CBSE Class 12 Physics Question Paper 2026 Solved Code 55-1-3

SECTION A

1. 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) Sustained 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 cannot interfere with them.

2. 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 \( n = \sqrt{\mu_r \varepsilon_r} = \sqrt{\frac{8}{3} \times \frac{3}{2}} = 2 \), so the speed becomes \( \frac{c}{2} \).
b) Frequency is fixed by the source and never changes on entering a new medium, so \( \lambda = \frac{v}{\nu} \) is halved.

3. 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 through an unbiased junction.
b) Both currents keep flowing, but they cancel each other exactly.

4. 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 \( \frac{q_1}{r_1} = \frac{q_2}{r_2} \), which gives \( q \propto r \).
b) Surface field \( E = \frac{kq}{r^2} \propto \frac{1}{r} \), so \( \frac{E_A}{E_B} = \frac{r_2}{r_1} \): the smaller sphere has the stronger field.

5. The electric potential for various points in x-y plane is given by V = 1.0 x² - 2.0 y², where V is in volts and x, y are in metres. The angle that the electric field at point (2.0 m, 1.0 m) makes with the positive x-axis is - [1 Mark]
(A) \( 45^{\circ} \)
(B) \( 90^{\circ} \)
(C) \( 135^{\circ} \)
(D) \( 315^{\circ} \)

Answer: (C) \( 135^{\circ} \)

Teacher's Note:
a) Use \( E_x = -\frac{\partial V}{\partial x} = -2x = -4 \) V/m and \( E_y = -\frac{\partial V}{\partial y} = 4y = 4 \) V/m at (2, 1).
b) A vector with a negative x-component and an equal positive y-component points at \( 135^{\circ} \) to the +x axis.

6. A current of 1.5 A is maintained in a copper wire of length 1 m with area of cross-section 1.7 \( \times \) 10^-7 m². The magnitude of electric field in the wire is [\( \rho_{Cu} = 1.7 \times 10^{-8} \) Ω m) [1 Mark]
(A) \( 0.15 \frac{V}{m} \)
(B) \( 0.30 \frac{V}{m} \)
(C) \( 1.5 \frac{V}{m} \)
(D) \( 3.0 \frac{V}{m} \)

Answer: (A) \( 0.15 \frac{V}{m} \)

Teacher's Note:
a) Use \( E = \rho J = \frac{\rho I}{A} = \frac{1.7 \times 10^{-8} \times 1.5}{1.7 \times 10^{-7}} = 0.15 \) V/m.
b) The length of the wire is not needed when you use \( E = \rho J \).

7. Light from a small object in air falls on a spherical glass surface (n = 1.5) of radius of curvature R. A real image of the object will be formed if the object distance u is related to R as : [1 Mark]
(A) \( u \lt \frac{R}{2} \)
(B) \( \frac{R}{2} \lt u \lt R \)
(C) \( R \lt u \lt 2R \)
(D) \( u \gt 2R \)

Answer: (D) \( u \gt 2R \)

Teacher's Note:
a) Use \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), which gives \( \frac{1.5}{v} = \frac{0.5}{R} - \frac{1}{|u|} \).
b) The image is real (v positive) only when \( \frac{0.5}{R} \gt \frac{1}{|u|} \), that is, when the object distance is greater than 2R.

8. The ratio of the potential energy to the kinetic energy of an electron in n^th orbit of Bohr model of hydrogen atom is [1 Mark]
(A) \( -\frac{1}{2} \)
(B) \( \frac{1}{2} \)
(C) 2
(D) -2

Answer: (D) -2

Teacher's Note:
a) In a Bohr orbit, \( K = \frac{ke^2}{2r} \) and \( U = -\frac{ke^2}{r} \), so U = -2K.
b) The ratio is the same for every orbit because it does not depend on n.

9. Welders wear special glass goggles or face masks with glass windows to protect their eyes from [1 Mark]
(A) Infrared rays
(B) Ultraviolet rays
(C) X-rays
(D) Microwaves

Answer: (B) Ultraviolet rays

Teacher's Note:
a) Welding arcs give out large amounts of UV radiation.
b) Special glass absorbs UV and protects the eyes; this is a standard NCERT fact about uses of UV rays.

