CBSE Class 12 Physics Question Paper 2026 Solved Code 55-1-2

SECTION A

1. In a region electric field is given by \( \vec{E} = 4x\hat{i} \) N/C. The potential difference between points A (x = 1 m) and B (x = 3 m), (V_A - V_B) is [1 Mark]
(A) -16 V
(B) 16 V
(C) - 8 V
(D) 8 V

Answer: (B) 16 V

Teacher's Note:
a) Use \( V_A - V_B = \int_{A}^{B} \vec{E}\cdot d\vec{x} = \int_{1}^{3} 4x\,dx \).
b) This gives \( 2(3^2 - 1^2) = 16 \) V; watch the order of limits to get the sign right.

2. 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, there is no net current across an unbiased junction.
b) So the diffusion current and the drift current must balance each other exactly.

3. A copper wire is stretched to increase its length by 1%. Then the change in its resistance is close to [1 Mark]
(A) 1%
(B) 4%
(C) -4%
(D) 2%

Answer: (D) 2%

Teacher's Note:
a) On stretching, the volume stays constant, so \( R = \frac{\rho l^2}{V} \) and \( R \propto l^2 \).
b) For small changes, \( \frac{\Delta R}{R} = 2\frac{\Delta l}{l} = 2 \times 1\% = 2\% \).

4. Four independent waves are expressed as [1 Mark]
(i) y₁ = A₁ sin ωt
(ii) y₂ = A₂ sin 2 ωt
(iii) y₃ = A₃ cos ωt
(iv) y₄ = A₄ sin (ωt + π/3)
The interference between two of these waves is possible in
(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 ω; wave (ii) has 2ω, so it cannot interfere with them.

5. A concave lens of focal length 40 cm is coaxially in contact with two convex lenses, each of focal length 20 cm, on each side. The focal length of the combination is [1 Mark]
(A) zero
(B) \( \frac{20}{3} \) cm
(C) \( -\frac{20}{3} \) cm
(D) \( \frac{40}{3} \) cm

Answer: (D) \( \frac{40}{3} \) cm

Teacher's Note:
a) For lenses in contact, \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \), with a negative sign for the concave lens.
b) \( \frac{1}{F} = \frac{1}{20} + \frac{1}{20} - \frac{1}{40} = \frac{3}{40} \), so \( F = \frac{40}{3} \) cm.

6. The distance-of-closest-approach for an alpha particle is 'd' when it moves head-on with speed v towards a target nucleus. If the alpha particle is replaced by a proton moving with the same speed, the new distance-of-closest-approach will be [1 Mark]
(A) 2d
(B) d
(C) \( \frac{d}{2} \)
(D) \( \frac{d}{\sqrt{2}} \)

Answer: (A) 2d

Teacher's Note:
a) At closest approach, \( \frac{1}{2}mv^2 = \frac{kZe \cdot q}{d} \), so \( d \propto \frac{q}{m} \) for the same speed.
b) For an alpha particle \( \frac{q}{m} = \frac{2e}{4m_p} \); for a proton it is \( \frac{e}{m_p} \), which is twice as large, so d doubles.

7. Electromagnetic waves used for purification of water are [1 Mark]
(A) X-rays
(B) Ultraviolet rays
(C) Infrared rays
(D) Ultrasonic rays

Answer: (B) Ultraviolet rays

Teacher's Note:
a) UV rays kill germs, so they are used in water purifiers.
b) Ultrasonic waves are sound waves, not electromagnetic waves, so option (D) is ruled out at once.

8. 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 \( \frac{kq_1}{r_1} = \frac{kq_2}{r_2} \).
b) Surface field \( E = \frac{kq}{r^2} = \frac{V}{r} \), so \( E \propto \frac{1}{r} \) and \( \frac{E_A}{E_B} = \frac{r_2}{r_1} \).

