CBSE Class 12 Physics Sample Paper 2026 27 with Solutions PDF Download

SECTION A

1. A point charge +Q is placed at the centre of a spherical Gaussian surface of radius R. If the radius of the sphere is doubled, the electric flux through the surface will [1 Mark]
A. become half
B. become double
C. remain the same
D. become four times

Answer: C. remain the same

Teacher's Note:
a) By Gauss's law, \( \phi = \frac{Q}{\varepsilon_0} \), which depends only on the charge enclosed.
b) The size or shape of the Gaussian surface does not change the flux.

2. If two identical heaters each rated as (1000 W, 220 V) are connected in parallel to 220 V, then the total power consumed is- [1 Mark]
A. 200 W
B. 250 W
C. 2000 W
D. 2500 W

Answer: C. 2000 W

Teacher's Note:
a) In parallel, each heater gets its rated 220 V, so each consumes 1000 W.
b) For a parallel combination, \( P = P_1 + P_2 = 1000 + 1000 = 2000 \) W.

3. An infinitely long cylinder is kept parallel to a uniform magnetic field B directed along negative y -axis. What will be the direction of induced current as seen from the y axis? [1 Mark]
A. clockwise of the negative y-axis
B. anticlockwise of the negative y-axis
C. no current will be induced
D. along the direction of the magnetic field

Answer: C. no current will be induced

Teacher's Note:
a) The cylinder is stationary and the field is uniform, so the flux linked with it does not change.
b) An emf is induced only when the magnetic flux changes with time.

4. If electric field associated with an electromagnetic wave is given by
E= 30 sin \( (0.2\pi \times 10^{-4}\ \text{m}^{-1} x - 0.6\pi \times 10^{4}\ \text{Hz}\ t)\ \hat{j} \) then the direction of propagation of the wave is along, [1 Mark]
A. \( \hat{i} \)
B. \( -\hat{i} \)
C. \( \hat{k} \)
D. \( -\hat{k} \)

Answer: A. \( \hat{i} \)

Teacher's Note:
a) The wave equation is a function of x, so the wave travels along the x-axis.
b) The form \( \sin(kx - \omega t) \) means the wave moves along +x; the form \( \sin(kx + \omega t) \) would mean -x.

5. Which one of the following phenomenon is not explained by Huygens' construction of wave front? [1 Mark]
A. Diffraction
B. Refraction
C. Reflection
D. Origin of spectra

Answer: D. Origin of spectra

Teacher's Note:
a) Huygens' construction explains reflection, refraction and diffraction.
b) The origin of spectra needs the quantum idea of energy levels, which Huygens' theory cannot give.

6. If wave front from star in nearby galaxy is detected by a telescope at the top of some mountain in India then the shape of the wavefront detected will be [1 Mark]
A. diverging spherical
B. converging spherical
C. plane
D. cylindrical

Answer: C. plane

Teacher's Note:
a) The star acts as a very distant point source.
b) A small part of a very large spherical wavefront is almost flat, so it is a plane wavefront.

7. The binding energy per nucleon curve shows a maximum around mass number A=56. What does this imply about nuclei near iron (Fe)? [1 Mark]
A. They are the least stable nuclei.
B. They are the most tightly bound nuclei.
C. They undergo spontaneous fission releasing maximum energy.
D. They cannot participate in nuclear reactions.

Answer: B. They are the most tightly bound nuclei.

Teacher's Note:
a) Maximum binding energy per nucleon means the nucleus is the most stable and most tightly bound.
b) Spontaneous fission is typical of very heavy nuclei like uranium, not iron.
c) Both fusion of lighter nuclei and fission of heavier nuclei release energy by moving towards iron.

8. A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. The direction of the movement of the ball and the magnetic moment would be- [1 Mark]
Direction of the movement of ball | Direction of the magnetic moment
A. Towards the bar magnet | Opposite to that of the bar magnet
B. Towards the bar magnet | Same direction as that of the bar magnet
C. Away from the bar magnet | Same direction as that of the bar magnet
D. Away from the bar magnet | Opposite to that of the bar magnet

Answer: D. Away from the bar magnet | Opposite to that of the bar magnet

Teacher's Note:
a) In liquid nitrogen the ball is below its critical temperature, so it becomes a superconductor.
b) A superconductor is perfectly diamagnetic: its induced magnetic moment is opposite to the field, so it is repelled.

9. A bridge circuit is shown in the following figure. The total current flowing in the circuit is- [1 Mark]
A. 18 A
B. 5 A
C. 5/18 A
D. 18/5A

[Figure: Bridge circuit between points X and Y. Upper arms: 2 Ω from X to the top junction and 4 Ω from the top junction to Y. Lower arms: 3 Ω from X to the bottom junction and 6 Ω from the bottom junction to Y. A 5 Ω resistor joins the top and bottom junctions. An 18 V battery is connected between X and Y.]

Answer: B. 5 A

Teacher's Note:
a) The bridge is balanced since \( \frac{2}{4} = \frac{3}{6} \), so no current flows through the 5 Ω resistor.
b) \( (2 + 4) = 6\ \Omega \) and \( (3 + 6) = 9\ \Omega \) are in parallel: \( R = \frac{6 \times 9}{6 + 9} = \frac{18}{5}\ \Omega \).
c) \( I = \frac{V}{R} = \frac{18}{18/5} = 5 \) A.

