CBSE Class 12 Physics: sample paper 2026-27

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQElectric Charges and Fields

   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. (A)become half
   2. (B)become double
   3. (C)remain the same
   4. (D)become four times

   Try it first, then show our solutionSolution

   1. By Gauss's law, $\Phi =\frac{Q_{enc}}{{\epsilon }_0}$, which depends only on the charge enclosed. It does not depend on the size of the surface.

      ½ mark

   2. Doubling $R$ leaves the enclosed charge $+Q$ unchanged, so the flux remains the same. Option (C).

      ½ mark

   Answer: Remains the same, option (C)

   Where the marks are

   • Uses Gauss's law: flux depends only on the enclosed charge, so it is unchanged (1 mark)

   Open Q1 on its own page

2. Q2 · 1 mark · MCQCurrent Electricity

   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. (A)200 W
   2. (B)250 W
   3. (C)2000 W
   4. (D)2500 W

   Try it first, then show our solutionSolution

   1. In parallel, each heater gets its rated 220 V, so each consumes its rated 1000 W.

      ½ mark

   2. Total power is $P=P_1+P_2=1000+1000=2000$ W. Option (C).

      ½ mark

   Answer: 2000 W, option (C)

   Where the marks are

   • Uses $P=P_1+P_2$ for a parallel combination to get 2000 W (1 mark)

3. Q3 · 1 mark · MCQElectromagnetic Induction

   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. (A)clockwise of the negative y-axis
   2. (B)anticlockwise of the negative y-axis
   3. (C)no current will be induced
   4. (D)along the direction of the magnetic field

   Try it first, then show our solutionSolution

   1. The field is uniform and constant, and the cylinder is stationary, so the magnetic flux linked with it does not change.

      ½ mark

   2. Since $\frac{d\Phi }{dt}=0$, no emf is induced and hence no current flows. Option (C).

      ½ mark

   Answer: No current will be induced, option (C)

   Where the marks are

   • States that the flux is constant, so no current is induced (1 mark)

   Open Q3 on its own page

4. Q4 · 1 mark · MCQElectromagnetic Waves

   If electric field associated with an electromagnetic wave is given by $\mathbf{E}=30\sin (0.2\pi \times 10^{-4}{\mathrm{m}}^{-1}x-0.6\pi \times 10^4\mathrm{Hz}t)\hat{j}$ then the direction of propagation of the wave is along,

   1. (A)$\hat{i}$
   2. (B)$-\hat{i}$
   3. (C)$\hat{k}$
   4. (D)$-\hat{k}$

   Try it first, then show our solutionSolution

   1. The wave has the form $\sin (kx-\omega t)$. A phase of this form describes a wave travelling in the $+x$ direction.

      ½ mark

   2. So the wave propagates along $\hat{i}$. Option (A).

      ½ mark

   Answer: $\hat{i}$, option (A)

   Where the marks are

   • Recognises the form $(kx-\omega t)$ as propagation along $+x$ (1 mark)

   Open Q4 on its own page

5. Q5 · 1 mark · MCQWave Optics

   Which one of the following phenomenon is not explained by Huygens’ construction of wave front?

   1. (A)Diffraction
   2. (B)Refraction
   3. (C)Reflection
   4. (D)Origin of spectra

   Try it first, then show our solutionSolution

   1. Huygens' construction explains reflection, refraction and diffraction, because these are wave propagation effects.

      ½ mark

   2. The origin of spectra comes from quantised atomic energy levels, which a wavefront construction cannot explain. Option (D).

      ½ mark

   Answer: Origin of spectra, option (D)

   Where the marks are

   • Identifies origin of spectra as not explained by Huygens' construction (1 mark)

   Open Q5 on its own page

6. Q6 · 1 mark · MCQWave Optics

   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. (A)diverging spherical
   2. (B)converging spherical
   3. (C)plane
   4. (D)cylindrical

   Try it first, then show our solutionSolution

   1. A star is a point source at an enormous distance from the Earth.

      ½ mark

   2. A spherical wavefront of very large radius is practically plane, so the telescope detects a plane wavefront. Option (C).

      ½ mark

   Answer: Plane, option (C)

   Where the marks are

   • Treats the distant star as a point source at infinity, giving a plane wavefront (1 mark)

   Open Q6 on its own page

7. Q7 · 1 mark · MCQNuclei

   The binding energy per nucleon curve shows a maximum around mass number A=56. What does this imply about nuclei near iron (Fe)?

   1. (A)They are the least stable nuclei.
   2. (B)They are the most tightly bound nuclei.
   3. (C)They undergo spontaneous fission releasing maximum energy.
   4. (D)They cannot participate in nuclear reactions.