9. A straight conductor lies along x-axis. It carries a current of 10 A along +x direction. The magnetic field \( \vec{B} \) due to 1 cm segment of this conductor, centred at the origin, at a point (0, 1 m, 0) is [1 Mark]
(A) \( (1\text{ nT})\hat{j} \)
(B) \( (10\text{ nT})\hat{k} \)
(C) \( -(10\text{ nT})\hat{k} \)
(D) \( -(1\text{ nT})\hat{j} \)

Answer: (B) \( (10\text{ nT})\hat{k} \)

Teacher's Note:
a) Biot-Savart law: \( d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l} \times \vec{r}}{r^3} \) with \( d\vec{l} = 0.01\hat{i} \) m and \( \vec{r} = 1\hat{j} \) m.
b) \( dB = 10^{-7} \times 10 \times 0.01 = 10^{-8} \) T = 10 nT, and \( \hat{i} \times \hat{j} = \hat{k} \) gives the direction.

10. The angular width of interference fringes in Young's double-slit experiment depends on [1 Mark]
(A) distance between the slits and the screen only
(B) wavelength of light used only
(C) both wavelength of light used and the slits separation
(D) slits separation only

Answer: (C) both wavelength of light used and the slits separation

Teacher's Note:
a) Angular fringe width \( \theta = \frac{\lambda}{d} \), where d is the slit separation.
b) It does not depend on the screen distance D; only the linear fringe width \( \beta = \frac{\lambda D}{d} \) does.

11. 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 \( n = \sqrt{\varepsilon_r \mu_r} = \sqrt{\frac{3}{2} \times \frac{8}{3}} = 2 \).
b) Frequency never changes on entering a new medium; speed and wavelength both become \( \frac{1}{n} \) times, that is, halved.

12. A resistor and an inductor of negligible resistance are connected in series to a 20 V ac source. If the voltage across the resistor is 12 V, the voltage across the inductor will be [1 Mark]
(A) 6 V
(B) 8 V
(C) 10 V
(D) 16 V

Answer: (D) 16 V

Teacher's Note:
a) In an LR circuit, \( V_R \) and \( V_L \) are \( 90^{\circ} \) out of phase, so \( V^2 = V_R^2 + V_L^2 \).
b) \( V_L = \sqrt{20^2 - 12^2} = \sqrt{256} = 16 \) V; do not simply subtract 12 from 20.

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. 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) Energy levels are discrete because of Bohr's quantisation condition on angular momentum, \( mvr = \frac{nh}{2\pi} \).
b) The Reason is true, but the electrostatic force only keeps the electron in orbit; it does not explain quantisation.

14. 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 difference in mass is called the mass defect, so the Assertion is true.
b) Energy is released (not absorbed) when nucleons bind to form a nucleus; this released energy is the binding energy.

15. 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.

16. 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 it is seen with electrons because of their wave nature.
b) The de Broglie wavelength \( \lambda = \frac{h}{mv} \) of the electrons decides the spread of the pattern.

SECTION B

17. (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 the dark fringes coincide: \( d\sin\theta = n\lambda_2 = (n+1)\lambda_1 \).
2. \( n \times 600 = (n+1) \times 400 \), which gives n = 2.
3. Position of this dark fringe from the central maximum: \( y = \frac{(n+1)\lambda_1 D}{d} \).
4. \( y = \frac{3 \times 400 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} = 1.8 \times 10^{-3} \) m = 1.8 mm.

Teacher's Note:
a) The smaller wavelength must have the higher order at the same point, so write \( (n+1)\lambda_1 = n\lambda_2 \).
b) Check with the other wavelength: \( y = \frac{2 \times 600 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} = 1.8 \) mm, the same result.

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

Teacher's Note:
a) Bright fringe positions in YDSE are \( y_n = \frac{n\lambda D}{d} \).
b) Convert 0.6 mm to \( 0.6 \times 10^{-3} \) m and nm to \( 10^{-9} \) m before substituting.

