CBSE Class 12 Physics Question Paper 2025 Solved Code 55-1-3

SECTION A

1. A charge Q is fixed in position. Another charge q is brought near charge Q and released from rest. Which of the following graphs is the correct representation of the acceleration of the charge q as a function of its distance r from charge Q ? [1 Mark]

[Figure: Four graphs of acceleration a (y-axis) versus distance r (x-axis). (A) a curve that starts very high at small r and falls steadily, flattening out at large r. (B) a curve that stays flat at a high value and then drops sharply to a low value. (C) a straight line falling from a high value at small r to zero. (D) a bell-shaped curve rising from zero to a peak and falling back.]

Answer: (A) the curve in which a is very large at small r and decreases steadily, flattening out as r increases

Teacher's Note:
a) By Coulomb's law, \( F = \frac{kQq}{r^2} \), so \( a = \frac{F}{m} \propto \frac{1}{r^2} \).
b) An inverse square graph falls steeply at first and then flattens out, which is graph (A).

2. Two conductors A and B of the same material have their lengths in the ratio 1:2 and radii in the ratio 2:3. If they are connected in parallel across a battery, the ratio \( \frac{v_A}{v_B} \) of the drift velocities of electrons in them will be - [1 Mark]
(A) 2
(B) \( \frac{1}{2} \)
(C) \( \frac{3}{2} \)
(D) \( \frac{8}{9} \)

Answer: (A) 2

Teacher's Note:
a) Drift velocity \( v_d = \frac{eV\tau}{ml} \), so for the same V and material, \( v_d \propto \frac{1}{l} \).
b) The radii do not matter here; \( \frac{v_A}{v_B} = \frac{l_B}{l_A} = \frac{2}{1} = 2 \).

3. A 1 cm segment of a wire lying along x-axis carries current of 0.5 A along +x direction. A magnetic field \( \vec{B} = (0.4\text{ mT})\hat{j} + (0.6\text{ mT})\hat{k} \) is switched on, in the region. The force acting on the segment is [1 Mark]
(A) \( (2\hat{j} + 3\hat{k}) \) mN
(B) \( (-3\hat{j} + 2\hat{k}) \) μN
(C) \( (6\hat{j} + 4\hat{k}) \) mN
(D) \( (-4\hat{j} + 6\hat{k}) \) μN

Answer: (B) \( (-3\hat{j} + 2\hat{k}) \) μN

Teacher's Note:
a) Use \( \vec{F} = I(\vec{L} \times \vec{B}) \) with \( \vec{L} = 0.01\hat{i} \) m.
b) Remember \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \times \hat{k} = -\hat{j} \) to get the correct signs.

4. The ratio of the number of turns of the primary to the secondary coils in an ideal transformer is 20 : 1. If 240 V ac is applied from a source to the primary coil of transformer and a 6.0 Ω resistor is connected across the output terminals, then current drawn by the transformer from the source will be - [1 Mark]
(A) 4.0 A
(B) 3.8 A
(C) 0.97 A
(D) 0.10 A

Answer: (D) 0.10 A

Teacher's Note:
a) Secondary voltage \( = \frac{240}{20} = 12 \) V, so secondary current \( = \frac{12}{6} = 2 \) A.
b) For an ideal transformer, \( V_pI_p = V_sI_s \), so \( I_p = \frac{2}{20} = 0.1 \) A.

5. You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be [1 Mark]
(A) 2592
(B) 2866
(C) 2976
(D) 3140

Answer: (B) 2866

Teacher's Note:
a) Use \( L = \frac{\mu_0 N^2 A}{l} \), so \( N^2 = \frac{Ll}{\mu_0 A} \) with \( A = \pi r^2 \).
b) Substituting the values gives \( N \approx 2866 \) turns.

6. A voltage \( v = v_0 \sin \omega t \) applied to a circuit drives a current \( i = i_0 \sin(\omega t + \phi) \) in the circuit. The average power consumed in the circuit over a cycle is [1 Mark]
(A) Zero
(B) \( i_0 v_0 \cos\phi \)
(C) \( \frac{i_0 v_0}{2} \)
(D) \( \frac{i_0 v_0}{2}\cos\phi \)

Answer: (D) \( \frac{i_0 v_0}{2}\cos\phi \)

Teacher's Note:
a) Average power in an AC circuit is \( P_{avg} = V_{rms}I_{rms}\cos\phi \).
b) Using \( V_{rms} = \frac{v_0}{\sqrt2} \) and \( I_{rms} = \frac{i_0}{\sqrt2} \) gives \( \frac{i_0v_0}{2}\cos\phi \).

7. X-rays are more harmful to human beings than ultraviolet radiations because X-rays - [1 Mark]
(A) have frequency lower than that of ultraviolet radiations.
(B) have wavelength smaller than that of ultraviolet radiations.
(C) move faster than ultraviolet radiations in air.
(D) are mechanical waves but ultraviolet radiations are electro-magnetic waves.

Answer: (B) have wavelength smaller than that of ultraviolet radiations.

Teacher's Note:
a) Smaller wavelength means higher frequency and higher photon energy \( \left(E = \frac{hc}{\lambda}\right) \).
b) Both are electromagnetic waves and travel at the same speed in air, so options (C) and (D) are wrong.

