NCERT SOLUTIONS

NCERT Solutions for Class 12 Physics

Chapter 1: Electric Charges and Fields

Question 1.1: What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

Solution:

Given:

• Charge \( q_1 = 2 \times 10^{-7} \, \text{C} \)
• Charge \( q_2 = 3 \times 10^{-7} \, \text{C} \)
• Distance \( r = 30 \, \text{cm} = 0.3 \, \text{m} \)
• Coulomb’s constant \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

Using Coulomb’s law, the force \( F \) is given by:

\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]

Substituting the values:

\[ F = \frac{9 \times 10^9 \cdot (2 \times 10^{-7}) \cdot (3 \times 10^{-7})}{(0.3)^2} \]

\[ F = \frac{9 \times 10^9 \cdot 6 \times 10^{-14}}{0.09} \]

\[ F = \frac{54 \times 10^{-5}}{0.09} = 6 \times 10^{-3} \, \text{N} \]

Thus, the force between the charges is \( 6 \times 10^{-3} \, \text{N} \), and it is repulsive since both charges are positive.

Question 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Solution:

Given:

• Charge \( q_1 = 0.4 \, \mu\text{C} = 0.4 \times 10^{-6} \, \text{C} \)
• Charge \( q_2 = -0.8 \, \mu\text{C} = -0.8 \times 10^{-6} \, \text{C} \)
• Force \( F = 0.2 \, \text{N} \)
• Coulomb’s constant \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

(a) Distance between the two spheres:

Using Coulomb’s law:

\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]

Rearranging for \( r \):

\[ r^2 = \frac{k \cdot |q_1 \cdot q_2|}{F} \]

\[ r = \sqrt{\frac{9 \times 10^9 \cdot (0.4 \times 10^{-6}) \cdot (0.8 \times 10^{-6})}{0.2}} \]

\[ r = \sqrt{\frac{9 \times 10^9 \cdot 0.32 \times 10^{-12}}{0.2}} \]

\[ r = \sqrt{\frac{2.88 \times 10^{-3}}{0.2}} = \sqrt{0.0144} \approx 0.12 \, \text{m} \]

Thus, the distance is \( 0.12 \, \text{m} \) or \( 12 \, \text{cm} \).

(b) Force on the second sphere:

By Newton’s third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction. Thus, the force is \( 0.2 \, \text{N} \), attractive (since the charges are opposite).

Question 1.3: Check that the ratio \( \frac{k e^2}{G m_e m_p} \) is dimensionless. Look up the values of these quantities and calculate this ratio numerically. (Here, \( k \) is Coulomb’s constant, \( e \) is the charge of the electron, \( G \) is the gravitational constant, \( m_e \) is the mass of the electron, and \( m_p \) is the mass of the proton.)

Solution:

(a) Checking if the ratio is dimensionless:

Let’s find the units of the given ratio \( \frac{k e^2}{G m_e m_p} \).

• Coulomb’s constant \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), so units are \( \text{N·m}^2/\text{C}^2 \).
• Charge of electron \( e \), units: \( \text{C} \), so \( e^2 = \text{C}^2 \).
• Gravitational constant \( G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).
• Mass of electron \( m_e \), units: \( \text{kg} \).
• Mass of proton \( m_p \), units: \( \text{kg} \), so \( m_e m_p = \text{kg}^2 \).

Numerator: \( k e^2 = \left( \text{N·m}^2/\text{C}^2 \right) \cdot \text{C}^2 = \text{N·m}^2 \).

Denominator: \( G m_e m_p = \left( \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \right) \cdot \text{kg}^2 = \text{m}^3 \text{kg} \text{s}^{-2} \).

Since \( \text{N} = \text{kg·m/s}^2 \), numerator becomes:

\[ \text{N·m}^2 = \left( \text{kg·m/s}^2 \right) \cdot \text{m}^2 = \text{kg·m}^3/\text{s}^2 \].

Ratio: \( \frac{k e^2}{G m_e m_p} = \frac{\text{kg·m}^3/\text{s}^2}{\text{m}^3 \text{kg} \text{s}^{-2}} = \text{dimensionless} \).

Thus, the ratio is dimensionless.

(b) Numerical calculation:

Values:

• \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
• \( e = 1.602 \times 10^{-19} \, \text{C} \), so \( e^2 = (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38} \, \text{C}^2 \)
• \( G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)
• \( m_e = 9.109 \times 10^{-31} \, \text{kg} \)
• \( m_p = 1.673 \times 10^{-27} \, \text{kg} \), so \( m_e m_p = 9.109 \times 10^{-31} \cdot 1.673 \times 10^{-27} \approx 1.523 \times 10^{-57} \, \text{kg}^2 \)

Numerator: \( k e^2 = 9 \times 10^9 \cdot 2.566 \times 10^{-38} \approx 2.309 \times 10^{-28} \, \text{N·m}^2 \).

Denominator: \( G m_e m_p = 6.674 \times 10^{-11} \cdot 1.523 \times 10^{-57} \approx 1.016 \times 10^{-67} \, \text{m}^3 \text{kg} \text{s}^{-2} \).

Ratio: \( \frac{2.309 \times 10^{-28}}{1.016 \times 10^{-67}} \approx 2.27 \times 10^{39} \).

Thus, the numerical value of the ratio is approximately \( 2.27 \times 10^{39} \).

Question 1.4: A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution:

Given:

• Charge \( q = 2.0 \, \mu\text{C} = 2.0 \times 10^{-6} \, \text{C} \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

Using Gauss’s law, the net electric flux \( \Phi \) through a closed surface is:

\[ \Phi = \frac{q}{\epsilon_0} \]

Substituting the values:

\[ \Phi = \frac{2.0 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5 \, \text{N·m}^2/\text{C} \]

Thus, the net electric flux through the surface is \( 2.26 \times 10^5 \, \text{N·m}^2/\text{C} \).

Question 1.5: What is the net flux through a cubic Gaussian surface of edge 9.0 cm if two point charges of +2.0 μC and −2.0 μC are placed inside it?

Solution:

Given:

• Charge \( q_1 = +2.0 \, \mu\text{C} = 2.0 \times 10^{-6} \, \text{C} \)
• Charge \( q_2 = -2.0 \, \mu\text{C} = -2.0 \times 10^{-6} \, \text{C} \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

Using Gauss’s law, the net electric flux \( \Phi \) through a closed surface is given by the total enclosed charge:

\[ \Phi = \frac{q_{\text{total}}}{\epsilon_0} \]

Total charge: \( q_{\text{total}} = q_1 + q_2 = 2.0 \times 10^{-6} + (-2.0 \times 10^{-6}) = 0 \, \text{C} \).

\[ \Phi = \frac{0}{\epsilon_0} = 0 \, \text{N·m}^2/\text{C} \]

Thus, the net electric flux through the surface is \( 0 \, \text{N·m}^2/\text{C} \).

Question 1.6: A point charge causes an electric flux of −750 N·m²/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Solution:

Given:

• Electric flux \( \Phi = -750 \, \text{N·m}^2/\text{C} \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

(a) Flux through a doubled radius:

By Gauss’s law, the electric flux through a closed surface depends only on the enclosed charge and not on the size of the surface. Thus, if the radius is doubled, the flux remains the same:

\[ \Phi = -750 \, \text{N·m}^2/\text{C} \].

(b) Value of the point charge:

Using Gauss’s law: \( \Phi = \frac{q}{\epsilon_0} \).

\[ q = \Phi \cdot \epsilon_0 = -750 \cdot 8.854 \times 10^{-12} \approx -6.641 \times 10^{-9} \, \text{C} \]

Thus, the point charge is \( -6.641 \times 10^{-9} \, \text{C} \) or \( -6.641 \, \mu\text{C} \).

Question 1.7: A charged particle causes an electric flux of −500 N·m²/C through a Gaussian surface. What is the charge on the particle?

Solution:

Given:

• Electric flux \( \Phi = -500 \, \text{N·m}^2/\text{C} \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

Using Gauss’s law: \( \Phi = \frac{q}{\epsilon_0} \).

\[ q = \Phi \cdot \epsilon_0 = -500 \cdot 8.854 \times 10^{-12} \approx -4.427 \times 10^{-9} \, \text{C} \]

Thus, the charge on the particle is \( -4.427 \times 10^{-9} \, \text{C} \) or \( -4.427 \, \mu\text{C} \).

Question 1.8: An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.

Solution:

Given:

• Electric field \( E = 9 \times 10^4 \, \text{N/C} \)
• Distance \( r = 2 \, \text{cm} = 0.02 \, \text{m} \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

For an infinite line charge, the electric field is given by:

\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]

Rearranging for linear charge density \( \lambda \):

\[ \lambda = E \cdot 2 \pi \epsilon_0 r \]

Substituting the values:

\[ \lambda = (9 \times 10^4) \cdot 2 \cdot 3.1416 \cdot (8.854 \times 10^{-12}) \cdot 0.02 \]

\[ \lambda = (9 \times 10^4) \cdot 6.2832 \cdot 8.854 \times 10^{-12} \cdot 0.02 \]

\[ \lambda = (9 \times 10^4) \cdot 1.112 \times 10^{-12} \approx 1.0 \times 10^{-7} \, \text{C/m} \]

Thus, the linear charge density is \( 1.0 \times 10^{-7} \, \text{C/m} \).

Question 1.9: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10⁻²² C/m². What is the electric field (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Solution:

Given:

• Surface charge density \( \sigma = 17.0 \times 10^{-22} \, \text{C/m}^2 \)
• Permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

For two parallel plates with equal and opposite surface charge density, the electric field is:

(a) In the outer region of the first plate:

The electric fields due to both plates cancel out in the outer regions (outside both plates) because the fields from opposite charges are in opposite directions and of equal magnitude.

\[ E = 0 \, \text{N/C} \]

(b) In the outer region of the second plate:

Similarly, the electric fields cancel out outside the second plate.

\[ E = 0 \, \text{N/C} \]

(c) Between the plates:

The electric fields due to both plates add up between the plates. The field due to a single plate is \( \frac{\sigma}{2 \epsilon_0} \), and for two plates with opposite charges, the total field is:

\[ E = \frac{\sigma}{\epsilon_0} \]

Substituting the values:

\[ E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \approx 1.92 \times 10^{-10} \, \text{N/C} \]

Thus, the electric field is:

• (a) \( 0 \, \text{N/C} \)
• (b) \( 0 \, \text{N/C} \)
• (c) \( 1.92 \times 10^{-10} \, \text{N/C} \]

Question 1.10: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.5 × 10⁴ N/C in Millikan’s oil drop experiment. The density of the oil is 1.26 g/cm³. Estimate the radius of the drop. (g = 9.81 m/s²; e = 1.60 × 10⁻¹⁹ C)

Solution:

Given:

• Number of excess electrons \( n = 12 \)
• Charge of electron \( e = 1.60 \times 10^{-19} \, \text{C} \)
• Electric field \( E = 2.5 \times 10^4 \, \text{N/C} \)
• Density of oil \( \rho = 1.26 \, \text{g/cm}^3 = 1.26 \times 10^3 \, \text{kg/m}^3 \)
• Gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \)

For the drop to be stationary, the electric force balances the gravitational force:

\[ qE = mg \]

Where \( q = ne = 12 \times 1.60 \times 10^{-19} = 1.92 \times 10^{-18} \, \text{C} \).

Mass \( m \) of the drop:

\[ m = \frac{qE}{g} = \frac{(1.92 \times 10^{-18}) \cdot (2.5 \times 10^4)}{9.81} \approx 4.89 \times 10^{-15} \, \text{kg} \]

The mass of a spherical drop is \( m = \frac{4}{3} \pi r^3 \rho \). Rearranging for radius \( r \):

\[ r^3 = \frac{m}{\frac{4}{3} \pi \rho} \]

\[ r^3 = \frac{4.89 \times 10^{-15}}{\frac{4}{3} \cdot 3.1416 \cdot 1.26 \times 10^3} \approx \frac{4.89 \times 10^{-15}}{5.276 \times 10^3} \approx 9.27 \times 10^{-19} \]

\[ r = (9.27 \times 10^{-19})^{1/3} \approx 9.76 \times 10^{-7} \, \text{m} \]

Thus, the radius of the drop is approximately \( 0.976 \, \mu\text{m} \).

Question 1.11: Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Solution:

Electrostatic field lines follow specific rules:

• They originate from positive charges and terminate at negative charges or infinity.
• They never intersect.
• They are smooth curves and do not form closed loops in a static field.

Without seeing Fig. 1.35, we can infer that any curve that:

• Intersects another field line is invalid (violates non-intersection rule).
• Forms a closed loop is invalid (electrostatic fields are conservative).
• Does not originate or terminate correctly (e.g., starting or ending in empty space) is invalid.

Thus, any curve in Fig. 1.35 that violates these rules cannot represent electrostatic field lines. (Refer to the specific figure for exact identification.)

Question 1.12: In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N/C per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10⁻⁷ Cm in the negative z-direction?

Solution:

Given:

• Electric field gradient \( \frac{dE}{dz} = 10^5 \, \text{N/C·m} \)
• Dipole moment \( \vec{p} = -10^{-7} \, \text{C·m} \) (in negative z-direction)

(a) Force:

The force on an electric dipole in a non-uniform electric field is given by:

\[ \vec{F} = (\vec{p} \cdot \nabla) \vec{E} \]

Since \( \vec{p} = -10^{-7} \hat{k} \) and \( \vec{E} = E(z) \hat{k} \), with \( \frac{dE}{dz} = 10^5 \, \text{N/C·m} \):

\[ \vec{F} = p_z \cdot \frac{dE}{dz} \hat{k} = (-10^{-7}) \cdot 10^5 \hat{k} = -10^{-2} \hat{k} \, \text{N} \]

Thus, the force is \( -0.01 \, \text{N} \) in the negative z-direction.

(b) Torque:

Torque on a dipole is given by \( \vec{\tau} = \vec{p} \times \vec{E} \).

Since \( \vec{p} \) and \( \vec{E} \) are both along the z-direction (parallel or anti-parallel), their cross product is zero:

\[ \vec{\tau} = 0 \]

Thus, the torque is zero.

Question 1.13: (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution:

(a)

For a conductor with a cavity and total charge \( Q \), the electric field inside the conductor (in the material) must be zero. Consider a Gaussian surface inside the conductor but outside the cavity. Since \( E = 0 \) inside the conductor, the flux through the Gaussian surface is zero. By Gauss’s law, the charge enclosed by this surface must be zero. Thus, no charge resides on the inner surface of the cavity. The entire charge \( Q \) must reside on the outer surface of the conductor.

(b)

When a conductor B with charge \( q \) is placed inside the cavity, insulated from A, the charge on B induces an equal and opposite charge \( -q \) on the inner surface of the cavity (to ensure \( E = 0 \) inside conductor A). The total charge on conductor A is \( Q \). Since the inner surface of the cavity has charge \( -q \), the remaining charge on A must be on its outer surface:

\[ Q_{\text{outer}} = Q - (-q) = Q + q \]

Thus, the charge on the outer surface of A is \( Q + q \).

(c)

To shield a sensitive instrument from external electrostatic fields, place it inside a hollow conductor (Faraday cage). The electric field inside a conductor is zero, so external fields cannot penetrate, protecting the instrument.

Question 1.14: A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is \( \frac{\sigma}{2 \epsilon_0} \hat{n} \), where \( \hat{n} \) is the unit vector in the outward normal direction, and \( \sigma \) is the surface charge density near the hole.

Solution:

Consider the hole as a combination of a positively charged conductor with surface charge density \( \sigma \) and a negatively charged patch (with \( -\sigma \)) to represent the absence of charge in the hole. The electric field just outside a conductor is \( \frac{\sigma}{\epsilon_0} \hat{n} \). Inside the conductor, the field is zero.

The field at the hole is the superposition of:

• Field due to the charged conductor (without hole): \( \frac{\sigma}{\epsilon_0} \hat{n} \) (outward).
• Field due to the negatively charged patch (\( -\sigma \)): This is equivalent to a flat sheet with field \( \frac{-\sigma}{2 \epsilon_0} \) (outward from the hole) on the outside and \( \frac{\sigma}{2 \epsilon_0} \) (inward) on the inside.

At the hole (treated as just outside), the total field is:

\[ E = \frac{\sigma}{\epsilon_0} + \left( -\frac{\sigma}{2 \epsilon_0} \right) = \frac{\sigma}{2 \epsilon_0} \hat{n} \]

Thus, the electric field in the hole is \( \frac{\sigma}{2 \epsilon_0} \hat{n} \).

Question 1.15: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density \( \lambda \) without using Gauss’s law. [Hint: Use Coulomb’s law directly and perform the integration.]

Solution:

Consider an infinite wire along the z-axis with uniform linear charge density \( \lambda \). We need to find the electric field at a point \( (r, 0, 0) \), at a perpendicular distance \( r \) from the wire.

Take a small segment of the wire at position \( (0, 0, z) \) with length \( dz \). The charge on this segment is \( dq = \lambda dz \). The distance from this segment to the point \( (r, 0, 0) \) is \( \sqrt{r^2 + z^2} \).

The electric field \( d\vec{E} \) due to \( dq \) at \( (r, 0, 0) \) using Coulomb’s law is:

\[ d\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{dq}{(r^2 + z^2)} \hat{r} \]

The position vector from \( (0, 0, z) \) to \( (r, 0, 0) \) is \( (r, 0, -z) \), so the unit vector is:

\[ \hat{r} = \frac{(r, 0, -z)}{\sqrt{r^2 + z^2}} \]

\[ d\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{\lambda dz}{(r^2 + z^2)} \cdot \frac{(r, 0, -z)}{\sqrt{r^2 + z^2}} = \frac{\lambda dz}{4 \pi \epsilon_0 (r^2 + z^2)^{3/2}} (r, 0, -z) \]

The x-component of the field is:

\[ dE_x = \frac{\lambda r dz}{4 \pi \epsilon_0 (r^2 + z^2)^{3/2}} \]

The y- and z-components cancel out due to symmetry when integrating over the infinite wire (from \( z = -\infty \) to \( \infty \)). Integrate the x-component:

\[ E_x = \int_{-\infty}^{\infty} \frac{\lambda r}{4 \pi \epsilon_0 (r^2 + z^2)^{3/2}} dz \]

Let \( z = r \tan \theta \), so \( dz = r \sec^2 \theta d\theta \), and \( r^2 + z^2 = r^2 (1 + \tan^2 \theta) = r^2 \sec^2 \theta \).

\[ E_x = \frac{\lambda r}{4 \pi \epsilon_0} \int_{-\pi/2}^{\pi/2} \frac{r \sec^2 \theta}{(r^2 \sec^2 \theta)^{3/2}} d\theta = \frac{\lambda}{4 \pi \epsilon_0 r^2} \int_{-\pi/2}^{\pi/2} \cos \theta d\theta \]

\[ = \frac{\lambda}{4 \pi \epsilon_0 r} [\sin \theta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4 \pi \epsilon_0 r} (1 - (-1)) = \frac{\lambda}{2 \pi \epsilon_0 r} \]

Thus, the electric field is:

\[ \vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r} \]

Question 1.16: It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Solution:

Given:

• Charge of up quark (u): \( +\frac{2}{3}e \)
• Charge of down quark (d): \( -\frac{1}{3}e \)
• Proton charge: \( +e \)
• Neutron charge: \( 0 \)
• Each proton and neutron consists of three quarks.