18. Find the ratio \( \left(\frac{\lambda_a}{\lambda_p}\right) \) of de Broglie wavelength λ_a associated with an alpha particle to de Broglie wavelength λ_p associated with a proton if both are moving with the (a) same velocity (b) same kinetic energy. [2 Marks]

Answer:
de Broglie wavelength \( \lambda = \frac{h}{mv} \).
(a) For the same velocity, \( \lambda \propto \frac{1}{m} \), so \( \frac{\lambda_a}{\lambda_p} = \frac{m_p}{m_a} = \frac{m_p}{4m_p} = \frac{1}{4} \).
(b) For the same kinetic energy, \( \lambda = \frac{h}{\sqrt{2mK}} \), so \( \lambda \propto \frac{1}{\sqrt{m}} \).
\( \frac{\lambda_a}{\lambda_p} = \sqrt{\frac{m_p}{4m_p}} = \frac{1}{2} \).

Teacher's Note:
a) Use mass of alpha particle = 4 times mass of proton.
b) Write the correct form of λ for each case: \( \frac{h}{mv} \) for same velocity and \( \frac{h}{\sqrt{2mK}} \) for same kinetic energy.

19. 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-sectional area A with n free electrons per unit volume. In time Δt, electrons move a distance \( \Delta x = v_d\Delta t \).
3. Total charge transported across the area in time Δt: \( Q = -neAv_d\Delta t \).
4. Electrons move opposite to the electric field, so the current in the direction of the field is given by \( I\Delta t = neAv_d\Delta t \), that is, \( I = neAv_d \).

Teacher's Note:
a) Half a mark is for the order of magnitude; the derivation carries the remaining marks.
b) Show the charge in the volume \( A v_d \Delta t \) clearly before dividing by Δt.

20. 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 square \( a = \frac{L}{4N} \) and radius of circular coil \( r = \frac{L}{2\pi N} \).
2. Maximum torque \( \tau = NIBA \), so \( \frac{\tau_1}{\tau_2} = \frac{NIBA_1}{NIBA_2} = \frac{A_1}{A_2} \).
3. \( \frac{\tau_1}{\tau_2} = \frac{\left(\frac{L}{4N}\right)^2}{\pi\left(\frac{L}{2\pi N}\right)^2} \).
4. \( \frac{\tau_1}{\tau_2} = \frac{\pi}{4} \).

Teacher's Note:
a) The length of wire per turn is \( \frac{L}{N} \); use it to find the side and the radius.
b) Since N, I and B are the same, the torque ratio is simply the ratio of the areas.

21. Draw a graph showing variation of binding energy per nucleon as a function of mass number A. The binding energy per nucleon for heavy nuclei \( (A \gt 170) \) decreases with increase in mass number. Explain its significance. [2 Marks]

Answer:
1. Graph: binding energy per nucleon (MeV) on the y-axis against mass number A (0 to 250) on the x-axis. It rises steeply for light nuclei (²H, ³H, ⁶Li), shows peaks for ⁴He, ¹²C, ¹⁶O, reaches a broad maximum of about 8.75 MeV near ⁵⁶Fe, and then falls slowly to about 7.6 MeV for ²³⁸U.
2. Significance: for A greater than 170, the decrease in binding energy per nucleon means these heavy nuclei are less stable.

Teacher's Note:
a) The marking scheme gives full marks for the correct shape of the graph alone.
b) Label both axes and mark the peak near Fe; the graph carries 1½ of the 2 marks.

SECTION C

22. 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 parallel horizontal plates, P1 P1' at the top and P2 P2' at the bottom, with a potential difference V applied between them. An electron (charge -e, mass m) enters from the left midway between the plates along a dashed horizontal line, and its dashed curved path bends downward to strike the right end P2' of the lower plate.]

Answer:
1. Acceleration of the electron: \( a = \frac{qE}{m} = \frac{qV}{mL} \), where L = 2 cm is the plate separation.
2. Time spent between the plates: \( t = \frac{x}{u_x} \), where x = 3 cm.
3. Vertical deflection: \( y = \frac{1}{2}\left(\frac{qV}{mL}\right)\left(\frac{x}{u_x}\right)^2 \), so \( V = \frac{2ymLu_x^2}{ex^2} \), with y = 1 cm.
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 vertical deflection needed is half the gap, y = 1 cm, not 2 cm.
b) The horizontal velocity stays constant; only the vertical motion is accelerated, like a projectile.
c) Keep all lengths in metres before substituting.