8. A point source is placed at the bottom of a tank containing a transparent liquid (refractive index n) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is [1 Mark]
(A) \( \frac{\pi H^2}{(n-1)} \)
(B) \( \frac{\pi H^2}{(n^2-1)} \)
(C) \( \frac{\pi H^2}{\sqrt{n^2-1}} \)
(D) \( \frac{\pi H^2}{(n^2+1)} \)

Answer: (B) \( \frac{\pi H^2}{(n^2-1)} \)

Teacher's Note:
a) Light escapes only within a circle where the angle of incidence is less than the critical angle, \( \sin C = \frac{1}{n} \).
b) Radius of the circle \( r = H\tan C = \frac{H}{\sqrt{n^2-1}} \), so area \( = \pi r^2 = \frac{\pi H^2}{n^2-1} \).

9. In a photoelectric experiment with a material of work function 2.1 eV, the stopping potential is found to be 2.5 V. The maximum kinetic energy of ejected photoelectrons is [1 Mark]
(A) 0.4 eV
(B) 2.1 eV
(C) 2.5 eV
(D) 4.6 eV

Answer: (C) 2.5 eV

Teacher's Note:
a) Maximum kinetic energy \( K_{max} = eV_0 \), where \( V_0 \) is the stopping potential.
b) The work function is not needed here; do not add or subtract it.

10. When a p-n junction diode is forward biased [1 Mark]
(A) the barrier height and the depletion layer width both increase.
(B) the barrier height increases and the depletion layer width decreases.
(C) the barrier height and the depletion layer width both decrease.
(D) the barrier height decreases and the depletion layer width increases.

Answer: (C) the barrier height and the depletion layer width both decrease.

Teacher's Note:
a) Forward bias opposes the built-in potential, so the barrier height falls.
b) A lower barrier means a thinner depletion layer.

11. Let \( \lambda_e \), \( \lambda_p \) and \( \lambda_d \) be the wavelengths associated with an electron, a proton and a deuteron, all moving with the same speed. Then the correct relation between them is [1 Mark]
(A) \( \lambda_d \gt \lambda_p \gt \lambda_e \)
(B) \( \lambda_e \gt \lambda_p \gt \lambda_d \)
(C) \( \lambda_p \gt \lambda_e \gt \lambda_d \)
(D) \( \lambda_e = \lambda_p = \lambda_d \)

Answer: (B) \( \lambda_e \gt \lambda_p \gt \lambda_d \)

Teacher's Note:
a) de Broglie wavelength \( \lambda = \frac{h}{mv} \), so for the same speed, \( \lambda \propto \frac{1}{m} \).
b) Mass order is electron less than proton less than deuteron, so the wavelength order is reversed.

12. Which of the following figures correctly represent the shape of curve of binding energy per nucleon as a function of mass number ? [1 Mark]

[Figure: Four graphs of B.E./A (y-axis) versus mass number A (x-axis). (A) rises steeply, peaks near A = 56 (dashed line) and then decreases gradually. (B) rises steeply up to about A = 56 and then stays almost flat. (C) is a symmetric hump peaking at A = 80. (D) rises steeply to a peak at A = 80 and then falls sharply before rising slightly.]

Answer: (A) the curve that rises steeply, peaks near A = 56 and then decreases gradually

Teacher's Note:
a) The real binding energy per nucleon curve has its maximum near A = 56 (iron region).
b) For heavier nuclei it falls slowly, not sharply, so curves peaking at A = 80 are wrong.

Note : Question numbers 13 to 16 are Assertion (A) and Reason (R) type questions. Two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

13. Assertion (A) : We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of a n-type semiconductor.
Reason (R) : In a p-type semiconductor \( \eta_e \gt\gt \eta_h \) while in a n-type semiconductor \( \eta_h \gt\gt \eta_e \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Physically joining two slabs leaves surface roughness and gaps, so a proper continuous junction is not formed.
b) The Reason is false: in p-type, holes are the majority carriers \( (\eta_h \gt\gt \eta_e) \), and in n-type, electrons are the majority carriers.

14. Assertion (A) : The potential energy of an electron revolving in any stationary orbit in a hydrogen atom is positive.
Reason (R) : The total energy of a charged particle is always positive. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (D) Assertion (A) is false and Reason (R) is also false.

Teacher's Note:
a) The electron-proton force is attractive, so the potential energy of the electron is negative.
b) The total energy of a bound electron is also negative, so the Reason is false too.

15. Assertion (A) : It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.
Reason (R) : The direction of induced current in a coil with its circuit closed, due to motion of a magnet, is such that it opposes the cause. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) This is Lenz's law: the induced current opposes the change that produces it.
b) More turns give a larger induced current, so the opposition to the moving magnet is larger.

16. Assertion (A) : The deflection in a galvanometer is directly proportional to the current passing through it.
Reason (R) : The coil of a galvanometer is suspended in a uniform radial magnetic field. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is also false.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) In a radial field the plane of the coil is always parallel to B, so the torque \( NIAB \) is proportional to I.
b) Balancing it with the restoring torque \( k\phi \) gives \( \phi \propto I \), which makes the scale linear.

