CBSE Class 12 Physics Question Paper 2026 Solved Code 55-1-1

SECTION A

1. In a region, the electric potential varies as V = 10 - 50x, where V is in volts and x in meters. The electric field in the region is : [1 Mark]
(A) 10 N/C along +x
(B) 10 N/C along -x
(C) 50 N/C along +x
(D) 50 N/C along -x

Answer: (C) 50 N/C along +x

Teacher's Note:
a) Use \( E = -\frac{dV}{dx} \); here \( \frac{dV}{dx} = -50 \), so \( E = +50 \) N/C.
b) The positive sign shows the field points along +x, the direction in which potential decreases.

2. A conducting wire connects two charged metallic spheres A and B of radii r₁ and r₂ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, (E_A/E_B) at the surfaces of spheres A and B will be [1 Mark]
(A) \( \frac{r_1}{r_2} \)
(B) \( \frac{r_2}{r_1} \)
(C) \( \frac{r_1^2}{r_2^2} \)
(D) \( \frac{r_2^2}{r_1^2} \)

Answer: (B) \( \frac{r_2}{r_1} \)

Teacher's Note:
a) Connected spheres reach the same potential, so \( \frac{q_1}{r_1} = \frac{q_2}{r_2} \).
b) Surface field \( E = \frac{kq}{r^2} = \frac{V}{r} \), so the field is inversely proportional to the radius.

3. A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance (a/2) from the axis of the wire will be [1 Mark]
(A) Zero
(B) \( \frac{\mu_0 I}{2\pi a} \)
(C) \( \frac{\mu_0 I}{4\pi a} \)
(D) \( \frac{\mu_0 I}{6\pi a} \)

Answer: (C) \( \frac{\mu_0 I}{4\pi a} \)

Teacher's Note:
a) Inside the wire, Ampere's law gives \( B = \frac{\mu_0 I r}{2\pi a^2} \), since only the current within radius r is enclosed.
b) Putting \( r = \frac{a}{2} \) gives \( B = \frac{\mu_0 I}{4\pi a} \).

4. The shape of the interference fringes in Young's double-slit experiment, when the distance between the slit and the screen is very large as compared to the slit-separation, is nearly [1 Mark]
(A) straight
(B) parabolic
(C) circular
(D) hyperbolic

Answer: (A) straight

Teacher's Note:
a) Strictly the fringes are hyperbolic, but when D is much greater than d they look like straight lines.
b) Read the condition in the question carefully; it decides between "straight" and "hyperbolic".

5. An electromagnetic wave passes from vacuum into a dielectric medium with relative electrical permittivity (3/2) and relative magnetic permeability (8/3). Then, its [1 Mark]
(A) wavelength is doubled and frequency remains unchanged.
(B) wavelength is doubled and frequency is halved.
(C) wavelength is halved and frequency remains unchanged.
(D) wavelength and frequency both will remain unchanged.

Answer: (C) wavelength is halved and frequency remains unchanged.

Teacher's Note:
a) Refractive index \( n = \sqrt{\mu_r \varepsilon_r} = \sqrt{\frac{8}{3} \times \frac{3}{2}} = 2 \).
b) Frequency never changes with medium, so speed and wavelength both become half.

6. In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short circuited, the voltage across the inductor will be [1 Mark]
(A) 10 V
(B) \( 5\sqrt{2} \) V
(C) \( \frac{5}{\sqrt{2}} \) V
(D) \( 10\sqrt{2} \) V

Answer: (B) \( 5\sqrt{2} \) V

Teacher's Note:
a) Equal voltages mean \( R = X_L = X_C \), so the circuit is at resonance and the source voltage is 10 V.
b) With C shorted, \( Z = \sqrt{2}R \), so \( V_L = \frac{10}{\sqrt{2}R} \times R = 5\sqrt{2} \) V.

7. Electromagnetic waves used in a diagnostic tool in medicine have a wavelength range [1 Mark]
(A) 1 nm to 10^-3 nm
(B) 400 nm to 1 nm
(C) 1 mm to 700 nm
(D) 0.1 m to 1 mm

Answer: (A) 1 nm to 10^-3 nm

Teacher's Note:
a) X-rays are used as a diagnostic tool in medicine.
b) Their wavelength range is about 1 nm to 10^-3 nm; do not confuse it with UV (400 nm to 1 nm).

8. The 'distance of closest approach' of an alpha-particle is 'd' when it moves with a velocity v head-on towards the target nucleus. If the velocity of alpha particle is halved, the new 'distance of closest approach' will be - [1 Mark]
(A) \( \frac{d}{2} \)
(B) 2d
(C) \( \frac{d}{4} \)
(D) 4d

Answer: (D) 4d

Teacher's Note:
a) At closest approach, \( \frac{1}{2}mv^2 = \frac{k(2e)(Ze)}{d} \), so \( d \propto \frac{1}{v^2} \).
b) Halving v makes \( v^2 \) one-fourth, so d becomes 4 times.

9. A concave lens of focal length 10 cm is cut into two identical plano-concave lenses. The focal length of each lens will be [1 Mark]
(A) 20 cm
(B) 30 cm
(C) 40 cm
(D) 5 cm

Answer: (A) 20 cm

Teacher's Note:
a) The original lens is two identical plano-concave lenses in contact, so \( \frac{1}{10} = \frac{1}{f} + \frac{1}{f} \).
b) This gives f = 20 cm in magnitude for each half (still a diverging lens).