10. A circular loop has radius R and carries current I as shown in figure. In order that the net magnetic field at the centre of the loop is zero, the current in wire AB should have magnitude [1 Mark]
(A) 2πI, along +X - axis
(B) 2πI along -X - axis
(C) πI along +X - axis
(D) πI along -X - axis

[Figure: A circular loop of radius R carrying current I in the clockwise sense (arrow at the top pointing right). A long straight wire AB lies horizontally below the loop; the distance from the level of the centre of the loop to the wire is marked 2R. Axes are shown: X to the right, Y upward and Z drawn diagonally towards the lower left.]

Answer: (A) 2πI, along +X - axis

Teacher's Note:
a) Equate magnitudes: \( \frac{\mu_0 I}{2R} = \frac{\mu_0 I'}{2\pi (2R)} \), which gives \( I' = 2\pi I \).
b) The clockwise loop gives a field into the page at the centre, so the wire current must be along +X to give a field out of the page above the wire.

11. The phenomenon of interference is shown by [1 Mark]
(A) longitudinal mechanical wave only
(B) transverse mechanical wave only
(C) electromagnetic waves only
(D) all these waves

Answer: (D) all these waves

Teacher's Note:
a) Interference follows from the principle of superposition, which holds for every kind of wave.
b) Do not confuse this with polarisation, which is shown only by transverse waves.

12. A series LCR circuit with R = 3 Ω, X_C = 4 Ω, X_L = 8 Ω is connected to a 220 V, 50 Hz ac source. The power factor for the circuit is [1 Mark]
(A) 0.30
(B) 0.45
(C) 0.50
(D) 0.60

Answer: (D) 0.60

Teacher's Note:
a) Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + 4^2} = 5 \) Ω.
b) Power factor \( \cos\phi = \frac{R}{Z} = \frac{3}{5} = 0.60 \); the supply voltage and frequency are not needed.

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) : 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 (printed as "Reason (A)") is the Assertion; it is true because of the mass defect.
b) Energy is released, not absorbed, when nucleons bind together; this released energy is the binding energy.

14. 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.

15. 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 true; the first one (printed as "Reason (A)") is the Assertion.
b) Discrete energy levels come from the quantisation of angular momentum, not from the electrostatic force acting as centripetal force.

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 electrons have a de Broglie wavelength \( \lambda = \frac{h}{p} \).
b) The wave nature of electrons directly explains the diffraction pattern, so R explains A.

SECTION B

17. 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 (about \( 10^{-3} \) m/s).
2. Consider a conductor of cross-sectional area A with n free electrons per unit volume. In time \( \Delta t \), the electrons drift a distance \( \Delta x = v_d \Delta t \).
3. The charge carried across the area in time \( \Delta t \) is \( Q = -neAv_d\Delta t \); electrons move opposite to the field, so in magnitude \( I\Delta t = neAv_d\Delta t \).
4. Hence \( I = neAv_d \).

Teacher's Note:
a) Half a mark is for stating the order of magnitude, so do not skip it.
b) Show the step "charge crossing area A in time \( \Delta t \)" clearly; this is where most marks lie.

18. 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 \( a = \frac{L}{4N} \); 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. \( \frac{\tau_1}{\tau_2} = \frac{\left(\frac{L}{4N}\right)^2}{\pi\left(\frac{L}{2\pi N}\right)^2} = \frac{\frac{1}{16}}{\frac{1}{4\pi}} \).
4. \( \frac{\tau_1}{\tau_2} = \frac{\pi}{4} \).

Teacher's Note:
a) Each coil has N turns, so the length of one turn is \( \frac{L}{N} \); do not forget to divide by N.
b) A circle encloses more area than a square of the same perimeter, so the circular coil gets the larger torque.

19. (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: \( d\sin\theta = n\lambda_2 = (n+1)\lambda_1 \).
2. \( n \times 600 = (n+1) \times 400 \), so n = 2.
3. Position of this dark fringe: \( y = \frac{n\lambda_2 D}{d} = \frac{2 \times 600 \times 10^{-9} \times 1.5}{1 \times 10^{-3}} \).
4. y = 1.8 mm (the same result comes from \( y = \frac{3\lambda_1 D}{d} \)).

