CBSE Class 12 2025 Physics 55-1-2 Question Paper Set-2: Download Solutions with Answer Key

The work done on a charge \( Q \) when it moves between two points is given by: \[ W = q \cdot \Delta V \]
where \( \Delta V \) is the potential difference between the initial and final positions.

The work done is maximum when the charge moves between points with the greatest potential difference. Path C is the one that passes through regions with the largest change in potential and will lead to maximum work done.

Thus, the correct answer is (C). Quick Tip: The work done on a charge depends on the potential difference. The larger the potential difference between two points, the greater the work done when the charge moves between them.

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. The area \( A \) of a wire with radius \( r \) is: \[ A = \pi r^2 \]

For the resistance to be \( \frac{R}{2} \), we need to adjust the length or radius.
If the radius \( r \) is kept the same, and the length \( L \) is halved, the new resistance becomes: \[ R' = \rho \frac{L/2}{A} = \frac{1}{2} \cdot R \]
Therefore, halving the length of the wire reduces the resistance to half of the original value.

Thus, the correct answer is (B). Quick Tip: To halve the resistance of a wire, you can either reduce its length by half or increase its radius by a factor of \( \sqrt{2} \).

The induced emf in a coil is given by Faraday's Law of Induction: \[ emf = - N \frac{d\Phi}{dt} \]
where:

\( N = 300 \) is the number of turns,

\( \Phi \) is the magnetic flux,

\( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux.

The flux \( \Phi = B \cdot A \), where \( B \) is the magnetic field and \( A \) is the area of the coil. The area of the coil is: \[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{15 \, mm}{2} \right)^2 = \pi (7.5 \times 10^{-3})^2 = 1.77 \times 10^{-4} \, m^2 \]

Now, consider the two cases for the magnetic field change:

1. In the first case, the magnetic field is reduced uniformly to zero in 20 ms: \[ emf_1 = - N \frac{d\Phi}{dt} = - 300 \times \frac{30 \times 10^{-3} \times 1.77 \times 10^{-4}}{20 \times 10^{-3}} = 0.80 \, V \]

2. In the second case, the magnetic field is increased back to 30 mT in 40 ms: \[ emf_2 = - N \frac{d\Phi}{dt} = - 300 \times \frac{30 \times 10^{-3} \times 1.77 \times 10^{-4}}{40 \times 10^{-3}} = 0.40 \, V \]

Thus, the ratio of the induced emfs is: \[ \frac{e_1}{e_2} = \frac{0.80}{0.40} = 2 \]

Therefore, the correct answer is (C) 2.
Quick Tip: The induced emf is proportional to the rate of change of magnetic flux. For the same change in magnetic field, the emf is inversely proportional to the time taken for the change.

Step 1: Understand the conditions for Total Internal Reflection (TIR).

Total Internal Reflection (TIR) is a phenomenon that occurs when a ray of light passes from a denser medium to a rarer medium. The two essential conditions for TIR are:

Light must travel from an optically denser medium to an optically rarer medium.
The angle of incidence in the denser medium must be greater than the critical angle (\(i_c\)).

Step 2: Determine the optical densities of the two media based on the given speeds.

The speed of light in a medium is inversely proportional to its refractive index. A medium with a lower speed of light is optically denser (higher refractive index), and a medium with a higher speed of light is optically rarer (lower refractive index).
Given:

Speed of light in medium '1' = \(v_1\)
Speed of light in medium '2' = \(v_2\)
Condition: \(v_2 > v_1\)

Since \(v_2 > v_1\), it means light travels faster in medium '2' than in medium '1'.
Therefore:

Medium '1' is optically denser.
Medium '2' is optically rarer.




Step 3: Determine the direction of light for TIR based on optical densities.

According to Condition 1 for TIR, the light must travel from the optically denser medium to the optically rarer medium.

Thus, the light must be incident from medium '1' (the denser medium) to medium '2' (the rarer medium).



Step 4: Calculate the critical angle (\(i_c\)).

The refractive index of a medium (\(\mu\)) is given by \(\mu = \frac{c}{v}\), where \(c\) is the speed of light in a vacuum.