23. (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 \( \sin\theta \geq \frac{2}{3} \). [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 enters from the lower left through layer A and strikes the A-B interface at angle θ with the normal (shown dashed).]

Answer:
(a) 1. Light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle \( (i \gt i_c) \).
(b) 3. 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)
4. 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)
5. From (1) and (2), \( \sin r = \frac{3}{2}\sin\theta \). For \( \sin\theta = \frac{2}{3} \), sin r = 1, that is, \( r = 90^{\circ} \) and the ray grazes the B-C surface.
6. Hence for \( \sin\theta \geq \frac{2}{3} \), the beam is totally reflected and does not enter region C at all.

Teacher's Note:
a) Part (a) carries ½ mark for each condition; part (b) carries 2 marks.
b) For parallel layers, \( n\sin\theta \) stays the same across every interface, which is a quick check: \( n\sin\theta = \frac{2}{3}n \sin r \).

24. (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:
(a) 1. Take a cylindrical Gaussian surface of cross-sectional area A, passing through the sheet, with its two flat faces 1 and 2 on either side of the sheet.
2. Only faces 1 and 2 contribute to the flux, since \( \vec{E} \) is parallel to the curved surface. The flux \( \vec{E}\cdot\Delta\vec{S} \) through both faces is equal and adds up, so the net flux is 2EA.
3. Charge enclosed = σA. 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.
(b) 4. (i) Inside the sheets: \( E_{in} = 0 \).
5. (ii) Outside the sheets: \( E_{out} = \frac{\sigma}{\varepsilon_0} \).

Teacher's Note:
a) Draw the Gaussian cylinder with the field shown on both faces; the diagram carries ½ mark.
b) Between two sheets with the same charge density, the fields of the two sheets are equal and opposite, so they cancel; outside they add up.

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. From A, a 2R resistor leads to point M. From M, a resistor R leads to point O and another resistor R (upper branch) leads to point P. A resistor R connects O and P. From P, a resistor R leads to point N, and from O another resistor R (lower branch) also leads to N. From N, a 3R resistor leads to B.]

Answer:
(a) 1. Consider a Wheatstone bridge ABCD with resistances R₁ (AD), R₂ (AB), R₃ (DC) and R₄ (BC), a galvanometer G between B and D, and a battery between A and C. At balance, \( I_g = 0 \) and \( V_B = V_D \).
2. Kirchhoff's loop rule for loop ADBA gives \( -I_1R_1 + 0 + I_2R_2 = 0 \) ... (1)
Loop CBDC gives \( I_4R_4 + 0 - I_3R_3 = 0 \). Since \( I_g = 0 \), \( I_1 = I_3 \) and \( I_2 = I_4 \), so \( I_2R_4 - I_1R_3 = 0 \) ... (2)
3. From (1) and (2): \( \frac{R_2}{R_1} = \frac{R_4}{R_3} \), which is the balance condition.
(b) 4. Between M and N, the resistors MO, MP, ON and PN (each R) with OP (R) in the middle form a balanced Wheatstone bridge, so no current flows through OP. Hence \( R_{MN} = \frac{2R \times 2R}{2R + 2R} = R \).
5. \( R_{AB} = R_{AM} + R_{MN} + R_{NB} = 2R + R + 3R = 6R \).

Teacher's Note:
a) Part (a) carries 2 marks, including ½ mark for the labelled bridge diagram.
b) In part (b), first spot the balanced bridge; then the middle resistor OP can be removed.

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 point A at the top, B at the bottom, P on the left and R on the right. Diode D1 is between P and A (pointing towards A) and diode D2 is between A and R (pointing 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 V_i is connected between A and B.]