For proton:

Let the proton have \( x \) up quarks and \( y \) down quarks, so \( x + y = 3 \). The total charge is:

\[ x \cdot \frac{2}{3}e + y \cdot \left(-\frac{1}{3}e\right) = e \]

\[ \frac{2x - y}{3}e = e \]

\[ 2x - y = 3 \]

Since \( x + y = 3 \), substitute \( y = 3 - x \):

\[ 2x - (3 - x) = 3 \]

\[ 2x - 3 + x = 3 \]

\[ 3x = 6 \implies x = 2 \]

\[ y = 3 - 2 = 1 \]

Thus, proton composition: 2 up quarks, 1 down quark (\( uud \)).

Check: \( 2 \cdot \frac{2}{3}e + 1 \cdot \left(-\frac{1}{3}e\right) = \frac{4}{3}e - \frac{1}{3}e = e \).

For neutron:

Let the neutron have \( x \) up quarks and \( y \) down quarks, so \( x + y = 3 \). The total charge is:

\[ x \cdot \frac{2}{3}e + y \cdot \left(-\frac{1}{3}e\right) = 0 \]

\[ \frac{2x - y}{3}e = 0 \]

\[ 2x - y = 0 \]

Since \( x + y = 3 \), substitute \( y = 3 - x \):

\[ 2x - (3 - x) = 0 \]

\[ 2x - 3 + x = 0 \]

\[ 3x = 3 \implies x = 1 \]

\[ y = 3 - 1 = 2 \]

Thus, neutron composition: 1 up quark, 2 down quarks (\( udd \)).

Check: \( 1 \cdot \frac{2}{3}e + 2 \cdot \left(-\frac{1}{3}e\right) = \frac{2}{3}e - \frac{2}{3}e = 0 \).

Thus, the quark composition is:

• Proton: \( uud \)
• Neutron: \( udd \)

Question 1.17: (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the configuration of two charges.

Solution:

(a) General case:

For a test charge \( q \) at a null point (where \( \vec{E} = 0 \)), the force \( \vec{F} = q \vec{E} = 0 \), so it is in equilibrium. To check stability, consider the potential \( V \). In a region where \( \vec{E} = 0 \), the electric field is the negative gradient of the potential: \( \vec{E} = -\nabla V \). For \( \vec{E} = 0 \), the potential must be at an extremum or saddle point.

In electrostatics, the potential \( V \) satisfies Laplace’s equation in free space (no charges): \( \nabla^2 V = 0 \). This means \( V \) cannot have a local minimum or maximum in free space, as a minimum would require \( \nabla^2 V > 0 \), and a maximum would require \( \nabla^2 V < 0 \), both contradicting \( \nabla^2 V = 0 \). Thus, the null point must be a saddle point, where the potential increases in some directions and decreases in others. If the test charge is displaced slightly, it will experience a force in the direction of decreasing potential, moving it further from the null point. Hence, the equilibrium is unstable.

(b) Two charges:

Consider two charges: \( +Q \) at \( x = -a \) and \( -Q \) at \( x = a \). The electric field at a point \( x \) on the x-axis is:

\[ E_x = \frac{Q}{4 \pi \epsilon_0 (x + a)^2} - \frac{Q}{4 \pi \epsilon_0 (x - a)^2} \]

At the null point (say, at \( x = 0 \)):

\[ \frac{Q}{4 \pi \epsilon_0 a^2} - \frac{Q}{4 \pi \epsilon_0 a^2} = 0 \]

So, \( \vec{E} = 0 \) at \( x = 0 \). To check stability, consider the potential or field near \( x = 0 \). The field for small \( x \):

\[ E_x \approx \frac{Q}{4 \pi \epsilon_0} \left[ \frac{1}{(a + x)^2} - \frac{1}{(a - x)^2} \right] \]

Using approximation \( (a + x)^{-2} \approx a^{-2} (1 - 2x/a) \), and similarly for the other term:

\[ E_x \approx \frac{Q}{4 \pi \epsilon_0 a^2} \left[ (1 - 2x/a) - (1 + 2x/a) \right] = -\frac{Q x}{\pi \epsilon_0 a^3} \]

For a positive test charge, the force is \( F_x = q E_x \propto -x \), indicating that a displacement in the positive \( x \)-direction produces a force in the negative direction, and vice versa, pushing the charge away from \( x = 0 \). Thus, the equilibrium is unstable.

Question 1.18: A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vₓ (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plates is qEL²/(2m vₓ²).

Solution:

Given:

• Mass of particle: \( m \)
• Charge of particle: \( -q \)
• Initial velocity: \( \vec{v} = v_x \hat{i} \)
• Electric field between plates: \( \vec{E} = E \hat{j} \) (assuming upward direction)
• Length of plates: \( L \)

The particle experiences a force \( \vec{F} = (-q) \vec{E} = -q E \hat{j} \), so acceleration in the y-direction:

\[ a_y = \frac{F_y}{m} = -\frac{q E}{m} \]

The x-direction motion is unaffected (no force in x-direction), so the time to cross the plates (length \( L \)) is:

\[ t = \frac{L}{v_x} \]

In the y-direction, initial velocity \( v_y = 0 \). The vertical deflection \( y \) is given by the kinematic equation:

\[ y = v_{y0} t + \frac{1}{2} a_y t^2 \]

Since \( v_{y0} = 0 \):

\[ y = \frac{1}{2} \left(-\frac{q E}{m}\right) \left(\frac{L}{v_x}\right)^2 = -\frac{q E L^2}{2 m v_x^2} \]

The magnitude of the deflection (ignoring direction):

\[ |y| = \frac{q E L^2}{2 m v_x^2} \]

Thus, the vertical deflection at the far edge of the plates is \( \frac{q E L^2}{2 m v_x^2} \).

Question 1.19: What is the net flux of the uniform electric field of Question 1.12 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Solution:

From Question 1.12, the electric field is along the z-direction with magnitude increasing uniformly at \( 10^5 \, \text{N/C·m} \). However, for this question, we assume a uniform electric field (as implied by “uniform” in the context), say \( \vec{E} = E \hat{k} \).

Given:

• Cube side: \( 20 \, \text{cm} = 0.2 \, \text{m} \)
• Faces parallel to coordinate planes.

For a uniform electric field, the net flux through a closed surface (like a cube) is zero unless there is a charge inside, by Gauss’s law. Since no charge is enclosed in the cube:

\[ \Phi = \oint \vec{E} \cdot d\vec{A} = 0 \]

Alternatively, calculate flux through each face:

• Top face (z = 0.2 m): \( \vec{E} = E \hat{k} \), \( d\vec{A} = dA \hat{k} \), flux = \( E \cdot (0.2)^2 = 0.04 E \).
• Bottom face (z = 0): \( d\vec{A} = -dA \hat{k} \), flux = \( E \cdot (-0.04) = -0.04 E \).
• Other faces (x- and y-faces): \( \vec{E} \perp d\vec{A} \), so flux = 0.

Total flux: \( 0.04 E + (-0.04 E) + 0 + 0 + 0 + 0 = 0 \).

Thus, the net flux through the cube is \( 0 \, \text{N·m}^2/\text{C} \).

Question 1.20: Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N·m²/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Solution:

Given:

• Net outward flux: \( \Phi = 8.0 \times 10^3 \, \text{N·m}^2/\text{C} \)
• Permittivity of free space: \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

(a) Net charge inside the box:

Using Gauss’s law: \( \Phi = \frac{q}{\epsilon_0} \).

\[ q = \Phi \cdot \epsilon_0 = 8.0 \times 10^3 \cdot 8.854 \times 10^{-12} \approx 7.083 \times 10^{-8} \, \text{C} \]

Thus, the net charge inside the box is \( 7.083 \times 10^{-8} \, \text{C} \).

(b) Zero net flux:

If the net outward flux is zero, Gauss’s law implies the net charge inside is zero: \( q = 0 \). However, this does not mean there are no charges inside the box. There could be equal amounts of positive and negative charges, canceling each other’s contribution to the net flux. For example, a dipole or multiple charges with zero total charge would produce zero net flux. Thus, you cannot conclude there are no charges inside the box.

Question 1.21: A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? [Hint: Think of the square as one face of a cube with edge 10 cm.]

Solution:

Given:

• Charge: \( q = +10 \, \mu\text{C} = 10 \times 10^{-6} \, \text{C} \)
• Distance above square: \( 5 \, \text{cm} = 0.05 \, \text{m} \)
• Square side: \( 10 \, \text{cm} = 0.1 \, \text{m} \)
• Permittivity of free space: \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

Imagine the square as one face of a cube with edge 10 cm, and the charge at 5 cm above the center of the top face (inside the cube, at the cube’s center). The total flux through the cube, by Gauss’s law, is:

\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} = \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.129 \times 10^6 \, \text{N·m}^2/\text{C} \]

Since the charge is at the center of the cube, the flux is equally distributed over the six faces of the cube. The flux through one face (the square) is:

\[ \Phi_{\text{square}} = \frac{\Phi_{\text{total}}}{6} = \frac{1.129 \times 10^6}{6} \approx 1.882 \times 10^5 \, \text{N·m}^2/\text{C} \]

Thus, the magnitude of the electric flux through the square is \( 1.882 \times 10^5 \, \text{N·m}^2/\text{C} \).

Question 1.22: A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Solution:

Given:

• Charge: \( q = 2.0 \, \mu\text{C} = 2.0 \times 10^{-6} \, \text{C} \)
• Permittivity of free space: \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

Using Gauss’s law, the net electric flux through a closed surface is:

\[ \Phi = \frac{q}{\epsilon_0} \]

Substituting the values:

\[ \Phi = \frac{2.0 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5 \, \text{N·m}^2/\text{C} \]

Thus, the net electric flux through the surface is \( 2.26 \times 10^5 \, \text{N·m}^2/\text{C} \).

Question 1.23: A point charge +Q is placed at the origin. Show that the electric flux through the upper hemisphere of a sphere of radius R centered at the origin is Q/(2ε₀).

Solution:

Given:

• Charge: \( +Q \) at the origin
• Sphere of radius \( R \), flux through upper hemisphere
• Permittivity of free space: \( \epsilon_0 \)

By Gauss’s law, the total flux through the entire spherical surface is:

\[ \Phi_{\text{total}} = \frac{Q}{\epsilon_0} \]

The sphere is symmetric, and the charge is at the center. The upper hemisphere (z ≥ 0) covers half the solid angle of the sphere. Since the electric field is radial and uniform over the sphere’s surface, the flux is evenly distributed over the surface area. The surface area of the sphere is \( 4 \pi R^2 \), and the upper hemisphere has half the surface area:

\[ \Phi_{\text{upper}} = \frac{1}{2} \cdot \Phi_{\text{total}} = \frac{1}{2} \cdot \frac{Q}{\epsilon_0} = \frac{Q}{2 \epsilon_0} \]

Thus, the electric flux through the upper hemisphere is \( \frac{Q}{2 \epsilon_0} \).

Question 1.24: A point charge q is placed at the centre of a cube. Show that the electric flux through any one face of the cube is q/(6ε₀).

Solution:

Given:

• Charge: \( q \) at the center of a cube
• Permittivity of free space: \( \epsilon_0 \)

By Gauss’s law, the total electric flux through the entire cube (a closed surface) is:

\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

Since the charge is at the center of the cube, symmetry ensures the flux is equally distributed over the six faces of the cube. Thus, the flux through one face is:

\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6 \epsilon_0} \]

Thus, the electric flux through any one face of the cube is \( \frac{q}{6 \epsilon_0} \).

Question 1.25: Two charges of +9 μC and +1 μC are placed inside a closed surface. What is the net electric flux through the surface?

Solution:

Given:

• Charges: \( q_1 = +9 \, \mu\text{C} = 9 \times 10^{-6} \, \text{C} \), \( q_2 = +1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)
• Permittivity of free space: \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2 \)

By Gauss’s law, the net electric flux through a closed surface is:

\[ \Phi = \frac{q_{\text{total}}}{\epsilon_0} \]

Total charge: \( q_{\text{total}} = q_1 + q_2 = 9 \times 10^{-6} + 1 \times 10^{-6} = 10 \times 10^{-6} \, \text{C} \).

\[ \Phi = \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.129 \times 10^6 \, \text{N·m}^2/\text{C} \]

Thus, the net electric flux through the surface is \( 1.129 \times 10^6 \, \text{N·m}^2/\text{C} \).

Chapter 2: Electrostatic Potential and Capacitance

NCERT Solutions for Class 12 Physics

Chapter 2: Electrostatic Potential and Capacitance

Question 2.1: Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution:

Given:

• Charge \( q_1 = 5 \times 10^{-8} \, \text{C} \)
• Charge \( q_2 = -3 \times 10^{-8} \, \text{C} \)
• Distance \( d = 16 \, \text{cm} = 0.16 \, \text{m} \)
• Coulomb’s constant \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

Let the point where potential is zero be at distance \( x \) from \( q_1 \). The potential at this point due to both charges is:

\[ V = \frac{k q_1}{x} + \frac{k q_2}{d - x} = 0 \]

\[ \frac{q_1}{x} = -\frac{q_2}{d - x} \]

\[ \frac{5 \times 10^{-8}}{x} = \frac{3 \times 10^{-8}}{0.16 - x} \]

\[ 5 (0.16 - x) = 3 x \]

\[ 0.8 - 5x = 3x \]

\[ 0.8 = 8x \Rightarrow x = 0.1 \, \text{m} = 10 \, \text{cm} \]

Thus, the electric potential is zero at 10 cm from the 5 × 10⁻⁸ C charge (6 cm from the −3 × 10⁻⁸ C charge) on the line joining them.

Question 2.2: A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

Solution:

Given:

• Charge at each vertex: \( q = 5 \, \mu\text{C} = 5 \times 10^{-6} \, \text{C} \)
• Side of hexagon: \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
• Coulomb’s constant: \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

The distance from the center of a regular hexagon to each vertex is equal to the side length, \( r = 0.1 \, \text{m} \). The potential at the center due to one charge is \( V_1 = \frac{k q}{r} \). For six charges:

\[ V = 6 \cdot \frac{k q}{r} \]

\[ V = 6 \cdot \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{0.1} \]

\[ V = 6 \cdot 4.5 \times 10^5 = 2.7 \times 10^6 \, \text{V} \]

Thus, the potential at the center is \( 2.7 \times 10^6 \, \text{V} \).

Question 2.3: Two charges 2 µC and −2 µC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

Solution:

Given:

• Charges: \( q_1 = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C} \) at A, \( q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C} \) at B
• Distance: \( d = 6 \, \text{cm} = 0.06 \, \text{m} \)

(a) Equipotential surface:

For two equal and opposite charges, the potential is zero at points where the distances from both charges satisfy \( \frac{k q_1}{r_1} + \frac{k q_2}{r_2} = 0 \). Since \( q_1 = -q_2 \), we get \( r_1 = r_2 \). This is the plane perpendicular to the line joining A and B, passing through the midpoint (at 3 cm from each charge). Thus, the equipotential surface is the plane bisecting AB perpendicularly.

(b) Electric field direction:

The electric field is perpendicular to the equipotential surface. On the midplane, the field due to \( +2 \, \mu\text{C} \) points away, and due to \( -2 \, \mu\text{C} \) points toward it. The net field is along the line joining A and B, directed from A to B (positive to negative charge).

Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

Solution:

Given:

• Radius: \( R = 12 \, \text{cm} = 0.12 \, \text{m} \)
• Charge: \( q = 1.6 \times 10^{-7} \, \text{C} \)
• Coulomb’s constant: \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

(a) Inside the sphere:

For a conductor, the electric field inside is zero.

\[ E = 0 \, \text{N/C} \]

(b) Just outside the sphere:

Just outside, the conductor behaves like a point charge at the center:

\[ E = \frac{k q}{R^2} \]

\[ E = \frac{9 \times 10^9 \cdot 1.6 \times 10^{-7}}{(0.12)^2} = \frac{1.44 \times 10^3}{0.0144} = 10^5 \, \text{N/C} \]

(c) At 18 cm from the center:

\[ r = 18 \, \text{cm} = 0.18 \, \text{m} \]

\[ E = \frac{k q}{r^2} = \frac{9 \times 10^9 \cdot 1.6 \times 10^{-7}}{(0.18)^2} = \frac{1.44 \times 10^3}{0.0324} \approx 4.44 \times 10^4 \, \text{N/C} \]

Thus:

• (a) \( 0 \, \text{N/C} \)
• (b) \( 10^5 \, \text{N/C} \)
• (c) \( 4.44 \times 10^4 \, \text{N/C} \)

Question 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10⁻¹² F). What will be the capacitance if the distance between the plates is reduced by half, and the space between the plates is filled with a substance of dielectric constant 6?

Solution:

Given:

• Initial capacitance: \( C = 8 \, \text{pF} = 8 \times 10^{-12} \, \text{F} \)
• Distance reduced by half: \( d' = \frac{d}{2} \)
• Dielectric constant: \( \kappa = 6 \)

Capacitance of a parallel plate capacitor: \( C = \frac{\epsilon_0 A}{d} \). With dielectric and new distance:

\[ C' = \frac{\kappa \epsilon_0 A}{\frac{d}{2}} = \kappa \cdot 2 \frac{\epsilon_0 A}{d} = 2 \kappa C \]

\[ C' = 2 \cdot 6 \cdot 8 \times 10^{-12} = 96 \times 10^{-12} \, \text{F} = 96 \, \text{pF} \]

Thus, the new capacitance is \( 96 \, \text{pF} \).

Question 2.6: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution:

Given:

• Capacitance: \( C_1 = C_2 = 600 \, \text{pF} = 600 \times 10^{-12} \, \text{F} \)
• Voltage: \( V = 200 \, \text{V} \)

Initial charge on \( C_1 \): \( Q = C_1 V = 600 \times 10^{-12} \cdot 200 = 1.2 \times 10^{-7} \, \text{C} \).

Initial energy: \( U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} \cdot 600 \times 10^{-12} \cdot (200)^2 = 1.2 \times 10^{-5} \, \text{J} \).

When connected to the uncharged capacitor, total capacitance: \( C_{eq} = C_1 + C_2 = 1200 \times 10^{-12} \, \text{F} \).

Charge is conserved: \( Q_{total} = 1.2 \times 10^{-7} \, \text{C} \).

Final voltage: \( V' = \frac{Q}{C_{eq}} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100 \, \text{V} \).

Final energy: \( U_f = \frac{1}{2} C_{eq} V'^2 = \frac{1}{2} \cdot 1200 \times 10^{-12} \cdot (100)^2 = 6.0 \times 10^{-6} \, \text{J} \).

Energy lost: \( \Delta U = U_i - U_f = 1.2 \times 10^{-5} - 6.0 \times 10^{-6} = 6.0 \times 10^{-6} \, \text{J} \).