Teacher's Note:
a) For a single slit, minima are at \( a\sin\theta = n\lambda \); set the two positions equal.
b) Convert nm to m and mm to m before substituting.

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. For coinciding bright fringes: \( \frac{n\lambda_2 D}{d} = \frac{(n+1)\lambda_1 D}{d} \), so \( n \times 660 = (n+1) \times 440 \), giving n = 2.
2. \( y = \frac{n\lambda_2 D}{d} = \frac{2 \times 660 \times 10^{-9} \times 1.5}{0.6 \times 10^{-3}} = 3.3 \) mm.

Teacher's Note:
a) The larger wavelength has the lower order at the coinciding point.
b) Check using the other wavelength: \( \frac{3 \times 440 \times 10^{-9} \times 1.5}{0.6 \times 10^{-3}} \) also gives 3.3 mm.

20. In an electron microscope, accelerated electrons have wavelength of 0.011 nm. Calculate the voltage through which electrons were accelerated to attain this wavelength.
(Take e = 1.6 \( \times \) 10^-19 C, m_e = 9 \( \times \) 10^-31 kg, h = 6.6 \( \times \) 10^-34 J.s) [2 Marks]

Answer:
1. de Broglie wavelength \( \lambda = \frac{h}{\sqrt{2meV}} \), so \( V = \frac{h^2}{2me\lambda^2} \).
2. \( V = \frac{(6.6 \times 10^{-34})^2}{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times (0.011 \times 10^{-9})^2} \).
3. V = 12500 volts.

Teacher's Note:
a) Rearranging the formula for V before substituting saves time and avoids mistakes.
b) Use the constants given in the question, not the general list, to get the exact answer.

21. Suppose a nucleus with mass number A = 240 and \( \frac{B.E.}{A} \) = 7.6 MeV, breaks into two nuclei, each of mass number A = 120 with \( \frac{B.E.}{A} \) = 8.5 MeV. Calculate the energy released in the process. [2 Marks]

Answer:
1. Binding energy of the nucleus (A = 240) = \( 240 \times 7.6 \) MeV.
2. Binding energy of the two fragments = \( 2 \times 120 \times 8.5 = 240 \times 8.5 \) MeV.
3. Energy released = \( 240 \times 8.5 - 240 \times 7.6 = 240 \times 0.9 \).
4. Energy released = 216 MeV.

Teacher's Note:
a) Energy released equals the gain in total binding energy.
b) Multiply B.E. per nucleon by the total number of nucleons; do not just subtract 8.5 - 7.6.

SECTION C

22. 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. [3 Marks]
(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.

[Figure: An ac source V_i connected between points A (top) and B (bottom) of a diamond-shaped network. Diode D₁ is between P and A, with its arrow pointing towards A. Diode D₂ is between A and R, with its arrow 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.]

Answer:
1. (a) Diode D₂ conducts, because in the positive half-cycle A is at higher potential, so D₂ (pointing from A to R) is forward biased while D₁ is reverse biased.
2. (b) Equivalent circuit: current flows from A through D₂ to R. From R it divides: one path goes through the 3 kΩ resistor to B, the other goes through the 1 kΩ resistor to P and then through the 2 kΩ resistor to B. D₁ is removed (open).
3. (c) At peak value, 12 V appears across R and B. The 3 kΩ resistor gets V_RB = 12 V.
4. In the branch R - P - B, the current is \( \frac{12}{3000} = 4 \) mA, so by V = IR, V_RP = 4 V and V_PB = 8 V.

Teacher's Note:
a) Look at the direction of the diode arrow: a diode conducts when its arrow points from higher to lower potential.
b) The marking scheme also accepts the alternative reading that D₁ conducts, with V_PB = 12 V, V_PR = 3 V and V_RB = 9 V, if justified consistently.
c) Use Ohm's law branch by branch; the two branches across R and B share the same 12 V.

23. Figure shows a narrow beam of electrons entering with a velocity of 3 \( \times \) 10⁷ 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, P₁P₁' on top and P₂P₂' below, with voltage V marked between them. An electron (charge -e, mass m) enters from the left midway between the plates; a dashed straight line shows the undeflected path and a dashed curve bends downward to reach the end P₂' of the lower plate.]