SECTION B

17. n identical cells, each of e.m.f. E and internal resistance r, are connected in series. Later on it was found out that two cells 'X' and 'Y' are connected in reverse polarities. Calculate the potential difference across the cell 'X'. [2 Marks]

Answer:
1. Each reversed cell cancels one correct cell, so net emf \( = (n-4)E \). Total internal resistance \( = nr \).
2. Current \( I = \frac{(n-4)E}{nr} \).
3. In cell X, current flows against its emf, so \( V = E + Ir = E + \frac{(n-4)E}{nr} \times r \).
4. \( V = \frac{(2n-4)E}{n} \).

Teacher's Note:
a) Two reversed cells reduce the net emf by 4E, not 2E: a common mistake.
b) For a cell being charged (current against its emf), use \( V = E + Ir \), not \( V = E - Ir \).

18. (a) In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at \( \theta = 30^{\circ} \). Calculate the width of the slit. [2 Marks]

Answer:
1. Condition for minima: \( a\sin\theta = n\lambda \). For the first minimum, n = 1.
2. \( a \sin 30^{\circ} = 600 \times 10^{-9} \) m, so \( a \times \frac{1}{2} = 600 \times 10^{-9} \) m.
3. \( a = 1200 \times 10^{-9} \) m \( = 1.2 \times 10^{-6} \) m.

Teacher's Note:
a) The minima condition \( a\sin\theta = n\lambda \) carries 1 mark; write it first.
b) Use \( \sin30^{\circ} = \frac{1}{2} \) and give the answer in metres.

OR

(b) In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point. [2 Marks]

Answer:
1. Phase difference \( \phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4} \).
2. \( I = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos\frac{\pi}{4} = 2I_0 + 2I_0 \times \frac{1}{\sqrt2} \).
3. \( I = I_0(2 + \sqrt2) = 3.414\, I_0 \).

Teacher's Note:
a) Always convert path difference to phase difference first.
b) The scheme also accepts \( I = 4I_0\cos^2\frac{\phi}{2} = 4I_0\cos^2\frac{\pi}{8} \), which gives the same value.

19. A double convex lens of glass has both faces of the same radius of curvature 17 cm. Find its focal length if it is immersed in water. The refractive indices of glass and water are 1.5 and 1.33 respectively. [2 Marks]

Answer:
1. Lens maker's formula in water: \( \frac{1}{f} = \left(\frac{n_g}{n_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).
2. For a double convex lens, \( R_1 = R = 17 \) cm and \( R_2 = -R \), so \( \frac{1}{f} = \left(\frac{1.5}{1.33} - 1\right)\frac{2}{R} \).
3. \( \frac{1}{f} = \left(\frac{1.5 - 1.33}{1.33}\right)\left(\frac{2}{17}\right) \).
4. \( f = 66.5 \) cm.

Teacher's Note:
a) In a liquid, use the relative refractive index \( \frac{n_g}{n_w} \), not \( n_g \) alone.
b) The focal length in water is much larger than in air (17 cm), which is a quick check.

20. An electron in Bohr model of hydrogen atom makes a transition from energy level -1.51 eV to -3.40 eV. Calculate the change in the radius of its orbit. The radius of orbit of electron in its ground state is 0.53 Å. [2 Marks]

Answer:
1. \( E_n = -\frac{13.6}{n^2} \) eV. For \( E_n = -1.51 \) eV, n = 3.
2. For \( E_n = -3.40 \) eV, n = 2.
3. \( r_n = 0.53\,n^2 \) Å, so change in radius \( \Delta r = 0.53\,(3^2 - 2^2) = 0.53 \times 5 \).
4. \( \Delta r = 2.65 \) Å.

Teacher's Note:
a) First find both quantum numbers from the energy values.
b) Remember \( r \propto n^2 \); the orbit becomes smaller in this transition.

21. A p-type Si semiconductor is made by doping an average of one dopant atom per \( 5 \times 10^7 \) silicon atoms. If the number density of silicon atoms in the specimen is \( 5 \times 10^{28} \) atoms m^-3, find the number of holes created per cubic centimetre in the specimen due to doping. Also give one example of such dopants. [2 Marks]

Answer:
1. Number of holes per m³ \( = \frac{5\times10^{28}}{5\times10^7} = 10^{21} \).
2. Number of holes per cm³ \( = \frac{10^{21}}{10^6} = 10^{15} \).
3. Example of dopant: Aluminium (or Indium / Gallium).

Teacher's Note:
a) Each trivalent dopant atom creates one hole.
b) Convert per m³ to per cm³ by dividing by 10⁶.

SECTION C

22. (a) Two batteries of emf's 3V & 6V and internal resistances 0.2 Ω & 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find :
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination [3 Marks]

Answer:
1. Equivalent emf \( E_{eq} = \frac{E_1r_2 + E_2r_1}{r_1 + r_2} = \frac{3 \times 0.4 + 6 \times 0.2}{0.6} = 4 \) V.
2. Equivalent internal resistance \( r_{eq} = \frac{r_1r_2}{r_1 + r_2} = \frac{0.2 \times 0.4}{0.2 + 0.4} = 0.133 \) Ω.
3. Current \( I = \frac{E_{eq}}{R + r_{eq}} = \frac{4}{4 + 0.133} = \frac{4}{4.133} \approx 0.97 \) A.

Teacher's Note:
a) Write each formula before substituting; the formula carries half a mark.
b) Treat the combination as one cell of emf \( E_{eq} \) and internal resistance \( r_{eq} \) in series with the 4 Ω resistor.
c) The marking scheme writes the final current as 0.9 A; the exact value of 4/4.133 is about 0.97 A.