10. Four independent waves are expressed as
(i) \( y_1 = A_1 \sin \omega t \)
(ii) \( y_2 = A_2 \sin 2\omega t \)
(iii) \( y_3 = A_3 \cos \omega t \)
(iv) \( y_4 = A_4 \sin (\omega t + \pi/3) \)
The interference between two of these waves is possible in [1 Mark]
(A) (i) and (iii) only
(B) (iii) and (iv) only
(C) (i), (iii) and (iv) only
(D) All of them

Answer: (C) (i), (iii) and (iv) only

Teacher's Note:
a) Interference needs waves of the same frequency with a constant phase difference.
b) Waves (i), (iii) and (iv) all have angular frequency \( \omega \); wave (ii) has \( 2\omega \), so it is left out.

11. Two heaters rated as (P₁, V) and (P₂, V) are connected in series across a dc source of \( \frac{V}{2} \) volt. The power consumed by the combination will be - [1 Mark]
(A) (P₁ + P₂)
(B) \( \frac{P_1 + P_2}{2} \)
(C) \( \frac{P_1 P_2}{2(P_1 + P_2)} \)
(D) \( \frac{P_1 P_2}{4(P_1 + P_2)} \)

Answer: (D) \( \frac{P_1 P_2}{4(P_1 + P_2)} \)

Teacher's Note:
a) Resistances are \( R_1 = \frac{V^2}{P_1} \) and \( R_2 = \frac{V^2}{P_2} \); in series they add.
b) Power \( P = \frac{(V/2)^2}{R_1 + R_2} = \frac{P_1 P_2}{4(P_1 + P_2)} \); do not forget the factor 4 from the halved voltage.

12. In an unbiased p-n junction, at equilibrium, which of the following statements is true ? [1 Mark]
(A) Diffusion current is zero but drift current exists.
(B) Diffusion current exists but drift current is zero.
(C) Diffusion and drift currents are equal and opposite.
(D) Both the diffusion and drift currents exist but are unequal.

Answer: (C) Diffusion and drift currents are equal and opposite.

Teacher's Note:
a) At equilibrium both currents flow, but they cancel, so the net current is zero.
b) Saying "both are zero" is a common mistake; the currents exist but balance each other.

For question number 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the options (A), (B), (C) and (D) as given below :

13. Assertion (A) : All atoms have a net magnetic moment.
Reason (R) : A current loop does not always behave as a magnetic dipole. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.

Answer: (D) Both Assertion (A) and Reason (R) are false.

Teacher's Note:
a) Atoms of diamagnetic substances have zero net magnetic moment, so the Assertion is false.
b) A current loop always behaves as a magnetic dipole, so the Reason is also false.

14. Assertion (A) : If accelerated electrons are passed through a narrow slit, a diffraction pattern is observed.
Reason (R) : Electrons behave as both particles and waves. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.

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

Teacher's Note:
a) Diffraction is a wave property, and electrons show it because of their wave nature.
b) The dual (particle and wave) nature of electrons directly explains the diffraction pattern.

15. Reason (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons.
Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.

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

Teacher's Note:
a) The first statement (labelled "Reason (A)" in the paper) is the Assertion; the mass defect makes it true.
b) Energy is released, not absorbed, when nucleons bind to form a nucleus, so the Reason is false.

16. Reason (A) : In Bohr model of hydrogen atom, the energy levels are discrete and quantised.
Reason (R) : In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) Both statements are correct facts of the Bohr model.
b) Quantised energy levels come from the quantisation of angular momentum, not from the electrostatic force being centripetal.

SECTION B

17. In a photoelectric experiment, the emitter plate is irradiated with radiation of 200 nm. The photocurrent becomes zero when the collector plate potential is - 0.80 V. Calculate the work function (in eV) of the emitter. [2 Marks]

Answer:
1. Einstein's equation: \( h\nu = \phi_0 + K_{max} \), so \( \frac{hc}{\lambda} = \phi_0 + eV_0 \) and \( \phi_0 = \frac{hc}{\lambda} - eV_0 \).
2. \( \phi_0 = \left[\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 200 \times 10^{-9}} - 0.80\right] \) eV \( = (6.2 - 0.80) \) eV.
3. Work function \( \phi_0 = 5.4 \) eV.

Teacher's Note:
a) Divide \( \frac{hc}{\lambda} \) by e to get the photon energy directly in eV.
b) The stopping potential of 0.80 V means \( K_{max} = 0.80 \) eV; use its magnitude only.

18. (a) A beam of light consisting of two wavelengths 400 nm and 600 nm is used to illuminate a single slit of width 1 mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed 1.5 m from the slit. [2 Marks]

Answer:
1. At the point where dark fringes coincide: \( d\sin\theta = n\lambda_2 = (n+1)\lambda_1 \), so \( n \times 600 = (n+1) \times 400 \), giving n = 2.
2. Position of this dark fringe from the central maximum: \( y = \frac{(n+1)\lambda_1 D}{d} = \frac{3 \times 400 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} \).
3. y = 1.8 mm (the same result comes from \( y = \frac{n\lambda_2 D}{d} = \frac{2 \times 600 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} \)).

Teacher's Note:
a) Find the smallest n for which the two minima fall at the same place before calculating y.
b) Convert nm to m and mm to m before substituting to avoid power-of-ten errors.

