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

SECTION A

1. Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), r being the separation between the two charges (q₁, q₂) and (q₂, q₃). If q₂ is positive and least in magnitude, then the magnitudes of q₁, q₂ and q₃ are such that [1 Mark]
(A) \( q_2 \lt q_3 \lt q_1 \)
(B) \( q_3 \lt q_1 \lt q_2 \)
(C) \( q_1 \lt q_2 \lt q_3 \)
(D) \( q_2 \lt q_1 \lt q_3 \)

[Figure: Graph of Coulomb force F (y-axis) versus \( \frac{1}{r^2} \) (x-axis) showing two straight lines through the origin, one for the pair (q1, q2) with a steeper slope and one for the pair (q2, q3) with a smaller slope.]

Answer: (A) \( q_2 \lt q_3 \lt q_1 \)

Teacher's Note:
a) The slope of the F versus \( \frac{1}{r^2} \) line equals \( k \) times the product of the two charges involved.
b) Since q2 is smallest and positive, compare the slopes of the two lines to arrange q1, q2, q3 in increasing order.

2. Two wires P and Q are made of the same material. The wire Q has twice the diameter and half the length as that of wire P. If the resistance of wire P is R, the resistance of wire Q will be [1 Mark]
(A) R
(B) \( \frac{R}{2} \)
(C) \( \frac{R}{8} \)
(D) 2R

Answer: (C) \( \frac{R}{8} \)

Teacher's Note:
a) Resistance \( R = \frac{\rho l}{A} \), and area \( A \propto d^2 \).
b) Doubling the diameter makes A four times larger, and halving the length divides R further by 2, giving \( \frac{R}{8} \).

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. A coil has 100 turns, each of area 0.05 m² and total resistance 1.5 Ω. It is inserted at an instant in a magnetic field of 90 mT, with its axis parallel to the field. The charge induced in the coil at that instant is : [1 Mark]
(A) 3.0 mC
(B) 0.30 C
(C) 0.45 C
(D) 1.5 C

Answer: (B) 0.30 C

Teacher's Note:
a) Induced charge \( q = \frac{N\Delta\phi}{R} = \frac{NBA}{R} \) since flux changes from zero to BA.
b) Substituting values gives \( q = \frac{100 \times 0.09 \times 0.05}{1.5} = 0.30 \) C.

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 \( L = \mu_0 n^2 A l \), where \( n = \frac{N}{l} \), so \( N^2 = \frac{Ll}{\mu_0 A} \).
b) Substituting 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 the \( \frac{i_0v_0}{2}\cos\phi \) formula.

7. The given diagram exhibits the relationship between the wavelength of the electromagnetic waves and the energy of photon associated with them. The three points P, Q and R marked on the diagram may correspond respectively to : [1 Mark]
(A) X-rays, microwaves, UV radiation
(B) X-rays, UV radiation, microwaves
(C) UV radiation, microwaves, X-rays
(D) Microwaves, UV radiation, X-rays

[Figure: A curve of Energy (y-axis) versus Wavelength λ (x-axis) decreasing steeply; point P lies high on the curve at small wavelength, point R lies in the middle, and point Q lies low on the curve at large wavelength.]

Answer: (A) X-rays, microwaves, UV radiation

Teacher's Note:
a) Photon energy \( E = \frac{hc}{\lambda} \) is inversely proportional to wavelength.
b) P (short wavelength, high energy) fits X-rays, Q (long wavelength, low energy) fits microwaves, and R (in between) fits UV radiation.

8. A beaker is filled with water (refractive index \( \frac{4}{3} \)) upto a height H. A coin is placed at its bottom. The depth of the coin, when viewed along the near normal direction, will be [1 Mark]
(A) \( \frac{H}{4} \)
(B) \( \frac{3H}{4} \)
(C) H
(D) \( \frac{4H}{3} \)

Answer: (B) \( \frac{3H}{4} \)

Teacher's Note:
a) Apparent depth \( = \frac{\text{real depth}}{n} \).
b) Substituting \( n = \frac{4}{3} \) gives apparent depth \( = \frac{3H}{4} \).