So, \(\mu_1 = \frac{c}{v_1}\) and \(\mu_2 = \frac{c}{v_2}\).
The critical angle \(i_c\) for light going from a denser medium (\(\mu_1\)) to a rarer medium (\(\mu_2\)) is given by:
\(\sin i_c = \frac{\mu_{rarer}}{\mu_{denser}} = \frac{\mu_2}{\mu_1}\)

Substitute the expressions for refractive indices in terms of speeds: \(\sin i_c = \frac{c/v_2}{c/v_1} = \frac{v_1}{v_2}\)

Therefore, the critical angle is \(i_c = \sin^{-1}\left(\frac{v_1}{v_2}\right)\).



Step 5: Combine all conditions to find the correct option.

For total internal reflection to occur, the ray of light must be incident from medium '1' (denser) and the angle of incidence must be greater than the critical angle, \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\).



Step 6: Evaluate the options.


(A) medium '1' and at an angle greater than \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\): This perfectly matches our derived conditions.
(B) medium '1' and at an angle greater than \(\cos^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect critical angle expression.
(C) medium '2' and at an angle greater than \(\sin^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect medium of incidence. Light must be incident from the denser medium ('1').
(D) medium '2' and at an angle greater than \(\cos^{-1}\left(\frac{v_1}{v_2}\right)\): Incorrect medium and critical angle expression.

Based on the physics principles and the given condition \(v_2 > v_1\), option (A) is the correct choice.
\[ \boxed{medium '1' and at an angle greater than \sin^{-1}\left(\frac{v_1}{v_2}\right)} \] Quick Tip: Remember that Total Internal Reflection (TIR) occurs only when light passes from an optically denser medium to an optically rarer medium. The critical angle for TIR is given by \(\sin i_c = \frac{\mu_{rarer}}{\mu_{denser}}\). If speeds are given, remember that the medium with lower speed is denser (\(v_{denser} < v_{rarer}\)), and thus \(\mu_{denser} > \mu_{rarer}\).

The energy of a photon is given by the equation: \[ E = h f \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, J \cdot s \)), and \( f \) is the frequency of the light. Substituting the given values: \[ E = 6.626 \times 10^{-34} \times 5.0 \times 10^{14} = 3.313 \times 10^{-19} \, J \]

Now, the number of photons \( n \) emitted per second can be calculated using the power \( P \) of the source and the energy \( E \) of one photon: \[ n = \frac{P}{E} \]
Given that the power emitted is 3.31 mW, or \( 3.31 \times 10^{-3} \) W, we have: \[ n = \frac{3.31 \times 10^{-3}}{3.313 \times 10^{-19}} = 10^{16} \, photons per second \]

Thus, the correct answer is (A). Quick Tip: To find the number of photons emitted by a light source, use the relationship between power, photon energy, and the number of photons per second: \( n = \frac{P}{E} \).

Step 1: Understanding Assertion (A):

In Bohr’s model, the potential energy of an electron in an orbit is negative and equal to \(-2\) times the kinetic energy. Hence, Assertion (A) is false.


Step 2: Understanding Reason (R):

The total energy (kinetic + potential) of a bound system like an electron in an atom is negative, not always positive. So Reason (R) is also false.


Step 3: Final evaluation:

Both statements are incorrect, so option (D) is the right choice.
Quick Tip: In bound systems, total energy is negative — indicating the electron is trapped in the atom.

Step 1: Evaluate Assertion (A):

You cannot create a working p-n junction diode just by physically joining p-type and n-type slabs because there will be no depletion region or electric field formed at the junction. Therefore, Assertion (A) is true.


Step 2: Evaluate Reason (R):

The Reason (R) is incorrect because the carrier concentrations are reversed: in a p-type semiconductor, hole concentration \(\eta_h \gg \eta_e\), and in an n-type semiconductor, electron concentration \(\eta_e \gg \eta_h\). So Reason (R) is false.


Step 3: Relationship between A and R:

Assertion is true, but Reason is false — hence, option (C) is correct.
Quick Tip: Always remember: p-n junctions must be chemically fabricated to allow diffusion and depletion region formation.

We start with the equation for the electric field and the current density in a material: \[ \vec{E} = \rho \vec{j} \]
where:
\( \vec{E} \) is the electric field (measured in volts per meter),

\( \rho \) is the resistivity of the material (measured in ohm meters),

\( \vec{j} \) is the current density (measured in amperes per square meter).

This equation describes the relationship between the electric field and the current density in a material. The resistivity \( \rho \) is a proportional constant that characterizes how strongly the material opposes the flow of current.

Now, let’s use Ohm's law, which states that: \[ \vec{E} = \frac{V}{L} \]
where:

\( V \) is the potential difference (in volts),

\( L \) is the length of the conductor (in meters).