OR

(b) (i) A conductor of length \( l \) is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If R and R' are initial and final values of resistance and \( v_d \) and \( v_d' \) are initial and final values of drift velocity, find the relation between (i) R' and R and (ii) \( v_d' \) and \( v_d \).
(ii) When electrons drift in a conductor from lower to higher potential, does it mean that all the 'free electrons' of the conductor are moving in the same direction ? [3 Marks]

Answer:
1. Volume stays the same: \( Al = A'(2l) \), so \( A' = \frac{A}{2} \). Then \( R' = \frac{\rho(2l)}{A/2} = 4\frac{\rho l}{A} \), so \( R' = 4R \).
2. \( v_d = \frac{eE\tau}{m} = \frac{eV\tau}{ml} \). With V the same, \( v_d' = \frac{eV\tau}{m(2l)} \), so \( \frac{v_d'}{v_d} = \frac{l}{2l} = \frac{1}{2} \), i.e. \( v_d' = \frac{v_d}{2} \).
3. No. Drift is only a small average velocity added to the random thermal motion; the free electrons still move in all directions.

Teacher's Note:
a) On stretching, volume is constant, so \( R \propto l^2 \) and doubling the length makes \( R' = 4R \).
b) The cell stays connected, so V is fixed and \( v_d \propto \frac{1}{l} \).

23. A particle of charge q is moving with a velocity \( \vec{v} \) at a distance 'd' from a long straight wire carrying a current 'I' as shown in figure. At this instant, it is subjected to a uniform electric field \( \vec{E} \) such that the particle keeps moving undeviated. In terms of unit vectors \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \), find -
(a) the magnetic field \( \vec{B} \),
(b) the magnetic force \( \vec{F}_m \), and
(c) the electric field \( \vec{E} \), acting on the charge. [3 Marks]

[Figure: A long straight wire along the x-axis carrying current I towards the left (-x direction). A charged particle is at distance d above the wire, moving with velocity v towards the left. Axes shown alongside: x to the right (unit vector i), y upwards (unit vector j) and z out of the page (unit vector k).]

Answer:
1. (a) \( \vec{B} = \frac{\mu_0 I}{2\pi d}(-\hat{k}) \).
2. (b) \( \vec{F}_m = q(\vec{v} \times \vec{B}) = \frac{qv\mu_0 I}{2\pi d}(-\hat{j}) \).
3. (c) For the particle to go undeviated, \( \vec{F}_e = -\vec{F}_m = \frac{qv\mu_0 I}{2\pi d}\hat{j} \).
4. Since \( \vec{F}_e = q\vec{E} \), \( \vec{E} = \frac{\mu_0 vI}{2\pi d}\hat{j} \).

Teacher's Note:
a) Use the right-hand thumb rule for the direction of B due to the wire at the position of the charge.
b) For undeviated motion, the electric force must be equal and opposite to the magnetic force.

24. An ac source of voltage \( v = v_m \sin \omega t \) is connected to a series combination of LCR circuit. Draw the phasor diagram. Using it obtain an expression for the impedance of the circuit and the phase difference between applied voltage and the current. [3 Marks]

Answer:
1. Phasor diagram: current phasor I is taken as reference. \( V_R \) is in phase with I, \( V_L \) leads I by \( 90^{\circ} \) and \( V_C \) lags I by \( 90^{\circ} \). \( V_L \) and \( V_C \) are opposite, so their resultant is \( (V_{Cm} - V_{Lm}) \), and the applied voltage \( V_m \) is the resultant of \( V_{Rm} \) and \( (V_{Cm} - V_{Lm}) \).
2. \( V_{Rm} = i_mR \), \( V_{Cm} = i_mX_C \), \( V_{Lm} = i_mX_L \).
3. From the diagram, \( V_m^2 = V_{Rm}^2 + (V_{Cm} - V_{Lm})^2 = i_m^2\left[R^2 + (X_C - X_L)^2\right] \).
4. \( i_m = \frac{V_m}{\sqrt{R^2 + (X_C - X_L)^2}} = \frac{V_m}{Z} \), so \( Z = \sqrt{R^2 + (X_C - X_L)^2} \).
5. Phase difference: \( \tan\phi = \frac{V_{Cm} - V_{Lm}}{V_{Rm}} = \frac{X_C - X_L}{R} \), so \( \phi = \tan^{-1}\left(\frac{X_C - X_L}{R}\right) \).

Teacher's Note:
a) The phasor diagram carries 1 mark; label I, \( V_R \), \( V_L \), \( V_C \) and the angle \( \phi \) clearly.
b) Always take current as the reference phasor in a series circuit, because it is the same in all elements.

25. (a) A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current \( (I_c) \) and the displacement current \( (I_d) \) has the same value at all points of the circuit.
(b) In case (a) above, is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor ? Explain. [3 Marks]

Answer:
1. Total current \( I = I_c + I_d \). Outside the capacitor, \( I_d = 0 \), so \( I = I_c \).
2. Inside the capacitor, \( I_c = 0 \), so \( I = I_d = \varepsilon_0\frac{d\phi_E}{dt} = \varepsilon_0\frac{d(EA)}{dt} \).
3. With \( E = \frac{Q}{\varepsilon_0 A} \), \( I_d = \varepsilon_0 A\frac{d}{dt}\left(\frac{Q}{\varepsilon_0 A}\right) = \frac{dQ}{dt} = I_c \). Hence \( I_c + I_d \) has the same value at all points of the circuit.
4. (b) Yes. The current entering the plate is \( I_c \) and the current between the plates is \( I_d \). Since \( I_c = I_d \), Kirchhoff's junction rule is valid at each plate.