9. The stopping potential \( V_0 \) measured in a photoelectric experiment for a metal surface is plotted against frequency ν of the incident radiation. Let m be the slope of the straight line so obtained. Then the value of charge of an electron is given by (h is the Planck's constant.) [1 Mark]
(A) mh
(B) \( \frac{m}{h} \)
(C) \( \frac{h}{m} \)
(D) \( \frac{1}{mh} \)

Answer: (C) \( \frac{h}{m} \)

Teacher's Note:
a) From \( eV_0 = h\nu - \phi \), \( V_0 = \frac{h}{e}\nu - \frac{\phi}{e} \), so slope \( m = \frac{h}{e} \).
b) Rearranging gives \( e = \frac{h}{m} \).

10. 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) Since mass order is electron less than proton less than deuteron, wavelength order is reversed.

11. Which of the following figures correctly represent the shape of curve of binding energy per nucleon as a function of mass number ? [1 Mark]
(A) option graph
(B) option graph
(C) option graph
(D) option graph

[Figure: Four graphs of Binding Energy per nucleon (B.E./A) versus mass number A. (A) rises steeply then gently curves down after a peak near A = 56. (B) rises steeply and stays almost flat after A = 56. (C) is a symmetric hump peaking near A = 80. (D) rises then falls sharply after peaking near A = 80.]

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

Teacher's Note:
a) The actual binding energy curve peaks near A = 56 (iron region) and falls off slowly for heavier nuclei.
b) A sharp symmetric hump or a peak near A = 80 does not match the real curve.

12. 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 field, reducing the barrier potential.
b) A reduced barrier means fewer uncompensated ions are needed, so the depletion width shrinks too.

13. 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 follows directly from Lenz's law, which states induced current opposes the change causing it.
b) More turns mean a larger opposing induced emf, making it harder to push the magnet in.

14. 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) A radial field keeps the plane of the coil always parallel to B, making torque directly proportional to current.
b) This linear relationship is what makes the galvanometer scale uniform.

15. 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) Physical joining cannot give the required continuous crystal structure at the junction, so Assertion is correct.
b) The Reason statement is wrong because in a p-type semiconductor holes (not electrons) are the majority carriers, and vice versa in n-type.

16. 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's potential energy in a hydrogen atom is negative because the electron-proton force is attractive.
b) The total energy of a charged particle can be positive, negative or zero, so the Reason statement is also false.

SECTION B

17. A battery of emf E and internal resistance r is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of E and r. [2 Marks]

Answer:
1. Using \( E = V + Ir \): for the first case, \( E = 5 + 2r \); for the second case, \( E = 4 + 4r \).
2. Solving these two equations simultaneously gives \( E = 6 \) V and \( r = 0.5 \) Ω.

Teacher's Note:
a) Always write \( E = V + Ir \) for each case separately before solving.
b) Subtracting the two equations quickly eliminates E to find r first.

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 the first minimum: \( a\sin\theta = n\lambda \), with n = 1.
2. \( a \sin 30^{\circ} = 600 \times 10^{-9} \), so \( a \times \frac{1}{2} = 600 \times 10^{-9} \), giving \( a = 1.2 \times 10^{-6} \) m.

Teacher's Note:
a) Remember the diffraction minima condition \( a\sin\theta = n\lambda \), not the interference maxima condition.
b) Substitute \( \sin30^{\circ} = 0.5 \) carefully to avoid calculation errors.

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 \( \Delta\phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{\lambda}\times\frac{\lambda}{8} = \frac{\pi}{4} \).
2. Using \( I = 4I_0\cos^2\left(\frac{\Delta\phi}{2}\right) = 4I_0\cos^2\left(\frac{\pi}{8}\right) \), which gives \( I \approx 3.414\, I_0 \).

Teacher's Note:
a) Convert path difference to phase difference before applying the intensity formula.
b) Use \( I = 4I_0\cos^2(\Delta\phi/2) \) for two equal-intensity coherent waves.

19. A transparent solid cylindrical rod (refractive index \( \frac{2}{\sqrt3} \)) is kept in air. A ray of light incident on its face travels along the surface of the rod, as shown in figure. Calculate the angle θ. [2 Marks]

[Figure: A cylindrical rod with a ray incident at angle θ on its flat end face, refracting inside and travelling along the curved surface of the rod, indicating grazing (critical angle) refraction at the surface.]