For Visually Impaired Candidates (in lieu of Q. 9)

In a Wheat-stone bridge, all the four arms have equal resistance R. If the resistance of the galvanometer arm is 2R, the equivalent resistance of the combination as seen by the battery is- [1 Mark]
A. 2R
B. R
C. R/2
D. R/4

Answer: B. R

Teacher's Note:
a) The bridge is balanced, so the galvanometer arm (2R) carries no current and is ineffective.
b) \( (R + R) \) and \( (R + R) \) are in parallel: \( R_{eq} = \frac{2R \times 2R}{2R + 2R} = R \).

10. Two electric bulbs, B₁ and B₂, are connected in series to an A.C. supply of 200 V. The power ratings printed on the bulbs are:
B₁: 200 V, 60 W
B₂: 200 V, 100 W
Using their rated voltage-power information, the ratio of their resistances R₁/R₂ is- [1 Mark]
A. 5:3
B. 3:5
C. 3:4
D. 4:3

[Figure: Two bulbs B₁ and B₂ connected in series across a 200 V A.C. supply. The voltage across B₁ is marked V₁ and the voltage across B₂ is marked V₂.]

Answer: A. 5:3

Teacher's Note:
a) From the ratings, \( R_1 = \frac{V^2}{P_1} \) and \( R_2 = \frac{V^2}{P_2} \), with the same rated V.
b) \( \frac{R_1}{R_2} = \frac{P_2}{P_1} = \frac{100}{60} = 5:3 \). The lower-wattage bulb has the higher resistance.

11. If nucleus scaled to football size is approximately 22cm, then the radius of the orbit of electron will be approximately: [1 Mark]
A. 22 cm
B. 22 m
C. 22 km
D. 22000 km

Answer: C. 22 km

Teacher's Note:
a) Nuclear radius is about \( 10^{-15} \) m and atomic radius is about \( 10^{-10} \) m, a ratio of about \( 10^{5} \).
b) \( 0.22\ \text{m} \times 10^{5} = 22000\ \text{m} = 22\ \text{km} \).

12. A current of 5A is passing through a long wire which has semicircular loop of radius 10 cm as shown in the figure-
The magnetic field produced at the center of the loop is- [1 Mark]
A. 2 π μ T
B. 4 π μ T
C. 5 π μ T
D. 8 π μ T

[Figure: A long straight wire carrying a current of 5A (arrow pointing right) rises into a semicircular loop of radius 10 cm and then continues as a straight wire. The radius of 10 cm is drawn from the centre of the semicircle.]

Answer: C. 5 π μ T

Teacher's Note:
a) For a semicircular loop, \( B = \frac{\mu_0 I}{4R} \); the straight parts pass through the centre line and add nothing.
b) \( B = \frac{4\pi \times 10^{-7} \times 5}{4 \times 0.10} = 5\pi \times 10^{-6}\ \text{T} = 5\pi\ \mu\text{T} \).

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

13. Assertion (A): In Rutherford's scattering experiment, most of the alpha particles passed through the gold foil undeflected.
Reason (R): The electrons in the atom are very light and do not affect the path of the alpha particles significantly. [1 Mark]
A. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false.
D. both Assertion and Reason are false.

Answer: B. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Teacher's Note:
a) Most alpha particles pass undeflected because most of the atom is empty space and the nucleus is tiny.
b) The light electrons do not deflect the energetic alpha particles, which is true but is not the reason asked for.

14. A convex lens of focal length 20 cm is used to form a real image of a point object placed 30 cm from the lens. The lens is then immersed in water of refractive index 1.33 (the lens material has refractive index 1.5).
Assertion (A): The focal length of the lens increases when it is immersed in water.
Reason (R): The focal length of a lens depends on the relative refractive index of the lens material with respect to the surrounding medium, as given by the lens maker's formula. [1 Mark]
A. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false.
D. both Assertion and Reason are false.

Answer: A. both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Teacher's Note:
a) Lens maker's formula uses \( \left( \frac{\mu_{lens}}{\mu_{medium}} - 1 \right) \).
b) In water this factor becomes smaller, so the focal length increases.

15. In Young's double slit experiment, the distance between the slits is 0.25mm and the screen is placed 1.0m away. A student uses monochromatic light of wavelength 500nm.
Assertion (A): The fringe width observed on the screen will be 2.0mm.
Reason (R): The fringe width in YDSE is directly proportional to the slit separation and inversely proportional to the wavelength of light. [1 Mark]
A. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false.
D. both Assertion and Reason are false.

Answer: C. Assertion is true but Reason is false.

Teacher's Note:
a) \( \beta = \frac{\lambda D}{d} = \frac{500 \times 10^{-9} \times 1.0}{0.25 \times 10^{-3}} = 2 \times 10^{-3}\ \text{m} = 2.0\ \text{mm} \), so A is true.
b) Fringe width is inversely proportional to slit separation and directly proportional to wavelength, so R is false.

16. Assertion (A): If a conductor is placed in an external electric field, the electric field inside the conductor becomes infinity in electrostatic equilibrium.
Reason (R): Free electrons inside the conductor rearrange themselves in such a way that the internal electric field is in the same direction as that of the external field. [1 Mark]
A. both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false.
D. both Assertion and Reason are false.

Answer: D. both Assertion and Reason are false.