   Try it first, then show our solutionSolution

   1. Binding energy per nucleon measures how strongly each nucleon is held in the nucleus.

      ½ mark

   2. Its maximum near $A=56$ means these nuclei are the most tightly bound and most stable. Option (B).

      ½ mark

   Answer: They are the most tightly bound nuclei, option (B)

   Where the marks are

   • Links maximum binding energy per nucleon to the greatest stability (1 mark)

   Open Q7 on its own page

8. Q8 · 1 mark · MCQMagnetism and Matter

   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. (A)Direction of movement of ball: Towards the bar magnet; magnetic moment: Opposite to that of the bar magnet
   2. (B)Direction of movement of ball: Towards the bar magnet; magnetic moment: Same direction as that of the bar magnet
   3. (C)Direction of movement of ball: Away from the bar magnet; magnetic moment: Same direction as that of the bar magnet
   4. (D)Direction of movement of ball: Away from the bar magnet; magnetic moment: Opposite to that of the bar magnet

   Try it first, then show our solutionSolution

   1. Liquid nitrogen cools the ball below its critical temperature, so it becomes a superconductor and behaves as a perfect diamagnet (Meissner effect).

      ½ mark

   2. A diamagnet acquires a magnetic moment opposite to the applied field and is repelled by the magnet. So it moves away with its moment opposite to the magnet's. Option (D).

      ½ mark

   Answer: Away from the magnet; moment opposite, option (D)

   Where the marks are

   • Identifies the superconductor as diamagnetic: it is repelled and has an opposite moment (1 mark)

   Open Q8 on its own page

9. Q9 · 1 mark · MCQCurrent Electricity

   A bridge circuit is shown in the following figure. The total current flowing in the circuit is-

   1. (A)18 A
   2. (B)5 A
   3. (C)5/18 A
   4. (D)18/5 A

   Figure in the paper: Bridge between points X and Y: 2 Ω from X to top junction, 4 Ω from top junction to Y, 3 Ω from X to bottom junction, 6 Ω from bottom junction to Y, 5 Ω connecting top and bottom junctions. A 18 V battery is connected between X and Y through the bottom junction path (battery connected across X and Y). See it in the official paper.

   Try it first, then show our solutionSolution

   1. Check the bridge: $\frac{2}{4}=\frac{3}{6}$, so the bridge is balanced and the 5 Ω resistor carries no current.

      ½ mark

   2. The arms $(2+4)=6$ Ω and $(3+6)=9$ Ω are in parallel: $R=\frac{6\times 9}{15}=\frac{18}{5}$ Ω. Then $I=\frac{V}{R}=\frac{18}{18/5}=5$ A. Option (B).

      ½ mark

   Answer: 5 A, option (B)

   Where the marks are

   • Recognises the balanced bridge, finds $R=18/5$ Ω and $I=5$ A (1 mark)

   Open Q9 on its own page

10. Q10 · 1 mark · MCQCurrent Electricity

    Two electric bulbs, $B_1$ and $B_2$, are connected in series to an A.C. supply of 200 V. The power ratings printed on the bulbs are:

    • $B_1$: 200 V, 60 W
    • $B_2$: 200 V, 100 W

    Using their rated voltage–power information, the ratio of their resistances $R_1/R_2$ is-

    1. (A)5:3
    2. (B)3:5
    3. (C)3:4
    4. (D)4:3

    Figure in the paper: Series circuit of two bulbs B1 and B2 connected to a 200 V A.C. source; voltage drops V1 across B1 and V2 across B2 marked. See it in the official paper.

    Try it first, then show our solutionSolution

    1. From the ratings, $R=\frac{V^2}{P}$, and both bulbs have the same rated voltage.

       ½ mark

    2. So $\frac{R_1}{R_2}=\frac{P_2}{P_1}=\frac{100}{60}=\frac{5}{3}$, i.e. 5:3. Option (A).

       ½ mark

    Answer: 5:3, option (A)

    Where the marks are

    • Uses $R=V^2/P$ to get the ratio 5:3 (1 mark)

    Open Q10 on its own page

11. Q11 · 1 mark · MCQAtoms

    If nucleus scaled to football size is approximately 22cm, then the radius of the orbit of electron will be approximately:

    1. (A)22 cm
    2. (B)22 m
    3. (C)22 km
    4. (D)22000 km

    Try it first, then show our solutionSolution

    1. The size of an atom is about $10^{-10}$ m and the size of a nucleus is about $10^{-15}$ m, so their ratio is about $10^5$.