Teacher's Note:
a) Treat the regions outside and inside the capacitor separately; each carries half a mark.
b) The key step is showing \( I_d = \frac{dQ}{dt} \), which is the conduction current in the wires.

26. (a) Mention any three features of results of experiment on photoelectric effect which cannot be explained using the wave theory of light.
(b) In his experiment on photoelectric effect, Robert A. Millikan found the slope of the cut-off voltage versus frequency of incident light plot to be \( 4.12 \times 10^{-15} \) Vs. Calculate the value of Planck's constant from it. [3 Marks]

Answer:
1. (a) (i) The existence of a threshold frequency \( (\nu_0) \).
2. (ii) The maximum kinetic energy of photoelectrons is independent of the intensity of incident radiation.
3. (iii) The photoelectric effect is instantaneous.
4. (b) Slope \( = \frac{h}{e} \), so \( h = e \times \text{slope} = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15} \).
5. \( h = 6.6 \times 10^{-34} \) Js.

Teacher's Note:
a) From \( eV_0 = h\nu - \phi_0 \), the graph of \( V_0 \) against \( \nu \) has slope \( \frac{h}{e} \).
b) Wave theory predicts that intensity should decide the kinetic energy and that emission may be delayed; both are wrong.

27. (a) Draw circuit arrangement for studying V-I characteristics of a p-n junction diode.
(b) Show the shape of the characteristics of a diode.
(c) Mention two information that you can get from these characteristics. [3 Marks]

Answer:
1. (a) Forward bias circuit: a battery with a rheostat (potential divider) and a switch connected to the diode with its p-side to the positive terminal, a milliammeter (mA) in series and a voltmeter across the diode.
2. Reverse bias circuit: the same arrangement with the diode reversed (p-side to the negative terminal) and a microammeter (μA) in place of the milliammeter.
3. (b) Characteristics: in forward bias the current stays very small up to the knee voltage (about 0.7 V for Si) and then rises sharply (current in mA). In reverse bias a very small, nearly constant current (in μA) flows until the breakdown voltage, after which the current rises suddenly.
4. (c) Two informations: knee voltage and reverse saturation current (others: breakdown voltage, very low resistance in forward bias, very high resistance in reverse bias).

Teacher's Note:
a) Use a milliammeter for forward bias and a microammeter for reverse bias; examiners check this.
b) As per the marking scheme, marks are not deducted for not writing values on the graph.

28. (a) Define 'Mass defect' and 'Binding energy' of a nucleus. Describe 'Fission process' on the basis of binding energy per nucleon.
(b) A deuteron contains a proton and a neutron and has a mass of 2.013553 u. Calculate the mass defect for it in u and its energy equivalence in MeV. \( (m_p = 1.007277 \text{ u}, m_n = 1.008665 \text{ u}, 1\text{u} = 931.5 \text{ MeV/c}^2) \) [3 Marks]

Answer:
1. Mass defect: the difference between the mass of the nucleus and the total mass of its constituent nucleons.
2. Binding energy: the energy required to separate the nucleons from the nucleus.
3. Fission: a heavy nucleus splits into lighter nuclei and energy is released, because the binding energy per nucleon increases.
4. (b) \( \Delta m = (m_p + m_n) - m_d = (1.007277 + 1.008665) - 2.013553 = 0.002389 \) u.
5. Energy \( = \Delta m \times c^2 = 0.002389 \times 931.5 = 2.2253 \) MeV \( \approx 2.22 \) MeV.

Teacher's Note:
a) Link fission to the rise in binding energy per nucleon of the products; this carries the mark.
b) Keep all six decimal places while subtracting masses to get the correct mass defect.

SECTION D

Question numbers 29 and 30 are case study based questions. Read the following paragraphs and answer the questions that follow.

29. A thin lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, one can obtain the 'lens maker formula' and then the 'lens formula'. A lens has two foci - called 'first focal point' and 'second focal point' of the lens, one on each side. [4 Marks]

(i) Consider the arrangement shown in figure. A black vertical arrow and a horizontal thick line with a ball are painted on a glass plate. It serves as the object. When the plate is illuminated, its real image is formed on the screen.
Which of the following correctly represents the image formed on the screen ? [1 Mark]

[Figure: An optical bench with a light box with plate at one end, a convex lens in the middle and a screen at the other end. An inset labelled "Plate on Hidden Side of Box" shows an upward vertical arrow with a short horizontal line ending in a ball on the right. Options: (A) upward arrow with the ball on the right; (B) upward arrow with the line tilted and the ball at lower left; (C) downward arrow with the ball on the left; (D) downward arrow with the line tilted and the ball at upper right.]

Answer: (C) the downward arrow with the ball on the left

Teacher's Note:
a) A real image formed by a convex lens is inverted, so the arrow points downward.
b) The image is also laterally inverted, so the ball moves from the right side to the left side.