OR

(b) In a Young's double-slit experimental set-up with slit separation 0.6 mm a beam of light consisting of two wavelengths 440 nm and 660 nm is used to obtain interference pattern on a screen kept 1.5 m in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide. [2 Marks]

Answer:
1. At the point where bright fringes coincide: \( \frac{n\lambda_2 D}{d} = \frac{(n+1)\lambda_1 D}{d} \), so \( n \times 660 = (n+1) \times 440 \), giving n = 2.
2. Position of this bright fringe: \( y = \frac{n\lambda_2 D}{d} = \frac{2 \times 660 \times 10^{-9} \times 1.5}{0.6 \times 10^{-3}} \).
3. y = 3.3 mm from the central maximum.

Teacher's Note:
a) The longer wavelength has the lower order at the point of coincidence.
b) You can check the answer with \( y = \frac{3 \times 440 \times 10^{-9} \times 1.5}{0.6 \times 10^{-3}} \), which also gives 3.3 mm.

19. A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil. [2 Marks]

Answer:
1. Side of the square \( a = \frac{L}{4N} \) and radius of the circular coil \( r = \frac{L}{2\pi N} \).
2. Maximum torque \( \tau = NIBA \), so with the same N, I and B, \( \frac{\tau_1}{\tau_2} = \frac{A_1}{A_2} \).
3. \( \frac{\tau_1}{\tau_2} = \frac{(L/4N)^2}{\pi (L/2\pi N)^2} = \frac{\pi}{4} \).

Teacher's Note:
a) The same wire length is shared by N turns, so each turn has perimeter \( \frac{L}{N} \).
b) Since N, I and B are the same, the torque ratio is just the ratio of areas.

20. What is the order of magnitude of drift velocity of electrons in a conductor ? Deduce the relation between the current flowing through a conductor and drift velocity of electrons in it. [2 Marks]

Answer:
1. The drift velocity of electrons in a conductor is of the order of a few mm/s.
2. Consider a conductor of cross-section area A with n free electrons per unit volume. In time \( \Delta t \), electrons move a distance \( \Delta x = v_d \Delta t \), so the charge crossing the area is \( Q = -neAv_d\Delta t \).
3. Electrons move opposite to the electric field, so the magnitude of charge crossing is \( I\Delta t = neAv_d\Delta t \).
4. Hence \( I = neAv_d \).

Teacher's Note:
a) Define n, A and \( v_d \) clearly before writing the charge crossing in time \( \Delta t \).
b) Half a mark is for the order of magnitude, so do not skip "a few mm/s".

21. Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot. [2 Marks]

Answer:
1. Graph: potential energy (MeV) on the y-axis and separation r (fm) on the x-axis. The curve falls steeply from high positive values at very small r, reaches a minimum (negative) at \( r_0 \approx 0.8 \) fm, and then rises gradually towards zero at larger r.
2. Conclusion (i): The potential energy is minimum at a distance \( r_0 \) of about 0.8 fm.
3. Conclusion (ii): The nuclear force is attractive for distances larger than 0.8 fm and repulsive for distances smaller than 0.8 fm.

Teacher's Note:
a) Label the axes (potential energy in MeV, r in fm) and mark \( r_0 \) at the minimum of the curve.
b) Any two conclusions are enough: minimum at about 0.8 fm, attractive beyond it, repulsive below it.

SECTION C

22. (a) Using Gauss's law, deduce an experession for electric field at a point due to a uniformly charged infinite plane thin sheet.
(b) Two large thin plane sheets, each having surface charge density σ, are held close and parallel to each other in air. What is the net electric field at a point (i) inside and (ii) outside, the sheets ? [3 Marks]

Answer:
1. (a) Take a cylindrical Gaussian surface of cross-section area A passing through the sheet, with its two flat faces 1 and 2 parallel to the sheet on either side. The field is normal to the sheet, so only faces 1 and 2 contribute to the flux; the curved surface gives zero flux.
2. Flux through each face is EA, and both add up, so the net flux through the Gaussian surface is 2EA. The charge enclosed is σA.
3. By Gauss's law, \( 2EA = \frac{\sigma A}{\varepsilon_0} \), so \( \vec{E} = \frac{\sigma}{2\varepsilon_0}\hat{n} \), where \( \hat{n} \) is the unit vector normal to the sheet, pointing away from it.
4. (b) (i) Inside, between the sheets, the fields of the two sheets are equal and opposite, so \( E_{in} = 0 \).
5. (b) (ii) Outside the sheets, the two fields add, so \( E_{out} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \).

Teacher's Note:
a) Draw the Gaussian cylinder and state why the curved surface gives no flux; this step carries marks.
b) For two sheets with the same sign of charge, the field cancels between them and doubles outside.

OR

(a) Obtain the condition of balance of a Wheatstone bridge.
(b) Find net resistance of the network of resistors connected between A and B, as shown in figure. [3 Marks]

[Figure: A network between terminals A and B: 2R from A to junction M; from M, a resistor R to junction O and another resistor R (upper branch) to junction P; resistor R from O to P; from O, a resistor R (lower branch) to junction N; resistor R from P to N; and 3R from N to B.]

Answer:
1. (a) In the bridge ABCD, let resistances \( R_1 \) (AB), \( R_2 \) (AD), \( R_3 \) (DC) and \( R_4 \) (BC) carry currents \( I_1, I_2, I_3, I_4 \), with the galvanometer G between B and D. At balance, \( V_B = V_D \) and \( I_g = 0 \).
2. Kirchhoff's loop rule for loop ADBA gives \( -I_1R_1 + 0 + I_2R_2 = 0 \) ... (1); for loop CBDC it gives \( I_4R_4 + 0 - I_3R_3 = 0 \).
3. Since \( I_g = 0 \), \( I_1 = I_3 \) and \( I_2 = I_4 \), so the second equation becomes \( I_2R_4 - I_1R_3 = 0 \) ... (2). From (1) and (2), \( \frac{R_2}{R_1} = \frac{R_4}{R_3} \), which is the balance condition.
4. (b) Between M and N, the resistors MO, MP, ON and PN (each R) with OP (R) form a balanced Wheatstone bridge, so no current flows through OP and \( R_{MN} = \frac{2R \times 2R}{2R + 2R} = R \).
5. \( R_{AB} = R_{AM} + R_{MN} + R_{NB} = 2R + R + 3R = 6R \) Ω.