Answer:
1. Acceleration \( a = \frac{qE}{m} = \frac{qV}{mL} \), where L = 2 cm is the plate separation; time spent between plates \( t = \frac{x}{u_x} \).
2. Vertical deflection \( y = \frac{1}{2}\left(\frac{qV}{mL}\right)\left(\frac{x}{u_x}\right)^2 \), so \( V = \frac{2ymLu_x^2}{ex^2} \).
3. With y = 1 cm, x = 3 cm: \( V = \frac{2 \times 1 \times 10^{-2} \times 9.1 \times 10^{-31} \times 2 \times 10^{-2} \times (3 \times 10^7)^2}{1.6 \times 10^{-19} \times (3 \times 10^{-2})^2} \).
4. V = 2275 volt.

Teacher's Note:
a) The beam enters midway, so the deflection needed is half the separation, y = 1 cm.
b) Treat the motion like a projectile: uniform speed along x and uniform acceleration along y.

24. (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 travels upward through A and meets the A-B interface at 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)
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), \( \sin r = \frac{3}{2}\sin\theta \). For \( \sin\theta = \frac{2}{3} \), sin r = 1, that is \( r = 90^{\circ} \): the ray just grazes the B-C surface.
5. For \( \sin\theta \gt \frac{2}{3} \), sin r would exceed 1, so the beam is totally reflected. Hence for \( \sin\theta \geq \frac{2}{3} \) the beam does not enter region C at all.

Teacher's Note:
a) Write Snell's law at both interfaces and eliminate the angle in layer B.
b) The key step is showing that sin r reaches 1 (grazing emergence) exactly at \( \sin\theta = \frac{2}{3} \).
c) Part (a) carries 1 mark; state both conditions separately.

25. (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 A passing through the sheet, with its two flat faces 1 and 2 on either side of the sheet (diagram of the sheet with the cylinder).
2. The field is perpendicular to the sheet, so only the two flat faces contribute flux; flux through each is EA, so total flux = 2EA. The charge enclosed is σA.
3. By Gauss's law, \( 2EA = \frac{\sigma A}{\varepsilon_0} \), so \( \vec{E} = \frac{\sigma}{2\varepsilon_0}\hat{n} \).
4. (b) (i) Inside, between the sheets: the fields of the two sheets are equal and opposite, so \( E_{in} = 0 \).
5. (ii) Outside the sheets: the fields add, so \( E_{out} = \frac{\sigma}{\varepsilon_0} \).

Teacher's Note:
a) Draw the Gaussian cylinder; the diagram itself carries half a mark.
b) The curved surface gives zero flux because E is parallel to it.
c) Both sheets have the same charge density σ, so their fields cancel between them.

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 goes to node M. Between nodes M and N: a resistor R from M to O, a resistor R from O to P, a resistor R from P to N, an upper resistor R joining M directly to P, and a lower resistor R joining O directly to N. From N, a 3R resistor goes to B.]

Answer:
1. (a) In a Wheatstone bridge with arms R₁, R₂, R₃, R₄ and galvanometer between B and D, apply Kirchhoff's loop rule to loop ADBA: \( -I_1R_1 + 0 + I_2R_2 = 0 \) (at balance \( I_g = 0 \), \( V_B = V_D \)) ... (1)
2. For loop CBDC: \( I_4R_4 + 0 - I_3R_3 = 0 \). 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), the balance condition is \( \frac{R_2}{R_1} = \frac{R_4}{R_3} \).
4. (b) The part between M and N is a balanced Wheatstone bridge (all arms R), so the resistor OP carries no current and \( R_{MN} = R \).
5. \( R_{AB} = R_{AM} + R_{MN} + R_{NB} = 2R + R + 3R = 6R \).

Teacher's Note:
a) Draw the bridge and mark the currents; the diagram carries half a mark.
b) Once all four arms are equal, the middle resistor can be removed; this gives \( R_{MN} = \frac{2R \times 2R}{4R} = R \).