Teacher's Note:
a) Free charges rearrange until their field exactly cancels the external field.
b) So the net electric field inside a conductor in electrostatic equilibrium is zero, not infinity.

SECTION B

17. The electric and magnetic fields associated with an electromagnetic radiation in a medium are given as follows:
E= 30 sin \( (2\pi \times 10^{18}\ \text{rad s}^{-1}\ t + \pi \times 10^{10}\ \text{m}^{-1}\ x) \) Vm^-1
B= 10^-7 sin \( (2\pi \times 10^{18}\ \text{rad}^{-1}\ t + \pi \times 10^{10}\ \text{m}^{-1}\ x) \) T
I. What is the refractive index of the medium through which the wave is propagating?
II. Find the frequency of the wave and identify the electromagnetic wave. [2 Marks]

Answer:
I. From the equation, \( \omega = 2\pi \times 10^{18}\ \text{rad s}^{-1} \) and \( k = \pi \times 10^{10}\ \text{m}^{-1} \).
Speed in the medium: \( v = \frac{\omega}{k} = \frac{2\pi \times 10^{18}}{\pi \times 10^{10}} = 2 \times 10^{8}\ \text{m/s} \)
Refractive index: \( \mu = \frac{c}{v} = \frac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5 \)
II. \( 2\pi\nu = 2\pi \times 10^{18}\ \text{rad s}^{-1} \)
\( \nu = 10^{18}\ \text{s}^{-1} = 10^{18}\ \text{Hz} \)
This frequency corresponds to X-rays.

Teacher's Note:
a) Read \( \omega \) (coefficient of t) and k (coefficient of x) directly from the equation; then \( v = \frac{\omega}{k} \).
b) Do not take \( 2\pi \times 10^{18} \) as the frequency; divide by \( 2\pi \) to get \( \nu \).
c) Remember the range: X-rays have frequencies of about \( 10^{16} \) to \( 10^{20} \) Hz.

18. Write the nature of path of free electrons in a conductor in the
a) presence of electric field,
b) absence of electric field. [2 Marks]

Answer:
a) In the presence of an electric field: Free electrons feel a force opposite to the field and drift slowly opposite to it. They still collide often with the atoms, so the path is zigzag (random motion) with a net drift superimposed on it. This net drift gives an electric current.
b) In the absence of an electric field: Electrons move randomly due to thermal energy, like gas molecules. There is no net movement in any direction, so no current flows.

Teacher's Note:
a) Keywords: "zigzag path with net drift opposite to E" and "random motion with no net drift".
b) Electrons drift opposite to the field because they are negatively charged.

19. Two like charges +q and +q are placed at a distance d. What work must be done by an external agent to bring another charge +q from infinity to the midpoint of the line joining the two given charges? [2 Marks]

Answer:
1. At the midpoint, the distance from each charge is \( \frac{d}{2} \).
2. Potential due to one charge: \( V_1 = \frac{kq}{d/2} = \frac{2kq}{d} \)
3. Potential due to both charges: \( V = 2 \times \frac{2kq}{d} = \frac{4kq}{d} \)
4. Work done by the external agent: \( W = qV = q \left( \frac{4kq}{d} \right) = \frac{4kq^{2}}{d} \), where \( k = \frac{1}{4\pi\varepsilon_0} \).

Teacher's Note:
a) Potential is a scalar, so the potentials of the two charges simply add.
b) The electric fields cancel at the midpoint, but the potential does not; do not write W = 0.

20. (A) A certain magnetic substance is found to have a relative permeability of 800. What type of magnetic material is it? Write any two characteristics that such a material exhibits. [2 Marks]

Answer:
Given \( \mu_r = 800 \). Since the relative permeability is much greater than 1, it is a ferromagnetic material.
Characteristics:
1. It is strongly attracted by a magnetic field because it has a very high magnetic permeability.
2. It can retain magnetism even after the external field is removed (it shows hysteresis), so it can be made into a permanent magnet.

Teacher's Note:
a) Diamagnetic: \( \mu_r \) slightly less than 1; paramagnetic: slightly more than 1; ferromagnetic: \( \mu_r \gg 1 \).
b) Identifying the material carries 1 mark; each characteristic carries half a mark.

OR

(B) State Gauss's law of magnetism. Also discuss its significance. [2 Marks]

Answer:
Gauss's law for magnetism: the net magnetic flux through any closed surface is always zero.
\( \phi_B = \oint \vec{B} \cdot d\vec{S} = 0 \)
This means that magnetic field lines neither begin nor end at any point.
Significance:
1. Magnetic monopoles do not exist.
2. Magnetic field lines always form continuous closed loops.

Teacher's Note:
a) Compare with Gauss's law in electrostatics, where the flux equals \( \frac{q}{\varepsilon_0} \); in magnetism the "enclosed pole" is always zero.
b) Write the integral form with the closed-surface symbol \( \oint \).

21. (A) I. Why hydrogen spectral lines are considered to be a fingerprint of the element?
II. If energy of an electron of the hydrogen atom in ground state is -13.6 eV, what should be the minimum energy of photon which can ionize the atom? [2 Marks]

Answer:
I. Each spectral line comes from a transition between the quantized energy levels of the hydrogen atom. This gives a unique set of lines (line spectrum) for hydrogen, and this distinct pattern identifies hydrogen wherever it is found. So the lines act like a fingerprint.
II. The minimum photon energy needed is 13.6 eV. This is the ionization energy needed to remove the electron from the ground state, assuming all the photon energy goes to the electron.