Answer:
1. For the ray to travel along the surface, the critical angle condition applies: \( \sin\theta_c = \frac{1}{n} = \frac{\sqrt3}{2} \), giving \( \theta_c = 60^{\circ} \), so \( r = 90^{\circ} - 60^{\circ} = 30^{\circ} \).
2. Applying Snell's law at the entry face, \( \sin\theta = n\sin r = \frac{2}{\sqrt3}\sin30^{\circ} = \frac{1}{\sqrt3} \), so \( \theta = \sin^{-1}\left(\frac{1}{\sqrt3}\right) \).

Teacher's Note:
a) Identify that grazing along the surface means the refraction angle at the curved surface equals the critical angle.
b) Use geometry to relate the refraction angle at the flat face to the critical angle at the curved surface.

20. Prove that, in Bohr model of hydrogen atom, the time period of revolution of an electron in n^th orbit is proportional to n³. [2 Marks]

Answer:
1. Time period \( T = \frac{2\pi r}{v} \). From Bohr's quantisation, \( v = \frac{nh}{2\pi mr} \).
2. Substituting v and using \( r \propto n^2 \) (from Bohr's radius formula) and \( v \propto \frac{1}{n} \), we get \( T = \frac{2\pi r}{v} \propto \frac{n^2}{1/n} = n^3 \), proving \( T \propto n^3 \).

Teacher's Note:
a) Recall the two key Bohr results: \( r \propto n^2 \) and \( v \propto \frac{1}{n} \).
b) Combine these directly in \( T = \frac{2\pi r}{v} \) instead of deriving r and v from scratch again.

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 \( = \frac{5\times10^{28}}{5\times10^7} = 10^{21} \), so per cm³ it is \( \frac{10^{21}}{10^6} = 10^{15} \).
2. One example of such a trivalent dopant is aluminium, indium or gallium.

Teacher's Note:
a) Divide the density of Si atoms by the doping ratio to get holes per unit volume.
b) Remember to convert from per m³ to per cm³ by dividing by 10⁶.

SECTION C

22. (a) 3V and 6V two batteries, whose internal resistances are 0.2 Ω and 0.4 Ω respectively, 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 \( E_{eq} = \frac{E_1r_2+E_2r_1}{r_1+r_2} = \frac{3(0.4)+6(0.2)}{0.6} = 4 \) V.
2. Equivalent internal resistance \( r_{eq} = \frac{r_1r_2}{r_1+r_2} = \frac{0.2\times0.4}{0.6} = 0.133 \) Ω.
3. Current drawn \( I = \frac{E_{eq}}{R+r_{eq}} = \frac{4}{4.133} \approx 0.9 \) A.

Teacher's Note:
a) Use the standard formulas for two cells in parallel to find equivalent emf and internal resistance.
b) Treat the combination as a single cell of \( E_{eq} \) and \( r_{eq} \) in series with the external resistor.

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. Since volume is constant, stretching to \( 2l \) halves the area, so \( R' = \frac{\rho(2l)}{A/2} = 4R \).
2. Drift velocity \( v_d = \frac{eE\tau}{m} = \frac{eV\tau}{ml} \), so \( v_d' = \frac{eV\tau}{m(2l)} = \frac{v_d}{2} \).
3. No, drift velocity is only the small net average velocity superimposed on the random thermal motion of electrons; individual free electrons do not all move in the same direction.

Teacher's Note:
a) Use \( R \propto \frac{l}{A} \) and constant volume \( (Al = \text{constant}) \) to find how R changes on stretching.
b) Remember drift velocity is an average quantity; it does not mean uniform directional motion of every electron.

23. Using Biot-Savart law, derive expression for the magnetic field \( (\vec{B}) \) due to a circular current carrying loop at a point on its axis and hence at its centre. [3 Marks]

Answer:
1. By Biot-Savart law, \( dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{(x^2+R^2)} \) for a current element at distance \( r = \sqrt{x^2+R^2} \) from the axial point.
2. Only the components along the axis survive after integrating around the loop; the perpendicular components cancel by symmetry, giving \( dB_x = dB\cos\theta \) where \( \cos\theta = \frac{R}{(R^2+x^2)^{1/2}} \).
3. Integrating over the loop (circumference \( 2\pi R \)) gives \( B = \frac{\mu_0 I R^2}{2(x^2+R^2)^{3/2}} \); at the centre (x = 0), this reduces to \( B = \frac{\mu_0 I}{2R} \).