(ii) Which of the following statements is incorrect ? [1 Mark]
(A) For a convex mirror magnification is always negative.
(B) For all virtual images formed by a mirror magnification is positive.
(C) For a concave lens magnification is always positive.
(D) For real and inverted images, magnification is always negative.

Answer: (A) For a convex mirror magnification is always negative.

Teacher's Note:
a) A convex mirror always forms a virtual, erect image, so its magnification is positive.
b) Positive magnification means an erect image; negative means an inverted image.

(iii) A convex lens of focal length 'f' is cut into two equal parts perpendicular to the principal axis. The focal length of each part will be : [1 Mark]
(A) f
(B) 2 f
(C) \( \frac{f}{2} \)
(D) \( \frac{f}{4} \)

Answer: (B) 2 f

Teacher's Note:
a) Each part is a plano-convex lens with only one curved surface, so its power is half the original power.
b) Half the power means double the focal length.

OR

(iii) If an object in case (i) above is 20 cm from the lens and the screen is 50 cm away from the object, the focal length of the lens used is [1 Mark]
(A) 10 cm
(B) 12 cm
(C) 16 cm
(D) 20 cm

Answer: (B) 12 cm

Teacher's Note:
a) Here u = -20 cm and v = 50 - 20 = 30 cm.
b) \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} + \frac{1}{20} = \frac{1}{12} \), so f = 12 cm.

(iv) The distance of an object from first focal point of a biconvex lens is \( X_1 \) and distance of the image from second focal point is \( X_2 \). The focal length of the lens is [1 Mark]
(A) \( X_1 X_2 \)
(B) \( \sqrt{X_1 + X_2} \)
(C) \( \sqrt{X_1 X_2} \)
(D) \( \sqrt{\frac{X_2}{X_1}} \)

Answer: (C) \( \sqrt{X_1 X_2} \)

Teacher's Note:
a) This is Newton's lens formula: \( X_1X_2 = f^2 \).
b) It follows from the lens formula by putting \( u = -(f + X_1) \) and \( v = f + X_2 \).

30. A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
As soon as the circuit is completed by closing key S₁ (keeping S₂ open) charges begin to flow between the capacitor plates and the battery terminals. The charge on the capacitor increases and consequently the potential difference \( V_c \) (= q/C) across the capacitor also increases with time. When this potential difference equals the potential difference across the battery, the capacitor is fully charged (Q = VC). During this process of charging, the charge q on the capacitor changes with time t as \( q = Q[1 - e^{-t/RC}] \)
The charging current can be obtained by differentiating it and using \( \frac{d}{dx}(e^{mx}) = me^{mx} \).
Consider the case when R = 20 kΩ, C = 500 μF and V = 10 V. [4 Marks]

[Figure: RC circuit with capacitor C in the top branch, resistor R on the left side, and a battery of emf V in the bottom branch with key S₁ in series with it. Key S₂ connects the lower end of R directly to the right side of the circuit, forming a path through R and C that bypasses the battery.]

(i) The final charge on the capacitor, when key S₁ is closed and S₂ is open, is [1 Mark]
(A) 5 μC
(B) 5 mC
(C) 25 mC
(D) 0.1 C

Answer: (B) 5 mC

Teacher's Note:
a) Final charge \( Q = VC = 10 \times 500 \times 10^{-6} = 5 \times 10^{-3} \) C.
b) Convert μF to F before multiplying.

(ii) For sufficient time the key S₁ is closed and S₂ is open. Now key S₂ is closed and S₁ is open. What is the final charge on the capacitor ? [1 Mark]
(A) Zero
(B) 5 mC
(C) 2.5 mC
(D) 5 μC

Answer: (A) Zero

Teacher's Note:
a) With S₁ open the battery is cut off, and S₂ lets the capacitor discharge through R.
b) After a long time the capacitor is fully discharged, so its charge is zero.

(iii) The dimensional formula for RC is [1 Mark]
(A) \( [M L^2 T^{-3} A^{-2}] \)
(B) \( [M^0 L^0 T^{-1} A^0] \)
(C) \( [M^{-1} L^{-2} T^4 A^2] \)
(D) \( [M^0 L^0 T A^0] \)

Answer: (D) \( [M^0 L^0 T A^0] \)

Teacher's Note:
a) RC is the time constant of the circuit, so it has the dimension of time.
b) Check: the power \( \frac{t}{RC} \) in the exponential must be dimensionless.

(iv) The key S₁ is closed and S₂ is open. The value of current in the resistor after 5 seconds, is [1 Mark]
(A) \( \frac{1}{2\sqrt{e}} \) mA
(B) \( \sqrt{e} \) mA
(C) \( \frac{1}{\sqrt{e}} \) mA
(D) \( \frac{1}{2e} \) mA

Answer: (A) \( \frac{1}{2\sqrt{e}} \) mA

Teacher's Note:
a) Charging current \( i = \frac{V}{R}e^{-t/RC} \), with \( \frac{V}{R} = 0.5 \) mA and RC = 10 s, so at t = 5 s, \( i = 0.5\,e^{-1/2} = \frac{1}{2\sqrt{e}} \) mA.
b) As per the CBSE marking scheme, 1 mark for this part may be given to all the students who have attempted other parts of the question.