Teacher's Note:
a) Ionization means taking the electron from n = 1 (-13.6 eV) to n = \( \infty \) (0 eV), so energy needed = 0 - (-13.6) = 13.6 eV.
b) Keyword for part I: every element has its own unique set of energy levels, so its own unique line spectrum.

OR

(B) State Bohr's quantization condition for defining stationary orbit. How does de Broglie hypothesis explain the stationary orbit? [2 Marks]

Answer:
Bohr's condition: Electrons revolve in certain discrete circular orbits without radiating energy, for which the angular momentum is an integral multiple of \( \frac{h}{2\pi} \):
\( mvr = \frac{nh}{2\pi} \), where m = mass of electron, v = its speed, r = orbit radius and n = principal quantum number.
de Broglie explanation: A moving electron behaves as a matter wave of wavelength \( \lambda = \frac{h}{mv} \). For a stable orbit, the electron wave must form a standing wave, so the circumference must be an integral multiple of the wavelength:
\( 2\pi r = n\lambda = \frac{nh}{mv} \)
or \( mvr = \frac{nh}{2\pi} \), which is exactly Bohr's quantization condition. So stationary orbits are those in which the electron wave forms a standing wave.

Teacher's Note:
a) Key steps that earn marks: \( \lambda = \frac{h}{mv} \), \( 2\pi r = n\lambda \), then \( mvr = \frac{nh}{2\pi} \).
b) Use the word "standing wave" in the explanation.

SECTION C

22. Compare the energy band diagrams of metals, insulators, and semiconductors.
I. Explain how the band gap determines conductivity.
II. Apply this reasoning to justify why diamond (C) is an insulator while tin (Sn) is a conductor, though both are the elements of group IV. [3 Marks]

Answer:
1. Metals: The valence and conduction bands overlap, or the conduction band is partly filled. Many free electrons are available, so conductivity is high.
2. Insulators: There is a large band gap. Almost no electrons are thermally excited at room temperature, so conductivity is very poor.
3. Semiconductors: The band gap is moderate. Carriers are excited thermally or optically, and conductivity can be changed by doping.
I. So the size of the band gap controls how many carriers are available at room temperature, and hence the conductivity.
II. Diamond (C) has a strong covalent network and a large band gap (about 5.4 eV), so it is an insulator at room temperature. Tin (Sn) has metallic bonding with overlapping bands, so electrons are freely available and it is a conductor.

Teacher's Note:
a) Remember: overlap (metal), small gap below about 3 eV (semiconductor), large gap above about 3 eV (insulator).
b) In the diamond vs tin part, quote the band gap value of diamond (about 5.4 eV).

23. Explain-
I. It is easier to start a car engine on a warm day than on a chilly day.
II. The resistance of our body is so large, even then one experiences a strong shock (sometimes even fatal)) when one accidentally touches the live wire of, say 240 V supply.
III. Heat is generated continuously in an electric heater but its temperature becomes constant after some time. [3 Marks]

Answer:
I. The internal resistance of a car battery decreases as temperature increases. So on a warm day the battery can give a larger current, which helps to start the engine.
II. Our body is sensitive to very small currents, even a few mA. Even with a large body resistance, 240 V can drive such a current through the body, so the shock is strong.
III. When the heater becomes hotter than its surroundings, it starts losing heat to them. Soon the rate of heat production becomes equal to the rate of heat loss, so the temperature becomes constant.

Teacher's Note:
a) Keywords: "internal resistance decreases with temperature", "body sensitive to a few mA", "rate of heat produced = rate of heat lost".
b) Each part carries 1 mark, so write one clear reason for each.

24. The atomic mass of \( {}^{16}_{8}\text{O} \) is 15.99493 u. Calculate the binding energy per nucleon of the oxygen nucleus.
(Given: mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/c²) [3 Marks]

Answer:
Given: Z = 8 protons, N = 16 - 8 = 8 neutrons.
Step 1: Sum of nucleon masses:
\( Zm_p + Nm_n = 8(1.00727) + 8(1.00866) = 8.05816 + 8.06928 = 16.12744 \) u
Step 2: Mass defect:
\( \Delta m = 16.12744 - 15.99493 = 0.13251 \) u
Step 3: Binding energy:
\( B = \Delta m \times 931.5 = 0.13251 \times 931.5 \approx 123.4 \) MeV
Step 4: Binding energy per nucleon:
\( \frac{B}{A} = \frac{123.4}{16} \approx 7.71 \) MeV per nucleon

Teacher's Note:
a) Marks are given for the mass defect, the total binding energy and the value per nucleon, 1 mark each.
b) Divide by the mass number A = 16, not by the atomic number Z = 8.
c) Quick check: about 7 to 9 MeV per nucleon is normal for most nuclei.

25. A diamond has a refractive index of 2.42.
I. A ray of light inside the diamond strikes the surface at an angle of incidence of \( 30^{\circ} \) .State whether it will undergo refraction or total internal reflection. [2 Marks]
II. Explain why diamonds sparkle when cut properly. [1 Mark]

Answer:
I. Critical angle: \( \sin\theta_c = \frac{1}{\mu} = \frac{1}{2.42} \approx 0.413 \)
\( \theta_c = \sin^{-1}\left( \frac{1}{2.42} \right) \approx 24.4^{\circ} \), so \( \theta_c \lt 30^{\circ} \).
The angle of incidence \( 30^{\circ} \) is greater than \( \theta_c \), so the ray will undergo total internal reflection (TIR) inside the diamond.
II. A well-cut diamond has facets angled so that light hitting them is totally internally reflected back inside instead of leaking out of the sides or bottom. The light is trapped and comes out only through the top, so the diamond sparkles.