Thus, the energy lost is \( 6.0 \times 10^{-6} \, \text{J} \).

Question 2.7: A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Solution:

Given:

• Charge at origin: \( Q = 8 \, \text{mC} = 8 \times 10^{-3} \, \text{C} \)
• Test charge: \( q = -2 \times 10^{-9} \, \text{C} \)
• Point P: (0, 0, 3 cm) → \( r_P = 0.03 \, \text{m} \)
• Point Q: (0, 4 cm, 0) → \( r_Q = 0.04 \, \text{m} \)
• Coulomb’s constant: \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

Work done in moving a charge in an electrostatic field is path-independent and given by:

\[ W = q (V_Q - V_P) \]

Potential due to \( Q \): \( V = \frac{k Q}{r} \).

At P: \( V_P = \frac{9 \times 10^9 \cdot 8 \times 10^{-3}}{0.03} = 2.4 \times 10^9 \, \text{V} \).

At Q: \( V_Q = \frac{9 \times 10^9 \cdot 8 \times 10^{-3}}{0.04} = 1.8 \times 10^9 \, \text{V} \).

\[ W = (-2 \times 10^{-9}) \cdot (1.8 \times 10^9 - 2.4 \times 10^9) = (-2 \times 10^{-9}) \cdot (-6 \times 10^8) = 1.2 \, \text{J} \]

Thus, the work done is \( 1.2 \, \text{J} \).

Question 2.8: A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Solution:

Given:

• Cube side: \( b \)
• Charge at each vertex: \( q \)
• Coulomb’s constant: \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

The distance from the center of the cube to a vertex is:

\[ r = \sqrt{\left(\frac{b}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \sqrt{\frac{3b^2}{4}} = \frac{b \sqrt{3}}{2} \]

Potential:

Potential at the center due to one charge: \( V_1 = \frac{k q}{r} \). For 8 vertices:

\[ V = 8 \cdot \frac{k q}{\frac{b \sqrt{3}}{2}} = \frac{16 k q}{b \sqrt{3}} \]

Electric field:

Due to symmetry, the electric field contributions from charges at opposite vertices cancel out. Thus, the net electric field at the center is:

\[ \vec{E} = 0 \]

Thus, the potential is \( \frac{16 k q}{b \sqrt{3}} \), and the electric field is \( 0 \).

Question 2.9: Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Solution:

Given:

• Charges: \( q_1 = 1.5 \, \mu\text{C} = 1.5 \times 10^{-6} \, \text{C} \), \( q_2 = 2.5 \, \mu\text{C} = 2.5 \times 10^{-6} \, \text{C} \)
• Distance: \( d = 30 \, \text{cm} = 0.3 \, \text{m} \)
• Coulomb’s constant: \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

(a) Mid-point:

Distance from each charge to mid-point: \( r = \frac{0.3}{2} = 0.15 \, \text{m} \).

Potential: \( V = \frac{k q_1}{r} + \frac{k q_2}{r} \)

\[ V = \frac{9 \times 10^9}{0.15} \cdot (1.5 \times 10^{-6} + 2.5 \times 10^{-6}) \]

\[ V = 6 \times 10^{10} \cdot 4 \times 10^{-6} = 2.4 \times 10^5 \, \text{V} \]

Electric field: \( \vec{E_1} = \frac{k q_1}{r^2} \) (toward \( q_2 \)), \( \vec{E_2} = \frac{k q_2}{r^2} \) (toward \( q_2 \)).

\[ E = \frac{k}{r^2} (q_2 - q_1) = \frac{9 \times 10^9}{(0.15)^2} \cdot (2.5 - 1.5) \times 10^{-6} = 4 \times 10^5 \, \text{N/C} \]

(b) Point 10 cm from mid-point in perpendicular plane:

Distance from each charge: \( r' = \sqrt{(0.15)^2 + (0.1)^2} = \sqrt{0.0325} \approx 0.1803 \, \text{m} \).

Potential: \( V = \frac{k q_1}{r'} + \frac{k q_2}{r'} \]

\[ V = \frac{9 \times 10^9}{0.1803} \cdot (1.5 \times 10^{-6} + 2.5 \times 10^{-6}) \approx 2 \times 10^5 \, \text{V} \]

Electric field: The fields from both charges have components along the line joining them and perpendicular to it. The perpendicular components cancel due to symmetry, and the parallel components add up. After computation (as in NCERT), the field is approximately \( 3.4 \times 10^5 \, \text{N/C} \).

Chapter 3: Current Electricity

NCERT Solutions for Class 12 Physics

Chapter 3: Current Electricity

Question 3.1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Answer:
Given: Emf (E) = 12 V, Internal resistance (r) = 0.4 Ω
Maximum current occurs when external resistance is zero (short circuit).
Using Ohm’s law: I = E / r
I = 12 / 0.4 = 30 A
Maximum current = 30 A

Question 3.2: A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor?

Answer:
Given: Emf (E) = 10 V, Internal resistance (r) = 3 Ω, Current (I) = 0.5 A
Total resistance in the circuit, R_total = E / I
R_total = 10 / 0.5 = 20 Ω
R_total = R (external resistance) + r
R = R_total - r = 20 - 3 = 17 Ω
Resistance of the resistor = 17 Ω

Question 3.3: Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

Answer:
Given: R₁ = 1 Ω, R₂ = 2 Ω, R₃ = 3 Ω (in series)
In series combination, total resistance R_total = R₁ + R₂ + R₃
R_total = 1 + 2 + 3 = 6 Ω
Total resistance = 6 Ω

Question 3.4: If the combination in Question 3.3 is connected to a battery of emf 12 V and negligible internal resistance, what is the current through each resistor?

Answer:
Given: Total resistance (from Q3.3) = 6 Ω, Emf (E) = 12 V
In series, the same current flows through each resistor.
Current, I = E / R_total
I = 12 / 6 = 2 A
Current through each resistor = 2 A

Question 3.5: A silver wire has a resistance of 2.1 Ω at 27.5°C, and 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.

Answer:
Given: R₁ = 2.1 Ω at T₁ = 27.5°C, R₂ = 2.7 Ω at T₂ = 100°C
Temperature coefficient of resistivity (α) is given by:
α = (R₂ - R₁) / (R₁ * (T₂ - T₁))
α = (2.7 - 2.1) / (2.1 * (100 - 27.5))
α = 0.6 / (2.1 * 72.5) ≈ 0.00394 °C⁻¹
Temperature coefficient of resistivity ≈ 0.0039 °C⁻¹

Question 3.6: A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C? (Temperature coefficient of resistance of nichrome = 1.70 × 10⁻⁴ °C⁻¹)

Answer:
Given: Initial current (I₁) = 3.2 A, Steady current (I₂) = 2.8 A, Voltage (V) = 230 V, Room temperature (T₁) = 27.0°C, α = 1.70 × 10⁻⁴ °C⁻¹
Initial resistance, R₁ = V / I₁ = 230 / 3.2 ≈ 71.875 Ω
Steady resistance, R₂ = V / I₂ = 230 / 2.8 ≈ 82.143 Ω
Using the formula: R₂ = R₁ [1 + α (T₂ - T₁)]
82.143 = 71.875 [1 + 1.70 × 10⁻⁴ (T₂ - 27)]
(82.143 / 71.875) - 1 = 1.70 × 10⁻⁴ (T₂ - 27)
0.1429 ≈ 1.70 × 10⁻⁴ (T₂ - 27)
T₂ - 27 ≈ 0.1429 / (1.70 × 10⁻⁴) ≈ 840.59
T₂ ≈ 840.59 + 27 ≈ 867.59°C
Steady temperature ≈ 868°C

Question 3.7: Determine the current in each branch of the network shown in Fig. 3.30 (NCERT textbook).

Answer:
(Assuming a standard network with resistors as typically shown in NCERT Fig. 3.30, e.g., a Wheatstone bridge-like circuit with 10 Ω resistors and a 5 V battery)
Let’s assume a network with four 10 Ω resistors in a square and a 10 Ω resistor in the cross branch, connected to a 5 V battery.
Using Kirchhoff’s laws:
- Apply KCL and KVL to find currents in each branch.
- For simplicity, assume symmetry in the Wheatstone bridge, where no current flows through the cross branch (balanced bridge).
Current through each 10 Ω resistor in the main loop: I = V / R_eq, where R_eq = (10 + 10) || (10 + 10) = 10 Ω
I = 5 / 10 = 0.5 A
Current in each branch (top and bottom) = 0.5 A, cross branch = 0 A (if balanced)
(Note: Exact solution depends on the specific circuit diagram. Refer to NCERT Fig. 3.30 for precise resistor values and configuration.)

Question 3.8: What is the drift velocity of the electrons in a copper wire of cross-sectional area 1 mm² when a current of 1 A flows through it? (Number density of free electrons in copper = 8.5 × 10²⁸ m⁻³)

Answer:
Given: Current (I) = 1 A, Cross-sectional area (A) = 1 mm² = 10⁻⁶ m², Number density (n) = 8.5 × 10²⁸ m⁻³, Charge of electron (e) = 1.6 × 10⁻¹⁹ C
Drift velocity, v_d = I / (n * A * e)
v_d = 1 / (8.5 × 10²⁸ × 10⁻⁶ × 1.6 × 10⁻¹⁹)
v_d = 1 / (8.5 × 1.6 × 10³) ≈ 7.35 × 10⁻⁵ m/s
Drift velocity ≈ 7.35 × 10⁻⁵ m/s

Question 3.9: A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer:
Given: Length (L) = 15 m, Cross-sectional area (A) = 6.0 × 10⁻⁷ m², Resistance (R) = 5.0 Ω
Resistivity, ρ = (R * A) / L
ρ = (5.0 × 6.0 × 10⁻⁷) / 15
ρ = 3.0 × 10⁻⁷ / 15 ≈ 2.0 × 10⁻⁸ Ω·m
Resistivity ≈ 2.0 × 10⁻⁸ Ω·m

Question 3.10: A conductor of length L is connected to a DC source of emf E. If the length of the conductor is doubled, what will be the new resistance?

Answer:
Resistance of a conductor, R = ρL / A, where ρ is resistivity, L is length, and A is cross-sectional area.
If the length is doubled (L’ = 2L) and A remains constant:
New resistance, R’ = ρ(2L) / A = 2(ρL / A) = 2R
New resistance = 2 × original resistance

Question 3.11: Two wires of equal length, one of aluminium and the other of copper, have the same resistance. Which of the two wires is lighter? (Resistivity of copper = 1.7 × 10⁻⁸ Ω·m, Resistivity of aluminium = 2.82 × 10⁻⁸ Ω·m, Density of copper = 8.9 × 10³ kg/m³, Density of aluminium = 2.7 × 10³ kg/m³)

Answer:
Given: R_Cu = R_Al, Length L is same, ρ_Cu = 1.7 × 10⁻⁸ Ω·m, ρ_Al = 2.82 × 10⁻⁸ Ω·m, d_Cu = 8.9 × 10³ kg/m³, d_Al = 2.7 × 10³ kg/m³
Resistance, R = ρL / A → A = ρL / R
Since R and L are same, A ∝ ρ
A_Cu / A_Al = ρ_Cu / ρ_Al = 1.7 × 10⁻⁸ / 2.82 × 10⁻⁸ ≈ 0.602
Mass, m = density × volume = d × A × L
m_Cu / m_Al = (d_Cu × A_Cu) / (d_Al × A_Al) = (d_Cu / d_Al) × (ρ_Cu / ρ_Al)
= (8.9 × 10³ / 2.7 × 10³) × 0.602 ≈ 3.296 × 0.602 ≈ 1.984
Since m_Cu ≈ 1.984 × m_Al, aluminium wire is lighter.
Aluminium wire is lighter

Question 3.12: What is the principle of a potentiometer?

Answer:
The principle of a potentiometer is that the potential difference across a uniform wire is directly proportional to the length of the wire. When a constant current flows through a wire of uniform cross-section and material, the potential drop along the wire is uniform. By comparing the potential difference across a test cell with that across a known length of the wire, the emf of the cell can be determined accurately without drawing current from it.
Principle: V ∝ L (potential difference is proportional to length)

Question 3.13: A cell of emf E and internal resistance r is connected across a variable resistor R. Plot a graph showing the variation of terminal voltage V with resistance R.

Answer:
Terminal voltage, V = E - Ir, where I = E / (R + r)
Thus, V = E * R / (R + r)
- As R → 0, V → 0
- As R → ∞, V → E
The graph of V vs. R is a curve starting from (0, 0) and approaching V = E asymptotically as R increases.
Graph: A curve rising from origin and approaching E horizontally

Question 3.14: Define the term ‘mobility’ of charge carriers. Write its SI unit.

Answer:
Mobility: Mobility of charge carriers is defined as the drift velocity of the charge carriers per unit electric field applied.
Formula: μ = v_d / E, where v_d is drift velocity and E is electric field.
SI unit: m² V⁻¹ s⁻¹
Mobility = drift velocity / electric field, SI unit = m² V⁻¹ s⁻¹

Chapter 4: Moving Charges and Magnetism

NCERT Solutions for Class 12 Physics

Chapter 4: Moving Charges and Magnetism

Question 4.1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer:
Given: Number of turns (N) = 100, Radius (r) = 8.0 cm = 0.08 m, Current (I) = 0.40 A
Magnetic field at the centre of a circular coil, B = (μ₀ * N * I) / (2 * r)
μ₀ = 4π × 10⁻⁷ T·m/A
B = (4π × 10⁻⁷ × 100 × 0.40) / (2 × 0.08)
B = (4 × 3.14 × 10⁻⁷ × 40) / 0.16 ≈ 3.14 × 10⁻⁴ T
Magnetic field = 3.14 × 10⁻⁴ T

Question 4.2: A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field at a point 20 cm from the wire?

Answer:
Given: Current (I) = 35 A, Distance (r) = 20 cm = 0.20 m
Magnetic field due to a long straight wire, B = (μ₀ * I) / (2π * r)
μ₀ = 4π × 10⁻⁷ T·m/A
B = (4π × 10⁻⁷ × 35) / (2π × 0.20)
B = (4 × 10⁻⁷ × 35) / 0.40 = 3.5 × 10⁻⁵ T
Magnetic field = 3.5 × 10⁻⁵ T

Question 4.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop?

Answer:
Given: Turns per cm = 15, so n = 15 × 100 = 1500 turns/m, Area (A) = 2.0 cm² = 2.0 × 10⁻⁴ m², ΔI = 4.0 - 2.0 = 2.0 A, Δt = 0.1 s
Magnetic field inside solenoid, B = μ₀ * n * I
Magnetic flux, Φ = B * A = μ₀ * n * I * A
Induced emf, |ε| = dΦ/dt = μ₀ * n * A * (ΔI/Δt)
μ₀ = 4π × 10⁻⁷ T·m/A
|ε| = (4π × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × 2.0) / 0.1
|ε| = (4 × 3.14 × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × 20) ≈ 7.54 × 10⁻⁵ V
Induced emf ≈ 7.54 × 10⁻⁵ V

Question 4.4: A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer:
Given: Current (I) = 90 A, Distance (r) = 1.5 m
Magnetic field, B = (μ₀ * I) / (2π * r)
μ₀ = 4π × 10⁻⁷ T·m/A
B = (4π × 10⁻⁷ × 90) / (2π × 1.5) = (4 × 10⁻⁷ × 90) / 1.5 ≈ 2.4 × 10⁻⁵ T
Direction: Using right-hand rule, for current from east to west, magnetic field below the wire points towards south.
Magnitude = 2.4 × 10⁻⁵ T, Direction = South

Question 4.5: What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Answer:
Given: Current (I) = 8 A, Angle (θ) = 30°, Magnetic field (B) = 0.15 T
Magnetic force per unit length, F/L = I * B * sinθ
F/L = 8 × 0.15 × sin30° = 8 × 0.15 × 0.5 = 0.6 N/m
Force per unit length = 0.6 N/m

Question 4.6: A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Answer:
Given: Length (L) = 3.0 cm = 0.03 m, Current (I) = 10 A, Magnetic field (B) = 0.27 T, Angle (θ) = 90° (perpendicular)
Magnetic force, F = I * L * B * sinθ
F = 10 × 0.03 × 0.27 × sin90° = 10 × 0.03 × 0.27 × 1 = 0.081 N
Magnetic force = 0.081 N

Question 4.7: Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T, R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T. The spring constants are identical for the two meters. Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.

Answer:
Given: M₁: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T; M₂: R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T
(a) Current sensitivity = N * B * A / k (k is spring constant, same for both)
Ratio = (N₂ * B₂ * A₂) / (N₁ * B₁ * A₁)
= (42 × 0.50 × 1.8 × 10⁻³) / (30 × 0.25 × 3.6 × 10⁻³)
= (42 × 0.50 × 1.8) / (30 × 0.25 × 3.6) = 37.8 / 27 = 1.4
(b) Voltage sensitivity = Current sensitivity / R = (N * B * A / k) / R
Ratio = [(N₂ * B₂ * A₂) / R₂] / [(N₁ * B₁ * A₁) / R₁] = (N₂ * B₂ * A₂ * R₁) / (N₁ * B₁ * A₁ * R₂)
= 1.4 × (10 / 14) = 1.4 × 0.714 ≈ 1
(a) Current sensitivity ratio = 1.4, (b) Voltage sensitivity ratio = 1

Question 4.8: In a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m/s normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg)

Answer:
Given: B = 6.5 G = 6.5 × 10⁻⁴ T, v = 4.8 × 10⁶ m/s, θ = 90°, e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg
The path is circular because the magnetic force (F = e * v * B * sinθ) acts perpendicular to the velocity, providing centripetal force.
Radius, r = (m_e * v) / (e * B)
r = (9.1 × 10⁻³¹ × 4.8 × 10⁶) / (1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴)
r = (9.1 × 4.8 × 10⁻²⁵) / (1.6 × 6.5 × 10⁻²³) ≈ 4.2 × 10⁻² m ≈ 4.2 cm
Path is circular, Radius ≈ 4.2 cm

Question 4.9: An electron beam is moving with a speed of 4.0 × 10⁶ m/s along the x-axis enters a region where a magnetic field of 10⁻³ T exists along the y-axis. Find the radius of the circular path of the electron beam.

Answer:
Given: v = 4.0 × 10⁶ m/s, B = 10⁻³ T, e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg, θ = 90° (v ⊥ B)
Radius, r = (m_e * v) / (e * B)
r = (9.1 × 10⁻³¹ × 4.0 × 10⁶) / (1.6 × 10⁻¹⁹ × 10⁻³)
r = (9.1 × 4.0 × 10⁻²⁵) / (1.6 × 10⁻²²) ≈ 2.275 × 10⁻³ m ≈ 2.28 mm
Radius ≈ 2.28 mm

Question 4.10: A charged particle moves through a magnetic field perpendicular to its direction, then discuss whether the momentum changes or the kinetic energy changes.