OR

(iv) The key S₁ is closed and S₂ is open. The initial value of charging current in the resistor, is [1 Mark]
(A) 5 mA
(B) 0.5 mA
(C) 2 mA
(D) 1 mA

Answer: (B) 0.5 mA

Teacher's Note:
a) At t = 0 the uncharged capacitor has no potential difference, so \( i_0 = \frac{V}{R} \).
b) \( i_0 = \frac{10}{20 \times 10^3} = 0.5 \times 10^{-3} \) A = 0.5 mA.

SECTION E

31. (a) (i) (1) What are coherent sources ? Why are they necessary for observing a sustained interference pattern ?
(2) Lights from two independent sources are not coherent. Explain.
(ii) Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.
(1) How far apart will adjacent bright interference fringes be on the screen ?
(2) Find the angular width (in degree) of the first bright fringe. [5 Marks]

Answer:
1. (i) (1) Two sources are coherent if they emit light of the same frequency (wavelength) and the phase difference between the waves from them is zero or does not change with time.
2. Coherent sources are needed because only then is the phase difference at each point constant, which gives a sustained (steady) interference pattern.
3. (2) Two independent sources are never coherent, because the phase difference between the light from them does not remain constant.
4. (ii) (1) Distance between adjacent bright fringes = fringe width \( \beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.2}{0.1 \times 10^{-3}} = 7.2 \) mm.
5. (2) Angular width \( \theta = \frac{\lambda}{d} = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6 \times 10^{-3} \) rad \( = 0.34^{\circ} \).

Teacher's Note:
a) Key words for the definition: same frequency and constant phase difference.
b) Fringe width uses \( \frac{\lambda D}{d} \) but angular width uses \( \frac{\lambda}{d} \); do not mix them up.
c) As per the marking scheme, full marks are given if the student writes the angular width in radians only.

OR

(b) (i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.
(ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed. [5 Marks]

Answer:
1. (i) A wavefront is the locus of all the points which oscillate in phase. Diagram: plane wavefronts (parallel straight lines) fall on the convex lens, and the refracted wavefronts are spherical, converging to the focus F on the other side.
2. (ii) Ray diagram: parallel rays from the distant source refract at the first surface of the ball and again at the second surface, and meet at a point beyond the ball.
3. First surface (rarer to denser, object at infinity): \( n_1 = 1 \), \( n_2 = 1.5 \), R = 15 cm, \( u = \infty \). Using \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \): \( \frac{1.5}{v} = \frac{0.5}{15} \), so v = 45 cm.
4. Second surface (denser to rarer): this image acts as the object at u = 15 cm from the second surface, with R = -15 cm. Using \( \frac{n_1}{v} - \frac{n_2}{u} = \frac{n_1 - n_2}{R} \): \( \frac{1}{v} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \).
5. \( \frac{1}{v} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30} \), so v = 7.5 cm. The final image is formed 7.5 cm beyond the second surface of the ball.

Teacher's Note:
a) The image from the first surface (45 cm) lies 15 cm beyond the second surface, since the ball's diameter is 30 cm.
b) Apply the single spherical surface formula twice and keep the sign of R correct for each surface.
c) The wavefront definition carries 1 mark and the two wavefront sketches carry half a mark each.

32. (a) (i) Two point charges 5 μC and -1 μC are placed at points (-3 cm, 0, 0) and (3 cm, 0, 0) respectively. An external electric field \( \vec{E} = \frac{A}{r^2}\hat{r} \) where \( A = 3 \times 10^5 \) Vm is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.
(ii) A system of two conductors is placed in air and they have net charge of +80μC and -80μC which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors ?
(3) If the charges on two conductors are changed to +160 μC and -160 μC, will the capacitance of the system change ? Give reason for your answer. [5 Marks]

Answer:
1. (i) Since \( \vec{E} = \frac{3 \times 10^5}{r^2}\hat{r} \) and \( dV = -\vec{E}\cdot d\vec{r} \), the potential is \( V = \frac{3 \times 10^5}{r} \).
2. Energy without the field \( U_i = \frac{kq_1q_2}{r_{12}} \); with the field \( U_f = \frac{kq_1q_2}{r_{12}} + q_1V(r_1) + q_2V(r_2) \). So \( \Delta U = U_f - U_i = q_1V(r_1) + q_2V(r_2) \).
3. \( \Delta U = \frac{5 \times 10^{-6} \times 3 \times 10^5}{3 \times 10^{-2}} - \frac{1 \times 10^{-6} \times 3 \times 10^5}{3 \times 10^{-2}} = 50 - 10 = 40 \) J.
4. (ii) (1) \( C = \frac{Q}{V} = \frac{80\ \mu C}{16\ V} = 5\ \mu F \). (2) \( C' = KC = 3 \times 5 = 15\ \mu F \), so \( V' = \frac{Q}{C'} = \frac{80\ \mu C}{15\ \mu F} = 5.33 \) V.
5. (3) No. The capacitance of the system depends only on its geometry, not on the charge given to it.

Teacher's Note:
a) Both charges are 3 cm from the origin, so both have the same external potential \( V = \frac{3 \times 10^5}{0.03} = 10^7 \) V.
b) The mutual energy \( \frac{kq_1q_2}{r_{12}} \) cancels in \( \Delta U \); only the energy in the external field changes.
c) Doubling Q doubles V, so C = Q/V stays the same.