Answer:
(a) 1. Diode D₂ conducts, because during the positive half-cycle point A is at higher potential, so D₂ is forward biased (D₁ is reverse biased).
(b) 2. Equivalent circuit: current flows from A through D₂ to R. From R, it divides: one part flows through the 3 kΩ resistor to B, the other flows through the 1 kΩ resistor to P and then through the 2 kΩ resistor to B. D₁ is removed (open).
(c) 3. At the peak, the voltage across R and B is 12 V, so \( V_{RB} = 12 \) V.
4. The branch R-P-B (1 kΩ + 2 kΩ) carries \( I = \frac{12}{3000} = 4 \) mA, so by Ohm's law, \( V_{RP} = 4 \times 10^{-3} \times 1000 = 4 \) V and \( V_{PB} = 4 \times 10^{-3} \times 2000 = 8 \) V.

Teacher's Note:
a) The marking scheme also gives credit if a student takes the positive cycle as reaching diode D₁: then D₁ conducts and \( V_{PB} = 12 \) V, \( V_{PR} = 3 \) V, \( V_{RB} = 9 \) V.
b) Check the diode symbol: current flows in the direction of the arrow head only.
c) Marks: ½ for (a), 1 for the circuit, and ½ each for the three voltage drops.

26. A 12.0 μF capacitor is charged to a potential difference of 150 V. The terminals of the charged capacitor are then connected to those of an uncharged 6.0 μF capacitor. Calculate final potential difference across and charge on, each capacitor. [3 Marks]

Answer:
1. Initial charge on the 12 μF capacitor: \( Q_i = C_1V_i = 12\,\mu F \times 150\,V = 1800\,\mu C \).
2. Common potential: \( V_f = \frac{C_1V_i}{C_1 + C_2} = \frac{1800\,\mu C}{(12 + 6)\,\mu F} = 100 \) V. So the final potential difference across each capacitor is 100 V.
3. Charge on the 12 μF capacitor: \( Q_1 = C_1V_f = 12\,\mu F \times 100\,V = 1200\,\mu C \).
4. Charge on the 6 μF capacitor: \( Q_2 = C_2V_f = 6\,\mu F \times 100\,V = 600\,\mu C \).

Teacher's Note:
a) Total charge is conserved: 1200 μC + 600 μC = 1800 μC, which is a quick check.
b) When connected, both capacitors are in parallel, so they share the same final potential.

27. A semiconductor has equal electron and hole concentration of \( 3 \times 10^8 \) m^-3. On doping with a certain impurity, the hole concentration increases to \( 6 \times 10^{10} \) m^-3.
(a) What type of semiconductor is obtained on doping ?
(b) Calculate the new electron concentration of the semiconductor.
(c) How does the energy gap of semiconductor change with doping ? Draw the energy band diagram for it. [3 Marks]

Answer:
(a) 1. A p-type semiconductor is obtained.
(b) 2. Using \( n_e n_h = n_i^2 \): \( n_e = \frac{n_i^2}{n_h} = \frac{(3 \times 10^8)^2}{6 \times 10^{10}} = 1.5 \times 10^6 \) m^-3.
(c) 3. The energy gap effectively decreases on doping.
4. Energy band diagram (p-type, T greater than 0 K): the conduction band (E_C) at the top and the valence band (E_V) at the bottom, separated by the gap E_g. The acceptor level E_A lies just above the valence band (about 0.01 - 0.05 eV above E_V), and holes are shown in the valence band.

Teacher's Note:
a) Hole concentration increased, so the impurity is trivalent (acceptor) and the material is p-type.
b) The mass-action law \( n_e n_h = n_i^2 \) holds for both pure and doped semiconductors.
c) The band diagram carries 1 mark; mark the acceptor level close to the valence band.

28. Draw a labelled ray diagram showing the formation of image by a compound microscope when final image is formed at least distance of distinct vision. Derive an expression for its magnifying power for this case. [3 Marks]

Answer:
1. Ray diagram: an object AB is placed just beyond the focus f_o of the objective. The objective forms a real, inverted, magnified image A'B' (height h') just within the focus f_e of the eyepiece. The eyepiece acts as a simple magnifier and forms the final virtual, magnified image A''B'' at distance D from the eye.
2. Magnification due to the objective: \( m_o = \frac{h'}{h} = \frac{L}{f_o} \), using \( \tan\beta = \frac{h}{f_o} = \frac{h'}{L} \), where L is the length of the tube.
3. When the final image is at the least distance of distinct vision, magnification by the eyepiece: \( m_e = \left(1 + \frac{D}{f_e}\right) \).
4. Total magnification: \( m = m_o \times m_e = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \).