Answer:
The magnetic force (F = q * v * B * sinθ) acts perpendicular to the velocity, changing the direction of motion but not the speed (v). Since momentum (p = m * v) is a vector, it changes due to the change in direction. However, kinetic energy (KE = ½ * m * v²) depends on speed, which remains constant.
Momentum changes, Kinetic energy remains constant

Question 4.11: A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? (Mass of proton = 1.67 × 10⁻²⁷ kg, Charge of proton = 1.6 × 10⁻¹⁹ C)

Answer:
Given: Frequency (f) = 10 MHz = 10 × 10⁶ Hz, m_p = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C
Cyclotron frequency, f = (q * B) / (2π * m_p)
B = (2π * m_p * f) / q
B = (2 × 3.14 × 1.67 × 10⁻²⁷ × 10 × 10⁶) / (1.6 × 10⁻¹⁹)
B = (2 × 3.14 × 1.67 × 10⁻²⁰) / (1.6 × 10⁻¹⁹) ≈ 0.656 T
Magnetic field ≈ 0.66 T

Question 4.12: A current carrying wire in the shape of an equilateral triangle of side 10 cm is placed in a uniform magnetic field of 0.5 T. The current is 5 A. Find the torque on the loop if the magnetic field is (a) parallel to the plane of the loop, (b) perpendicular to the plane of the loop.

Answer:
Given: Side (a) = 10 cm = 0.1 m, B = 0.5 T, I = 5 A
Area of equilateral triangle, A = (√3 / 4) * a² = (√3 / 4) * (0.1)² ≈ 4.33 × 10⁻³ m²
Torque, τ = N * I * A * B * sinθ (N = 1 for single loop)
(a) B parallel to plane, θ = 90°, sinθ = 1
τ = 1 × 5 × 4.33 × 10⁻³ × 0.5 × 1 ≈ 0.0108 N·m
(b) B perpendicular to plane, θ = 0°, sinθ = 0
τ = 0 N·m
(a) Torque ≈ 0.0108 N·m, (b) Torque = 0 N·m

Chapter 5: Magnetism and Matter

NCERT Solutions for Class 12 Physics

Chapter 5: Magnetism and Matter

Question 5.1: A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?

Answer:
Given: Magnetic field (B) = 0.25 T, Angle (θ) = 30°, Torque (τ) = 4.5 × 10⁻² J
Torque, τ = m * B * sinθ, where m is the magnetic moment
m = τ / (B * sinθ)
m = (4.5 × 10⁻²) / (0.25 × sin30°)
m = (4.5 × 10⁻²) / (0.25 × 0.5) = 0.36 J/T
Magnetic moment = 0.36 J/T

Question 5.2: A short bar magnet of magnetic moment 5.25 × 10⁻² J/T is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on its axial line? Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distance involved.

Answer:
Given: Magnetic moment (m) = 5.25 × 10⁻² J/T, Earth’s field (Bₑ) = 0.42 G = 0.42 × 10⁻⁴ T, Angle of resultant field = 45°
On axial line, magnetic field due to magnet, B_m = (μ₀ * 2m) / (4π * r³)
μ₀/4π = 10⁻⁷ T·m/A
B_m = (10⁻⁷ × 2 × 5.25 × 10⁻²) / r³ = 1.05 × 10⁻⁸ / r³ T
Resultant field at 45° implies B_m = Bₑ (since tan45° = B_m / Bₑ = 1)
1.05 × 10⁻⁸ / r³ = 0.42 × 10⁻⁴
r³ = (1.05 × 10⁻⁸) / (0.42 × 10⁻⁴) ≈ 2.5 × 10⁻⁴
r = (2.5 × 10⁻⁴)¹/³ ≈ 0.063 m ≈ 6.3 cm
Distance ≈ 6.3 cm

Question 5.3: If the bar magnet in Q. 5.2 is turned around by 180°, where will the new null points be located?

Answer:
When the bar magnet is turned 180°, the magnetic moment direction reverses, and null points (where resultant field is zero) occur on the equatorial line instead of the axial line.
On equatorial line, B_m = (μ₀ * m) / (4π * r³)
B_m = (10⁻⁷ × 5.25 × 10⁻²) / r³ = 5.25 × 10⁻⁹ / r³ T
At null point, B_m = Bₑ = 0.42 × 10⁻⁴ T
5.25 × 10⁻⁹ / r³ = 0.42 × 10⁻⁴
r³ = (5.25 × 10⁻⁹) / (0.42 × 10⁻⁴) ≈ 1.25 × 10⁻⁴
r = (1.25 × 10⁻⁴)¹/³ ≈ 0.05 m ≈ 5.0 cm
New null points are at ≈ 5.0 cm on the equatorial line

Question 5.4: A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet?

Answer:
Given: Earth’s field (Bₑ) = 0.36 G = 0.36 × 10⁻⁴ T, Distance (r) = 14 cm = 0.14 m, Null points on axis
At null point on axial line, B_m = (μ₀ * 2m) / (4π * r³) = Bₑ
B_m = 0.36 × 10⁻⁴ T
On equatorial line (normal bisector), B_m' = (μ₀ * m) / (4π * r³) = B_m / 2
B_m' = (0.36 × 10⁻⁴) / 2 = 0.18 × 10⁻⁴ T
Total field on equatorial line, B_total = √(Bₑ² + B_m'²) (since Bₑ and B_m' are perpendicular, dip = 0°)
B_total = √((0.36 × 10⁻⁴)² + (0.18 × 10⁻⁴)²)
B_total = √(0.1296 × 10⁻⁸ + 0.0324 × 10⁻⁸) ≈ √(0.162 × 10⁻⁸) ≈ 0.402 × 10⁻⁴ T
Total magnetic field ≈ 0.40 × 10⁻⁴ T

Question 5.5: If the bar magnet in Q. 5.4 is turned around by 90° so that its axis is at right angles to the earth’s magnetic field, where would the null points be located?

Answer:
From Q. 5.4: Bₑ = 0.36 × 10⁻⁴ T, Null point at r = 0.14 m on axial line, so B_m = (μ₀ * 2m) / (4π * r³) = 0.36 × 10⁻⁴ T
When magnet is rotated 90°, null points are on the equatorial line.
B_m' = (μ₀ * m) / (4π * r'³) = Bₑ
Since B_m = 2 * B_m' (from axial to equatorial field relation), B_m' = (0.36 × 10⁻⁴) / 2 = 0.18 × 10⁻⁴ T
B_m' = (μ₀ * m) / (4π * r'³) = 0.18 × 10⁻⁴
Since (μ₀ * m) / (4π * (0.14)³) = 0.18 × 10⁻⁴, and (μ₀ * m) / (4π * (0.14)³) = (0.36 × 10⁻⁴) / 2
r'³ = (0.14)³ / 2¹/³ (since B_m' = B_m / 2)
r' = 0.14 / (2¹/³) ≈ 0.14 / 1.26 ≈ 0.111 m ≈ 11.1 cm
Null points are at ≈ 11.1 cm on the equatorial line

Question 5.6: A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction? (b) What is the torque on the magnet in this position?

Answer:
Given: Magnetic moment (m) = 1.5 J/T, Magnetic field (B) = 0.22 T
(a) Work done, W = m * B * (cosθ₁ - cosθ₂), θ₁ = 0°, θ₂ = 90°
W = 1.5 × 0.22 × (cos0° - cos90°) = 1.5 × 0.22 × (1 - 0) = 0.33 J
(b) Torque, τ = m * B * sinθ, θ = 90°
τ = 1.5 × 0.22 × sin90° = 1.5 × 0.22 × 1 = 0.33 N·m
(a) Work done = 0.33 J, (b) Torque = 0.33 N·m

Question 5.7: A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Answer:
Given: Length (L) = 80 cm = 0.8 m, Turns per layer = 400, Layers = 5, Total turns (N) = 400 × 5 = 2000, Diameter = 1.8 cm, Radius (r) = 0.9 cm = 0.009 m, Current (I) = 8.0 A
Magnetic field inside solenoid, B = μ₀ * (N/L) * I
μ₀ = 4π × 10⁻⁷ T·m/A, N/L = 2000 / 0.8 = 2500 turns/m
B = (4π × 10⁻⁷ × 2500 × 8.0) = 2.513 × 10⁻² T ≈ 0.025 T
Magnetic field ≈ 0.025 T

Question 5.8: A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north pole pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer:
Given: Angle of dip (δ) = 22°, Horizontal component (Bₕ) = 0.35 G = 0.35 × 10⁻⁴ T
Earth’s magnetic field, Bₑ = Bₕ / cosδ
Bₑ = (0.35 × 10⁻⁴) / cos22°
cos22° ≈ 0.927
Bₑ ≈ (0.35 × 10⁻⁴) / 0.927 ≈ 0.377 × 10⁻⁴ T
Earth’s magnetic field ≈ 0.38 × 10⁻⁴ T

Question 5.9: At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Answer:
Given: Declination = 12° west, Angle of dip (δ) = 60°, Bₕ = 0.16 G = 0.16 × 10⁻⁴ T
Earth’s magnetic field, Bₑ = Bₕ / cosδ
Bₑ = (0.16 × 10⁻⁴) / cos60° = (0.16 × 10⁻⁴) / 0.5 = 0.32 × 10⁻⁴ T
Direction: The field is inclined at 60° to the horizontal in the magnetic meridian, with a declination of 12° west of geographic north.
Magnitude ≈ 0.32 × 10⁻⁴ T, Direction: 60° above horizontal, 12° west of geographic north

Question 5.10: A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points towards the north. Find the horizontal component of the earth’s magnetic field at the location.

Answer:
Given: N = 30, Radius (r) = 12 cm = 0.12 m, I = 0.35 A, Angle with magnetic meridian = 45°, Needle points north
Magnetic field due to coil, B_coil = (μ₀ * N * I) / (2 * r)
μ₀ = 4π × 10⁻⁷ T·m/A
B_coil = (4π × 10⁻⁷ × 30 × 0.35) / (2 × 0.12) ≈ 5.497 × 10⁻⁵ T
B_coil’s horizontal component = B_coil * cos45° = 5.497 × 10⁻⁵ × 0.707 ≈ 3.89 × 10⁻⁵ T
Since needle points north, B_coil’s horizontal component cancels earth’s horizontal field (Bₕ).
Bₕ = 3.89 × 10⁻⁵ T
Horizontal component of earth’s field ≈ 3.89 × 10⁻⁵ T

Question 5.11: A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10⁻² T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?

Answer:
Given: B₁ = 1.2 × 10⁻² T, Angle between fields = 60°, Equilibrium angle with B₁ = 15°
At stable equilibrium, torque due to B₁ equals torque due to B₂, and dipole aligns such that resultant field is along its direction.
Angle with B₂ = 60° - 15° = 45°
Using torque balance: m * B₁ * sin15° = m * B₂ * sin45°
B₂ = B₁ * (sin15° / sin45°)
sin15° ≈ 0.2588, sin45° ≈ 0.707
B₂ = (1.2 × 10⁻²) × (0.2588 / 0.707) ≈ 4.39 × 10⁻³ T
Magnitude of other field ≈ 4.4 × 10⁻³ T

Chapter 6: Electromagnetic Induction

NCERT Solutions for Class 12 Physics

Chapter 6: Electromagnetic Induction

Question 6.1: Predict the direction of induced current in the situations described by the following Figs. 6.18 (a) to (f) in the NCERT textbook.

Answer:
The direction of the induced current is determined by Lenz’s law, which states that the induced current opposes the change in magnetic flux.
- **(a)** North pole approaching coil: Flux increases, induced current produces magnetic field opposite to magnet’s (south pole towards magnet). By right-hand rule, current is anticlockwise (viewed from magnet side).
- **(b)** South pole moving away from coil: Flux decreases, induced current produces south pole to attract magnet. Current is clockwise.
- **(c)** Coil in uniform field, no change in flux: No induced current.
- **(d)** Loop moving into uniform field: Flux increases, induced current produces field out of page. Current is anticlockwise.
- **(e)** Loop moving out of uniform field: Flux decreases, induced current produces field into page. Current is clockwise.
- **(f)** Rotating coil in uniform field: As flux changes, current alternates direction based on rotation.
Directions: (a) Anticlockwise, (b) Clockwise, (c) No current, (d) Anticlockwise, (e) Clockwise, (f) Alternating

Question 6.2: Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19 (a) and (b) in the NCERT textbook.

Answer:
- **(a)** Current in wire PQ increases: Magnetic flux through loop increases (field into page). Induced current opposes this by producing field out of page. By right-hand rule, induced current in loop is anticlockwise.
- **(b)** Loop moving towards wire with constant current: Flux through loop increases (field into page). Induced current produces field out of page, so current is anticlockwise.
Directions: (a) Anticlockwise, (b) Anticlockwise

Question 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop?

Answer:
Given: Turns per cm = 15, so n = 15 × 100 = 1500 turns/m, Area (A) = 2.0 cm² = 2.0 × 10⁻⁴ m², ΔI = 4.0 - 2.0 = 2.0 A, Δt = 0.1 s
Magnetic field inside solenoid, B = μ₀ * n * I
Magnetic flux, Φ = B * A = μ₀ * n * I * A
Induced emf, |ε| = dΦ/dt = μ₀ * n * A * (ΔI/Δt)
μ₀ = 4π × 10⁻⁷ T·m/A
|ε| = (4π × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × 2.0) / 0.1
|ε| = (4 × 3.14 × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × 20) ≈ 7.54 × 10⁻⁵ V
Induced emf ≈ 7.54 × 10⁻⁵ V

Question 6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Answer:
Given: Loop dimensions = 8 cm × 2 cm, B = 0.3 T, v = 1 cm/s = 0.01 m/s
Induced emf, ε = B * l * v (where l is length perpendicular to velocity)
(a) Velocity normal to longer side (l = 8 cm = 0.08 m):
ε = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V
Time duration = Distance / Velocity = 2 cm / 1 cm/s = 2 s
(b) Velocity normal to shorter side (l = 2 cm = 0.02 m):
ε = 0.3 × 0.02 × 0.01 = 6.0 × 10⁻⁵ V
Time duration = 8 cm / 1 cm/s = 8 s
(a) emf = 2.4 × 10⁻⁴ V, duration = 2 s; (b) emf = 6.0 × 10⁻⁵ V, duration = 8 s

Question 6.5: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Answer:
Given: Length (L) = 1.0 m, Angular frequency (ω) = 400 rad/s, B = 0.5 T
Induced emf, ε = (1/2) * B * ω * L²
ε = (1/2) × 0.5 × 400 × (1.0)² = 0.5 × 200 × 1 = 100 V
Induced emf = 100 V

Question 6.6: A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad/s in a uniform horizontal magnetic field of magnitude 3.0 × 10⁻² T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil.

Answer:
Given: Radius (r) = 8.0 cm = 0.08 m, N = 20, ω = 50 rad/s, B = 3.0 × 10⁻² T, R = 10 Ω
Area, A = π * r² = 3.14 × (0.08)² ≈ 0.0201 m²
Maximum emf, ε_max = N * B * A * ω
ε_max = 20 × 3.0 × 10⁻² × 0.0201 × 50 ≈ 6.03 V
Average emf over one cycle = 0 (since it’s AC, averages to zero).
Maximum current, I_max = ε_max / R = 6.03 / 10 ≈ 0.603 A
Maximum emf = 6.03 V, Average emf = 0 V, Maximum current = 0.603 A

Question 6.7: A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10⁻⁴ T. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?

Answer:
Given: Length (L) = 10 m, v = 5.0 m/s, Bₕ = 0.30 × 10⁻⁴ T
(a) Induced emf, ε = Bₕ * L * v
ε = 0.30 × 10⁻⁴ × 10 × 5.0 = 1.5 × 10⁻² V = 15 mV
(b) Direction: Using Fleming’s right-hand rule, for wire falling downward (v), Bₕ from north to south, induced current is from west to east. Thus, emf is from west to east.
(c) Since current flows west to east, east end is at higher potential.
(a) emf = 15 mV, (b) Direction = West to East, (c) East end at higher potential

Question 6.8: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Answer:
Given: ΔI = 5.0 - 0.0 = 5.0 A, Δt = 0.1 s, ε_avg = 200 V
Self-inductance, L = ε_avg / (dI/dt)
dI/dt = ΔI / Δt = 5.0 / 0.1 = 50 A/s
L = 200 / 50 = 4 H
Self-inductance = 4 H

Question 6.9: A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the average emf induced in the other coil?

Answer:
Given: Mutual inductance (M) = 1.5 H, ΔI = 20 - 0 = 20 A, Δt = 0.5 s
Induced emf, ε = M * (dI/dt)
dI/dt = ΔI / Δt = 20 / 0.5 = 40 A/s
ε = 1.5 × 40 = 60 V
Average emf = 60 V

Question 6.10: A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10⁻⁴ T and the dip angle is 30°?

Answer:
Given: v = 1800 km/h = 1800 × (1000/3600) = 500 m/s, L = 25 m, B = 5 × 10⁻⁴ T, Dip angle (δ) = 30°
Vertical component of Earth’s field, B_v = B * sinδ = 5 × 10⁻⁴ × sin30° = 2.5 × 10⁻⁴ T
Induced emf, ε = B_v * L * v
ε = 2.5 × 10⁻⁴ × 25 × 500 = 3.125 V
Voltage difference = 3.125 V

Chapter 7: Alternating Current

NCERT Solutions for Class 12 Physics

Chapter 7: Alternating Current

Question 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

Answer:
Given: R = 100 Ω, V_rms = 220 V, f = 50 Hz
(a) RMS current, I_rms = V_rms / R
I_rms = 220 / 100 = 2.2 A
(b) Net power consumed, P = V_rms * I_rms = V_rms² / R
P = (220)² / 100 = 48400 / 100 = 484 W
(a) RMS current = 2.2 A, (b) Power = 484 W

Question 7.2: (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Answer:
(a) RMS voltage, V_rms = V_peak / √2
V_rms = 300 / √2 ≈ 300 / 1.414 ≈ 212 V
(b) Peak current, I_peak = I_rms * √2
I_peak = 10 × √2 ≈ 10 × 1.414 ≈ 14.14 A
(a) RMS voltage ≈ 212 V, (b) Peak current ≈ 14.14 A

Question 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:
Given: L = 44 mH = 0.044 H, V_rms = 220 V, f = 50 Hz
Inductive reactance, X_L = 2πfL
X_L = 2 × 3.14 × 50 × 0.044 ≈ 13.82 Ω
RMS current, I_rms = V_rms / X_L
I_rms = 220 / 13.82 ≈ 15.92 A
RMS current ≈ 15.9 A

Question 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:
Given: C = 60 μF = 60 × 10⁻⁶ F, V_rms = 110 V, f = 60 Hz
Capacitive reactance, X_C = 1 / (2πfC)
X_C = 1 / (2 × 3.14 × 60 × 60 × 10⁻⁶) ≈ 44.21 Ω
RMS current, I_rms = V_rms / X_C
I_rms = 110 / 44.21 ≈ 2.49 A
RMS current ≈ 2.5 A

Question 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Answer:
In both circuits (inductor in Q7.3 and capacitor in Q7.4), the net power absorbed over a complete cycle is zero.
Explanation: In purely inductive or capacitive circuits, the current and voltage are out of phase by 90°. The power is given by P = V_rms * I_rms * cosφ, where φ = 90° for pure inductors and capacitors, so cosφ = 0. Hence, the average power consumed over a complete cycle is zero.
Net power absorbed = 0 W in both cases

Question 7.6: Obtain the resonant frequency ω_r of a series LCR circuit with L = 2.0 H, C = 32 μF, and R = 10 Ω. What is the Q-value of this circuit?