Teacher's Note:
a) Draw the bridge with the galvanometer and write both loop equations using \( I_g = 0 \).
b) In (b), spot the balanced bridge first; the middle resistor OP can then be ignored.

23. A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?
(a) Capacitance of the capacitor,
(b) Energy stored in the capacitor, and
(c) The potential difference between the plates of the capacitor.
Justify your answer in each case. [3 Marks]

Answer:
1. (a) With the slab, \( C = \frac{K\varepsilon_0 A}{d} \); without it, \( C' = \frac{\varepsilon_0 A}{d} = \frac{C}{K} \). So the capacitance decreases.
2. (b) The battery is disconnected, so the charge Q stays constant. \( U = \frac{Q^2}{2C} \) becomes \( U' = \frac{Q^2}{2C'} = KU \). So the energy stored increases.
3. (c) \( V = \frac{Q}{C} \) becomes \( V' = \frac{Q}{C'} = KV \). So the potential difference between the plates increases.

Teacher's Note:
a) First state that Q is constant because the battery is disconnected; every justification depends on it.
b) Use \( U = \frac{Q^2}{2C} \), not \( \frac{1}{2}CV^2 \), since V also changes here.

24. Figure shows a narrow beam of electrons entering with a velocity of \( 3 \times 10^7 \) m/s, symmetrically through the space between two parallel horizontal plates P₁ P₁' and P₂ P₂' kept 2 cm apart.
If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end P₂'. [3 Marks]

[Figure: Two horizontal parallel plates, upper P1P1' and lower P2P2', with potential difference V marked on the upper plate. An electron (-e, mass m) enters midway between the plates moving horizontally; a dashed straight line shows its original direction and a dashed curve shows it bending down to strike the end P2' of the lower plate.]

Answer:
1. Given: \( u_x = 3 \times 10^7 \) m/s, plate separation L = 2 cm, plate length x = 3 cm, vertical deflection y = 1 cm (half the separation).
2. Acceleration \( a = \frac{qE}{m} = \frac{qV}{mL} \); time between the plates \( t = \frac{x}{u_x} \).
3. Deflection \( y = \frac{1}{2}\left(\frac{qV}{mL}\right)\left(\frac{x}{u_x}\right)^2 \), so \( V = \frac{2ymLu_x^2}{ex^2} \).
4. \( V = \frac{2 \times 1 \times 10^{-2} \times 9.1 \times 10^{-31} \times 2 \times 10^{-2} \times (3 \times 10^7)^2}{1.6 \times 10^{-19} \times (3 \times 10^{-2})^2} \).
5. V = 2275 V.

Teacher's Note:
a) The beam enters midway, so the deflection needed is half the plate separation (1 cm).
b) Horizontal velocity stays constant; only the vertical motion is accelerated, like a projectile.

25. An ac voltage V_i = 12 sin (100 πt)V is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure.
During the positive half-cycle of the input voltage V_i supplied to the network.
(a) Identify which of the two diodes will conduct and why ?
(b) Redraw an equivalent circuit diagram to show the flow of current.
(c) Calculate the output voltage drops V₀ across the three resistors when the input voltage attains its peak value. [3 Marks]

[Figure: A diamond-shaped network with top vertex A, bottom vertex B, left vertex P and right vertex R. Diode D1 is between P and A with its arrow pointing from P towards A; diode D2 is between A and R with its arrow pointing from A towards R. A 1 kΩ resistor joins P and R, a 2 kΩ resistor joins P and B, and a 3 kΩ resistor joins R and B. The ac source Vi is connected between A and B.]

Answer:
1. (a) Diode D₂ conducts, because in the positive half-cycle A is at higher potential and D₂ (pointing from A to R) is forward biased; D₁ is reverse biased.
2. (b) Equivalent circuit: current flows from A through D₂ to R. From R it divides: one path through the 3 kΩ resistor to B, and the other through the 1 kΩ resistor to P and then through the 2 kΩ resistor to B. D₁ is left out as an open switch.
3. (c) At the peak, V_i = 12 V appears between R and B, so \( V_{RB} = 12 \) V across the 3 kΩ resistor.
4. The 1 kΩ and 2 kΩ resistors in series (3 kΩ) also have 12 V across them, so current = 4 mA, giving \( V_{RP} = 4 \) V and \( V_{PB} = 8 \) V.

Teacher's Note:
a) Check the arrow direction of each diode to decide which one is forward biased.
b) The marking scheme also accepts the case where a student takes D₁ as conducting, with \( V_{PB} = 12 \) V, \( V_{PR} = 3 \) V and \( V_{RB} = 9 \) V.

