitting.

Question 10: Nuclear Fission (2024) - Why is energy release in fission significant for power generation?

Answer 10: Fission releases ~200 MeV per event, far more than chemical reactions, enabling efficient energy production in nuclear reactors.

Question 1: Nuclear Fusion (2015) - What is nuclear fusion?

Answer 1: Nuclear fusion is the process where two light nuclei (e.g., hydrogen isotopes) combine to form a heavier nucleus, releasing energy.

Question 2: Nuclear Fusion (2016) - Why does fusion release energy?

Answer 2: Fusion products (e.g., helium) have higher binding energy per nucleon (~7–8 MeV) than reactants (~1 MeV), releasing energy via E = mc².

Question 3: Nuclear Fusion (2017) - Why is fusion the primary energy source in stars?

Answer 3: Stars provide high temperatures (~10⁷ K) and pressures to overcome Coulomb repulsion, enabling fusion of light nuclei, releasing vast energy.

Question 4: Nuclear Fusion (2018) - Calculate the energy released in a fusion reaction with a mass defect of 0.018 u.

Answer 4: Energy = 0.018 × 931.5 = 16.767 MeV.

Question 5: Nuclear Fusion (2019) - Why is fusion difficult to achieve on Earth?

Answer 5: Fusion requires high temperatures (~10⁷ K) and pressures to overcome Coulomb repulsion, necessitating advanced confinement (e.g., magnetic or inertial).

Question 6: Nuclear Fusion (2020) - What is the role of plasma in fusion reactors?

Answer 6: Plasma, a hot ionized gas, allows nuclei to move freely at high energies, enabling fusion while being confined by magnetic fields in reactors like tokamaks.

Question 7: Nuclear Fusion (2021) - Name a common fusion reaction in stars.

Answer 7: The proton-proton chain, where four protons fuse to form helium-4, releasing energy, is a primary fusion reaction in stars like the Sun.

Question 8: Nuclear Fusion (2022) - Why is fusion considered a clean energy source?

Answer 8: Fusion produces minimal radioactive waste, uses abundant fuels (e.g., deuterium), and has no greenhouse gas emissions, making it environmentally friendly.

Question 9: Nuclear Fusion (2023) - How does fusion differ from fission in terms of reactants?

Answer 9: Fusion involves combining light nuclei (e.g., hydrogen), while fission involves splitting heavy nuclei (e.g., uranium), both increasing binding energy per nucleon.

Question 10: Nuclear Fusion (2024) - Why is deuterium-tritium fusion preferred in reactors?

Answer 10: Deuterium-tritium fusion has a high reaction cross-section and releases significant energy (~17.6 MeV), making it efficient for controlled fusion.

Chapter 14: Semiconductor Electronics

Question 1: Energy Bands (2020) - How do energy bands differ in conductors, semiconductors, and insulators?

Answer 1: In conductors, the valence and conduction bands overlap, allowing free electron movement. In semiconductors, a small energy gap (~1 eV) separates the bands, enabling limited conduction. In insulators, a large energy gap (>5 eV) prevents electron flow.

Question 2: Energy Bands (2019) - Why do insulators not conduct electricity?

Answer 2: Insulators have a large energy gap between the valence and conduction bands, preventing electrons from gaining enough energy to jump to the conduction band, thus inhibiting electrical conduction.

Question 3: Energy Bands (2018) - What is the significance of the energy gap in semiconductors?

Answer 3: The energy gap in semiconductors (~1 eV) allows electrons to move to the conduction band with moderate energy input (e.g., thermal energy), enabling controlled conductivity, which is crucial for electronic devices.

Question 4: Energy Bands (2017) - Explain the role of the conduction band in electrical conductivity.

Answer 4: The conduction band contains free electrons that can move and conduct electricity. In semiconductors, electrons excited to the conduction band from the valence band contribute to conductivity.

Question 5: Energy Bands (2016) - Why does the conductivity of a semiconductor increase with temperature?

Answer 5: As temperature rises, more electrons gain energy to cross the energy gap from the valence band to the conduction band, increasing the number of charge carriers and thus conductivity.