Teacher's Note:
a) TIR needs two conditions: light going from denser to rarer medium and angle of incidence greater than the critical angle.
b) Always find \( \theta_c \) first and compare it with the given angle.

26. Derive Snell's law of refraction using Huygens' wave theory. What prediction of the wave theory was later experimentally verified? [3 Marks]

Answer:
Huygens' principle: Every point on a wavefront acts as a source of secondary wavelets. The new wavefront at a later time is the common tangent (envelope) to these wavelets.
Diagram (in words): A horizontal boundary separates medium 1 (above, speed \( v_1 \)) from medium 2 (below, speed \( v_2 \)). A plane incident wavefront AB meets the boundary at A, making angle i with it. B reaches C on the boundary after time t. From A, a secondary wavelet of radius AE spreads into medium 2. CE is the refracted wavefront, making angle r with the boundary.
1. Let A strike the boundary first. In time t, B moves to C: \( BC = v_1 t \). In the same time, the wavelet from A grows to radius \( AE = v_2 t \).
2. From the geometry: \( \sin i = \frac{BC}{AC} \) and \( \sin r = \frac{AE}{AC} \)
\( \frac{\sin i}{\sin r} = \frac{BC}{AE} = \frac{v_1 t}{v_2 t} = \frac{v_1}{v_2} \)
3. Since \( n = \frac{c}{v} \), \( \frac{v_1}{v_2} = \frac{n_2}{n_1} \). Therefore \( n_1 \sin i = n_2 \sin r \). This is Snell's law of refraction.
Prediction verified: Wave theory predicted that light travels slower in an optically denser medium (\( v \propto \frac{1}{n} \)). Foucault (and later Fizeau) showed that the speed of light in water is less than in air, which contradicted Newton's corpuscular theory.

Teacher's Note:
a) Draw and label the diagram: AB, C, E, angles i and r, \( v_1 t \) and \( v_2 t \).
b) Common mistake: writing \( \frac{\sin i}{\sin r} = \frac{v_2}{v_1} \); the faster medium has the larger angle.

27. (A) I. An ammeter has an internal resistance of 13 Ω and is originally designed to measure currents only up to 100 A. To extend its range for high-current measurements, a shunt resistor is connected in parallel with it. After adding the shunt, the same meter can now read currents as high as 750 A without damage. Using the principle of current division in parallel circuits, determine the value of the shunt resistance required to increase the measuring range of the ammeter.
II. The coil of a moving coil galvanometer is wound over a metal frame. Why? [3 Marks]

Answer:
I. Treat the original ammeter as a galvanometer: G = 13 Ω, \( I_g = 100 \) A, I = 750 A.
The shunt carries \( I - I_g \), and the meter and shunt have the same voltage:
\( S = \frac{I_g G}{I - I_g} \)
\( S = \frac{100 \times 13}{750 - 100} = \frac{1300}{650} = 2\ \Omega \)
II. To provide electromagnetic damping. When the coil moves, eddy currents are induced in the metal frame. These currents oppose the motion and bring the coil to rest quickly.

Teacher's Note:
a) Write the formula first, then substitute and give the answer with the unit Ω.
b) Keyword for part II: "electromagnetic damping" by induced (eddy) currents in the frame.

OR

(B) I. A small current loop behaves like a tiny magnet when placed in a magnetic field. Using this idea, explain what physical quantity determines the strength and orientation of such a magnetic dipole. Also state the SI unit of this quantity.
II. A steel wire of length l has a magnetic moment m. It is bent into a semicircular arc. What is its new magnetic moment? [3 Marks]

Answer:
I. The magnetic dipole moment determines the strength and orientation of the dipole. SI unit: ampere metre² (A m²) or joule per tesla (J T^-1).
II. Pole strength: \( q_m = \frac{m}{l} \)
When bent into a semicircle, the distance between the poles becomes the diameter 2r.
Length of arc: \( l = \pi r \), so \( r = \frac{l}{\pi} \)
New magnetic moment: \( m' = q_m \times 2r = \frac{m}{l} \times \frac{2l}{\pi} = \frac{2m}{\pi} \)

Teacher's Note:
a) Pole strength does not change on bending; only the distance between the poles changes.
b) The new moment \( \frac{2m}{\pi} \) is less than m, which is a quick check.

28. I. A bar magnet falls from a height through a metal ring. Its acceleration will be less than 'g'. Explain.
II. A coil of copper wire is wound uniformly on a wooden equilateral triangular frame. If each side of the triangle is increased to 5 times its original length, while keeping the number of turns per unit length unchanged, how does the self-inductance of the coil change? [3 Marks]

Answer:
I. As the magnet falls, the magnetic flux linked with the ring increases. By Lenz's law, the current induced in the ring opposes the downward motion of the magnet. So its acceleration is less than g.
II. Self-inductance: \( L \propto \frac{N^{2}A}{l} \)
With turns per unit length fixed, N becomes 5N.
Area varies as the square of the side: A becomes \( 5^{2}A = 25A \).
Magnetic path length: l becomes 5l.
\( \frac{L'}{L} = \frac{(5N)^{2} \times 25A / 5l}{N^{2}A / l} = \frac{25 \times 25}{5} = 125 \)
So \( L' = 125L \); the self-inductance becomes 125 times.