26. Briefly explain the two important processes that occur during the formation of a p-n junction. [3 Marks]

Answer:
1. The two important processes are (i) diffusion and (ii) drift.
2. Diffusion: Due to the concentration gradient across the p- and n-sides, holes diffuse from the p-side to the n-side \( (p \to n) \) and electrons diffuse from the n-side to the p-side \( (n \to p) \). This motion of charge carriers gives rise to the diffusion current across the junction.
3. Drift: Diffusion leaves a positive space charge region on the n-side and a negative space charge region on the p-side. This sets up an electric field from the positive charge towards the negative charge. Due to this field, an electron on the p-side moves to the n-side and a hole on the n-side moves to the p-side. This motion due to the field is called drift, and it constitutes the drift current.

Teacher's Note:
a) Name both processes first; each name carries half a mark.
b) Mention the cause of each: concentration gradient for diffusion, and the junction electric field for drift.

27. (a) Draw the ray diagram to show the image formation by a refracting telescope and write the expression for angular magnification for the telescope in normal adjustment.
(b) Give two reasons to explain why a reflecting telescope is preferred over a refracting telescope. [3 Marks]

Answer:
1. (a) Ray diagram: parallel rays from a distant object pass through the objective (focal length \( f_o \)) and form a real, inverted image A'B' at its focus. This image lies at the focus of the eyepiece (focal length \( f_e \)), so the rays emerge parallel and the final image is formed at infinity.
2. Angular magnification in normal adjustment: \( |m| = \frac{f_o}{f_e} \).
3. (b) Reason (i): There is no chromatic aberration in a reflecting telescope, as the objective is a mirror.
4. Reason (ii): The image formed is brighter and clear.

Teacher's Note:
a) Label the objective, eyepiece, \( f_o \), \( f_e \), the intermediate image and the final image at infinity.
b) Other accepted reasons: less mechanical support is needed, it is cost efficient, and there is no or less spherical aberration.

28. (a) State the two conditions under which total internal reflection occurs.
(b) A transparent container contains layers of three immiscible transparent liquids A, B and C of refractive indices n, \( \frac{3}{4} \)n and \( \frac{2}{3} \)n, respectively. A laser beam is incident at the interface between A and B at an angle θ as shown in figure. Prove that the beam does not enter region C at all for sin θ \( \ge \frac{2}{3} \). [3 Marks]

[Figure: A container with three horizontal liquid layers: A (refractive index n) at the bottom, B (\( \frac{3}{4} \)n) in the middle and C (\( \frac{2}{3} \)n) at the top. A laser beam travels upward through A and strikes the A-B interface at an angle θ with the normal.]

Answer:
1. (a) (i) Light must travel from a denser medium to a rarer medium. (ii) The angle of incidence must be greater than the critical angle \( (i \gt i_c) \).
2. (b) Refraction at the A-B interface: \( \frac{\sin\theta}{\sin\theta_1} = \frac{3n/4}{n} \), so \( \sin\theta = \frac{3}{4}\sin\theta_1 \) ... (1), where \( \theta_1 \) is the angle in B.
3. Refraction at the B-C interface: \( \frac{\sin\theta_1}{\sin r} = \frac{2n/3}{3n/4} \), so \( \sin\theta_1 = \frac{8}{9}\sin r \) ... (2).
4. From (1) and (2), \( \frac{3}{2}\sin\theta = \sin r \). For \( \sin\theta = \frac{2}{3} \), \( \sin r = 1 \), i.e. \( r = 90^{\circ} \) and the ray grazes the B-C surface.
5. For larger values of \( \sin\theta \), the angle in B exceeds the critical angle and the ray is totally reflected. Hence for \( \sin\theta \ge \frac{2}{3} \), the beam does not enter region C at all.

Teacher's Note:
a) Apply Snell's law at each interface separately and then combine the two equations.
b) The boundary case \( r = 90^{\circ} \) (grazing emergence) gives the limiting value \( \sin\theta = \frac{2}{3} \).

SECTION D

29. A galvanometer is used to detect or/and measure small currents in an electrical circuit. It essentially works on the fact that a current-carrying coil experiences a deflecting torque when placed in a magnetic field. This deflection in the coil can be measured and it is related to the current flowing in the coil, the number of turns in the coil, area of the coil and the magnetic field. A hair spring attached to the coil provides a counter torque and helps in measuring the deflection. A galvanometer can be converted to an ammeter or a voltmeter of desired range by using suitable resistances. [4 Marks]

(I) The torque on the coil remains constant irrespective of the coil's orientation during rotation due to [1 Mark]
(A) use of soft iron core which increases the magnetic field.
(B) radial magnetic field
(C) hair spring which provides the counter torque
(D) eddy current in the iron core which causes damping.

Answer: (B) radial magnetic field

Teacher's Note:
a) In a radial field, the plane of the coil is always parallel to the field lines.
b) So the angle factor stays at its maximum and the torque NIAB does not depend on orientation.

(II) The best way to increase current sensitivity of a galvanometer is by [1 Mark]
(A) increasing number of turns of the coil
(B) increasing area of coil and magnitic field strength
(C) decreasing area of coil and magnetic field strength
(D) increasing torsional constant of the hair spring

Answer: (A) increasing number of turns of the coil

Teacher's Note:
a) Current sensitivity \( \frac{\phi}{I} = \frac{NAB}{k} \), so it increases with N.
b) Increasing the spring's torsional constant k would reduce the sensitivity.

(III) A moving coil galvanometer has a coil with area of cross-section \( 4.0 \times 10^{-3} \) m² and number of turns 50. The coil is rotating in a magnetic field of 0.25 T. The torque acting on the coil when a current of 5 A passes through it is [1 Mark]
(A) 1.0 N m
(B) 2.0 N m
(C) 0.50 N m
(D) 0.25 N m

Answer: (D) 0.25 N m

Teacher's Note:
a) Use \( \tau = NIAB = 50 \times 5 \times 4.0 \times 10^{-3} \times 0.25 \).
b) This gives 0.25 N m; multiply step by step to avoid power-of-ten slips.