Teacher's Note:
a) The labelled ray diagram carries 1½ marks; mark objective, eyepiece, f_o, f_e, L and D.
b) The marking scheme gives full marks for any other correct alternative method of derivation.

SECTION D

29. 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 V_S (y-axis) against frequency ν (x-axis) with origin O. Two parallel straight lines, A and B, cut the frequency axis at ν₁ (line A) and ν₂ (line B), with ν₂ greater than ν₁. Their dashed extensions below the frequency axis meet the V_S-axis below O at points marked V₁ (line A) and V₂ (line B).]

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 frequency axis is the threshold frequency \( \nu_0 \), and work function \( \phi_0 = h\nu_0 \).
b) V₁ and V₂ are potentials, so they would have to be multiplied by e to give energies; check units before choosing.

(II) For radiation of frequency \( \nu \gt \nu_2 \) 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) Einstein's equation: \( K_{max} = h\nu - \phi_0 \).
b) For the same ν, the metal with the smaller work function (A, since \( \nu_1 \lt \nu_2 \)) gives the larger \( K_{max} \).

(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 of the \( V_s \)-ν line is \( \frac{h}{e} \), a constant that does not depend on intensity.
b) Higher intensity means more photons per second, so more photoelectrons (a larger photocurrent), but the same stopping potential.

(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 \( \frac{E_1}{E_2} = \frac{2}{5} \); do not take the simple ratio of frequencies \( \frac{3}{6} \).

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 \( eV_s = h\nu - \phi_0 \), we get \( V_s = \frac{h}{e}\nu - \frac{\phi_0}{e} \).
b) The slope is \( m = \frac{h}{e} \), so \( h = me \).

30. 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, so \( \sin\theta = 1 \) in every position.
b) Then the torque \( \tau = NIAB \) depends only on the current, not on the 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, A and B and decreases with k.
b) Increasing the torsional constant k lowers sensitivity, so option (D) is wrong.

(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 \( \tau = NIAB \) for the maximum (radial field) torque.
b) \( \tau = 50 \times 5 \times 4.0 \times 10^{-3} \times 0.25 = 0.25 \) N m.

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) For a voltmeter, a high resistance R is joined in series: \( R = \frac{V}{I_g} - G \).
b) \( R = \frac{12}{3 \times 10^{-3}} - 15 = 4000 - 15 = 3985 \) Ω.

(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) For an ammeter, a low resistance (shunt) S is joined in parallel: \( S = \frac{I_gG}{I - I_g} \).
b) \( S = \frac{5 \times 10^{-3} \times 20}{10 - 0.005} \approx 0.01 \) Ω.

SECTION E

31. (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:
(a) 1. The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit: \( \varepsilon = -\frac{d\phi_B}{dt} \).
(b) 2. Magnetic field inside a long solenoid with n turns per unit length carrying current I: \( B = \mu_0 nI \).
3. Total flux linked with the solenoid: \( N\phi_B = (nl)(\mu_0 nI)(A) = \mu_0 n^2 AlI \), where nl = N is the total number of turns.
4. Self-inductance \( L = \frac{N\phi_B}{I} = \mu_0 n^2 Al \) (or \( L = \frac{\mu_0 N^2 A}{l} \)).
(c) 5. Induced emf \( \varepsilon = \frac{1}{2}Bl^2\omega \), with ω = 60 rpm = 2π × 1 rad/s.
\( \varepsilon = \frac{1}{2} \times 4 \times 10^{-3} \times (50 \times 10^{-2})^2 \times (2\pi \times 1) = 3.14 \times 10^{-3} \) V = 3.14 mV.

Teacher's Note:
a) Marks: 1 for Faraday's law, 2 for the derivation and 2 for the numerical; full marks are given for any other correct derivation.
b) Convert 60 rpm to 1 revolution per second, that is, ω = 2π rad/s.
c) Convert 50 cm to 0.5 m and 4.0 mT to \( 4 \times 10^{-3} \) T before substituting.