At the same time, the current density \( \vec{j} \) is given by: \[ \vec{j} = \frac{I}{A} \]
where:

\( I \) is the current (in amperes),

\( A \) is the cross-sectional area of the conductor (in square meters).

Substituting \( \vec{j} \) into the first equation, we get: \[ \vec{E} = \rho \frac{I}{A} \]
This is the form of Ohm's law, where \( \vec{E} \) is proportional to \( I \), and the proportionality constant is \( \frac{\rho}{A} \), which can also be interpreted as the resistance \( R \) of the material: \[ R = \frac{\rho L}{A} \]
Thus, we have: \[ \vec{E} = I R \]
This is the familiar form of Ohm's law, \( \vec{E} = I R \), where \( R \) is the resistance of the material.

However, Ohm's law does not hold for materials where the resistivity \( \rho \) is not constant. For example:

Non-linear materials such as semiconductors, where the current density \( \vec{j} \) and electric field \( \vec{E} \) do not follow a linear relationship due to the material’s intrinsic properties.

Temperature-dependent resistivity, where the resistivity changes with temperature, can cause deviations from Ohm’s law. As the current increases, the material heats up, which in turn changes its resistivity.

Therefore, Ohm's law is not valid when the resistivity \( \rho \) depends on factors such as temperature, current, or the electric field. Quick Tip: Ohm's law applies to materials with constant resistivity. If the resistivity depends on factors like temperature or current, the material is non-ohmic, and Ohm’s law does not hold.

We start with the equation for the electric field and the current density in a material: \[ \vec{E} = \rho \vec{j} \]
where:
\( \vec{E} \) is the electric field (measured in volts per meter),

\( \rho \) is the resistivity of the material (measured in ohm meters),

\( \vec{j} \) is the current density (measured in amperes per square meter).

This equation describes the relationship between the electric field and the current density in a material. The resistivity \( \rho \) is a proportional constant that characterizes how strongly the material opposes the flow of current.

Now, let’s use Ohm's law, which states that: \[ \vec{E} = \frac{V}{L} \]
where:

\( V \) is the potential difference (in volts),

\( L \) is the length of the conductor (in meters).

At the same time, the current density \( \vec{j} \) is given by: \[ \vec{j} = \frac{I}{A} \]
where:

\( I \) is the current (in amperes),

\( A \) is the cross-sectional area of the conductor (in square meters).

Substituting \( \vec{j} \) into the first equation, we get: \[ \vec{E} = \rho \frac{I}{A} \]
This is the form of Ohm's law, where \( \vec{E} \) is proportional to \( I \), and the proportionality constant is \( \frac{\rho}{A} \), which can also be interpreted as the resistance \( R \) of the material: \[ R = \frac{\rho L}{A} \]
Thus, we have: \[ \vec{E} = I R \]
This is the familiar form of Ohm's law, \( \vec{E} = I R \), where \( R \) is the resistance of the material.

However, Ohm's law does not hold for materials where the resistivity \( \rho \) is not constant. For example:

Non-linear materials such as semiconductors, where the current density \( \vec{j} \) and electric field \( \vec{E} \) do not follow a linear relationship due to the material’s intrinsic properties.

Temperature-dependent resistivity, where the resistivity changes with temperature, can cause deviations from Ohm’s law. As the current increases, the material heats up, which in turn changes its resistivity.

Therefore, Ohm's law is not valid when the resistivity \( \rho \) depends on factors such as temperature, current, or the electric field. Quick Tip: Ohm's law applies to materials with constant resistivity. If the resistivity depends on factors like temperature or current, the material is non-ohmic, and Ohm’s law does not hold.

Let:

\( E_1 = 3\,V, \quad r_1 = 0.2\,\Omega \)

\( E_2 = 6\,V, \quad r_2 = 0.4\,\Omega \)


Step 1: Equivalent EMF of the parallel combination: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{\frac{3}{0.2} + \frac{6}{0.4}}{\frac{1}{0.2} + \frac{1}{0.4}} = \frac{15 + 15}{5 + 2.5} = \frac{30}{7.5} = 4\,V \]

Step 2: Equivalent internal resistance of the combination: \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{0.2 \times 0.4}{0.2 + 0.4} = \frac{0.08}{0.6} = \frac{2}{15}\,\Omega \approx 0.133\,\Omega \]

Step 3: Total resistance in the circuit: \[ R_{total} = r_{eq} + R = 0.133 + 4 = 4.133\,\Omega \]

Step 4: Current drawn from the combination: \[ I = \frac{E_{eq}}{R_{total}} = \frac{4}{4.133} \approx 0.968\,A \] Quick Tip: To combine batteries in parallel, use the formula: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}, \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \] Then apply Ohm’s law with total external resistance.