       ½ mark

    2. Scaled up, the orbit radius is $0.22\text{ m}\times 10^5=2.2\times 10^4$ m $=22$ km. Option (C).

       ½ mark

    Answer: 22 km, option (C)

    Where the marks are

    • Uses the atom-to-nucleus size ratio of about $10^5$ to get 22 km (1 mark)

    Open Q11 on its own page

12. Q12 · 1 mark · MCQMoving Charges and Magnetism

    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. (A)$2\pi \mu$T
    2. (B)$4\pi \mu$T
    3. (C)$5\pi \mu$T
    4. (D)$8\pi \mu$T

    Figure in the paper: A long straight wire carrying 5 A from the left, with a semicircular loop (radius 10 cm marked by an arrow from centre to arc) in the middle, then continuing as a straight wire to the right. See it in the official paper.

    Try it first, then show our solutionSolution

    1. The straight parts point towards the centre (their lines pass through it), so they give no field there. Only the semicircle contributes: $B=\frac{{\mu }_{0}I}{4R}$.

       ½ mark

    2. Substituting: $B=\frac{4\pi \times 10^{-7}\times 5}{4\times 0.10}=5\pi \times 10^{-6}$ T $=5\pi \mu$T. Option (C).

       ½ mark

    Answer: $5\pi \mu$T, option (C)

    Where the marks are

    • Uses $B={\mu }_{0}I/4R$ for the semicircle and gets $5\pi \mu$T (1 mark)

    Open Q12 on its own page

13. Q13 · 1 mark · Assertion-reasonCompetency-basedAtoms

    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. (A)both Assertion and Reason are true and Reason is the correct explanation of Assertion.
    2. (B)both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)Assertion is true but Reason is false.
    4. (D)both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. Assertion: true. Most alpha particles pass undeflected because most of the atom is empty space.

       ½ mark

    2. Reason: true, since light electrons barely deflect alpha particles. However, the empty-space explanation is the real cause, so R does not explain A. Option (B).

       ½ mark

    Answer: Option (B)

    Where the marks are

    • Both statements true, but R is not the correct explanation (1 mark)

    Open Q13 on its own page

14. Q14 · 1 mark · Assertion-reasonCompetency-basedRay Optics and Optical Instruments

    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. (A)both Assertion and Reason are true and Reason is the correct explanation of Assertion.
    2. (B)both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)Assertion is true but Reason is false.
    4. (D)both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. By the lens maker's formula, $\frac{1}{f}=\left(\frac{{\mu }_l}{{\mu }_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.

       ½ mark

    2. In air, ${\mu }_l/{\mu }_m-1=0.5$. In water, $\frac{1.5}{1.33}-1\approx 0.128$. This factor is smaller, so $f$ increases. Both statements are true and R explains A. Option (A).

       ½ mark

    Answer: Option (A)

    Where the marks are

    • Applies the lens maker's formula with relative refractive index to show $f$ increases (1 mark)

    Open Q14 on its own page

15. Q15 · 1 mark · Assertion-reasonCompetency-basedWave Optics

    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. (A)both Assertion and Reason are true and Reason is the correct explanation of Assertion.
    2. (B)both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)Assertion is true but Reason is false.
    4. (D)both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. Fringe width: $\beta =\frac{\lambda D}{d}=\frac{500\times 10^{-9}\times 1.0}{0.25\times 10^{-3}}=2.0\times 10^{-3}$ m $=2.0$ mm, so A is true.

       ½ mark

    2. From the formula, $\beta \propto \lambda$ and $\beta \propto 1/d$. R states the reverse, so R is false. Option (C).

       ½ mark

    Answer: Option (C)

    Where the marks are

    • Computes $\beta =2.0$ mm and notes that R reverses the proportionality (1 mark)

    Open Q15 on its own page

16. Q16 · 1 mark · Assertion-reasonCompetency-basedElectric Charges and Fields

    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. (A)both Assertion and Reason are true and Reason is the correct explanation of Assertion.
    2. (B)both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
    3. (C)Assertion is true but Reason is false.
    4. (D)both Assertion and Reason are false.

    Try it first, then show our solutionSolution

    1. In electrostatic equilibrium the field inside a conductor is zero, not infinite, so A is false.

       ½ mark

    2. Free electrons rearrange to produce an internal field opposite to the external field, cancelling it. R says it is in the same direction, so R is false. Option (D).