OR

A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 3 mA. The value of resistance required to convert it into a voltmeter of range (0 - 12 V) is [1 Mark]
(A) 4015 Ω
(B) 3985 Ω
(C) 415 Ω
(D) 385 Ω

Answer: (B) 3985 Ω

Teacher's Note:
a) Series resistance \( R = \frac{V}{I_g} - G = \frac{12}{3 \times 10^{-3}} - 15 \).
b) This gives 4000 - 15 = 3985 Ω; do not forget to subtract the coil resistance.

(IV) A galvanometer with coil of resistance 20 Ω shows full scale deflection for a current of 5 mA. To convert it into an ammeter of range (0 - 10 A), a resistance of [1 Mark]
(A) 0.05 Ω should be connected in series with it.
(B) 0.05 Ω should be connected in parallel with it.
(C) 0.01 Ω should be connected in parallel with it.
(D) 0.01 Ω should be connected in series with it.

Answer: (C) 0.01 Ω should be connected in parallel with it.

Teacher's Note:
a) Shunt \( S = \frac{I_gG}{I - I_g} = \frac{5 \times 10^{-3} \times 20}{10 - 0.005} \approx 0.01 \) Ω.
b) An ammeter needs a low resistance in parallel (shunt); a voltmeter needs a high resistance in series.

30. A researcher performs an experiment on photo-electric effect using two metals A and B with unknown work functions. She illuminates the surfaces of A and B with monochromatic radiation of various frequencies and records the value of corrosponding stopping potentials (V_s). The graph shows the variation of stopping potential (V_s) with the frequency of incident radiation (ν) for metals A and B. [4 Marks]

[Figure: Graph of stopping potential Vs (y-axis) versus frequency ν (x-axis) showing two parallel straight lines A and B. Line A meets the ν-axis at ν1 and line B at ν2, with ν2 greater than ν1. Dashed extensions of the lines below the ν-axis meet the Vs-axis at points marked V1 (for A) and V2 (for B, lower).]

Answer the following questions :

(I) From the graph, the work functions of A and B are (h is Planck's constant and e value of charge on an electron) [1 Mark]
(A) ν₁ and ν₂
(B) V₁ and V₂
(C) hν₁ and hν₂
(D) \( \frac{h\nu_1}{e} \) and \( \frac{h\nu_2}{e} \)

Answer: (C) hν₁ and hν₂

Teacher's Note:
a) The intercept on the ν-axis is the threshold frequency \( \nu_0 \) of each metal.
b) Work function \( \phi_0 = h\nu_0 \), so the values are \( h\nu_1 \) and \( h\nu_2 \).

(II) For radiation of frequency ν \( \gt \) ν₂ incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is [1 Mark]
(A) greater for metal A because it has a smaller work function.
(B) greater for metal B because it has a larger work function.
(C) greater for metal B because it has higher threshold frequency.
(D) the same for both metal A and metal B because it is independent of work functions of metals.

Answer: (A) greater for metal A because it has a smaller work function.

Teacher's Note:
a) \( K_{max} = h\nu - \phi_0 \); for the same ν, a smaller \( \phi_0 \) gives a larger \( K_{max} \).
b) From the graph, A has the lower threshold frequency, so it has the smaller work function.

(III) If the intensity of the incident radiation for both metals A and B, is doubled keeping its frequency constant, then [1 Mark]
(A) the slope of the parallel lines will increase.
(B) the slope of the parallel lines will decrease.
(C) the threshold frequencies for both A and B will decrease.
(D) the slope of the parallel lines will not change but more electrons will be emitted per second.

Answer: (D) the slope of the parallel lines will not change but more electrons will be emitted per second.

Teacher's Note:
a) The slope is \( \frac{h}{e} \), a constant, so intensity cannot change it.
b) Higher intensity means more photons per second, so more photoelectrons are emitted per second.

(IV) The threshold frequency for a metal surface is ν₀. If the radiation of frequency 3ν₀ illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is E₁. If the frequency were increased to 6ν₀, the maximum KE of the photoelectrons becomes E₂. Then \( \left(\frac{E_1}{E_2}\right) \) equals [1 Mark]
(A) 1/3
(B) 1/2
(C) 2/5
(D) 3/4

Answer: (C) 2/5

Teacher's Note:
a) \( E_1 = h(3\nu_0) - h\nu_0 = 2h\nu_0 \) and \( E_2 = h(6\nu_0) - h\nu_0 = 5h\nu_0 \).
b) So \( \frac{E_1}{E_2} = \frac{2}{5} \); it is not simply \( \frac{3}{6} \), because the work function must be subtracted.

OR

Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant 'h' is given by [1 Mark]
(A) me
(B) \( \frac{1}{me} \)
(C) \( \frac{m}{e} \)
(D) \( \frac{e}{m} \)

Answer: (A) me

Teacher's Note:
a) From \( eV_s = h\nu - \phi_0 \), \( V_s = \frac{h}{e}\nu - \frac{\phi_0}{e} \), so the slope \( m = \frac{h}{e} \).
b) Rearranging gives \( h = me \).