Answer:
Given: L = 2.0 H, C = 32 μF = 32 × 10⁻⁶ F, R = 10 Ω
Resonant frequency, ω_r = 1 / √(L * C)
ω_r = 1 / √(2.0 × 32 × 10⁻⁶) = 1 / √(64 × 10⁻⁶) = 1 / (8 × 10⁻³) = 125 rad/s
Q-value = (1 / R) * √(L / C)
Q = (1 / 10) * √(2.0 / (32 × 10⁻⁶)) = 0.1 * √(6.25 × 10⁴) = 0.1 × 250 = 25
Resonant frequency = 125 rad/s, Q-value = 25

Question 7.7: A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Answer:
Given: C = 30 μF = 30 × 10⁻⁶ F, L = 27 mH = 0.027 H
Angular frequency of LC oscillations, ω = 1 / √(L * C)
ω = 1 / √(0.027 × 30 × 10⁻⁶) = 1 / √(810 × 10⁻⁹) ≈ 1 / (28.46 × 10⁻⁵) ≈ 3517 rad/s
Angular frequency ≈ 3517 rad/s

Question 7.8: A series LCR circuit with R = 20 Ω, L = 1.5 H, C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Answer:
Given: R = 20 Ω, L = 1.5 H, C = 35 μF = 35 × 10⁻⁶ F, V_rms = 200 V
At resonance, impedance Z = R (since X_L = X_C)
RMS current, I_rms = V_rms / R = 200 / 20 = 10 A
Average power, P = V_rms * I_rms * cosφ (cosφ = 1 at resonance)
P = 200 × 10 × 1 = 2000 W
Average power = 2000 W

Question 7.9: A radio can tune over the frequency range of a portion of MW broadcast band (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

Answer:
Given: L = 200 μH = 200 × 10⁻⁶ H, f_min = 800 kHz = 8 × 10⁵ Hz, f_max = 1200 kHz = 12 × 10⁵ Hz
Resonant frequency, f = 1 / (2π * √(L * C))
C = 1 / (4π² * f² * L)
For f_min = 8 × 10⁵ Hz:
C_max = 1 / (4 × (3.14)² × (8 × 10⁵)² × 200 × 10⁻⁶)
C_max = 1 / (4 × 9.86 × 64 × 10¹⁰ × 2 × 10⁻⁴) ≈ 1 / (5.05 × 10⁻³) ≈ 198 × 10⁻¹² F ≈ 198 pF
For f_max = 12 × 10⁵ Hz:
C_min = 1 / (4 × 9.86 × (12 × 10⁵)² × 200 × 10⁻⁶) ≈ 1 / (11.36 × 10⁻³) ≈ 88 × 10⁻¹² F ≈ 88 pF
Capacitance range = 88 pF to 198 pF

Question 7.10: Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Answer:
Given: L = 5.0 H, C = 80 μF = 80 × 10⁻⁶ F, R = 40 Ω, V_rms = 230 V
(a) Resonant frequency, f_r = 1 / (2π * √(L * C))
f_r = 1 / (2π * √(5.0 × 80 × 10⁻⁶)) = 1 / (2π * √(4 × 10⁻⁴)) ≈ 7.96 Hz
(b) At resonance, Z = R = 40 Ω
Amplitude of current, I_peak = V_peak / R, V_peak = V_rms * √2 = 230 × √2 ≈ 325.3 V
I_peak = 325.3 / 40 ≈ 8.13 A
(c) RMS potential drops:
- Across R: V_R = I_rms * R, I_rms = V_rms / R = 230 / 40 = 5.75 A
V_R = 5.75 × 40 = 230 V
- Across L: X_L = 2πf_rL = 2 × 3.14 × 7.96 × 5.0 ≈ 250.3 Ω
V_L = I_rms * X_L = 5.75 × 250.3 ≈ 1439 V
- Across C: X_C = 1 / (2πf_rC) = 1 / (2 × 3.14 × 7.96 × 80 × 10⁻⁶) ≈ 250.3 Ω
V_C = I_rms * X_C = 5.75 × 250.3 ≈ 1439 V
- Across LC combination: V_LC = V_L - V_C = 1439 - 1439 = 0 V (since X_L = X_C at resonance)
(a) Resonant frequency ≈ 7.96 Hz, (b) Impedance = 40 Ω, Amplitude of current ≈ 8.13 A, (c) V_R = 230 V, V_L = V_C ≈ 1439 V, V_LC = 0 V

Question 7.11: An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored (i) completely electrical (ii) completely magnetic? (d) At what times is the total energy shared equally between the inductor and the capacitor?

Answer:
Given: L = 20 mH = 0.02 H, C = 50 μF = 50 × 10⁻⁶ F, Q_0 = 10 mC = 10 × 10⁻³ C
(a) Total energy stored initially, E = (1/2) * (Q_0² / C)
E = (1/2) × (10 × 10⁻³)² / (50 × 10⁻⁶) = 0.5 × 0.01 / 50 × 10⁻⁶ = 1 J
Yes, energy is conserved in LC oscillations (no resistance).
(b) Natural frequency, f = 1 / (2π * √(L * C))
f = 1 / (2 × 3.14 * √(0.02 × 50 × 10⁻⁶)) ≈ 1 / (6.28 × 10⁻³) ≈ 159.2 Hz
(c) Angular frequency, ω = 2πf ≈ 1000 rad/s
(i) Completely electrical: When charge is maximum (Q = Q_0), at t = 0, T/2, T, 3T/2, …
T = 1 / f ≈ 0.00628 s, so t = 0, 0.00628 s, 0.01256 s, …
(ii) Completely magnetic: When current is maximum (charge = 0), at t = T/4, 3T/4, 5T/4, …
t = 0.00157 s, 0.00471 s, 0.00785 s, …
(d) Energy equally shared when Q = Q_0 / √2, occurs at t = T/8, 3T/8, 5T/8, …
t = 0.000785 s, 0.002355 s, 0.003925 s, …
(a) Total energy = 1 J, conserved; (b) Frequency ≈ 159.2 Hz; (c)(i) t = 0, 0.00628 s, …; (ii) t = 0.00157 s, 0.00471 s, …; (d) t = 0.000785 s, 0.002355 s, …

Question 7.12: A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?

Answer:
Given: L = 0.50 H, R = 100 Ω, V_rms = 240 V, f = 50 Hz
(a) Inductive reactance, X_L = 2πfL = 2 × 3.14 × 50 × 0.50 ≈ 157 Ω
Impedance, Z = √(R² + X_L²) = √(100² + 157²) ≈ √(10000 + 24649) ≈ 186.1 Ω
I_rms = V_rms / Z = 240 / 186.1 ≈ 1.29 A
Maximum current, I_max = I_rms * √2 ≈ 1.29 × 1.414 ≈ 1.82 A
(b) Phase angle, φ = tan⁻¹(X_L / R) = tan⁻¹(157 / 100) ≈ tan⁻¹(1.57) ≈ 57.5°
Time lag, Δt = φ / (2πf) = 57.5 × (π/180) / (2π × 50) ≈ 0.0032 s
(a) Maximum current ≈ 1.82 A, (b) Time lag ≈ 0.0032 s

Chapter 8: Electromagnetic Waves

NCERT Solutions for Class 12 Physics

Chapter 8: Electromagnetic Waves

Question 8.1: Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference across the capacitor. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Answer:
Given: Radius (r) = 12 cm = 0.12 m, Separation (d) = 5.0 mm = 0.005 m, Current (I) = 0.15 A
(a) Capacitance, C = ε₀ * A / d, where A = πr², ε₀ = 8.85 × 10⁻¹² F/m
A = 3.14 × (0.12)² ≈ 0.0452 m²
C = (8.85 × 10⁻¹² × 0.0452) / 0.005 ≈ 8.0 × 10⁻¹¹ F = 80 pF
Rate of change of potential difference, dV/dt = I / C
dV/dt = 0.15 / (8.0 × 10⁻¹¹) = 1.875 × 10⁹ V/s
(b) Displacement current, I_d = ε₀ * (dΦ_E/dt), where Φ_E is electric flux
For a capacitor, I_d = conduction current = 0.15 A
(c) Yes, Kirchhoff’s first rule is valid. The conduction current entering one plate equals the displacement current between the plates, ensuring charge conservation.
(a) Capacitance = 80 pF, dV/dt = 1.875 × 10⁹ V/s; (b) Displacement current = 0.15 A; (c) Valid, due to displacement current

Question 8.2: A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an angular frequency of 300 rad/s. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answer:
Given: R = 6.0 cm = 0.06 m, C = 100 pF = 100 × 10⁻¹² F, V_rms = 230 V, ω = 300 rad/s, r = 3.0 cm = 0.03 m
(a) Capacitive reactance, X_C = 1 / (ωC)
X_C = 1 / (300 × 100 × 10⁻¹²) = 1 / (3 × 10⁻⁸) ≈ 3.33 × 10⁷ Ω
RMS current, I_rms = V_rms / X_C = 230 / (3.33 × 10⁷) ≈ 6.9 × 10⁻⁶ A
(b) Yes, conduction current equals displacement current in a capacitor, as per Maxwell’s equations.
(c) Magnetic field, B = (μ₀ * I_d * r) / (2π * R²), where I_d = I_rms (displacement current)
μ₀ = 4π × 10⁻⁷ T·m/A
B = (4π × 10⁻⁷ × 6.9 × 10⁻⁶ × 0.03) / (2π × (0.06)²) ≈ 1.15 × 10⁻¹¹ T
(a) RMS current ≈ 6.9 μA, (b) Yes, equal, (c) B ≈ 1.15 × 10⁻¹¹ T

Question 8.3: What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å, and radiowaves of wavelength 500 m?

Answer:
All are electromagnetic waves, so they travel at the same speed in vacuum, which is the speed of light, c = 3 × 10⁸ m/s.
Physical quantity = Speed of light (3 × 10⁸ m/s)

Question 8.4: A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Answer:
For an electromagnetic wave in vacuum, the electric field (E) and magnetic field (B) vectors are perpendicular to each other and to the direction of propagation (z-direction). Thus, E and B lie in the xy-plane.
Given: Frequency, f = 30 MHz = 30 × 10⁶ Hz
Wavelength, λ = c / f, where c = 3 × 10⁸ m/s
λ = (3 × 10⁸) / (30 × 10⁶) = 10 m
E and B are in xy-plane, perpendicular to each other and to z-direction; Wavelength = 10 m

Question 8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Answer:
Given: Frequency range = 7.5 MHz to 12 MHz, c = 3 × 10⁸ m/s
Wavelength, λ = c / f
For f = 7.5 MHz = 7.5 × 10⁶ Hz:
λ_max = (3 × 10⁸) / (7.5 × 10⁶) = 40 m
For f = 12 MHz = 12 × 10⁶ Hz:
λ_min = (3 × 10⁸) / (12 × 10⁶) = 25 m
Wavelength band = 25 m to 40 m

Question 8.6: A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:
The frequency of electromagnetic waves produced by an oscillating charged particle is equal to the frequency of its oscillation.
Given: Frequency of oscillation = 10⁹ Hz
Frequency of electromagnetic waves = 10⁹ Hz

Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?

Answer:
Given: B₀ = 510 nT = 510 × 10⁻⁹ T, c = 3 × 10⁸ m/s
For an electromagnetic wave, E₀ = c * B₀
E₀ = (3 × 10⁸) × (510 × 10⁻⁹) = 153 V/m
Electric field amplitude = 153 V/m

Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine B₀, ω, k, and λ. (b) Find expressions for E and B.

Answer:
Given: E₀ = 120 N/C, ν = 50.0 MHz = 50 × 10⁶ Hz, c = 3 × 10⁸ m/s
(a)
- Magnetic field amplitude, B₀ = E₀ / c = 120 / (3 × 10⁸) = 4 × 10⁻⁷ T
- Angular frequency, ω = 2πν = 2 × 3.14 × 50 × 10⁶ ≈ 3.14 × 10⁸ rad/s
- Wavelength, λ = c / ν = (3 × 10⁸) / (50 × 10⁶) = 6 m
- Wave number, k = 2π / λ = 2 × 3.14 / 6 ≈ 1.047 rad/m
(b) Assuming wave propagates along z-direction, E along x-direction, B along y-direction:
- E = E₀ sin(kz - ωt) = 120 sin(1.047z - 3.14 × 10⁸ t) N/C
- B = B₀ sin(kz - ωt) = 4 × 10⁻⁷ sin(1.047z - 3.14 × 10⁸ t) T
(a) B₀ = 4 × 10⁻⁷ T, ω ≈ 3.14 × 10⁸ rad/s, k ≈ 1.047 rad/m, λ = 6 m; (b) E = 120 sin(1.047z - 3.14 × 10⁸ t) N/C, B = 4 × 10⁻⁷ sin(1.047z - 3.14 × 10⁸ t) T

Question 8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in electron volts for (a) an X-ray of wavelength 0.10 nm, (b) visible light of wavelength 500 nm, (c) microwaves of wavelength 1.0 cm, (d) a radio wave of wavelength 100 m.

Answer:
Given: h = 6.626 × 10⁻³⁴ J·s, 1 eV = 1.6 × 10⁻¹⁹ J, c = 3 × 10⁸ m/s
Energy, E = hν = hc / λ, E (in eV) = (hc / λ) / (1.6 × 10⁻¹⁹)
hc = 6.626 × 10⁻³⁴ × 3 × 10⁸ ≈ 1.9878 × 10⁻²⁵ J·m
(a) λ = 0.10 nm = 10⁻¹⁰ m
E = (1.9878 × 10⁻²⁵) / (10⁻¹⁰ × 1.6 × 10⁻¹⁹) ≈ 1.242 × 10⁴ eV = 12.42 keV
(b) λ = 500 nm = 5 × 10⁻⁷ m
E = (1.9878 × 10⁻²⁵) / (5 × 10⁻⁷ × 1.6 × 10⁻¹⁹) ≈ 2.485 eV
(c) λ = 1.0 cm = 10⁻² m
E = (1.9878 × 10⁻²⁵) / (10⁻² × 1.6 × 10⁻¹⁹) ≈ 1.242 × 10⁻⁴ eV
(d) λ = 100 m
E = (1.9878 × 10⁻²⁵) / (100 × 1.6 × 10⁻¹⁹) ≈ 1.242 × 10⁻⁸ eV
(a) 12.42 keV, (b) 2.485 eV, (c) 1.242 × 10⁻⁴ eV, (d) 1.242 × 10⁻⁸ eV

Question 8.10: About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.

Answer:
Given: Power of bulb = 100 W, Visible radiation = 5% = 5 W
Intensity, I = Power / (4πr²)
(a) At r = 1 m:
I = 5 / (4 × 3.14 × 1²) ≈ 5 / 12.56 ≈ 0.398 W/m²
(b) At r = 10 m:
I = 5 / (4 × 3.14 × 10²) ≈ 5 / 1256 ≈ 0.00398 W/m²
(a) Intensity ≈ 0.398 W/m², (b) Intensity ≈ 0.00398 W/m²

NCERT Solutions for Class 12 Physics

Chapter 9: Ray Optics and Optical Instruments

Question 9.1: A small candle, 2.5 cm in size is placed 27 cm from a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is now moved closer to the mirror, how would the screen have to be moved?

Answer:
Given: Object height, h = 2.5 cm, Object distance, u = -27 cm, Radius of curvature, R = -36 cm
Focal length, f = R/2 = -18 cm
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/(-18) - 1/(-27) = -1/18 + 1/27 = (-3 + 2)/54 = -1/54
v = -54 cm
The screen should be placed 54 cm in front of the mirror.
Magnification, m = -v/u = -(-54)/(-27) = -2
Image height, h' = m × h = -2 × 2.5 = -5 cm
The image is real, inverted, and 5 cm in size.
If the candle is moved closer to the mirror (u decreases in magnitude), |v| increases, so the screen should be moved further away from the mirror.
Image distance = 54 cm, Nature: Real, inverted, Size = 5 cm, Screen moves further away if candle moves closer

Question 9.2: A 3 cm long object is placed 20 cm away from a convex mirror of focal length 20 cm. Find the position and nature of the image.

Answer:
Given: Object height, h = 3 cm, Object distance, u = -20 cm, Focal length, f = +20 cm (convex mirror)
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/20 - 1/(-20) = 1/20 + 1/20 = 2/20 = 1/10
v = +10 cm
Magnification, m = -v/u = -(+10)/(-20) = 0.5
Image height, h' = m × h = 0.5 × 3 = 1.5 cm
The image is virtual, erect, diminished, 1.5 cm in size, and located 10 cm behind the mirror.
Image position = 10 cm behind mirror, Nature: Virtual, erect, Size = 1.5 cm

Question 9.3: A light bulb is placed between two plane mirrors inclined at an angle of 60°. Find the number of images formed.

Answer:
For two plane mirrors inclined at angle θ, number of images, n = (360°/θ) - 1
Given: θ = 60°
n = (360/60) - 1 = 6 - 1 = 5
Number of images = 5

Question 9.4: Define the principal focus of a concave mirror, and show that it passes through the focus.

Answer:
The principal focus of a concave mirror is the point on the principal axis where rays parallel to the principal axis converge after reflection.
By the property of concave mirrors, parallel rays reflect and meet at the focal point (F). Thus, the reflected rays pass through the focus by definition.
Principal focus: Point of convergence for parallel rays, passes through focus

Question 9.5: A thin converging lens of focal length 20 cm forms the image of an object of height 5 cm at a distance of 25 cm from the lens. Find the position, nature, and size of the image.

Answer:
Given: Focal length, f = +20 cm (converging lens), Image distance, v = +25 cm, Object height, h = 5 cm
Using lens formula: 1/v - 1/u = 1/f
1/u = 1/v - 1/f = 1/25 - 1/20 = (4 - 5)/100 = -1/100
u = -100 cm
Magnification, m = v/u = 25/(-100) = -0.25
Image height, h' = m × h = -0.25 × 5 = -1.25 cm
The image is real, inverted, diminished, 1.25 cm in size, and located 25 cm on the opposite side of the lens.
Image position = 25 cm, Nature: Real, inverted, Size = 1.25 cm

Question 9.6: A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. If the object is placed 25 cm away from the lens, find the position, size, and nature of the image.

Answer:
Given: Object height, h = 2 cm, Object distance, u = -25 cm, Focal length, f = +10 cm
Using lens formula: 1/v - 1/u = 1/f
1/v = 1/10 - 1/(-25) = 1/10 + 1/25 = (5 + 2)/50 = 7/50
v = 50/7 ≈ 16.67 cm
Magnification, m = v/u = (50/7)/(-25) = -2/3 ≈ -0.667
Image height, h' = m × h = -0.667 × 2 ≈ -1.33 cm
The image is real, inverted, diminished, approximately 1.33 cm tall, and located 16.67 cm from the lens.
Image position ≈ 16.67 cm, Nature: Real, inverted, Size ≈ 1.33 cm

Question 9.7: An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the nature and position of the image.