Teacher's Note:
a) Part I: name Lenz's law and say the induced current opposes the cause (the fall of the magnet).
b) Part II: change N, A and l together; forgetting any one of them gives a wrong factor.

SECTION D

29.

In 2023, researchers developed flexible semiconductor films using organic polymers that could be printed like newspaper sheets. These films can bend, fold, and still conduct electricity efficiently. Unlike traditional silicon chips, polymer semiconductors are lightweight, biodegradable, and can be integrated into clothing or medical patches. Imagine a shirt that monitors your heartbeat or a bandage that tracks wound healing in real time. Such innovations extend the idea of semiconductors beyond rigid circuits, showing how manipulating band gaps and carrier mobility in new materials can revolutionize electronics for healthcare, sustainability, and wearable technology.

I. Which advantage of semiconductors makes them ideal for wearable devices like smart shirts? [1 Mark]
A. High voltage operation (~100 V)
B. Bulky design and short life
C. Low power consumption and reliability
D. Requirement of external heating

Answer: C. Low power consumption and reliability

Teacher's Note:
a) Wearable devices must work continuously on very little energy and perform dependably.
b) The other options are disadvantages, not advantages.

II. Carbon (diamond) has a band gap of 5.4 eV, while silicon has 1.1 eV. Why is silicon used in electronic devices but not the diamond? [1 Mark]
A. Diamond has too many free electrons.
B. Diamond's band gap is too large for thermal excitation at room temperature.
C. Silicon has weaker covalent bonds than diamond.
D. Diamond cannot form covalent bonds.

Answer: B. Diamond's band gap is too large for thermal excitation at room temperature.

Teacher's Note:
a) With a 5.4 eV gap, almost no charge carriers are produced at room temperature.
b) Silicon's small gap (1.1 eV) lets enough electrons cross into the conduction band.

III. In an intrinsic semiconductor at room temperature [1 Mark]
A. only electrons contribute to conduction.
B. only holes contribute to conduction.
C. number of electrons equals number of holes.
D. no conduction occurs at all.

Answer: C. number of electrons equals number of holes.

Teacher's Note:
a) In a pure semiconductor, electrons and holes are created in pairs, so \( n_e = n_h = n_i \).
b) Both electrons and holes contribute to conduction.

IV. How will you use the concept of majority and minority carriers to explain the efficiency of polymer semiconductors in wearable technology? [1 Mark]

Answer: In polymer semiconductors, conduction is mainly due to majority carriers (electrons in n-type, holes in p-type). By controlling their concentration and mobility, the polymer conducts efficiently even when bent or stretched. Minority carriers take part in recombination and stability. This balance lets polymer semiconductors work reliably at low power, which suits flexible, lightweight wearables such as smart clothing and medical patches.

Teacher's Note:
a) Keywords: majority carriers carry the current; minority carriers affect recombination and stability.
b) Link the answer to low power and flexibility, as the passage asks about wearable technology.

30.

In 1905, Albert Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. Each photon carries an energy \( E = h\nu \). When a photon strikes a metal surface, it can transfer its energy to an electron. If the photon's energy exceeds the work function of the metal, the electron can be emitted with maximum kinetic energy \( K_{max} = h\nu - \phi_0 \). This discovery challenged the classical wave theory, which predicted that electron emission should depend only on the intensity of light. Instead, experiments showed that emission of electrons depends on frequency too, proving the particle nature of light.

I. Define the term work function. [1 Mark]

Answer: The work function (\( \phi_0 \)) of a metal is the minimum energy required to remove an electron from the metal surface so that it just escapes with zero kinetic energy.

Teacher's Note:
a) Keywords: "minimum energy" and "from the surface of the metal".
b) It is usually given in eV and depends on the metal.

II. Why the increase in the intensity of light does not increase the maximum kinetic energy of photoelectrons? [1 Mark]

Answer: Increasing intensity increases only the number of photons, not the energy of each photon. Photon energy depends only on frequency (\( E = h\nu \)), so the maximum kinetic energy depends only on frequency. Higher intensity only increases the number of emitted electrons (photoelectric current).

Teacher's Note:
a) One photon is absorbed by one electron, so the energy per electron is fixed by \( h\nu \).
b) Intensity affects the current; frequency affects \( K_{max} \).

III. A metal surface has a work function of 2 eV. The light of frequency 1.2 ×10¹⁵Hz falls on it. Which statement is correct? [1 Mark]
A. No electron is emitted because the intensity is low.
B. Electrons are emitted with maximum kinetic energy less than 2 eV.
C. Electrons are emitted with maximum kinetic energy greater than 0 eV.
D. Electron emission depends only on the intensity of light and not on the frequency.

Answer: C. Electrons are emitted with maximum kinetic energy greater than 0 eV.

Teacher's Note:
a) Photon energy: \( E = h\nu \approx 4.14 \times 10^{-15}\ \text{eV s} \times 1.2 \times 10^{15}\ \text{Hz} \approx 5 \) eV.
b) \( K_{max} = E - \phi_0 = 5 - 2 = 3 \) eV, which is greater than 0 eV.