The torque \( \tau \) experienced by a current-carrying coil in a magnetic field is given by the formula: \[ \tau = N I A B \sin \theta \]
where:
- \( \tau \) is the torque,
- \( N \) is the number of turns of the coil,
- \( I \) is the current flowing through the coil,
- \( A \) is the area of the coil,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the normal to the coil's plane and the magnetic field.

Step 1: Use the initial torque to find the magnetic field.

Given:
- Number of turns: \( N = 60 \),
- Area of the coil: \( A = 1.5 \times 10^{-3} \, m^2 \),
- Current: \( I = 2 \, A \),
- Initial torque \( \tau_1 = 0.12 \, Nm \) (when the angle between the normal to the coil's plane and the magnetic field is \( \theta = 0^\circ \)).

When \( \theta = 0^\circ \), \( \sin 0^\circ = 0 \), so the formula becomes: \[ \tau_1 = N I A B \]
Solving for \( B \): \[ B = \frac{\tau_1}{N I A} = \frac{0.12}{60 \times 2 \times 1.5 \times 10^{-3}} = \frac{0.12}{0.18} = 0.6667 \, T \]

Thus, the magnetic field strength is \( B = 0.6667 \, T \).

Step 2: Use the final torque to find the magnetic field.

Now, the coil is rotated by \( 90^\circ \), so the angle \( \theta = 90^\circ \), and the torque changes to \( \tau_2 = 0.05 \, Nm \). When \( \theta = 90^\circ \), \( \sin 90^\circ = 1 \), so the torque formula becomes: \[ \tau_2 = N I A B \sin 90^\circ = N I A B \]
Using the values: \[ B = \frac{\tau_2}{N I A} = \frac{0.05}{60 \times 2 \times 1.5 \times 10^{-3}} = \frac{0.05}{0.18} = 0.2778 \, T \]

Step 3: Calculate the resultant magnetic field strength.

Now, we need to use the relationship between the two torques, \( \tau_1 \) and \( \tau_2 \), to find the magnitude of the magnetic field. Given: \[ \tau_1 = N I A B \sin \theta_1 \quad and \quad \tau_2 = N I A B \sin \theta_2 \]
For \( \tau_1 = 0.12 \, Nm \), \( \theta_1 = 90^\circ \) (maximum torque), and for \( \tau_2 = 0.05 \, Nm \), the coil is rotated by 90°.

The total magnetic field is given by: \[ B = \sqrt{B_1^2 \sin^2 \theta + B_2^2 \cos^2 \theta} \]

Substituting the values: \[ B = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{5}{18}\right)^2} = \frac{13}{18} \, T \]

Conclusion:
- The resultant magnetic field strength is \( B = \frac{13}{18} \, T \). Quick Tip: The torque on a coil in a magnetic field depends on the number of turns, current, area of the coil, magnetic field strength, and the angle between the normal to the coil's plane and the magnetic field. For a coil rotated by 90°, the torque equation simplifies to \( \tau = N I A B \).

The mutual inductance \( M \) between two solenoids depends on the flux linkage and the induced EMF. For two solenoids with magnetic fields interacting, the mutual inductance is given by:
\[ M_{12} = \frac{\mu_0 n_1 n_2 A l}{l} \]
Where:
\( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \)),

\( n_1 \) and \( n_2 \) are the number of turns per unit length of solenoid 1 and 2 respectively,

\( A \) is the cross-sectional area of the solenoid,

\( l \) is the length of the solenoid.

The magnetic flux through solenoid \( S_2 \) due to solenoid \( S_1 \) is given by:
\[ \Phi_{B2} = B_1 \cdot A = \left( \frac{\mu_0 n_1 I_1}{l} \right) \cdot A \]

For the induced EMF in solenoid \( S_2 \), the formula for mutual inductance becomes:
\[ M_{12} = \frac{\mu_0 n_1 n_2 A l}{l} \]

Similarly, the mutual inductance of solenoid \( S_1 \) with respect to solenoid \( S_2 \) is:
\[ M_{21} = \frac{\mu_0 n_1 n_2 A l}{l} \]

Thus, we can see that:
\[ M_{21} = M_{12} \]

This shows that the mutual inductance is symmetric. Quick Tip: The mutual inductance between two solenoids depends on their turns per unit length, cross-sectional area, and the permeability of free space. The mutual inductance is symmetric, meaning \( M_{12} = M_{21} \).