SECTION E

31. (a) An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression for the electric field \( \vec{E} \) due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. r \( \gt\gt \) a.
(b) A dipole is placed in x-y plane such that charges q and -q are located at x = a and x = b respectively. There exists an electric field \( \vec{E} = 2\hat{i}\ \frac{N}{C} \) in the region. Calculate the force \( \vec{F} \) and torque \( \vec{\tau} \) experienced by the dipole. [5 Marks]

Answer:
1. (a) Let P be a point on the equatorial plane at distance r from the centre O of the dipole. Each charge is at distance \( \sqrt{r^2 + a^2} \) from P, so the magnitudes of the fields are \( E_{+q} = E_{-q} = \frac{q}{4\pi\varepsilon_0}\frac{1}{(r^2 + a^2)} \).
2. The components of the two fields normal to the dipole axis cancel. The components along the dipole axis add up, and the total field is opposite to \( \vec{p} \): \( \vec{E} = -(E_{+q} + E_{-q})\cos\theta\ \hat{p} \), where \( \cos\theta = \frac{a}{\sqrt{r^2 + a^2}} \).
3. So \( \vec{E} = -\frac{1}{4\pi\varepsilon_0}\frac{2qa}{(r^2 + a^2)^{3/2}}\hat{p} \), where p = q(2a).
4. At a far off point (r \( \gt\gt \) a): \( \vec{E} = \frac{-2qa}{4\pi\varepsilon_0 r^3}\hat{p} = -\frac{\vec{p}}{4\pi\varepsilon_0 r^3} \).
5. (b) Net force \( \vec{F} = \vec{F}_{+q} + \vec{F}_{-q} = q(2\hat{i}) - q(2\hat{i}) = 0 \) N. The dipole moment points from -q to +q, i.e. along \( -\hat{i} \), so torque \( \vec{\tau} = \vec{p} \times \vec{E} = p(-\hat{i}) \times 2\hat{i} = 0 \) (the angle between \( \vec{p} \) and \( \vec{E} \) is π, so \( \tau = pE\sin\pi = 0 \)).

Teacher's Note:
a) Draw the figure showing \( E_{+q} \), \( E_{-q} \) and their resultant at P; the diagram carries half a mark.
b) State clearly that the perpendicular components cancel and the axial components add.
c) In a uniform field the net force on a dipole is always zero; torque is zero here because \( \vec{p} \) and \( \vec{E} \) are antiparallel.

OR

(a) Two cells of emf E₁ and E₂ with internal resistances r₁ and r₂ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.
(b) A parallel combination, as stated in (a) above, of two cells of emfs E and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R. [5 Marks]

Answer:
1. (a) Let \( I_1 \) and \( I_2 \) be the currents leaving the positive terminals of the cells, so \( I = I_1 + I_2 \). If V is the common terminal potential difference, then \( V = E_1 - I_1r_1 \) and \( V = E_2 - I_2r_2 \).
2. \( I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2} = \left(\frac{E_1}{r_1} + \frac{E_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right) \).
3. Rearranging, \( V = \frac{E_1r_2 + E_2r_1}{r_1 + r_2} - I\left(\frac{r_1r_2}{r_1 + r_2}\right) \). Comparing with \( V = E_{eq} - Ir_{eq} \): \( E_{eq} = \frac{E_1r_2 + E_2r_1}{r_1 + r_2} \) and \( r_{eq} = \frac{r_1r_2}{r_1 + r_2} \).
4. (b) \( E_{eq} = \frac{E \times R + 3E \times R}{R + R} = \frac{4ER}{2R} = 2E \), and \( r_{eq} = \frac{R \times R}{R + R} = \frac{R}{2} \).
5. Current through 2R: \( I = \frac{E_{eq}}{2R + r_{eq}} = \frac{2E}{2R + \frac{R}{2}} = \frac{4E}{5R} \) A.

Teacher's Note:
a) Draw the parallel combination and its single-cell equivalent before writing the equations.
b) The key step is comparing your result with \( V = E_{eq} - Ir_{eq} \).
c) In (b), treat the combination as one cell of emf 2E and internal resistance \( \frac{R}{2} \) in series with 2R.

32. (a) Using the relation for refraction at a curved spherical surface, derive the expression for lens maker's formula.
(b) Three lenses L₁, L₂ and L₃, each of focal length 40 cm, are placed coaxially. The distance between L₁ and L₂ and between L₂ and L₃ are 120 cm and 20 cm respectively. An object is kept at a distance of 80 cm to the left of lens L₁.
Find the distance of the final image formed from the object. [5 Marks]

Answer:
1. (a) Consider a thin lens of refractive index \( n_2 \) in a medium of index \( n_1 \), with surfaces ABC (centre \( C_1 \), radius \( R_1 \)) and ADC (centre \( C_2 \), radius \( R_2 \)). The first surface forms the image \( I_1 \) of object O: \( \frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{BC_1} \) ... (1).
2. \( I_1 \) acts as a virtual object for the second surface ADC, which forms the final image I: \( -\frac{n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_2 - n_1}{DC_2} \) ... (2). For a thin lens, \( BI_1 = DI_1 \).
3. Adding (1) and (2): \( \frac{n_1}{OB} + \frac{n_1}{DI} = (n_2 - n_1)\left[\frac{1}{BC_1} + \frac{1}{DC_2}\right] \). Using the sign convention, \( -\frac{n_1}{u} + \frac{n_1}{v} = (n_2 - n_1)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \), so \( \frac{1}{v} - \frac{1}{u} = \left(\frac{n_2}{n_1} - 1\right)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \).
4. If the object is at infinity, the image forms at the focus (v = f), so \( \frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \), which is the lens maker's formula.
5. (b) Lens L₁: \( \frac{1}{40} = \frac{1}{v_1} + \frac{1}{80} \), so \( v_1 = 80 \) cm. Lens L₂: \( u_2 = 120 - 80 = 40 \) cm (object at its focus), so \( \frac{1}{v_2} = 0 \) and \( v_2 = \infty \). Lens L₃: \( u_3 = \infty \), so \( v_3 = 40 \) cm. Distance of final image from the object = 80 + 120 + 20 + 40 = 260 cm.