IV. Which observation/s directly contradicts the classical wave theory of light? [1 Mark]
A. Photoelectric current increases with intensity
B. The maximum kinetic energy depends only on the frequency and not on the intensity.
C. The electrons are emitted instantaneously.
D. Both B and C.

Answer: D. Both B and C.

Teacher's Note:
a) Wave theory predicts K_max should depend on intensity and that emission should take time; both are wrong.
b) Option A (current increases with intensity) does not contradict wave theory.

SECTION E

31. (A) I. An isolated conductor cannot have a large capacitance. Why? [2 Marks]
II. How would you connect 6 μF, 9 μF and 18 μF capacitors to obtain- [2 Marks]
i. the minimum capacitance
ii. the maximum capacitance
III. The capacitance of a parallel plate capacitor is increased as the area of the plates increases. What are the other two factors on which the capacitance depends? [1 Mark]

Answer:
I. The capacitance of an isolated conductor depends directly on its size. For an isolated sphere of radius R:
\( C = 4\pi\varepsilon_0 R \)
Since \( \varepsilon_0 \) is a constant, capacitance can be increased only by increasing R. For example, even a sphere of radius 1 m has \( C \approx 1.1 \times 10^{-10} \) F. A very large capacitance would need an extremely large conductor, which is not practical. So an isolated conductor always has a small capacitance.
II. i. For minimum capacitance, connect all three in series:
\( \frac{1}{C_{min}} = \frac{1}{6} + \frac{1}{9} + \frac{1}{18} = \frac{3 + 2 + 1}{18} = \frac{6}{18} = \frac{1}{3} \)
\( C_{min} = 3\ \mu\text{F} \)
ii. For maximum capacitance, connect all three in parallel:
\( C_{max} = 6 + 9 + 18 = 33\ \mu\text{F} \)
III. The other two factors are:
1. Distance between the plates, d (\( C \propto \frac{1}{d} \)).
2. Permittivity of the medium between the plates, \( \varepsilon \) (\( C \propto \varepsilon \)).

Teacher's Note:
a) Series gives a value smaller than the smallest capacitor; parallel gives the sum.
b) Use \( C = \frac{\varepsilon A}{d} \) to recall all three factors of a parallel plate capacitor.

OR

(B) I. Show that the electric field at any point is equal to the negative of the potential gradient at that point. [3 Marks]
II. A small metallic sphere is gradually charged in open air. It is observed that electrical breakdown of air occurs when the electric field at its surface reaches 2.5 x 10⁶ V m^-1 .If the sphere is allowed to attain a maximum potential of 5 x 10⁵ V without causing any discharge, determine the minimum radius the insulated sphere must have. [2 Marks]

Answer:
I. 1. Consider a small positive charge +q placed in an electric field \( \vec{E} \), displaced through a very small distance dr along the field.
2. Force on the charge: \( \vec{F} = q\vec{E} \)
3. Work done by the electric field: \( dW = \vec{F} \cdot d\vec{r} = qE\,dr \) ...(1)
4. Change in electric potential: \( dV = -\frac{dW}{q} \)
5. From (1): \( dV = -\frac{qE\,dr}{q} = -E\,dr \)
6. So \( E = -\frac{dV}{dr} \), i.e. the electric field at any point equals the negative of the potential gradient at that point.
II. Given: maximum field (potential gradient) \( E = \frac{dV}{dr} = 2.5 \times 10^{6}\ \text{V m}^{-1} \) and V = \( 5 \times 10^{5} \) V.
For the sphere: \( V = E\,r \)
\( r = \frac{V}{E} = \frac{5 \times 10^{5}}{2.5 \times 10^{6}} \)
\( r = 2 \times 10^{-1}\ \text{m} = 0.2\ \text{m} \)

Teacher's Note:
a) The negative sign shows that the field points in the direction of decreasing potential.
b) For a charged sphere, \( E = \frac{kQ}{r^{2}} \) and \( V = \frac{kQ}{r} \), so \( V = Er \); this gives the radius in one step.

32. (A) A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece lens of focal length 5 cm. The object is placed 2.2 cm from the objective lens.
I. Calculate the position of the image formed by the objective lens. [2 Marks]
II. The eyepiece is so placed that the final image is formed at infinity. Find the magnifying power of the microscope. [2 Marks]
III. Explain briefly why the compound microscope provides much higher magnification compared to a simple magnifier. [1 Mark]

Answer:
Given: \( f_o = 2 \) cm, \( f_e = 5 \) cm, \( u_o = -2.2 \) cm.
I. Lens formula: \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \)
\( \frac{1}{2} = \frac{1}{v_o} - \frac{1}{(-2.2)} \)
\( \frac{1}{v_o} = \frac{1}{2} - \frac{1}{2.2} = \frac{2.2 - 2}{4.4} = \frac{0.2}{4.4} \)
\( v_o = 22 \) cm
The image formed by the objective is 22 cm from it (on the other side, real).
II. For final image at infinity: \( M = \left( \frac{v_o}{|u_o|} \right) \left( \frac{D}{f_e} \right) \), with D = 25 cm.
\( M = \left( \frac{22}{2.2} \right) \left( \frac{25}{5} \right) = 10 \times 5 = 50 \)
Magnifying power of the microscope = 50.
III. A compound microscope uses two lenses. The objective forms a highly magnified real image of the tiny object, and the eyepiece magnifies this image further. The total magnification is the product of the two magnifications, so it is much greater than that of a simple magnifier, which uses only one lens.