The plot of frequency \( \nu \) of incident radiation versus the stopping potential \( V_0 \) is a linear relationship. According to Einstein's photoelectric equation: \[ E_k = h\nu - \phi \]
where:

\( E_k \) is the kinetic energy of the emitted electrons,

\( h \) is Planck's constant,

\( \nu \) is the frequency of the incident radiation,

\( \phi \) is the work function of the material.


The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electrons by: \[ E_k = eV_0 \]
where \( e \) is the charge of the electron.

Therefore, combining these two equations: \[ eV_0 = h\nu - \phi \]
This gives the equation of a straight line: \[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]
Thus, the plot of \( V_0 \) versus \( \nu \) will be a straight line with slope \( \frac{h}{e} \) and intercept \( -\frac{\phi}{e} \) on the \( V_0 \)-axis. The intercept on the stopping potential axis gives information about the work function \( \phi \) of the photo-emissive material.

Quick Tip: The slope of the plot \( V_0 \) vs \( \nu \) gives \( \frac{h}{e} \), where \( h \) is Planck's constant and \( e \) is the charge of the electron. The intercept on the \( V_0 \)-axis corresponds to the work function \( \phi \) of the material.

(i) Momentum:

The momentum of an electron is given by: \[ p = \sqrt{2mK} \]
where:

\( m = 9.1 \times 10^{-31} \, kg \) is the mass of the electron,

\( K = 80 \, eV = 80 \times 1.6 \times 10^{-19} \, J = 1.28 \times 10^{-17} \, J \) is the kinetic energy of the electron.

Now, substituting the known values: \[ p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.28 \times 10^{-17}} \, kg \cdot m/s \] \[ p = 4.8 \times 10^{-24} \, kg \cdot m/s \]

Thus, the momentum \( p \approx 4.8 \times 10^{-24} \, kg \cdot m/s \).


(ii) de Broglie Wavelength:

The de Broglie wavelength \( \lambda \) is given by the formula: \[ \lambda = \frac{h}{p} \]
where:

\( h = 6.63 \times 10^{-34} \, J \cdot s \) is Planck’s constant,

\( p = 4.8 \times 10^{-24} \, kg \cdot m/s \) is the momentum.

Substituting the known values: \[ \lambda = \frac{6.63 \times 10^{-34}}{4.8 \times 10^{-24}} \, m \] \[ \lambda \approx 1.38 \times 10^{-10} \, m \]

Thus, the de Broglie wavelength \( \lambda \approx 1.38 \times 10^{-10} \, m \). Quick Tip: To find the momentum, use \( p = \sqrt{2mK} \) and to find the de Broglie wavelength, use \( \lambda = \frac{h}{p} \).

Step 1: Understanding the setup:

The lens is convex and the screen is placed on the opposite side of the object. A real image is formed when object is beyond the focal length.


Step 2: Nature of real image:

A real image formed by a convex lens is always inverted and appears on the side opposite to the object (on the screen).


Step 3: Image matching:

Option (C) shows an inverted image on the correct side — the screen. Hence, it is correct.
Quick Tip: Convex lenses form real, inverted images on the opposite side of the screen when the object is beyond focus.

Coherent sources are sources of light that emit waves with a constant phase relationship. In other words, the phase difference between the waves from coherent sources remains constant over time. For sustained interference patterns, the waves need to maintain a fixed phase difference, which results in a stable constructive or destructive interference.

This constant phase relationship is necessary to observe sustained and stable interference patterns. Without coherence, the interference pattern would fade or change over time, making it impossible to observe clear and stable fringes.

(2) Lights from two independent sources are not coherent. Explain.


% Solution
Solution:

When lights come from two independent sources, the phase difference between the waves from each source is random. These sources do not have a fixed phase relationship and therefore are not coherent. As a result, the interference pattern formed by two such sources would be irregular and would not sustain itself, as the random phase differences lead to fluctuating constructive and destructive interference. Quick Tip: For sustained interference, coherence is key, ensuring that the phase difference between the waves remains constant.