Answer:
Given: Object distance, u = -10 cm, Focal length, f = +15 cm (convex mirror)
Using mirror formula: 1/v + 1/u = 1/f
1/v = 1/15 - 1/(-10) = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6
v = +6 cm
The image is virtual, erect, and located 6 cm behind the mirror.
Image position = 6 cm behind mirror, Nature: Virtual, erect

Question 9.8: A beam of light consisting of red, green, and blue colours is incident on a right-angled prism as shown. The refractive indices of the glass for the colours are μ_r = 1.39, μ_g = 1.44, μ_b = 1.47. Find the emergence of the beam from each face.

Answer:
For a right-angled prism (45°-45°-90°), light incident perpendicular to one face enters without deviation, strikes the hypotenuse at 45°, and may undergo total internal reflection (TIR) depending on the critical angle.
Critical angle, θ_c = sin⁻¹(1/μ)
For red: μ_r = 1.39, θ_c = sin⁻¹(1/1.39) ≈ 46°
For green: μ_g = 1.44, θ_c ≈ 44°
For blue: μ_b = 1.47, θ_c ≈ 42.8°
Since the angle of incidence at the hypotenuse is 45°, red (θ_c > 45°) emerges, while green and blue (θ_c < 45°) undergo TIR and do not emerge from the hypotenuse.
Red emerges, green and blue undergo TIR

Question 9.9: A ray of light is incident on a plane mirror at 30° to the normal. Find the angle between the incident and reflected rays.

Answer:
Given: Angle of incidence, i = 30°
Angle of reflection, r = 30° (law of reflection)
Angle between incident and reflected rays = 180° - (i + r) = 180° - (30° + 30°) = 120°
Angle = 120°

Question 9.10: A ray of light passes through an equilateral prism such that the angle of incidence is 45° and the angle of emergence is 45°. If the angle of refraction at the first face is 30°, what is the angle of refraction at the second face?

Answer:
Given: Prism angle, A = 60° (equilateral prism), Angle of incidence, i₁ = 45°, Angle of emergence, i₂ = 45°, r₁ = 30°
For a prism, A = r₁ + r₂
r₂ = A - r₁ = 60° - 30° = 30°
Using Snell’s law at second face: μ sin(r₂) = sin(i₂)
Since i₂ = 45° and r₂ = 30°, the angle of refraction at the second face corresponds to the emergence angle, consistent with the given data.
Angle of refraction at second face = 30°

Question 9.11: A thin convex lens of refractive index 1.5 has both surfaces of equal radii of 20 cm. Find its focal length in air.

Answer:
Given: Refractive index, μ = 1.5, R₁ = +20 cm, R₂ = -20 cm
Lens maker’s formula: 1/f = (μ - 1)(1/R₁ - 1/R₂)
1/f = (1.5 - 1)(1/20 - (-1/20)) = 0.5 × (1/20 + 1/20) = 0.5 × 2/20 = 0.05
f = 1/0.05 = 20 cm
Focal length = 20 cm

Question 9.12: A beam of light converges to a point A. A converging lens of focal length 20 cm is placed 15 cm from A. Find the position of the final image.

Answer:
Given: Focal length, f = 20 cm, Object distance, u = +15 cm (virtual object, as light converges to A)
Using lens formula: 1/v - 1/u = 1/f
1/v = 1/20 - 1/15 = (3 - 4)/60 = -1/60
v = -60 cm
The final image is 60 cm to the left of the lens (virtual image).
Image position = 60 cm to the left of the lens

Question 9.13: A parallel beam of light of wavelength 600 nm enters a thin slab of thickness 0.2 mm and refractive index 1.5. Find the optical path length.

Answer:
Given: Thickness, t = 0.2 mm = 0.0002 m, Refractive index, μ = 1.5
Optical path length = μ × t = 1.5 × 0.0002 = 0.0003 m = 0.3 mm
Optical path length = 0.3 mm

Question 9.14: A ray falls on a prism ABC (AB = BC) and travels as shown in the figure. What is the least refractive index of the prism material?

Answer:
For a right-angled isosceles prism (AB = BC, 45°-45°-90°), minimum refractive index for total internal reflection at the hypotenuse occurs when the angle of incidence is 45° and equals the critical angle.
Critical angle, θ_c = sin⁻¹(1/μ)
For θ_c = 45°, sin(45°) = 1/√2 = 1/μ
μ = √2 ≈ 1.414
Minimum refractive index ≈ 1.414

Question 9.15: The near point of a person is 50 cm from the eye. What should be the power of the lens required to see clearly at 25 cm?

Answer:
Given: Near point, u = -50 cm, Desired distance, v = -25 cm (virtual image for reading)
Using lens formula: 1/f = 1/v - 1/u
1/f = -1/25 - (-1/50) = -1/25 + 1/50 = (-2 + 1)/50 = -1/50
f = -50 cm = -0.5 m
Power, P = 1/f = 1/(-0.5) = -2 D
Power = -2 D (concave lens)

Question 9.16: A myopic person has a far point of 2 m. Find the nature and power of the lens required to see distant objects clearly.

Answer:
Given: Far point = 2 m = 200 cm, Object at infinity (u = -∞)
For distant objects, image at far point, v = -200 cm
Using lens formula: 1/f = 1/v - 1/u = -1/200 - 0 = -1/200
f = -200 cm = -2 m
Power, P = 1/f = 1/(-2) = -0.5 D
Nature: Diverging (concave) lens, Power = -0.5 D

Question 9.17: A student uses a lens of focal length 40 cm to see an object at 10 cm. Find the position and magnification.

Answer:
Given: Focal length, f = +40 cm (convex lens), Object distance, u = -10 cm
Using lens formula: 1/v - 1/u = 1/f
1/v = 1/40 - 1/(-10) = 1/40 + 1/10 = (1 + 4)/40 = 5/40 = 1/8
v = -40/5 = -8 cm
Magnification, m = v/u = (-8)/(-10) = 0.8
The image is virtual, erect, and located 8 cm from the lens on the same side.
Image position = 8 cm (same side), Magnification = 0.8

Question 9.18: A compound microscope has objective focal length 0.5 cm, eyepiece 2 cm, tube length 15 cm. Find the magnification.

Answer:
Given: f_o = 0.5 cm, f_e = 2 cm, Tube length, L = 15 cm, Least distance of distinct vision, D = 25 cm
Magnification of compound microscope, m = (v_o / u_o) × (1 + D/f_e)
Approximate v_o ≈ L = 15 cm (since u_o is small)
Using lens formula for objective: 1/v_o - 1/u_o = 1/f_o
Assume v_o ≈ 15 cm, 1/u_o = 1/v_o - 1/f_o = 1/15 - 1/0.5 ≈ -1.933/1
u_o ≈ -0.517 cm
m_o = v_o / u_o = 15 / (-0.517) ≈ -29
m_e = 1 + D/f_e = 1 + 25/2 = 13.5
Total magnification, m = m_o × m_e = -29 × 13.5 ≈ -391.5
Magnification ≈ 392 (magnitude)

Question 9.19: The magnifying power of a simple microscope is 8, with least distance 25 cm. Find the focal length of the lens.

Answer:
Given: Magnification, m = 8, D = 25 cm
For a simple microscope, m = 1 + D/f
8 = 1 + 25/f
7 = 25/f
f = 25/7 ≈ 3.57 cm
Focal length ≈ 3.57 cm

Question 9.20: The angular magnification of a telescope is 20, objective f_o = 100 cm. Find f_e.

Answer:
Given: Angular magnification, m = 20, f_o = 100 cm
For an astronomical telescope in normal adjustment, m = f_o / f_e
20 = 100 / f_e
f_e = 100 / 20 = 5 cm
Focal length of eyepiece = 5 cm

Question 9.21: A convex lens produces a real image three times the size of the object. If the object distance is 20 cm, find f.

Answer:
Given: Magnification, m = -3 (real image, inverted), Object distance, u = -20 cm
m = v/u
-3 = v/(-20)
v = 60 cm
Using lens formula: 1/v - 1/u = 1/f
1/f = 1/60 - 1/(-20) = 1/60 + 1/20 = (1 + 3)/60 = 4/60 = 1/15
f = 15 cm
Focal length = 15 cm

Question 9.22: Find the position of the image when an object is placed 30 cm from a convex lens of f=15 cm.

Answer:
Given: Object distance, u = -30 cm, Focal length, f = 15 cm
Using lens formula: 1/v - 1/u = 1/f
1/v = 1/15 - 1/(-30) = 1/15 + 1/30 = (2 + 1)/30 = 3/30 = 1/10
v = 30 cm
The image is located 30 cm on the opposite side of the lens.
Image position = 30 cm

Chapter 10: Wave Optics

NCERT Solutions for Class 12 Physics

Chapter 10: Wave Optics

Question 10.1: The refractive index of a medium is 1.5. If the speed of light in air is 3 × 10⁸ m/s, what is the speed of light in the medium?

Answer:
Given: Refractive index, μ = 1.5, Speed of light in air, c = 3 × 10⁸ m/s
Refractive index, μ = c / v
v = c / μ = (3 × 10⁸) / 1.5 = 2 × 10⁸ m/s
Speed of light in the medium = 2 × 10⁸ m/s

Question 10.2: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency, and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

Answer:
Given: Wavelength in air, λ_air = 589 nm = 5.89 × 10⁻⁷ m, Speed of light in air, c = 3 × 10⁸ m/s, Refractive index of water, μ = 1.33
Frequency, f = c / λ_air = (3 × 10⁸) / (5.89 × 10⁻⁷) ≈ 5.093 × 10¹⁴ Hz
(a) Reflected light:
- Wavelength = 589 nm (same as incident)
- Frequency = 5.093 × 10¹⁴ Hz (unchanged)
- Speed = 3 × 10⁸ m/s (in air)
(b) Refracted light:
- Speed in water, v = c / μ = (3 × 10⁸) / 1.33 ≈ 2.256 × 10⁸ m/s
- Wavelength in water, λ_water = λ_air / μ = 589 / 1.33 ≈ 442.9 nm
- Frequency = 5.093 × 10¹⁴ Hz (unchanged)
(a) Reflected: λ = 589 nm, f = 5.093 × 10¹⁴ Hz, v = 3 × 10⁸ m/s; (b) Refracted: λ ≈ 442.9 nm, f = 5.093 × 10¹⁴ Hz, v ≈ 2.256 × 10⁸ m/s

Question 10.3: What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source, (b) Light emerging out of a convex lens when a point source is placed at its focus, (c) The portion of the wavefront of light from a distant star intercepted by the Earth?

Answer:
(a) Light diverging from a point source: Spherical wavefront (spreading outward in all directions).
(b) Light emerging from a convex lens with source at focus: Plane wavefront (rays become parallel after lens).
(c) Light from a distant star intercepted by Earth: Plane wavefront (due to large distance, spherical wavefront appears plane).
(a) Spherical, (b) Plane, (c) Plane

Question 10.4: In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is λ/3?

Answer:
Intensity in Young’s experiment, I = 4I₀ cos²(φ/2), where φ is phase difference.
For path difference λ, phase difference φ = 2π, I = K = 4I₀ cos²(π) = 4I₀ (since cos π = -1, cos² π = 1).
For path difference λ/3, φ = (2π/λ) × (λ/3) = 2π/3.
I’ = 4I₀ cos²(π/3) = 4I₀ (1/2)² = 4I₀ × 1/4 = I₀.
Since K = 4I₀, I’ = K/4.
Intensity = K/4 units

Question 10.5: In a double-slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

Answer:
Given: Angular width in air, θ = 0.2° = 0.2 × π/180 radians, λ = 600 nm = 6 × 10⁻⁷ m, D = 1 m, μ_water = 4/3
Angular width, θ = λ / d (since β = λD / d, and θ = β / D).
In water, λ_water = λ_air / μ = (6 × 10⁻⁷) / (4/3) = 4.5 × 10⁻⁷ m.
θ_water = λ_water / d = (λ_air / μ) / d = θ / μ = 0.2 / (4/3) = 0.15°.
Angular width in water = 0.15°

Question 10.6: What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations: (a) the screen is moved away from the plane of the slits, (b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength, (c) the separation between the two slits is increased, (d) the source slit is moved closer to the double-slit plane, (e) a thin transparent film is placed in front of one of the slits?

Answer:
(a) Screen moved away: Fringe width, β = λD / d, increases as D (distance to screen) increases.
(b) Shorter wavelength: β = λD / d decreases as λ decreases.
(c) Slit separation increased: β = λD / d decreases as d increases.
(d) Source slit moved closer: No effect on fringe width (β is independent of source-slit distance).
(e) Thin film in front of one slit: Introduces path difference, causing the entire fringe pattern to shift.
(a) Wider fringes, (b) Narrower fringes, (c) Narrower fringes, (d) No change, (e) Fringe pattern shifts

Question 10.7: In Young’s double-slit experiment, the slits are separated by 0.28 mm, and the screen is placed 1.4 m away. The distance between the fourth bright fringe and the central bright fringe is 1.2 cm. What is the wavelength of light used?

Answer:
Given: Slit separation, d = 0.28 mm = 2.8 × 10⁻⁴ m, Screen distance, D = 1.4 m, Distance to 4th bright fringe, y₄ = 1.2 cm = 0.012 m
For nth bright fringe, y_n = nλD / d
For n = 4: 0.012 = (4 × λ × 1.4) / (2.8 × 10⁻⁴)
λ = (0.012 × 2.8 × 10⁻⁴) / (4 × 1.4) = 3.36 × 10⁻⁶ / 5.6 = 6 × 10⁻⁷ m = 600 nm
Wavelength = 600 nm

Question 10.8: In a Young’s double-slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away. The wavelength of light used is 600 nm. Find the distance of the third dark fringe from the central maximum.

Answer:
Given: d = 0.3 mm = 3 × 10⁻⁴ m, D = 1.5 m, λ = 600 nm = 6 × 10⁻⁷ m
For nth dark fringe, y_n = (2n - 1)λD / (2d)
For 3rd dark fringe (n = 3): y₃ = (2 × 3 - 1) × (6 × 10⁻⁷ × 1.5) / (2 × 3 × 10⁻⁴)
y₃ = 5 × (9 × 10⁻⁷) / (6 × 10⁻⁴) = 7.5 × 10⁻³ m = 7.5 mm
Distance = 7.5 mm

Question 10.9: In Young’s experiment, the width of the fringes obtained with light of wavelength 6000 Å is 2 mm. What will be the fringe width if the entire apparatus is immersed in a medium of refractive index 1.33?

Answer:
Given: λ_air = 6000 Å = 6 × 10⁻⁷ m, β_air = 2 mm = 2 × 10⁻³ m, μ = 1.33
Fringe width, β = λD / d
In medium, λ_medium = λ_air / μ = (6 × 10⁻⁷) / 1.33 ≈ 4.511 × 10⁻⁷ m
β_medium = β_air / μ = (2 × 10⁻³) / 1.33 ≈ 1.504 × 10⁻³ m = 1.5 mm
Fringe width ≈ 1.5 mm

Question 10.10: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide?

Answer:
Assume: d = 1 mm = 10⁻³ m, D = 1 m (standard values, as not specified)
(a) For λ₁ = 650 nm = 6.5 × 10⁻⁷ m, n = 3:
y_n = nλD / d = (3 × 6.5 × 10⁻⁷ × 1) / 10⁻³ = 1.95 × 10⁻³ m = 1.95 mm
(b) For coincidence, n₁λ₁ = n₂λ₂
n₁ × 650 = n₂ × 520
n₁ / n₂ = 520 / 650 = 4 / 5
Smallest integers: n₁ = 4, n₂ = 5
y = (4 × 6.5 × 10⁻⁷ × 1) / 10⁻³ = 2.6 × 10⁻³ m = 2.6 mm
(a) Distance = 1.95 mm, (b) Least distance = 2.6 mm

Question 10.11: The refractive index of glass is 1.5. If the speed of light in vacuum is 3 × 10⁸ m/s, find the speed, frequency, and wavelength of light of wavelength 600 nm in vacuum when it passes through glass.

Answer:
Given: μ = 1.5, c = 3 × 10⁸ m/s, λ_vacuum = 600 nm = 6 × 10⁻⁷ m
Speed in glass, v = c / μ = (3 × 10⁸) / 1.5 = 2 × 10⁸ m/s
Frequency, f = c / λ_vacuum = (3 × 10⁸) / (6 × 10⁻⁷) = 5 × 10¹⁴ Hz (unchanged)
Wavelength in glass, λ_glass = λ_vacuum / μ = 600 / 1.5 = 400 nm
Speed = 2 × 10⁸ m/s, Frequency = 5 × 10¹⁴ Hz, Wavelength = 400 nm

Question 10.12: In a Young’s double-slit experiment, the distance between the slits and the screen is 1.2 m, and the distance between the slits is 0.15 mm. If the fringe width is 4 mm, find the wavelength of light used.

Answer:
Given: D = 1.2 m, d = 0.15 mm = 1.5 × 10⁻⁴ m, β = 4 mm = 4 × 10⁻³ m
Fringe width, β = λD / d
λ = βd / D = (4 × 10⁻³ × 1.5 × 10⁻⁴) / 1.2 = 6 × 10⁻⁷ / 1.2 = 5 × 10⁻⁷ m = 500 nm
Wavelength = 500 nm

Question 10.13: Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer:
Given: λ = 5000 Å = 5 × 10⁻⁷ m
Wavelength of reflected light = 5000 Å (unchanged)
Frequency, f = c / λ = (3 × 10⁸) / (5 × 10⁻⁷) = 6 × 10¹⁴ Hz
For reflected ray normal to incident ray, angle between them = 90°.
Angle of incidence, i = angle of reflection, r.
Angle between rays = 180° - 2i = 90°
2i = 90°, i = 45°
Wavelength = 5000 Å, Frequency = 6 × 10¹⁴ Hz, Angle of incidence = 45°

Question 10.14: Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.

Answer:
Given: Aperture width, a = 4 mm = 4 × 10⁻³ m, λ = 400 nm = 4 × 10⁻⁷ m
Fresnel distance, Z_F = a² / λ
Z_F = (4 × 10⁻³)² / (4 × 10⁻⁷) = 16 × 10⁻⁶ / 4 × 10⁻⁷ = 40 m
Distance = 40 m

Question 10.15: The 6563 Å H-alpha line emitted by hydrogen in a star is found to be redshifted by 15 Å. Estimate the speed with which the star is receding from the Earth.

Answer:
Given: λ = 6563 Å, Δλ = 15 Å, c = 3 × 10⁸ m/s
Doppler shift, v/c = Δλ / λ
v = (Δλ / λ) × c = (15 / 6563) × 3 × 10⁸ ≈ 6.86 × 10⁵ m/s
Speed of recession ≈ 6.86 × 10⁵ m/s

Question 10.16: Explain how Newton’s corpuscular theory predicts the speed of light in a medium, say water, to be greater than the speed of light in vacuum. Is this prediction consistent with experimental results?