Teacher's Note:
a) Draw the ray diagram with both centres of curvature and the intermediate image \( I_1 \); it carries 1 mark.
b) In (b), notice that the image by L₁ falls at the focus of L₂, so parallel rays reach L₃.
c) Add all the distances from the object to the final image, not just from L₃.

OR

(a) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the centre of curvature. Using this diagram, derive the mirror formula.
(b) A concave mirror produces a two times magnified virtual image of an object kept 10 cm in front of it. Calculate the focal length of the mirror. [5 Marks]

Answer:
1. (a) Ray diagram: object AB is placed between F and C of a concave mirror with pole P. A ray from B parallel to the axis reflects through F, and a ray from B towards the pole reflects symmetrically. They meet beyond C to form a real, inverted, magnified image A'B'. Let M be the point where the parallel ray strikes the mirror and N the foot of the perpendicular from M on the axis.
2. Triangles BAP and B'A'P are similar, so \( \frac{BA}{B'A'} = \frac{AP}{A'P} \) ... (1).
3. Triangles MNF and B'A'F are similar, so \( \frac{MN}{B'A'} = \frac{NF}{A'F} \). Since BA = MN and N is very close to P, \( \frac{BA}{B'A'} = \frac{PF}{A'F} \) ... (2).
4. From (1) and (2), \( \frac{AP}{A'P} = \frac{PF}{A'F} \). Using the sign convention, \( \frac{-u}{-v} = \frac{-f}{-v + f} \), so \( -uv + uf = -vf \). Dividing by uvf gives \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
5. (b) \( m = -\frac{v}{u} = 2 \), so \( v = -2u = -2(-10) = 20 \) cm. Then \( \frac{1}{f} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} \), giving f = -20 cm (focal length 20 cm).

Teacher's Note:
a) Mark the pole, focus, centre of curvature, object and image clearly; the diagram carries 1 mark.
b) The marking scheme awards full marks for deriving the mirror formula by any other justified method.
c) A virtual image is erect, so m = +2 and v is positive (behind the mirror).

33. (a) State Faraday's law of electromagnetic induction.
(b) Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns.
(c) A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod. [5 Marks]

Answer:
1. (a) The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit, \( \varepsilon = -\frac{d\phi_B}{dt} \).
2. (b) For a long solenoid with n turns per unit length \( \left(n = \frac{N}{l}\right) \) carrying current I, the field inside is \( B = \mu_0 nI \).
3. Total flux linked \( N\phi_B = (nl)(\mu_0 nI)(A) = \mu_0 n^2 AlI \). So self-inductance \( L = \frac{N\phi_B}{I} = \mu_0 n^2 Al = \frac{\mu_0 N^2 A}{l} \).
4. (c) Induced emf \( \varepsilon = \frac{1}{2}Bl^2\omega \), with B = \( 4 \times 10^{-3} \) T, l = 0.5 m and \( \omega = 2\pi \times 1 \) rad/s (60 rpm = 1 rotation per second).
5. \( \varepsilon = \frac{1}{2} \times 4 \times 10^{-3} \times (50 \times 10^{-2})^2 \times (2\pi \times 1) = 3.14 \times 10^{-3} \) V = 3.14 mV.

Teacher's Note:
a) Convert rpm to rad/s using \( \omega = 2\pi \times \) (revolutions per second).
b) The marking scheme awards full marks in (b) for any other correct alternative method.
c) Remember that total number of turns N = nl when writing the flux linkage.

OR

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.
(b) The ratio of the number of turns in the primary to the secondary of an ideal transformer is 1 : 5. If 5 kW power at 200 V is supplied to the primary, find
(i) current in the primary, and
(ii) output voltage. [5 Marks]

Answer:
1. (a) Diagram: a soft iron core with the primary coil (fewer turns) wound on one limb and the secondary coil (more turns) on the other limb, labelled primary, secondary and soft iron core.
2. Principle: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it (mutual induction).
3. If φ is the flux in each turn, the emf induced in the secondary is \( \varepsilon_s = -N_s\frac{d\phi}{dt} \) and the back emf in the primary is \( \varepsilon_p = -N_p\frac{d\phi}{dt} \). With \( \varepsilon_p = v_p \) and \( \varepsilon_s = v_s \), \( \frac{v_s}{v_p} = \frac{N_s}{N_p} \).
4. For an ideal transformer, input power = output power, so \( I_pV_p = I_sV_s \) and \( \frac{V_s}{V_p} = \frac{I_p}{I_s} \). Thus \( \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \).
5. (b) (i) \( P = V_pI_p \), so \( 5000 = 200I_p \) and \( I_p = 25 \) A. (ii) \( \frac{N_p}{N_s} = \frac{V_p}{V_s} \), so \( \frac{1}{5} = \frac{200}{V_s} \) and \( V_s = 1000 \) V.

Teacher's Note:
a) In a step-up transformer the secondary has more turns than the primary; show this clearly in the diagram.
b) Write both the turns ratio and the current ratio, as the question asks for both.
c) Convert 5 kW to 5000 W before finding the primary current.

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