       ½ mark

    Answer: Option (D)

    Where the marks are

    • Both statements false: the internal field is zero because induced charges oppose the external field (1 mark)

    Open Q16 on its own page

Section B

1. Q17 · 2 marks · Very short answerElectromagnetic Waves

   The electric and magnetic fields associated with an electromagnetic radiation in a medium are given as follows:

   $E=30\sin (2\pi \times 10^{18}\mathrm{rad}{\mathrm{s}}^{-1}t+\pi \times 10^{10}{\mathrm{m}}^{-1}x)\mathrm{V}{\mathrm{m}}^{-1}$
   $B=10^{-7}\sin (2\pi \times 10^{18}\mathrm{rad}{\mathrm{s}}^{-1}t+\pi \times 10^{10}{\mathrm{m}}^{-1}x)\mathrm{T}$

   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.

   Try it first, then show our solutionSolution

   1. Compare with $E=E_0\sin (\omega t+kx)$: $\omega =2\pi \times 10^{18}$ rad s${}^{-1}$ and $k=\pi \times 10^{10}$ m${}^{-1}$. Speed in the medium: $v=\frac{\omega }{k}=\frac{2\pi \times 10^{18}}{\pi \times 10^{10}}=2\times 10^8$ m s${}^{-1}$.

      ½ mark

   2. Refractive index: $\mu =\frac{c}{v}=\frac{3\times 10^8}{2\times 10^8}=1.5$.

      ½ mark

   3. II. $\omega =2\pi \nu =2\pi \times 10^{18}$, so $\nu =10^{18}$ Hz.

      ½ mark

   4. A frequency of $10^{18}$ Hz (wavelength $\approx 10^{-10}$ m) lies in the X-ray region.

      ½ mark

   Answer: I. μ = 1.5; II. ν = 10^18 Hz, X-rays

   Where the marks are

   • Finds wave speed $v=\omega /k=2\times 10^8$ m/s (½ mark)
   • Gets $\mu =c/v=1.5$ (½ mark)
   • Finds $\nu =10^{18}$ Hz from $\omega =2\pi \nu$ (½ mark)
   • Identifies the wave as X-rays (½ mark)

   Open Q17 on its own page

2. Q18 · 2 marks · Very short answerCurrent Electricity

   Write the nature of path of free electrons in a conductor in the

   a) presence of electric field,
   b) absence of electric field.

   Try it first, then show our solutionSolution

   1. (a) With an electric field: each free electron still moves randomly between collisions, but between collisions it is accelerated opposite to $\vec{E}$. Its path becomes a curved zig-zag path with a slow net drift opposite to the field (drift velocity), which constitutes the current.

      1 mark

   2. (b) Without an electric field: free electrons move randomly in straight lines between collisions with the ions due to thermal energy, changing direction at each collision; the path is a random zig-zag, the average velocity is zero, so there is no net flow and no current.

      1 mark

   Answer: (a) Curved zig-zag path with net drift opposite to E; (b) random straight-line zig-zag path with zero net displacement (no current).

   Where the marks are

   • Presence of field: zig-zag (curved) path with net drift opposite to field (1 mark)
   • Absence of field: random zig-zag straight-line paths, zero average velocity, no current (1 mark)

   Open Q18 on its own page

3. Q19 · 2 marks · Very short answerElectrostatic Potential and Capacitance

   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?

   Try it first, then show our solutionSolution

   1. The midpoint is at distance $d/2$ from each charge.

      ½ mark

   2. Potential due to one charge at the midpoint: $V_1=\frac{kq}{d/2}=\frac{2kq}{d}$, where $k=\frac{1}{4\pi {\epsilon }_0}$.

      ½ mark

   3. Net potential (scalar sum): $V=2\times \frac{2kq}{d}=\frac{4kq}{d}$.

      ½ mark

   4. Work done by external agent (potential at infinity = 0): $W=qV=\frac{4k{q}^2}{d}=\frac{q^2}{\pi {\epsilon }_{0}d}$.

      ½ mark

   Answer: W = 4kq²/d = q²/(πε₀d)

   Where the marks are

   • Distance of midpoint from each charge = d/2 (½ mark)
   • Potential of one charge = 2kq/d (½ mark)
   • Total potential = 4kq/d (½ mark)
   • W = qV = 4kq²/d (½ mark)

   Open Q19 on its own page

4. Answer one of these 2 alternatives.

   Q20(A) · 2 marks · Very short answerMagnetism and Matter

   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.

   Try it first, then show our solutionSolution

   1. Since ${\mu }_r=800\gg 1$, the substance is ferromagnetic (e.g., iron, cobalt, nickel).

      1 mark

   2. Characteristic 1: It is strongly attracted by a magnet and moves from weaker to stronger regions of a non-uniform field; its susceptibility is large and positive.