26. A parallel plate capacitor of capacitance C is charged to V volt by a battery. After sometime the battery is disconnected and the distance between the plates is doubled. A slab of dielectric constant k = 1.8 is then introduced to completely fill the space between the plates. How will the following be affected ? [3 Marks]
(a) The capacitance of the capacitor.
(b) The electric field between the plates of the capacitor.
(c) The energy stored in the capacitor.
Justify your answer in each case.

Answer:
1. (a) Initially \( C_0 = \frac{\varepsilon_0 A}{d} \). New capacitance \( C = \frac{K\varepsilon_0 A}{2d} = \frac{1.8}{2}C_0 = 0.9C_0 \). The capacitance decreases.
2. (b) The battery is disconnected, so the charge Q stays the same. New potential \( V' = \frac{Q}{0.9C_0} \).
3. New field \( E' = \frac{V'}{2d} = \frac{Q}{0.9C_0 \times 2d} = \frac{1}{1.8}\frac{Q}{C_0 d} = \frac{E}{1.8} \). The electric field decreases.
4. (c) Energy \( U = \frac{1}{2}\frac{Q^2}{C} \); with the same Q, \( U' = \frac{1}{2}\frac{Q^2}{0.9C_0} = \frac{U}{0.9} \). The energy stored increases.

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

27. (a) Draw the V-I characteristics of silicon diode.
(b) Explain the following terms :
(i) minority carrier injection in forward bias
(ii) breakdown voltage in reverse bias [3 Marks]

Answer:
1. (a) V-I graph: V (V) on the x-axis, I (mA) on the positive y-axis and I (μA) on the negative y-axis. In forward bias the current stays almost zero up to the threshold (knee) voltage and then rises sharply. In reverse bias a very small, nearly constant current flows until the breakdown voltage V_Br, where the reverse current rises sharply.
2. (b) (i) Under forward bias, electrons from the n-side cross the depletion region and reach the p-side, and holes from the p-side cross the junction and reach the n-side, where they are minority carriers. This process is called minority carrier injection.
3. (ii) Under reverse bias the current is almost independent of voltage up to a critical reverse voltage; at this voltage the reverse current starts increasing sharply. This critical voltage is called the breakdown voltage.

Teacher's Note:
a) Label the axes with different units (mA for forward, μA for reverse) and mark the breakdown voltage on the graph.
b) Each term in (b) carries 1 mark; use the words "minority carriers" and "sharp increase in reverse current".

28. (a) Why does one prefer to view the image formed at infinity than that formed at near point in microscope/telescope ?
(b) Consider lenses L₁, L₂ and L₃ as specified in the following table. Which of them will you select as objective and eyepiece for constructing best possible (i) telescope (ii) compound microscope ? Give reason for your answer.
Lens: L₁ | L₂ | L₃
Power: 6 D | 3 D | 10 D
Aperture: 1 cm | 8 cm | 1 cm [3 Marks]

Answer:
1. (a) In normal adjustment (image at infinity) the eye muscles remain relaxed, while they are strained when the image is at the near point. Hence the image at infinity is preferred.
2. (b) (i) Telescope: L₂ as objective and L₃ as eyepiece.
3. (ii) Compound microscope: L₃ as objective and L₁ as eyepiece.
4. Reason: a telescope objective should have a large aperture and a large focal length (L₂). For a microscope, the objective and eye lens should have moderate aperture, but the objective should have large power (L₃).

Teacher's Note:
a) Large aperture gathers more light and gives better resolution in a telescope.
b) Remember: small power means large focal length (\( f = \frac{1}{P} \)).

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, so \( \sin\theta = 1 \) in every position.
b) The radial field is produced by concave pole pieces and the soft iron core together.

(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 is \( \frac{\phi}{I} = \frac{NAB}{k} \).
b) Decreasing A and B or increasing the spring constant k would reduce the sensitivity, so (C) and (D) are 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) In a radial field, \( \tau = NIAB = 50 \times 5 \times 4.0 \times 10^{-3} \times 0.25 = 0.25 \) N m.
b) Multiply step by step and keep track of the power of 10.