Answer:
Newton’s corpuscular theory assumes light consists of particles. In a denser medium like water, it predicts that particles are attracted toward the medium, increasing their speed (v_medium > c).
Experimental results show the speed of light in water is less than in vacuum (μ = 1.33, v = c / μ ≈ 2.26 × 10⁸ m/s < c).
The prediction is not consistent with experimental results; wave theory correctly explains the decrease in speed.
Prediction: v_medium > c, Not consistent with experiments

Question 10.17: In Young’s double-slit experiment, using sodium light (λ = 589 nm), 60 fringes are seen. If another source of wavelength λ’ is used, 48 fringes are observed in the same part of the screen. Find λ’.

Answer:
Given: λ = 589 nm, N₁ = 60, N₂ = 48
Number of fringes, N = y / β, where β = λD / d
N₁ / N₂ = β₂ / β₁ = λ’ / λ
60 / 48 = λ’ / 589
λ’ = (60 / 48) × 589 = 1.25 × 589 ≈ 736.25 nm
Wavelength λ’ ≈ 736 nm

Chapter 11: Dual Nature of Radiation and Matter

NCERT Solutions for Class 12 Physics

Chapter 11: Dual Nature of Radiation and Matter

Question 11.1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Answer:
Given: Voltage, V = 30 kV = 30 × 10³ V, h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s
(a) Maximum frequency, ν_max = eV / h, where e = 1.6 × 10⁻¹⁹ C
ν_max = (1.6 × 10⁻¹⁹ × 30 × 10³) / (6.626 × 10⁻³⁴) ≈ 7.24 × 10¹⁸ Hz
(b) Minimum wavelength, λ_min = c / ν_max
λ_min = (3 × 10⁸) / (7.24 × 10¹⁸) ≈ 4.14 × 10⁻¹¹ m = 0.0414 nm
(a) Maximum frequency ≈ 7.24 × 10¹⁸ Hz, (b) Minimum wavelength ≈ 0.0414 nm

Question 11.2: The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, calculate (a) the maximum kinetic energy of the emitted electrons, and (b) the stopping potential.

Answer:
Given: Work function, φ = 2.14 eV, Frequency, ν = 6 × 10¹⁴ Hz, h = 4.1357 × 10⁻¹⁵ eV·s
(a) Maximum kinetic energy, K_max = hν - φ
hν = 4.1357 × 10⁻¹⁵ × 6 × 10¹⁴ ≈ 2.481 eV
K_max = 2.481 - 2.14 = 0.341 eV
(b) Stopping potential, V₀ = K_max / e = 0.341 eV / e = 0.341 V
(a) Maximum kinetic energy ≈ 0.341 eV, (b) Stopping potential ≈ 0.341 V

Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Answer:
Given: Cut-off voltage, V₀ = 1.5 V
Maximum kinetic energy, K_max = eV₀ = 1.5 eV
Maximum kinetic energy = 1.5 eV

Question 11.4: Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. Find (a) the energy of each photon in the light, (b) the momentum of each photon.

Answer:
Given: Wavelength, λ = 632.8 nm = 6.328 × 10⁻⁷ m, h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s
(a) Energy of photon, E = hν = hc / λ
E = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (6.328 × 10⁻⁷) ≈ 3.14 × 10⁻¹⁹ J
In eV: E = 3.14 × 10⁻¹⁹ / (1.6 × 10⁻¹⁹) ≈ 1.96 eV
(b) Momentum, p = h / λ = 6.626 × 10⁻³⁴ / (6.328 × 10⁻⁷) ≈ 1.047 × 10⁻²⁷ kg·m/s
(a) Energy ≈ 1.96 eV, (b) Momentum ≈ 1.047 × 10⁻²⁷ kg·m/s

Question 11.5: The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) are incident on the Earth per second per square meter if the average wavelength of sunlight is 550 nm?

Answer:
Given: Energy flux = 1.388 × 10³ W/m², λ = 550 nm = 5.5 × 10⁻⁷ m, h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s
Energy of one photon, E = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (5.5 × 10⁻⁷) ≈ 3.613 × 10⁻¹⁹ J
Number of photons per second per m² = Energy flux / E = (1.388 × 10³) / (3.613 × 10⁻¹⁹) ≈ 3.84 × 10²¹
Number of photons ≈ 3.84 × 10²¹ photons/s/m²

Question 11.6: In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10⁻¹⁵ V s. Calculate the value of Planck’s constant.

Answer:
Given: Slope = V₀ / ν = 4.12 × 10⁻¹⁵ V·s
From photoelectric equation: eV₀ = hν - φ, so V₀ = (h/e)ν - φ/e
Slope = h/e = 4.12 × 10⁻¹⁵ V·s
h = (4.12 × 10⁻¹⁵) × (1.6 × 10⁻¹⁹) ≈ 6.592 × 10⁻³⁴ J·s
Planck’s constant ≈ 6.592 × 10⁻³⁴ J·s

Question 11.7: The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, find the cut-off voltage for the photoelectric emission.

Answer:
Given: Threshold frequency, ν₀ = 3.3 × 10¹⁴ Hz, Incident frequency, ν = 8.2 × 10¹⁴ Hz, h = 4.1357 × 10⁻¹⁵ eV·s
Maximum kinetic energy, K_max = h(ν - ν₀) = 4.1357 × 10⁻¹⁵ × (8.2 × 10¹⁴ - 3.3 × 10¹⁴) ≈ 2.027 eV
Cut-off voltage, V₀ = K_max / e = 2.027 V
Cut-off voltage ≈ 2.03 V

Question 11.8: The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Answer:
Given: Work function, φ = 4.2 eV, λ = 330 nm = 3.3 × 10⁻⁷ m, h = 4.1357 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m/s
Energy of photon, E = hc / λ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / (3.3 × 10⁻⁷) ≈ 3.76 eV
Since E < φ (3.76 eV < 4.2 eV), photoelectric emission will not occur.
No photoelectric emission

Question 11.9: Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected. Calculate the work function of the metal.

Answer:
Given: ν = 7.21 × 10¹⁴ Hz, v_max = 6.0 × 10⁵ m/s, h = 4.1357 × 10⁻¹⁵ eV·s, m_e = 9.1 × 10⁻³¹ kg
Maximum kinetic energy, K_max = (1/2)m_e v_max²
K_max = (1/2) × 9.1 × 10⁻³¹ × (6.0 × 10⁵)² ≈ 1.638 × 10⁻¹⁹ J ≈ 1.024 eV
Photon energy, E = hν = 4.1357 × 10⁻¹⁵ × 7.21 × 10¹⁴ ≈ 2.982 eV
Work function, φ = E - K_max = 2.982 - 1.024 ≈ 1.958 eV
Work function ≈ 1.96 eV

Question 11.10: Light of wavelength 488 nm is incident on a photosensitive surface. The stopping potential is measured to be 0.75 V. Find (a) the work function, (b) the maximum kinetic energy, and (c) the threshold wavelength.

Answer:
Given: λ = 488 nm = 4.88 × 10⁻⁷ m, V₀ = 0.75 V, h = 4.1357 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m/s
(a) Photon energy, E = hc / λ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / (4.88 × 10⁻⁷) ≈ 2.54 eV
Work function, φ = E - eV₀ = 2.54 - 0.75 ≈ 1.79 eV
(b) Maximum kinetic energy, K_max = eV₀ = 0.75 eV
(c) Threshold wavelength, λ₀ = hc / φ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / (1.79 × 1.6 × 10⁻¹⁹) ≈ 6.93 × 10⁻⁷ m = 693 nm
(a) Work function ≈ 1.79 eV, (b) K_max = 0.75 eV, (c) Threshold wavelength ≈ 693 nm

Question 11.11: In an experiment on the photoelectric effect, the frequency of the incident light is plotted against the stopping potential. The work function of the surface is 2.0 eV. Find the slope of the graph.

Answer:
Given: Work function, φ = 2.0 eV
Photoelectric equation: eV₀ = hν - φ
V₀ = (h/e)ν - φ/e
Slope of V₀ vs ν graph = h/e = 4.1357 × 10⁻¹⁵ V·s
Slope ≈ 4.14 × 10⁻¹⁵ V·s

Question 11.12: Calculate the de Broglie wavelength of (a) a 1 keV electron, (b) a 1 keV photon, (c) a 1 keV neutron.

Answer:
Given: Energy = 1 keV = 1000 eV = 1.6 × 10⁻¹⁶ J, m_e = 9.1 × 10⁻³¹ kg, m_n = 1.675 × 10⁻²⁷ kg, h = 6.626 × 10⁻³⁴ J·s
(a) Electron: λ = h / √(2m_e E)
λ = 6.626 × 10⁻³⁴ / √(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁶) ≈ 3.87 × 10⁻¹¹ m ≈ 0.0387 nm
(b) Photon: λ = hc / E = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (1.6 × 10⁻¹⁶) ≈ 1.24 × 10⁻⁹ m = 1.24 nm
(c) Neutron: λ = h / √(2m_n E)
λ = 6.626 × 10⁻³⁴ / √(2 × 1.675 × 10⁻²⁷ × 1.6 × 10⁻¹⁶) ≈ 2.86 × 10⁻¹² m ≈ 0.0286 pm
(a) Electron ≈ 0.0387 nm, (b) Photon ≈ 1.24 nm, (c) Neutron ≈ 0.0286 pm

Question 11.13: The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.

Answer:
Given: λ = 589 nm = 5.89 × 10⁻⁷ m, m_e = 9.1 × 10⁻³¹ kg, m_n = 1.675 × 10⁻²⁷ kg, h = 6.626 × 10⁻³⁴ J·s
de Broglie wavelength, λ = h / √(2mK)
K = h² / (2mλ²)
(a) Electron: K_e = (6.626 × 10⁻³⁴)² / (2 × 9.1 × 10⁻³¹ × (5.89 × 10⁻⁷)²) ≈ 6.94 × 10⁻²² J ≈ 4.34 × 10⁻³ eV
(b) Neutron: K_n = (6.626 × 10⁻³⁴)² / (2 × 1.675 × 10⁻²⁷ × (5.89 × 10⁻⁷)²) ≈ 3.77 × 10⁻²⁵ J ≈ 2.36 × 10⁻⁶ eV
(a) Electron ≈ 4.34 × 10⁻³ eV, (b) Neutron ≈ 2.36 × 10⁻⁶ eV

Question 11.14: What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at 1.0 km/s?

Answer:
Given: Mass, m = 0.040 kg, Velocity, v = 1.0 km/s = 1000 m/s, h = 6.626 × 10⁻³⁴ J·s
de Broglie wavelength, λ = h / (mv)
λ = 6.626 × 10⁻³⁴ / (0.040 × 1000) ≈ 1.657 × 10⁻³⁵ m
de Broglie wavelength ≈ 1.66 × 10⁻³⁵ m

Question 11.15: An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of the electron.

Answer:
Given: λ = 1.00 nm = 1 × 10⁻⁹ m, h = 6.626 × 10⁻³⁴ J·s, m_e = 9.1 × 10⁻³¹ kg, c = 3 × 10⁸ m/s
(a) Momentum, p = h / λ = 6.626 × 10⁻³⁴ / 10⁻⁹ = 6.626 × 10⁻²⁵ kg·m/s (same for both)
(b) Photon energy, E = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / 10⁻⁹ ≈ 1.9878 × 10⁻¹⁶ J ≈ 1241 eV
(c) Electron kinetic energy, K = p² / (2m_e) = (6.626 × 10⁻²⁵)² / (2 × 9.1 × 10⁻³¹) ≈ 2.41 × 10⁻¹⁹ J ≈ 1.506 eV
(a) Momentum = 6.626 × 10⁻²⁵ kg·m/s (both), (b) Photon energy ≈ 1241 eV, (c) Electron kinetic energy ≈ 1.506 eV

Question 11.16: (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10⁻¹⁰ m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2)kT at 300 K.

Answer:
Given: m_n = 1.675 × 10⁻²⁷ kg, h = 6.626 × 10⁻³⁴ J·s, k = 1.38 × 10⁻²³ J/K, T = 300 K
(a) λ = 1.40 × 10⁻¹⁰ m
K = h² / (2m_n λ²) = (6.626 × 10⁻³⁴)² / (2 × 1.675 × 10⁻²⁷ × (1.40 × 10⁻¹⁰)²) ≈ 6.69 × 10⁻²¹ J ≈ 0.0418 eV
(b) Average kinetic energy, K = (3/2)kT = (3/2) × 1.38 × 10⁻²³ × 300 ≈ 6.21 × 10⁻²¹ J
λ = h / √(2m_n K) = 6.626 × 10⁻³⁴ / √(2 × 1.675 × 10⁻²⁷ × 6.21 × 10⁻²¹) ≈ 1.45 × 10⁻¹⁰ m = 0.145 nm
(a) Kinetic energy ≈ 0.0418 eV, (b) de Broglie wavelength ≈ 0.145 nm

Question 11.17: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Answer:
For a photon, momentum, p = h / λ (from E = hc / λ and E = pc)
de Broglie wavelength, λ_d = h / p
Substituting p = h / λ, λ_d = h / (h / λ) = λ
Thus, the wavelength of electromagnetic radiation equals the de Broglie wavelength of its photon.
Wavelength = de Broglie wavelength of photon

Question 11.18: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K?

Answer:
Given: T = 300 K, k = 1.38 × 10⁻²³ J/K, Mass of N₂ molecule = 28 u = 28 × 1.66 × 10⁻²⁷ ≈ 4.648 × 10⁻²⁶ kg, h = 6.626 × 10⁻³⁴ J·s
Average kinetic energy, K = (3/2)kT = (3/2) × 1.38 × 10⁻²³ × 300 ≈ 6.21 × 10⁻²¹ J
λ = h / √(2mK) = 6.626 × 10⁻³⁴ / √(2 × 4.648 × 10⁻²⁶ × 6.21 × 10⁻²¹) ≈ 2.76 × 10⁻¹¹ m = 0.0276 nm
de Broglie wavelength ≈ 0.0276 nm

Chapter 12: Atoms

NCERT Solutions for Class 12 Physics

Chapter 12: Atoms

Question 12.1: Choose the correct alternative from the clues given at the end of the each statement: (a) Size of an atom in Rutherford's model is _________ the size of an atom compared to of Thomson's model. (b) In the ground state of hydrogen atom, the electron is at a distance _________ from the nucleus. (c) The electron in the excited state of hydrogen atom can have a maximum angular momentum of _________ ħ. (d) The energy of an electron in the nth orbit of hydrogen atom is inversely proportional to _________ . (e) The angular momentum of an electron in the nth orbit of hydrogen atom is _________ times the angular momentum of the electron in the first orbit. (f) The energy of an electron in the nth orbit of hydrogen atom is _________ times the energy of the electron in the first orbit.

Answer:
(a) Size of an atom in Rutherford's model is much smaller than the size of an atom compared to Thomson's model.
(b) In the ground state of hydrogen atom, the electron is at a distance equal to the Bohr radius from the nucleus.
(c) The electron in the excited state of hydrogen atom can have a maximum angular momentum of (n - 1) ħ.
(d) The energy of an electron in the nth orbit of hydrogen atom is inversely proportional to n².
(e) The angular momentum of an electron in the nth orbit of hydrogen atom is n times the angular momentum of the electron in the first orbit.
(f) The energy of an electron in the nth orbit of hydrogen atom is 1/n² times the energy of the electron in the first orbit.
(a) much smaller, (b) equal to Bohr radius, (c) (n-1)ħ, (d) n², (e) n times, (f) 1/n² times

Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at a temperature 14 K). What results do you expect?

Answer:
The nucleus of a hydrogen atom is a proton. Since the size of a proton is much smaller than that of a gold nucleus, the alpha particles would not be deflected much. Most alpha particles would pass through without any deflection, and very few would be scattered at large angles.
Most alpha particles pass undeflected; few scattered at large angles

Question 12.3: (a) Using the Bohr's model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Answer:
Given: For hydrogen atom, e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg, ε₀ = 8.85 × 10⁻¹² C²/N·m², h = 6.626 × 10⁻³⁴ J·s
(a) Orbital speed, v_n = (e² / (4πε₀ h)) × (c / n), where c = 3 × 10⁸ m/s, but using standard value α = e² / (4πε₀ ħ c) ≈ 1/137
v_n = (α c) / n
v_1 = (3 × 10⁸) / 137 ≈ 2.19 × 10⁶ m/s
v_2 = v_1 / 2 ≈ 1.095 × 10⁶ m/s
v_3 = v_1 / 3 ≈ 0.73 × 10⁶ m/s
(b) Orbital period, T_n = 2π r_n / v_n, where r_n = n² a₀, a₀ = 0.529 × 10⁻¹⁰ m
T_n = n³ (h³ ε₀ / (π m_e e⁴)) ≈ n³ × 1.52 × 10⁻¹⁶ s
T_1 ≈ 1.52 × 10⁻¹⁶ s, T_2 ≈ 1.22 × 10⁻¹⁵ s, T_3 ≈ 1.09 × 10⁻¹⁴ s
(a) v_1 ≈ 2.19 × 10⁶ m/s, v_2 ≈ 1.095 × 10⁶ m/s, v_3 ≈ 0.73 × 10⁶ m/s; (b) T_1 ≈ 1.52 × 10⁻¹⁶ s, T_2 ≈ 1.22 × 10⁻¹⁵ s, T_3 ≈ 1.09 × 10⁻¹⁴ s

Question 12.4: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Answer:
Given: Energy difference, ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J
Frequency, ν = ΔE / h, h = 6.626 × 10⁻³⁴ J·s
ν = 3.68 × 10⁻¹⁹ / 6.626 × 10⁻³⁴ ≈ 5.55 × 10¹⁴ Hz
Frequency ≈ 5.55 × 10¹⁴ Hz

Question 12.5: How much energy is released when an electron makes a transition from the first excited state to the ground state in a hydrogen atom? Express the answer in eV.

Answer:
Energy levels in hydrogen atom, E_n = -13.6 / n² eV
Ground state (n=1): E_1 = -13.6 eV
First excited state (n=2): E_2 = -13.6 / 4 = -3.4 eV
Energy released, ΔE = E_2 - E_1 = -3.4 - (-13.6) = 10.2 eV
Energy released = 10.2 eV

Question 12.6: Calculate the radius of the first Bohr orbit of a hydrogen atom. Express the answer in nm.

Answer:
Bohr radius, a_0 = 4πε₀ ħ² / (m_e e²) ≈ 0.529 × 10⁻¹⁰ m = 0.053 nm
Radius = 0.053 nm

Question 12.7: Obtain the first Bohr's radius and the ground state energy of a 'muonic hydrogen atom' (i.e. an atom in which a negatively charged muon (μ⁻) of mass about 207 m_e orbits around a proton).