Teacher's Note:
a) Use the sign convention: the object distance is negative, \( u_o = -2.2 \) cm.
b) \( m_o = \frac{v_o}{|u_o|} = 10 \) and \( m_e = \frac{D}{f_e} = 5 \) for normal adjustment; multiply them.

OR

(B) An astronomical telescope uses an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm.
I. Calculate the magnifying power of the telescope when the final image is formed at infinity. [2 Marks]
II. If the least distance of distinct vision of the observer is 25 cm, calculate the magnifying power when the final image is formed at this distance. [2 Marks]
III. Explain why the objective lens of an astronomical telescope is made with a large focal length and aperture, while the eyepiece has a small focal length. [1 Mark]

Answer:
Given: \( f_o = 100 \) cm, \( f_e = 5 \) cm, D = 25 cm.
I. Normal adjustment (final image at infinity):
\( M = \frac{f_o}{f_e} = \frac{100}{5} = 20 \)
Magnifying power = 20.
II. Final image at the least distance of distinct vision:
\( M = \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right) \)
\( M = \frac{100}{5} \left( 1 + \frac{5}{25} \right) = 20 \times 1.2 = 24 \)
Magnifying power = 24.
III. The objective has a large focal length to give a large magnified real image of distant objects, and a large aperture to collect more light, giving a brighter image with better resolution. The eyepiece has a small focal length so that it magnifies the image formed by the objective further. Together they give high magnifying power and a bright, clear image.

Teacher's Note:
a) Magnifying power is larger when the final image is at D than at infinity.
b) Large aperture improves brightness and resolving power; it does not change the magnifying power.

33. (A) A certain electrical device is used to convert an alternating voltage of smaller magnitude into a much larger alternating voltage without violating the law of conservation of energy.
I. With the help of a neat labelled diagram, describe the principle on which this device operates and explain how it increases the voltage. [2 Marks]
II. State the basic working mechanism of the device and mention the condition required in its input signal for proper functioning. [2 Marks]
III. Give one reason why this device does not operate with 100% efficiency in practical situations. [1 Mark]

Answer:
The device is a step-up transformer.
I. Diagram (in words): A rectangular laminated soft iron core. The primary coil (fewer turns) is wound on one arm and the secondary coil (more turns) on the opposite arm. The input a.c. is connected to the primary, and the output is taken from the secondary.
Principle - mutual induction: An alternating current in the primary produces a time-varying magnetic flux in the core. This changing flux links the secondary, and by Faraday's law an emf is induced in the secondary.
For an ideal transformer: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). If \( N_s \gt N_p \), then \( V_s \gt V_p \), so the voltage is stepped up.
Energy is conserved because, in an ideal transformer, \( P_{in} = P_{out} \Rightarrow V_p I_p = V_s I_s \); when the voltage increases, the current decreases.
II. Working mechanism: The primary is supplied with an alternating voltage, which produces an alternating magnetic field in the core. The changing flux linked with the secondary induces an alternating emf in it. So energy passes from primary to secondary through the changing magnetic field, without any electrical contact between the windings.
Condition: The input must be alternating current (a.c.), not d.c. With d.c. the flux does not change, so no emf is induced in the secondary.
III. Copper losses (heating of the windings due to their resistance). Because of such losses, the output power is less than the input power, so efficiency is less than 100%.

Teacher's Note:
a) Label the primary coil, secondary coil and laminated core in the diagram.
b) Other accepted losses: eddy current loss, hysteresis loss and flux leakage; any one is enough.
c) A transformer steps up voltage, not power; always mention that current is stepped down.

OR

(B) I. What do you mean by the resonance condition of a series LCR-circuit? [3 Marks]
II. Write an expression for the resonant frequency.
Describe the use of a series resonant circuit in the tuning of a radio receiver. [2 Marks]

Answer:
I. A series LCR circuit has a resistor R, an inductor L and a capacitor C in series with an a.c. source. The circuit is in resonance when the inductive reactance equals the capacitive reactance:
\( X_L = X_C \) or \( \omega L = \frac{1}{\omega C} \), where \( \omega \) is the angular frequency of the a.c. supply.
At resonance:
1. Net reactance \( X = X_L - X_C = 0 \).
2. Impedance Z = R, which is the minimum.
3. Current is maximum for the given applied voltage, and the circuit behaves as purely resistive.
II. Resonant frequency: \( f_0 = \frac{\omega_0}{2\pi} = \frac{1}{2\pi\sqrt{LC}} \)
Use in tuning: Each radio station transmits at a definite frequency. The input stage of the receiver has a series LCR (or LC) circuit with a fixed inductor and a variable capacitor. By adjusting C, \( f_0 \) is changed until it equals the frequency of the wanted station. Then the circuit is at resonance for that station: impedance is minimum, so the current (and signal) is maximum. Other stations are not at resonance, so their currents are small and they are rejected. In this way the receiver selects one station.

Teacher's Note:
a) Keywords for resonance: \( X_L = X_C \), Z = R (minimum), current maximum.
b) In tuning, mention "variable capacitor" and "resonant frequency equals station frequency".