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 \( R = \frac{V}{I_g} - G = \frac{12}{3 \times 10^{-3}} - 15 = 4000 - 15 = 3985 \) Ω.
b) A common mistake is to forget to subtract the coil resistance, which gives 4000 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) Shunt \( S = \frac{I_g G}{I - I_g} = \frac{5 \times 10^{-3} \times 20}{10 - 0.005} \approx 0.01 \) Ω.
b) An ammeter needs a small resistance in parallel (shunt); a voltmeter needs a large 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 V_S (y-axis) versus 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 meet the negative V_S axis at V₁ (line A) and V₂ (line B), with V₂ further below O.]

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, and work function \( \phi_0 = h\nu_0 \).
b) Option (D) has the unit of volt, so it cannot be a work function.

(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) \( K_{max} = h\nu - \phi_0 \), so a smaller work function gives a larger kinetic energy.
b) From the graph, line A has the higher stopping potential at any frequency.

(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, and threshold frequency depends only on the metal.
b) Intensity affects only the number of photoelectrons (photocurrent), not their maximum energy.

(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 \( E_2 = 6h\nu_0 - h\nu_0 = 5h\nu_0 \).
b) Do not simply take the ratio of frequencies (1/2); subtract the work function first.

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) Therefore h = me; the slope is the same for every metal.

SECTION E

31. (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) Diagram: a thin lens with surfaces ABC (centre of curvature C₁) and ADC (centre of curvature C₂); object O, intermediate image I₁ and final image I on the principal axis.
2. Refraction at 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)
3. 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)
4. For a thin lens \( BI_1 = DI_1 \). Adding (1) and (2) and using sign convention (OB = -u, DI = v, BC₁ = R₁, DC₂ = -R₂): \( \frac{1}{v} - \frac{1}{u} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).
5. For an object at infinity the image is at the focus (v = f), so \( \frac{1}{f} = \left(\frac{n_2}{n_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \). This is the lens maker's formula.
6. (b) Lens L₁: \( \frac{1}{40} = \frac{1}{v_1} + \frac{1}{80} \), so v₁ = 80 cm to the right of L₁.
7. Lens L₂: object distance u₂ = 120 - 80 = 40 cm, which is at its focus, so \( \frac{1}{v_2} = 0 \) and v₂ = ∞ (parallel beam).
8. Lens L₃: u₃ = ∞, so v₃ = f = 40 cm to the right of L₃.
9. Distance of the final image from the object = 80 + 120 + 20 + 40 = 260 cm.

Teacher's Note:
a) Part (a) carries 3 marks including 1 mark for the ray diagram; show both refracting surfaces clearly.
b) In part (b), notice that the image by L₁ falls exactly at the focus of L₂, so L₂ gives a parallel beam.
c) Add all the distances along the axis carefully to find the object-to-image distance.

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 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 reflects symmetrically. They meet at B', forming a real, inverted, enlarged image A'B' beyond C. MN is the perpendicular from M to the axis.
2. Triangles BAP and B'A'P are similar: \( \frac{BA}{B'A'} = \frac{AP}{A'P} \) ... (1)
3. Triangles MNF and B'A'F are similar, and BA = MN with N very close to P: \( \frac{BA}{B'A'} = \frac{NF}{A'F} = \frac{PF}{A'F} \) ... (2)
4. From (1) and (2): \( \frac{AP}{A'P} = \frac{PF}{A'F} \). With sign convention, \( \frac{-u}{-v} = \frac{-f}{-v + f} \), so \( -uv + uf = -vf \).
5. Dividing by uvf: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), the mirror formula.
6. (b) Magnification \( m = -\frac{v}{u} = 2 \), so \( v = -2u = -2(-10) = 20 \) cm (virtual image behind the mirror).
7. \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \): \( \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = \frac{1}{f} \), so f = -20 cm.

Teacher's Note:
a) The ray diagram carries 1 mark and the derivation 2 marks; the scheme gives full credit for any other justified method of derivation.
b) For a virtual, erect image the magnification is positive, so v comes out positive (behind the mirror).
c) The negative focal length confirms the mirror is concave.

32. (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) 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 \( L = \frac{\mu_0 N^2 A}{l} \)).
5. (c) Induced emf \( \varepsilon = \frac{1}{2}Bl^2\omega \), with 60 rpm = 1 rev/s, so \( \omega = 2\pi \times 1 \) rad/s.
6. \( \varepsilon = \frac{1}{2} \times 4 \times 10^{-3} \times (50 \times 10^{-2})^2 \times (2\pi \times 1) = 3.14 \) mV.