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:
(a) 1. Diagram: a soft iron core with the primary coil (fewer turns) wound on one limb and the secondary coil (more turns) wound on the other limb (or both wound on the same limb); input terminals on the primary and output terminals on the secondary.
2. Principle: mutual induction. 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.
3. If φ is the flux in each turn of the core, the emf induced in the secondary is \( \varepsilon_s = -N_s\frac{d\phi}{dt} \) and the back emf in the primary is \( \varepsilon_p = -N_p\frac{d\phi}{dt} \). For an ideal transformer, \( \varepsilon_p = v_p \) and \( \varepsilon_s = v_s \), so \( v_s = -N_s\frac{d\phi}{dt} \) and \( v_p = -N_p\frac{d\phi}{dt} \).
4. Dividing, \( \frac{v_s}{v_p} = \frac{N_s}{N_p} \). For an ideal transformer, input power = output power: \( I_pV_p = I_sV_s \), so \( \frac{V_s}{V_p} = \frac{I_p}{I_s} \). Hence \( \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \).
(b) 5. (i) \( P = V_pI_p \): \( 5000 = 200 I_p \), so \( I_p = 25 \) A.
(ii) \( \frac{N_p}{N_s} = \frac{V_p}{V_s} \): \( \frac{1}{5} = \frac{200}{V_s} \), so \( V_s = 1000 \) V.

Teacher's Note:
a) Marks: 1 for the labelled diagram, ½ for the principle, 2 for the ratio, ½ for the primary current and 1 for the output voltage.
b) In a step-up transformer the secondary has more turns, so the voltage rises and the current falls in the same ratio.
c) Use P = VI with the primary values only to find the primary current.

32. (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:
(a) 1. Consider a point P on the equatorial plane at distance r from the centre of the dipole. The magnitudes of the fields due to +q and -q are equal: \( E_{+q} = E_{-q} = \frac{q}{4\pi\varepsilon_0}\frac{1}{(r^2 + a^2)} \).
2. The components 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. \( \vec{E} = -\frac{1}{4\pi\varepsilon_0}\frac{2qa}{(r^2 + a^2)^{3/2}}\hat{p} \).
4. At a large distance \( (r \gt\gt a) \): \( \vec{E} = \frac{-2qa}{4\pi\varepsilon_0 r^3}\hat{p} \).
(b) 5. Force: \( \vec{F} = \vec{F}_{+q} + \vec{F}_{-q} = q(2\hat{i}) - q(2\hat{i}) = 0 \) N.
Torque: \( \vec{\tau} = \vec{p} \times \vec{E} = p(-\hat{i}) \times 2\hat{i} = 0 \). (Alternatively, \( \tau = pE\sin\theta \) with the angle between \( \vec{p} \) and \( \vec{E} \) equal to π, so τ = 0.)

Teacher's Note:
a) Marks: 2½ for the equatorial field derivation (including a labelled diagram), ½ for the far-field expression and 2 for force and torque.
b) A dipole in a uniform field always has zero net force; the torque is zero here because \( \vec{p} \) lies along the x-axis, parallel or antiparallel to \( \vec{E} \).
c) Remember the equatorial field points opposite to \( \vec{p} \) (from +q towards -q direction).

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:
(a) 1. Let I₁ and I₂ be the currents leaving the positive electrodes of the cells \( \varepsilon_1 \) and \( \varepsilon_2 \), so \( I = I_1 + I_2 \). The potential difference V across the combination is \( V = \varepsilon_1 - I_1r_1 = \varepsilon_2 - I_2r_2 \).
2. \( I = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2} = \left(\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right) \).
3. Rearranging: \( V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\left(\frac{r_1r_2}{r_1 + r_2}\right) \). Comparing with \( V = \varepsilon_{eq} - Ir_{eq} \): \( \varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \) and \( r_{eq} = \frac{r_1r_2}{r_1 + r_2} \).
(b) 4. \( E_{eq} = \frac{E \times R + 3E \times R}{R + R} = \frac{4ER}{2R} = 2E \) and \( r_{eq} = \frac{R \times R}{R + R} = \frac{R}{2} \).
5. Current through 2R: \( I = \frac{E_{eq}}{2R + \frac{R}{2}} = \frac{2E}{5R/2} = \frac{4E}{5R} \).