Teacher's Note:
a) Draw the loop with a current element and clearly mark the perpendicular and axial components of dB.
b) Symmetry argument (cancellation of perpendicular components) is a key step examiners look for.
c) Setting x = 0 directly gives the centre field formula, a common follow-up mistake is forgetting this substitution.

24. (a) Show that the energy required to build up the current I in a coil of inductance L is \( \frac{1}{2}LI^2 \). [3 Marks]

Answer:
1. The induced emf in the coil is \( |\varepsilon| = L\frac{dI}{dt} \), so the rate of doing work is \( \frac{dW}{dt} = |\varepsilon|I \).
2. Total work done in establishing current I is \( W = \int_0^I LI\,dI \), which gives \( W = \frac{1}{2}LI^2 \), the energy stored in the inductor.

Teacher's Note:
a) Start from the induced emf expression before integrating for work done.
b) Remember the limits of integration are from 0 to I, the final steady current.

(b) Considering the case of magnetic field produced by air-filled current carrying solenoid, show that the magnetic energy density of a magnetic field B is \( \frac{B^2}{2\mu_0} \). [3 Marks]

Answer:
1. Using \( U_B = \frac{1}{2}LI^2 \) and \( L = \mu_0n^2Al \), with \( B = \mu_0nI \) so \( I = \frac{B}{\mu_0n} \).
2. Substituting gives \( U_B = \frac{1}{2}(\mu_0n^2Al)\left(\frac{B}{\mu_0n}\right)^2 = \frac{1}{2}\frac{B^2}{\mu_0}Al \).
3. Dividing by the volume Al of the solenoid gives energy density \( u_B = \frac{B^2}{2\mu_0} \).

Teacher's Note:
a) Express both L and B in terms of the solenoid's turns per unit length n before substituting.
b) Dividing total energy by volume (Al) is the key step to reach energy density.

25. (a) A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current \( (I_c) \) and displacement current \( (I_d) \) has the same value at all points of the circuit. [3 Marks]

Answer:
1. Outside the capacitor, \( I_d = 0 \), so total current \( I = I_c \); inside the capacitor (between the plates), \( I_c = 0 \), so \( I = I_d \).
2. Inside, \( I_d = \varepsilon_0\frac{d\phi_E}{dt} = \varepsilon_0\frac{d}{dt}(EA) = \frac{d}{dt}\left(\frac{Q}{A}\times A\right) = \frac{dQ}{dt} = I_c \).
3. Since both expressions equal \( \frac{dQ}{dt} \), the sum \( I_c + I_d \) is the same at every point of the circuit.

Teacher's Note:
a) Consider the two regions separately: outside the capacitor only conduction current flows, inside only displacement current flows.
b) Show both reduce to \( \frac{dQ}{dt} \) to prove continuity of total current.

(b) In case (a) above, is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor ? Explain. [Marks included above]

Answer: Yes, because the conduction current entering a plate exactly equals the displacement current leaving between the plates \( (I_c = I_d) \), so the total current in equals the total current out, validating Kirchhoff's junction rule at each plate.

Teacher's Note:
a) Displacement current is what "completes" the circuit inside the capacitor gap.
b) This idea was Maxwell's key contribution to extend Ampere's law consistently.

26. Answer the following giving reason :
(a) All the photo electrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensities.
(c) If one goes on increasing the wavelength of light incident on a metal surface, keeping its intensity constant, emission of photoelectrons stop at a certain wavelength on that metal surface. [3 Marks]

Answer:
1. Electrons in a metal are bound with varying strengths, so more tightly bound electrons emerge with less kinetic energy and less tightly bound ones with more, giving a spread in kinetic energies.
2. The number of photoelectrons ejected per second (saturation current) is directly proportional to the intensity of incident radiation, so different intensities give different saturation currents.
3. As wavelength increases, frequency and photon energy decrease; when photon energy falls below the work function (\(\lambda \gt \lambda_0\)), no photoelectrons can be emitted, so emission stops.

Teacher's Note:
a) Link kinetic energy spread to the different binding energies of electrons within the metal.
b) Remember saturation current depends on intensity, while stopping potential depends on frequency, not intensity.
c) The threshold wavelength condition \( \lambda_0 = \frac{hc}{\phi} \) explains why emission stops beyond a certain wavelength.