OR

(b) (i) Consider three metal spherical shells A, B and C, each of radius R. Each shell is having a concentric metal ball of radius R/10. The spherical shells A, B and C are given charges +6q, -4q, and 14q respectively. Their inner metal balls are also given charges -2q, +8q and -10q respectively. Compare the magnitude of the electric fields due to shells A, B and C at a distance 3R from their centres.
(ii) A charge -6 μC is placed at the centre B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge +5 μC is moved from point 'C' to point 'A' along the circumference. Calculate the work done on the charge. [5 Marks]

[Figure: A horizontal line with points D, C, B and A from left to right. A charge marked +6q is at D and a charge marked -6q is at B. A semicircle centred at B is drawn above the line from C to A.]

Answer:
1. (i) Total charge for A = 6q - 2q = +4q; for B = -4q + 8q = +4q; for C = 14q - 10q = +4q.
2. Field outside at r = 3R: \( E = \frac{kQ}{r^2} = \frac{k(4q)}{9R^2} = \frac{4kq}{9R^2} \) for each, so \( E_A = E_B = E_C \).
3. (ii) C is 5 cm from B and 5 cm from D: \( V_C = \frac{k(-6 \times 10^{-6})}{5 \times 10^{-2}} + \frac{k(6 \times 10^{-6})}{5 \times 10^{-2}} = 0 \).
4. A is 5 cm from B and 15 cm from D: \( V_A = \frac{k(6 \times 10^{-6})}{15 \times 10^{-2}} + \frac{k(-6 \times 10^{-6})}{5 \times 10^{-2}} = \frac{9 \times 10^9 \times 6 \times 10^{-6} \times (-2)}{15 \times 10^{-2}} = -7.2 \times 10^5 \) V.
5. Work done \( W = q(V_A - V_C) = 5 \times 10^{-6} \times (-7.2 \times 10^5 - 0) = -3.6 \) J.

Teacher's Note:
a) Outside a spherical charge distribution, the field depends only on the total charge enclosed.
b) Work done in an electrostatic field depends only on the end points, so the semicircular path does not matter.
c) Keep the sign of the potential; a negative answer means the field does positive work on the charge.

33. (a) (i) A proton moving with velocity \( \vec{V} \) in a non-uniform magnetic field traces a path as shown in the figure.
The path followed by the proton is always in the plane of the paper. What is the direction of the magnetic field in the region near points P, Q and R ? What can you say about relative magnitude of magnetic fields at these points ?
(ii) A current carrying circular loop of area A produces a magnetic field B at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} \). [5 Marks]

[Figure: A proton (+) moving to the right with velocity V along a straight line up to point P. After P the path curves upward through point Q to a crest at point R, and then curves downward.]

Answer:
1. (i) Near P, the force on the proton acts upwards, so the magnetic field is into the plane of the paper. Near Q, the force is also upwards, so the field is into the plane of the paper.
2. Near R, the force acts downwards, so the magnetic field is out of the plane of the paper.
3. Since \( B \propto \frac{1}{r} \) (r = radius of curvature of the path): near P, B is small; near Q, B is smaller than at P; near R, B is larger than at P. So \( B_Q \lt B_P \lt B_R \).
4. (ii) Let r be the radius and I the current. \( B = \frac{\mu_0 I}{2r} \), so \( I = \frac{2Br}{\mu_0} \). Also \( A = \pi r^2 \), so \( r = \sqrt{\frac{A}{\pi}} \).
5. \( M = IA = \frac{2Br}{\mu_0}A = \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} \).

Teacher's Note:
a) Find the direction of the force from the way the path bends, then use \( \vec{F} = q(\vec{v} \times \vec{B}) \) for the direction of B.
b) From \( r = \frac{mv}{qB} \), a sharper bend (smaller r) means a stronger field.
c) In part (ii), write r in terms of A before substituting into M = IA.

OR

(b) (i) Derive an expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field.
(ii) A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path (2) its time period of revolution change ? [5 Marks]

Answer:
1. (i) Consider a rectangular coil of sides a and b carrying current I, with its normal at angle \( \theta \) to B. The forces on the two sides of length b are \( F_1 = F_2 = IbB \), equal and opposite, and they form a couple.
2. Torque \( \tau = F_1\frac{a}{2}\sin\theta + F_2\frac{a}{2}\sin\theta = IabB\sin\theta = IAB\sin\theta \). In vector form, \( \vec{\tau} = I\vec{A} \times \vec{B} \). (The forces on the other two sides cancel.)
3. (ii) (1) \( r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \), so \( r \propto \sqrt{K} \).
4. \( \frac{r'}{r} = \sqrt{\frac{K/2}{K}} = \frac{1}{\sqrt2} \), so \( r' = \frac{r}{\sqrt2} \); the radius decreases.
5. (2) \( T = \frac{2\pi m}{qB} \) does not depend on kinetic energy, so the time period will not change.

Teacher's Note:
a) Draw the coil with the forces on each arm; the diagram and the couple idea carry marks.
b) Halving K reduces r by a factor \( \sqrt2 \), not by half: a common mistake.
c) The time period in a magnetic field is independent of speed and kinetic energy.

Please click the link below to download pdf file of CBSE Class 12 Physics Question Paper 2025 Solved Code 55-1-3