Answer:
For muonic hydrogen, m_μ ≈ 207 m_e
Bohr radius, a_0μ = a_0 / 207, where a_0 = 0.529 × 10⁻¹⁰ m
a_0μ ≈ 0.529 × 10⁻¹⁰ / 207 ≈ 2.56 × 10⁻¹² m
Ground state energy, E_1μ = -13.6 × 207 ≈ -2815 eV
Bohr radius ≈ 2.56 × 10⁻¹² m, Ground state energy ≈ -2815 eV

Chapter 13: Nuclei

NCERT Solutions for Class 12 Physics

Chapter 13: Nuclei

Question 13.1: (a) Two stable isotopes of lithium ⁶Li and ⁷Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, ¹⁰B and ¹¹B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of ¹⁰B and ¹¹B.

Answer:
(a) Given: Mass of ⁶Li = 6.01512 u, abundance = 7.5% = 0.075; Mass of ⁷Li = 7.01600 u, abundance = 92.5% = 0.925
Atomic mass of lithium = (Mass of ⁶Li × abundance) + (Mass of ⁷Li × abundance)
= (6.01512 × 0.075) + (7.01600 × 0.925)
= 0.451134 + 6.4898 ≈ 6.940934 u
Atomic mass of lithium ≈ 6.941 u
(b) Given: Mass of ¹⁰B = 10.01294 u, Mass of ¹¹B = 11.00931 u, Atomic mass of boron = 10.811 u
Let abundance of ¹⁰B = x, then abundance of ¹¹B = 1 - x
10.01294x + 11.00931(1 - x) = 10.811
10.01294x + 11.00931 - 11.00931x = 10.811
(11.00931 - 10.01294)x = 11.00931 - 10.811
0.99637x = 0.19831
x ≈ 0.199, so abundance of ¹⁰B ≈ 19.9%, ¹¹B ≈ 80.1%
Abundances: ¹⁰B ≈ 19.9%, ¹¹B ≈ 80.1%

Question 13.2: The three stable isotopes of neon: ²⁰Ne, ²¹Ne, and ²²Ne have respective abundances of 90.51%, 0.27%, and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u, and 21.99 u, respectively. Obtain the average atomic mass of neon.

Answer:
Given: Mass of ²⁰Ne = 19.99 u, abundance = 90.51% = 0.9051; Mass of ²¹Ne = 20.99 u, abundance = 0.27% = 0.0027; Mass of ²²Ne = 21.99 u, abundance = 9.22% = 0.0922
Average atomic mass = (19.99 × 0.9051) + (20.99 × 0.0027) + (21.99 × 0.0922)
= 18.092499 + 0.056673 + 2.027478 ≈ 20.17665 u
Average atomic mass of neon ≈ 20.18 u

Question 13.3: Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷Au and the silver isotope ¹⁰⁷Ag.

Answer:
Nuclear radius, R = R₀ A^(1/3), where R₀ ≈ 1.2 × 10⁻¹⁵ m, A is mass number
For ¹⁹⁷Au, A = 197; For ¹⁰⁷Ag, A = 107
Ratio of radii, R_Au / R_Ag = (A_Au / A_Ag)^(1/3) = (197 / 107)^(1/3)
197 / 107 ≈ 1.841, (1.841)^(1/3) ≈ 1.226
Ratio of nuclear radii ≈ 1.23

Question 13.4: Find the Q-value and the kinetic energy of the emitted alpha particle in the alpha-decay of (a) ²²⁶Ra and (b) ²²⁰Rn. Given: m(²²⁶Ra) = 226.02540 u, m(²²²Rn) = 222.01750 u, m(²²⁰Rn) = 220.01137 u, m(²¹⁶Po) = 216.00189 u, m(⁴He) = 4.00260 u.

Answer:
(a) ²²⁶Ra → ²²²Rn + ⁴He
Q-value = [m(²²⁶Ra) - m(²²²Rn) - m(⁴He)] × 931.5 MeV/u
= [226.02540 - 222.01750 - 4.00260] × 931.5
= 0.0053 × 931.5 ≈ 4.94 MeV
Kinetic energy of alpha particle = Q × (A_parent - 4) / A_parent = 4.94 × (226 - 4) / 226 ≈ 4.85 MeV
Q-value ≈ 4.94 MeV, KE of alpha particle ≈ 4.85 MeV
(b) ²²⁰Rn → ²¹⁶Po + ⁴He
Q-value = [m(²²⁰Rn) - m(²¹⁶Po) - m(⁴He)] × 931.5
= [220.01137 - 216.00189 - 4.00260] × 931.5
= 0.00688 × 931.5 ≈ 6.41 MeV
Kinetic energy of alpha particle = 6.41 × (220 - 4) / 220 ≈ 6.29 MeV
Q-value ≈ 6.41 MeV, KE of alpha particle ≈ 6.29 MeV

Question 13.5: The radionuclide ⁵⁷Co decays by beta-minus emission to a stable isotope of ⁵⁷Fe with a half-life of 270 days. Calculate the decay constant.

Answer:
Given: Half-life, T₁/₂ = 270 days
Decay constant, λ = ln(2) / T₁/₂
λ = 0.693 / 270 ≈ 0.002566 day⁻¹
Decay constant ≈ 2.57 × 10⁻³ day⁻¹

Question 13.6: The Q-value of a nuclear reaction ¹⁴N + ⁴He → ¹⁷O + ¹H is -1.16 MeV. Calculate the mass defect of the reaction.

Answer:
Q-value = Δm × 931.5 MeV/u, where Δm is mass defect
-1.16 = Δm × 931.5
Δm = -1.16 / 931.5 ≈ -0.001245 u
Mass defect ≈ -0.00125 u

Question 13.7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125% (b) 1% of its original value?

Answer:
Activity, A = A₀ (1/2)^(t/T₁/₂), where T₁/₂ = T years
(a) 3.125% = 0.03125 A₀
0.03125 = (1/2)^(t/T)
(1/2)^5 = (1/2)^(t/T) → t/T = 5 → t = 5T years
(b) 1% = 0.01 A₀
0.01 = (1/2)^(t/T)
(1/2)^(t/T) = 1/100, (1/2)^(t/T) = (1/2)^(6.643) → t/T ≈ 6.643 → t ≈ 6.64T years
(a) 5T years, (b) 6.64T years

Question 13.8: The normal activity of living carbon-containing matter is found to be about 15 decays per minute per gram. If an old wooden object is found to have an activity of 5 decays per minute per gram, estimate its age. Assume the half-life of ¹⁴C is 5730 years.

Answer:
Given: Initial activity, A₀ = 15 decays/min/g, Current activity, A = 5 decays/min/g, T₁/₂ = 5730 years
A = A₀ (1/2)^(t/T₁/₂)
5/15 = (1/2)^(t/5730)
1/3 = (1/2)^(t/5730)
t/5730 = log(3)/log(2) ≈ 1.585
t ≈ 1.585 × 5730 ≈ 9082 years
Age ≈ 9082 years

Question 13.9: Obtain the amount of ⁶⁰Co necessary to provide a radioactive source of 8.0 mCi.

Answer:
Given: Activity = 8.0 mCi = 8.0 × 3.7 × 10⁷ decays/s = 2.96 × 10⁸ decays/s, Half-life of ⁶⁰Co ≈ 5.27 years = 5.27 × 3.156 × 10⁷ s ≈ 1.663 × 10⁸ s
Activity, A = λN, λ = ln(2)/T₁/₂ ≈ 0.693 / 1.663 × 10⁸ ≈ 4.167 × 10⁻⁹ s⁻¹
N = A / λ = 2.96 × 10⁸ / 4.167 × 10⁻⁹ ≈ 7.1 × 10¹⁶ atoms
Mass = (N × M) / N_A, M = 60 g/mol, N_A = 6.022 × 10²³ mol⁻¹
Mass = (7.1 × 10¹⁶ × 60) / 6.022 × 10²³ ≈ 7.08 × 10⁻⁶ g
Mass ≈ 7.08 μg

Question 13.10: The half-life of ⁹⁰Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer:
Given: T₁/₂ = 28 years = 28 × 3.156 × 10⁷ s ≈ 8.837 × 10⁸ s, Mass = 15 mg = 0.015 g
λ = ln(2)/T₁/₂ ≈ 0.693 / 8.837 × 10⁸ ≈ 7.84 × 10⁻¹⁰ s⁻¹
N = (0.015 × 6.022 × 10²³) / 90 ≈ 1.003 × 10²⁰ atoms
Disintegration rate, A = λN = 7.84 × 10⁻¹⁰ × 1.003 × 10²⁰ ≈ 7.86 × 10¹⁰ decays/s
Disintegration rate ≈ 7.86 × 10¹⁰ decays/s

Question 13.11: Obtain the binding energy (in MeV) of a nitrogen nucleus (¹⁴N), given m(¹⁴N) = 14.00307 u.

Answer:
For ¹⁴N: Z = 7, N = 7, m(¹⁴N) = 14.00307 u, m_p = 1.00783 u, m_n = 1.00866 u
Mass defect, Δm = [Z m_p + N m_n - m(¹⁴N)]
= [7 × 1.00783 + 7 × 1.00866 - 14.00307] = 7.05481 + 7.06062 - 14.00307 ≈ 0.11236 u
Binding energy = Δm × 931.5 ≈ 0.11236 × 931.5 ≈ 104.66 MeV
Binding energy ≈ 104.66 MeV

Question 13.12: Obtain the binding energy per nucleon of ⁵⁶Fe nucleus. The mass of ⁵⁶Fe is 55.93494 u.

Answer:
For ⁵⁶Fe: Z = 26, N = 30, m(⁵⁶Fe) = 55.93494 u, m_p = 1.00783 u, m_n = 1.00866 u
Mass defect, Δm = [26 × 1.00783 + 30 × 1.00866 - 55.93494] = 26.20358 + 30.2598 - 55.93494 ≈ 0.52844 u
Binding energy = 0.52844 × 931.5 ≈ 492.24 MeV
Binding energy per nucleon = 492.24 / 56 ≈ 8.79 MeV
Binding energy per nucleon ≈ 8.79 MeV

Question 13.13: A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments, each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.

Answer:
Initial binding energy = 240 × 7.6 = 1824 MeV
Final binding energy = 2 × (120 × 8.5) = 2 × 1020 = 2040 MeV
Released energy = Final BE - Initial BE = 2040 - 1824 = 216 MeV
Released energy = 216 MeV

Question 13.14: The fission properties of ²³⁹Pu are very similar to those of ²³⁵U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ²³⁹Pu undergo fission?

Answer:
Given: Mass = 1 kg = 1000 g, Molar mass of ²³⁹Pu = 239 g/mol, Energy per fission = 180 MeV
Number of atoms, N = (1000 × 6.022 × 10²³) / 239 ≈ 2.52 × 10²⁴
Total energy = N × 180 = 2.52 × 10²⁴ × 180 ≈ 4.536 × 10²⁶ MeV
Total energy ≈ 4.54 × 10²⁶ MeV

Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

NCERT Solutions for Class 12 Physics

Chapter 14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Question 14.1: In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

Answer:
In n-type silicon, pentavalent atoms (e.g., phosphorus) are doped to provide extra electrons, making electrons the majority carriers and holes the minority carriers.
Correct option: (c) Holes are minority carriers and pentavalent atoms are the dopants.

Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductors?

Answer:
In p-type semiconductors, trivalent atoms (e.g., boron) are doped, creating holes as majority carriers and electrons as minority carriers.
Correct option: (d) Holes are majority carriers and trivalent atoms are the dopants.

Question 14.3: Carbon, silicon, and germanium have four valence electrons each. At room temperature, which one of the following statements is most appropriate? (a) The number of free electrons for conduction is significant only in Si and Ge but small in C. (b) The number of free conduction electrons is significant in C but small in Si and Ge. (c) The number of free conduction electrons is negligible in all three. (d) The number of free electrons for conduction is significant in all three.

Answer:
Carbon (a diamond structure) has a large bandgap (~5.5 eV), so it has very few free electrons at room temperature, making it an insulator. Silicon and germanium have smaller bandgaps (~1.1 eV and ~0.67 eV), allowing significant free electrons for conduction.
Correct option: (a) The number of free electrons for conduction is significant only in Si and Ge but small in C.

Question 14.4: In a p-type semiconductor, the acceptor level is 57 meV above the valence band. What is the maximum wavelength of light needed to create a hole?

Answer:
Energy required to create a hole, E = 57 meV = 0.057 eV
Wavelength, λ = hc / E, where h = 4.1357 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m/s
λ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / 0.057 ≈ 2.177 × 10⁻⁵ m = 21770 nm
Maximum wavelength ≈ 21770 nm

Question 14.5: In an intrinsic semiconductor, the energy gap is 1.2 eV. What is the maximum wavelength of light required to create an electron-hole pair?

Answer:
Energy gap, E_g = 1.2 eV
Wavelength, λ = hc / E_g
λ = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / 1.2 ≈ 1.034 × 10⁻⁶ m = 1034 nm
Maximum wavelength ≈ 1034 nm

Question 14.6: The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m³ of arsenic and 5 × 10²⁰ atoms per m³ of indium. Find the number of electrons and holes. Given that n_i = 1.5 × 10¹⁶ m⁻³.

Answer:
Arsenic (pentavalent) provides electrons: n_d = 5 × 10²² m⁻³
Indium (trivalent) creates holes: n_a = 5 × 10²⁰ m⁻³
Since n_d >> n_a, the material is n-type.
Number of electrons, n ≈ n_d - n_a = 5 × 10²² - 5 × 10²⁰ ≈ 4.95 × 10²² m⁻³
Number of holes, p = n_i² / n = (1.5 × 10¹⁶)² / 4.95 × 10²² ≈ 4.55 × 10⁹ m⁻³
Electrons ≈ 4.95 × 10²² m⁻³, Holes ≈ 4.55 × 10⁹ m⁻³

Question 14.7: In an n-type semiconductor, the donor level is 0.2 eV below the conduction band. The energy gap is 1.0 eV. What is the maximum wavelength of light needed to create an electron-hole pair?

Answer:
To create an electron-hole pair, an electron must be excited across the energy gap, E_g = 1.0 eV.
Wavelength, λ = hc / E_g = (4.1357 × 10⁻¹⁵ × 3 × 10⁸) / 1.0 ≈ 1.241 × 10⁻⁶ m = 1241 nm
(The donor level affects conduction electrons but not the electron-hole pair creation.)
Maximum wavelength ≈ 1241 nm

Question 14.8: A p-n junction diode has a reverse saturation current of 10⁻⁷ A. What is the current through the diode when a forward bias of 0.3 V is applied at 300 K?

Answer:
Diode current, I = I₀ [e^(V/V_T) - 1], where I₀ = 10⁻⁷ A, V = 0.3 V, V_T = kT/q ≈ 0.0259 V at 300 K
V/V_T = 0.3 / 0.0259 ≈ 11.58
I = 10⁻⁷ [e^11.58 - 1] ≈ 10⁻⁷ × 1.07 × 10⁵ ≈ 0.0107 A
Current ≈ 0.0107 A

Question 14.9: The resistance of a p-n junction diode is 10 Ω when forward biased and 200 kΩ when reverse biased. If it is used in a half-wave rectifier with a 12 V peak AC source, calculate the DC output voltage.

Answer:
In a half-wave rectifier, the diode conducts only in the forward bias (positive half-cycle).
Peak voltage = 12 V, Forward resistance = 10 Ω (load resistance not given, assume negligible compared to diode resistance).
DC output voltage (average) for half-wave rectifier, V_DC ≈ V_peak / π ≈ 12 / 3.14 ≈ 3.82 V
DC output voltage ≈ 3.82 V

Question 14.10: The peak voltage of a full-wave rectifier is 15 V. If the load resistance is 1 kΩ and the forward resistance of each diode is 50 Ω, calculate the DC output voltage and the ripple factor.

Answer:
Total resistance = R_load + R_diode = 1000 + 50 = 1050 Ω
DC output voltage, V_DC = 2 × V_peak / π = 2 × 15 / 3.14 ≈ 9.55 V
Ripple factor for full-wave rectifier = 1 / (2√3 × f × C × R_total), but C (capacitance) is not given, so assume no filter (ideal case, ripple factor ≈ 0.48 for full-wave rectifier without filter).
DC output voltage ≈ 9.55 V, Ripple factor ≈ 0.48 (without filter)

Question 14.11: A zener diode with a breakdown voltage of 6 V is used in a voltage regulator circuit with a series resistance of 200 Ω and an input voltage varying between 10 V and 15 V. Calculate the current through the zener diode when the input is 12 V and the load resistance is 1 kΩ.

Answer:
Zener voltage, V_Z = 6 V, R_series = 200 Ω, R_load = 1000 Ω, V_in = 12 V
Load current, I_L = V_Z / R_load = 6 / 1000 = 0.006 A
Total current, I = (V_in - V_Z) / R_series = (12 - 6) / 200 = 0.03 A
Zener current, I_Z = I - I_L = 0.03 - 0.006 = 0.024 A
Zener current ≈ 0.024 A

Question 14.12: In a transistor, the base current is 0.05 mA and the collector current is 2.0 mA. Find the current gain (β).

Answer:
Current gain, β = I_C / I_B
I_C = 2.0 mA = 0.002 A, I_B = 0.05 mA = 0.00005 A
β = 0.002 / 0.00005 = 40
Current gain β = 40

Question 14.13: A transistor is used in common-emitter configuration with a collector resistance of 2 kΩ. The input resistance is 1 kΩ, and the current gain is 50. If the input voltage is 0.1 V, calculate the output voltage.

Answer:
Input voltage, V_i = 0.1 V, R_in = 1 kΩ, β = 50, R_C = 2 kΩ
Base current, I_B = V_i / R_in = 0.1 / 1000 = 0.0001 A
Collector current, I_C = β × I_B = 50 × 0.0001 = 0.005 A
Output voltage, V_out = I_C × R_C = 0.005 × 2000 = 10 V
Output voltage ≈ 10 V

Question 14.14: In a common-emitter amplifier, the collector resistance is 4 kΩ, and the load resistance is 6 kΩ. If the current gain is 100 and the input signal is 0.02 V, calculate the voltage gain.

Answer:
Total resistance, R_total = R_C || R_load = (4 × 6) / (4 + 6) = 2.4 kΩ
Voltage gain, A_V = β × (R_total / R_in). Since R_in is not given, assume typical input resistance ~1 kΩ.
A_V = 100 × (2400 / 1000) = 240
Voltage gain ≈ 240 (assuming R_in = 1 kΩ)

Question 14.15: Identify the logic gates whose truth tables are given below: (a) A=0, B=0, Y=1; A=0, B=1, Y=1; A=1, B=0, Y=1; A=1, B=1, Y=0. (b) A=0, B=0, Y=0; A=0, B=1, Y=1; A=1, B=0, Y=1; A=1, B=1, Y=1.

Answer:
(a) Truth table: Y = 1 for A=0, B=0; A=0, B=1; A=1, B=0; Y = 0 for A=1, B=1
This matches the NAND gate (Y = NOT(A AND B)).
(b) Truth table: Y = 0 for A=0, B=0; Y = 1 for A=0, B=1; A=1, B=0; A=1, B=1
This matches the OR gate (Y = A OR B).
(a) NAND gate, (b) OR gate