Teacher's Note:
a) Marks: 2 for the equivalent emf, 1 for the equivalent internal resistance and 2 for part (b).
b) The equivalent internal resistance of cells in parallel is like resistors in parallel.
c) In part (b), add the external 2R to \( r_{eq} \) before finding the current.

33. (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:
(a) 1. Diagram: a thin convex lens (medium n₂) in a medium n₁, with surfaces ABC (centre C₁) and ADC (centre C₂); the object O, the intermediate image I₁ and the final image I on the axis. The first surface ABC forms the image of O at I₁: \( \frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{BC_1} \) ... (1)
2. I₁ acts as a virtual object for the second surface ADC: \( -\frac{n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_2 - n_1}{DC_2} \) ... (2)
3. For a thin lens, \( BI_1 = DI_1 \). 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 \( \frac{1}{v} - \frac{1}{u} = \left(\frac{n_2}{n_1} - 1\right)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \).
If the object is at infinity, the image forms at the focus, so \( \frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \).
(b) 4. Lens L₁: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \): \( \frac{1}{40} = \frac{1}{v_1} + \frac{1}{80} \), so \( v_1 = 80 \) cm. Lens L₂: \( u_2 = -(120 - 80) = -40 \) cm, so \( \frac{1}{40} = \frac{1}{v_2} + \frac{1}{40} \), giving \( v_2 = \infty \).
5. Lens L₃: \( u_3 = \infty \), so \( v_3 = 40 \) cm. Distance between final image and object = 80 + 120 + 20 + 40 = 260 cm.

Teacher's Note:
a) Marks: 3 for the derivation (including 1 for the ray diagram) and 2 for the numerical.
b) The image by L₁ falls at the focus of L₂, so the rays leave L₂ parallel and L₃ focuses them at its focus.
c) Add all the distances from the object to the final image, not just the distance 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:
(a) 1. Ray diagram: object AB placed between F and C of a concave mirror with pole P. A ray from B parallel to the axis strikes the mirror at M and passes through F; a ray from B to the pole P is reflected symmetrically. They meet at B', forming a real, inverted, magnified image A'B' beyond C. MN is the perpendicular from M to the axis.
2. ΔBAP and ΔB'A'P are similar, so \( \frac{BA}{B'A'} = \frac{AP}{A'P} \) ... (1). Since BA = MN, and ΔMNF and ΔB'A'F are similar, \( \frac{MN}{B'A'} = \frac{NF}{A'F} \), so \( \frac{BA}{B'A'} = \frac{NF}{A'F} = \frac{PF}{A'F} \) (N is very close to P) ... (2)
3. From (1) and (2): \( \frac{AP}{A'P} = \frac{PF}{A'F} \), so \( \frac{-u}{-v} = \frac{-f}{-v + f} \). This gives \( -uv + uf = -vf \). Dividing by uvf: \( -\frac{1}{f} + \frac{1}{v} = -\frac{1}{u} \), so \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \).
(b) 4. \( m = -\frac{v}{u} = 2 \), so \( v = -2u = -2(-10) = 20 \) cm (virtual image behind the mirror).
5. \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \): \( \frac{1}{20} - \frac{1}{10} = \frac{1}{f} \), so \( \frac{1 - 2}{20} = \frac{1}{f} \) and f = -20 cm. The focal length is 20 cm (concave mirror).

Teacher's Note:
a) Marks: 1 for the ray diagram, 2 for the derivation and 2 for the numerical; the scheme awards the two derivation marks for any other justified method.
b) A virtual image from a concave mirror is erect, so m is positive (+2) and v is positive.
c) Apply the sign convention consistently: u = -10 cm because the object is in front of the mirror.

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