27. (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. \( (m_p = 1.007277 \text{ u}; m_n = 1.008665 \text{ u}, 1u = 931.5 \text{ MeV/c}^2) \) [3 Marks]

Answer:
1. Mass defect is the difference between the sum of masses of the individual nucleons and the actual mass of the nucleus, and binding energy is the energy needed to separate the nucleus into its free nucleons.
2. In fission, a heavy nucleus splits into two lighter nuclei, releasing energy since the binding energy per nucleon of the products is higher than that of the parent nucleus.
3. Mass defect \( \Delta m = (m_p+m_n) - m_d = 1.007277+1.008665-2.013553 = 0.002389 \) u; energy equivalent \( = 0.002389 \times 931.5 \approx 2.22 \) MeV.

Teacher's Note:
a) Keep the definitions of mass defect and binding energy distinct but connected.
b) In fission, always relate energy release to the increase in binding energy per nucleon.
c) Do the subtraction to at least six decimal places to avoid rounding errors in mass defect.

28. (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]

[Figure: Two circuit diagrams - one for forward bias with a milliammeter and voltmeter connected to a forward-biased p-n diode with a variable battery, and one for reverse bias with a microammeter and voltmeter connected to a reverse-biased diode.]

Answer:
1. The forward-bias circuit connects the diode's p-side to the positive terminal through a rheostat, milliammeter and voltmeter; the reverse-bias circuit reverses the diode polarity and uses a microammeter.
2. The V-I graph shows a small forward current until the knee voltage is crossed, after which current rises sharply; in reverse bias, a very small nearly constant reverse saturation current flows until breakdown.
3. From the characteristics we can find the knee (threshold) voltage and the reverse saturation current or breakdown voltage of the diode.

Teacher's Note:
a) Use different current meters (mA for forward, μA for reverse) since the current scales are very different.
b) The knee voltage and breakdown voltage are the two most commonly asked pieces of information from this graph.

SECTION D

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

29. 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: An RC series circuit with a resistor R, a capacitor C, an ideal battery of emf V, and two switches S1 and S2 arranged so that S1 connects the battery to charge the capacitor and S2 provides a separate discharge path through R alone.]

(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 (fully charged) value is simply \( Q = VC = 10 \times 500\times10^{-6} = 5\times10^{-3} \) C.
b) Remember to convert μF correctly to get charge in the right unit.

(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 disconnected, so the capacitor discharges completely through R via S2.
b) A fully discharged capacitor always ends up with zero charge, regardless of R and C values.

(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 represents a time constant, so its dimension must be that of time alone.
b) A quick check: the exponential \( e^{-t/RC} \) must be dimensionless, so RC must have the same dimension as t.

(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 \( i = i_0 e^{-t/RC} \), with \( i_0 = \frac{V}{R} = 0.5 \) mA and RC = 10 s.
b) At t = 5 s, \( t/RC = 0.5 \), so \( i = 0.5\,e^{-0.5} = \frac{1}{2\sqrt e} \) mA.

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 capacitor behaves like a plain wire, so initial current \( i_0 = \frac{V}{R} \).
b) Substituting values gives \( i_0 = \frac{10}{20000} = 0.5 \) mA.

30. 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 focus points, called "first focal point" and "second focal point" and in these one is on one side of the lens and the other is on the other side of the lens. [4 Marks]

[Figure: An optical bench with a light box (with a plate hidden on the far side of the box), a convex lens mounted on the bench, and a screen on which the image is formed.]

(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]
(A) upward arrow with dot to its upper right
(B) downward arrow with dot to its upper left
(C) downward arrow with dot on the opposite (inverted and reversed) side compared to the object
(D) downward arrow with dot to its upper right

[Figure: Four small diagrams (A)-(D), each showing a short vertical arrow together with a dot, arranged in different relative positions to represent possible orientations of the real image.]

Answer: (C) the image with the arrow inverted and the dot also inverted relative to the object

Teacher's Note:
a) A real image formed by a convex lens is always inverted both vertically and laterally.
b) Check that both the arrow and the dot flip position relative to each other, not just the arrow.

(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, diminished image, so its magnification is always positive, not negative.
b) Remember: positive magnification means erect image, negative means inverted image.

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

Answer: (B) 2f

Teacher's Note:
a) Cutting perpendicular to the principal axis reduces the curvature contribution of each surface, effectively halving the lens's power.
b) Halved power means doubled focal length for each part.