Teacher's Note:
a) Convert rpm to rad/s before substituting: \( \omega = \frac{2\pi \times 60}{60} \).
b) In (b), the scheme gives full marks for any other correct method.
c) Remember that N = nl when changing between the two forms of L.

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, both labelled.
2. Principle: mutual induction. An alternating voltage in the primary produces an alternating magnetic flux which links the secondary and induces an emf in it.
3. If \( \phi \) 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 negligible resistance, \( \varepsilon_p = v_p \) and \( \varepsilon_s = v_s \).
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 \( \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \).
5. (b) (i) \( P = V_pI_p \): \( 5000 = 200 I_p \), so I_p = 25 A.
6. (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) In a step-up transformer the secondary must have more turns than the primary; show this in the diagram.
b) Write both the turns ratio and the current ratio; the question asks for both.
c) Convert 5 kW to 5000 W before finding the current.

33. (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) Diagram: dipole -q and +q separated by 2a, point P on the equatorial line at distance r from the centre; fields E_+q and E_-q at P, each making angle θ with the dipole axis.
2. Magnitudes: \( E_{+q} = E_{-q} = \frac{q}{4\pi\varepsilon_0}\frac{1}{(r^2 + a^2)} \).
3. The components normal to the dipole axis cancel; the components along the axis add, and the total field is opposite to \( \vec{p} \): \( \vec{E} = -(E_{+q} + E_{-q})\cos\theta\,\hat{p} \), with \( \cos\theta = \frac{a}{(r^2 + a^2)^{1/2}} \).
4. \( \vec{E} = -\frac{1}{4\pi\varepsilon_0}\frac{2qa}{(r^2 + a^2)^{3/2}}\hat{p} \).
5. For r \( \gt\gt \) a: \( \vec{E} = \frac{-2qa}{4\pi\varepsilon_0 r^3}\hat{p} \).
6. (b) Net force \( \vec{F} = \vec{F}_{+q} + \vec{F}_{-q} = q(2\hat{i}) - q(2\hat{i}) = 0 \) N.
7. Torque \( \vec{\tau} = \vec{p} \times \vec{E} = p(-\hat{i}) \times 2\hat{i} = 0 \), since the angle between \( \vec{p} \) and \( \vec{E} \) is π (\( \tau = pE\sin\theta = 0 \)).

Teacher's Note:
a) Draw the diagram showing both fields and their components; only the components parallel to the axis survive.
b) In a uniform field the net force on any dipole is always zero.
c) Torque is zero whenever \( \vec{p} \) is parallel or antiparallel to \( \vec{E} \).

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 I₁ and I₂ be the currents leaving the positive terminals of the cells \( \varepsilon_1 \) and \( \varepsilon_2 \), so \( I = I_1 + I_2 \).
2. Terminal potential difference: \( V = \varepsilon_1 - I_1r_1 = \varepsilon_2 - I_2r_2 \), so \( I = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2} = \left(\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right) \).
3. Rearranging: \( V = \frac{\varepsilon_1r_2 + \varepsilon_2r_1}{r_1 + r_2} - I\left(\frac{r_1r_2}{r_1 + r_2}\right) \). Comparing with \( V = \varepsilon_{eq} - Ir_{eq} \):
4. \( \varepsilon_{eq} = \frac{\varepsilon_1r_2 + \varepsilon_2r_1}{r_1 + r_2} \) and \( r_{eq} = \frac{r_1r_2}{r_1 + r_2} \).
5. (b) \( E_{eq} = \frac{E \times R + 3E \times R}{R + R} = \frac{4ER}{2R} = 2E \); \( r_{eq} = \frac{R \times R}{R + R} = \frac{R}{2} \).
6. Current through 2R: \( I = \frac{E_{eq}}{2R + \frac{R}{2}} = \frac{2E}{\frac{5R}{2}} = \frac{4E}{5R} \) A.

Teacher's Note:
a) Write the terminal voltage for each cell first; the equivalent emf carries 2 marks and the internal resistance 1 mark.
b) In (b), add the equivalent internal resistance to the external 2R before applying Ohm's law.

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