Question 1: P-N Junction (2020) - What is a p-n junction, and how is it formed?

Answer 1: A p-n junction is formed by joining p-type (hole-rich) and n-type (electron-rich) semiconductors. Diffusion of charge carriers creates a depletion region with a potential barrier at the junction.

Question 2: P-N Junction (2019) - Explain the effect of forward bias on a p-n junction.

Answer 2: In forward bias, the external voltage reduces the potential barrier at the p-n junction, allowing electrons and holes to recombine, enabling significant current flow.

Question 3: P-N Junction (2018) - What happens to the depletion region in reverse bias?

Answer 3: In reverse bias, the depletion region widens as the external voltage increases the potential barrier, preventing significant current flow except for a small reverse saturation current.

Question 4: P-N Junction (2017) - Why does a p-n junction conduct in one direction only?

Answer 4: A p-n junction conducts in forward bias as the barrier is reduced, allowing current flow. In reverse bias, the barrier increases, blocking significant current, thus acting as a one-way conductor.

Question 5: P-N Junction (2016) - What is the significance of the depletion region in a p-n junction?

Answer 5: The depletion region acts as a barrier to charge flow, controlling the behavior of the p-n junction in forward and reverse bias, enabling applications like rectification.

Question 1: Diode (2021) - Draw the I-V characteristics of a p-n junction diode in forward and reverse bias.

Answer 1: In forward bias, the I-V curve shows negligible current below the threshold voltage (~0.7 V for silicon), then a sharp exponential rise. In reverse bias, a small saturation current flows until breakdown voltage, where current increases suddenly.

Question 2: Diode (2019) - How does a p-n junction diode function as a rectifier?

Answer 2: A p-n junction diode allows current in forward bias (positive AC half-cycle) and blocks it in reverse bias (negative half-cycle), converting AC to pulsating DC in a rectifier circuit.

Question 3: Diode (2018) - What is meant by the breakdown voltage of a diode?

Answer 3: Breakdown voltage is the reverse voltage at which a p-n junction diode starts conducting significantly due to mechanisms like Zener effect or avalanche breakdown.

Question 4: Diode (2017) - Explain the term 'reverse saturation current' in a diode.

Answer 4: Reverse saturation current is the small current in a reverse-biased p-n junction diode due to minority carriers, remaining nearly constant until the breakdown voltage is reached.

Question 5: Diode (2016) - Why is a diode used in rectification circuits?

Answer 5: A diode is used in rectification circuits because it allows current to flow in one direction (forward bias) while blocking it in the opposite direction (reverse bias), converting AC to DC.

Question 1: Semiconductors (2021) - Differentiate between intrinsic and extrinsic semiconductors.

Answer 1: Intrinsic semiconductors are pure, with equal numbers of electrons and holes, and conductivity depends on temperature. Extrinsic semiconductors are doped with impurities, creating n-type (electron-rich) or p-type (hole-rich) materials with enhanced conductivity.

Question 2: Semiconductors (2019) - What is meant by doping in semiconductors?

Answer 2: Doping is the process of adding impurities to a semiconductor to increase its conductivity. Pentavalent impurities (e.g., phosphorus) create n-type, while trivalent impurities (e.g., boron) create p-type semiconductors.

Question 3: Semiconductors (2018) - What are donor and acceptor impurities?

Answer 3: Donor impurities (e.g., phosphorus) have five valence electrons and donate extra electrons to create n-type semiconductors. Acceptor impurities (e.g., boron) have three valence electrons and accept electrons, creating holes in p-type semiconductors.

Question 4: Semiconductors (2017) - Why are extrinsic semiconductors preferred over intrinsic ones?

Answer 4: Extrinsic semiconductors have higher conductivity due to increased charge carriers from doping, making them more suitable for electronic devices like diodes and transistors compared to intrinsic semiconductors.

Question 5: Semiconductors (2016) - Draw the energy band diagram of an n-type semiconductor.

Answer 5: The energy band diagram of an n-type semiconductor shows the Fermi level closer to the conduction band, with donor energy levels just below it, indicating extra electrons from donor impurities.