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 (image distance from the lens).
b) Using \( \frac{1}{f} = \frac{1}{v}-\frac{1}{u} = \frac{1}{30}+\frac{1}{20} = \frac{1}{12} \), 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 the lens's 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) \( \sqrt{X_1X_2} \)
(D) \( \sqrt{\frac{X_2}{X_1}} \)

Answer: (C) \( \sqrt{X_1X_2} \)

Teacher's Note:
a) This is Newton's lens formula, relating object and image distances measured from the respective focal points.
b) Remember the formula as \( f = \sqrt{X_1X_2} \), a useful shortcut different from the usual lens formula.

SECTION E

31. (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 \( A = 3\times10^5 \) 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. Since \( \vec{E} = \frac{A}{r^2}\hat r \), the potential is \( V(r) = \frac{A}{r} \). The change in electrostatic energy is \( \Delta U = q_1V(r_1)+q_2V(r_2) \).
2. Substituting \( r_1=r_2=0.03 \) m, \( \Delta U = \frac{3\times10^5}{0.03}(5\times10^{-6}-1\times10^{-6}) = 40 \) J.
3. Capacitance of the system \( C = \frac{Q}{V} = \frac{80\,\mu C}{16\,V} = 5\, \mu F \).
4. With dielectric constant K = 3, \( C' = KC = 15\, \mu F \), so new potential difference \( V' = \frac{Q}{C'} = \frac{80\,\mu C}{15\,\mu F} = 5.33 \) V.
5. No, the capacitance will not change on changing the charge, because capacitance depends only on the geometry (size, shape, separation) of the conductors, not on the charge or potential.

Teacher's Note:
a) Find the potential function by integrating \( \vec E \) before calculating the energy change.
b) Remember capacitance is a purely geometric property; changing Q and V together keeps C fixed.
c) Introducing a dielectric always increases capacitance by a factor K for the same charge.

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. These spherical shells are given charges respectively +6q, -4q and 14q. Their inner balls are also given charges respectively -2q, +8q and -10q. Compare the electric fields due to these 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 along the circumference of this semicircle from point C to point A. Calculate the work done on the charge. [5 Marks]

[Figure: A horizontal line with points D, C, B, A marked from left to right; a semicircle of radius 5 cm is drawn above the line with centre B, on which points C and A lie at the two ends of the diameter through B; a charge +6q is at D and -6q is at B.]

Answer:
1. Total enclosed charge (shell + inner ball) is +4q for A, +4q for B, and +4q for C, all equal, so at 3R from the centre, \( E_A = E_B = E_C \) since field outside depends only on total enclosed charge.
2. \( V_C = k\left(\frac{-6\times10^{-6}}{0.05}\right) + k\left(\frac{6\times10^{-6}}{0.05}\right) = 0 \), since C is equidistant (5 cm) from both B and D.
3. \( V_A = k\left(\frac{-6\times10^{-6}}{0.05}\right) + k\left(\frac{6\times10^{-6}}{0.15}\right) = -7.2\times10^5 \) V (using distance BA = 5 cm and DA = 15 cm).
4. Work done \( W = q(V_A - V_C) = 5\times10^{-6}\times(-7.2\times10^5 - 0) = -3.6 \) J.

Teacher's Note:
a) For a spherical shell, the field outside depends only on the total charge enclosed, so add shell and inner ball charges first.
b) Calculate potentials at start and end points carefully using distances from both charges before finding work done.
c) Work done by an external agent moving a charge equals \( q\Delta V \), regardless of the path taken.

32. (a) (i) A proton moving with velocity \( \vec{V} \) in a non-uniform magnetic field traces a path as shown in 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 curved path traced by a positively charged proton moving with initial velocity V, curving progressively more sharply from point P through Q to R, all lying in the plane of the paper.]

Answer:
1. Near P and Q, the magnetic force needed to curve the path upward means the field points into the plane of the paper; near R, since the force is downward, the field points out of the plane of the paper.
2. Since the radius of curvature is smaller where the field is stronger \( \left(B \propto \frac{1}{r}\right) \), the field is smallest near P, larger near R, giving \( B_Q \lt B_P \lt B_R \).
3. Using \( B = \frac{\mu_0I}{2r} \), we get \( I = \frac{2Br}{\mu_0} \); with \( A = \pi r^2 \), so \( r = \sqrt{A/\pi} \).
4. Magnetic moment \( M = IA = \frac{2Br}{\mu_0}A = \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} \), as required.

Teacher's Note:
a) Use the right-hand rule at each point separately, based on the direction the path is curving.
b) Smaller radius of curvature corresponds to a stronger local magnetic field.
c) Express r in terms of A before substituting into M = IA to reach the required form.

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 loses 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. For a rectangular loop of sides a and b carrying current I in field B, the forces on the two arms of length b are equal and opposite, each of magnitude \( F = IbB \), forming a couple.
2. The net torque is \( \tau = F_1\frac{a}{2}\sin\theta+F_2\frac{a}{2}\sin\theta = IabB\sin\theta = IAB\sin\theta \), so in vector form \( \vec\tau = I\vec A\times\vec B \).
3. Since \( r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} \), and \( r\propto\sqrt{K} \), when K becomes K/2, the new radius \( r' = \frac{r}{\sqrt2} \), so the radius decreases by a factor of \( \sqrt2 \).
4. Time period \( T = \frac{2\pi m}{qB} \) does not depend on kinetic energy, so the time period remains unchanged.

Teacher's Note:
a) Identify the pair of forces on the loop's arms that form the couple producing torque.
b) Radius depends on \( \sqrt{K} \), so halving K changes the radius by a factor of \( \frac{1}{\sqrt2} \), not by half.
c) Time period of circular motion in a magnetic field is independent of speed or kinetic energy - a key exam point.

33. (a) (i) (1) What are coherent sources ? Why are they necessary for observing a sustained interference pattern ?
(2) Two independent sources of light are not coherent. Explain.
(ii) In a Young's double slit experiment two slits are arranged 0.1 mm apart from each other, 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 the nearby bright interference fringes be on the screen ?
(2) Find the angular width of the first bright fringe (in degrees). [5 Marks]

Answer:
1. Coherent sources are those that emit waves of the same frequency with a constant (or zero) phase difference; they are necessary because only then does the phase difference at any point stay fixed with time, giving a stable, sustained interference pattern.
2. Two independent sources are never coherent, as the phase difference between the light emitted by their independently excited atoms keeps changing randomly and rapidly with time.
3. Fringe width \( \beta = \frac{\lambda D}{d} = \frac{600\times10^{-9}\times1.2}{0.1\times10^{-3}} = 7.2 \) mm.
4. Angular width \( \theta = \frac{\lambda}{d} = \frac{600\times10^{-9}}{0.1\times10^{-3}} = 6\times10^{-3} \) rad \( \approx 0.34^{\circ} \).

Teacher's Note:
a) A sustained pattern requires the phase difference to stay constant with time, which only coherent sources give.
b) Always use \( \beta = \frac{\lambda D}{d} \) for fringe spacing and \( \theta = \frac{\lambda}{d} \) for angular fringe width.
c) Convert the final angle from radians to degrees if the question specifically asks for degrees.

OR

(b) (i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted from it. Draw the incident and refracted wavefronts to show this.
(ii) A light beam 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]

[Figure: A plane wavefront (shown as parallel lines with arrows) incident on a convex lens on the left, and a converging spherical wavefront (shown as curved arcs converging to a focal point F) emerging on the right after refraction.]

Answer:
1. A wavefront is the locus of all points in a medium that oscillate in the same phase; a plane wavefront incident on a convex lens emerges as a converging spherical wavefront.
2. For refraction at the first (near) surface of the glass ball: \( n_1=1, n_2=1.5, R=15 \) cm, \( u=\infty \); using \( \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \), we get \( v = 45 \) cm (measured from the first surface).
3. For refraction at the second (far) surface, this image acts as an object at u = +15 cm (since it lies 15 cm beyond that surface), with \( n_1=1.5, n_2=1, R=-15 \) cm; solving \( \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \) gives \( v = 7.5 \) cm from the second surface, which is the position of the final image.

Teacher's Note:
a) A wavefront is always perpendicular to the direction of propagation (rays) at every point.
b) Apply the single spherical surface refraction formula twice, once at each surface of the ball, carrying the image from the first surface as the object for the second.
c) Be careful with sign conventions for R, since it changes sign for the second (diverging away) surface.

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