      ½ mark

   3. Characteristic 2: It retains magnetism after the external field is removed (shows hysteresis / retentivity), so it can be made into a permanent magnet; it becomes paramagnetic above the Curie temperature.

      ½ mark

   Answer: Ferromagnetic material

   Where the marks are

   • Identifies material as ferromagnetic since μr ≫ 1 (1 mark)
   • Any two correct characteristics (½ + ½ mark)

   Open Q20(A) on its own page

   OR

   Q20(B) · 2 marks · Very short answerMagnetism and Matter

   State Gauss’s law of magnetism. Also discuss its significance.

   Try it first, then show our solutionSolution

   1. Gauss's law of magnetism: the net magnetic flux through any closed surface is always zero: $\oint \vec{B}\cdot d\vec{S}=0$.

      1 mark

   2. Significance 1: Isolated magnetic poles (magnetic monopoles) do not exist; poles always occur in pairs.

      ½ mark

   3. Significance 2: Magnetic field lines form continuous closed loops — as many lines enter a closed surface as leave it; they neither start nor end anywhere.

      ½ mark

   Answer: ∮B·dS = 0; monopoles do not exist and field lines form closed loops.

   Where the marks are

   • Correct statement with $\oint \vec{B}\cdot d\vec{S}=0$ (1 mark)
   • Monopoles do not exist (½ mark)
   • Field lines form closed loops (½ mark)

   Open Q20(B) on its own page

5. Answer one of these 2 alternatives.

   Q21(A) · 2 marks · Very short answerAtoms

   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?

   Try it first, then show our solutionSolution

   1. I. Hydrogen has its own unique set of quantised energy levels; each line corresponds to a definite transition $h\nu =E_i-E_f$. No other element has the same set of wavelengths, so the pattern of lines identifies hydrogen uniquely, like a fingerprint.

      1 mark

   2. II. To ionise, the electron must be raised from $E_1=-13.6$ eV to $E_{\infty }=0$. Minimum photon energy $=0-(-13.6)=13.6$ eV.

      1 mark

   Answer: I. Unique characteristic line spectrum; II. 13.6 eV

   Where the marks are

   • Unique discrete energy levels give a characteristic line spectrum (1 mark)
   • Minimum ionising photon energy = 13.6 eV (1 mark)

   Open Q21(A) on its own page

   OR

   Q21(B) · 2 marks · Very short answerAtoms

   State Bohr’s quantization condition for defining stationary orbit. How does de Broglie hypothesis explain the stationary orbit?

   Try it first, then show our solutionSolution

   1. Bohr's quantisation condition: an electron revolves without radiating only in those orbits for which its angular momentum is an integral multiple of $h/2\pi$: $mvr=\frac{nh}{2\pi }$, $n=1,2,3,\dots$

      1 mark

   2. de Broglie explanation: the electron has wavelength $\lambda =h/mv$. A stable orbit is one where this wave forms a standing wave, i.e. the circumference holds a whole number of wavelengths: $2\pi r=n\lambda =\frac{nh}{mv}\Rightarrow mvr=\frac{nh}{2\pi }$, which is exactly Bohr's condition.

      1 mark

   Answer: mvr = nh/2π; stationary orbits are those whose circumference fits a whole number of de Broglie wavelengths (2πr = nλ).

   Where the marks are

   • States $mvr=nh/2\pi$ (1 mark)
   • Uses $2\pi r=n\lambda$ with $\lambda =h/mv$ to obtain the condition (1 mark)

   Open Q21(B) on its own page

Section C

1. Q22 · 3 marks · Short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

   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.

   Try it first, then show our solutionSolution

   1. Band diagrams: Metals – conduction band partly filled or overlapping the valence band (no gap). Insulators – full valence band and empty conduction band separated by a large gap ($E_g>3$ eV). Semiconductors – similar but with a small gap ($E_g<3$ eV, e.g. Si 1.1 eV).

      1 mark

   2. I. Conductivity needs electrons in the conduction band. With no gap (metal) many free electrons exist → high conductivity. With a large gap, thermal energy at room temperature ($kT\approx 0.025$ eV) cannot excite electrons across → negligible conductivity. With a small gap, some electrons are thermally excited, leaving holes → moderate conductivity that rises with temperature and doping.

      1 mark

   3. II. Diamond (C) has a large gap ($\approx 5.4$ eV), so practically no electrons reach the conduction band at room temperature → insulator. In tin (Sn), the bands overlap (zero gap), so free electrons are always available → conductor. Thus, though both are group IV, the size of the band gap decides conductivity.

      1 mark

   Answer: Metals: overlapping bands; semiconductors: small gap; insulators: large gap. Diamond (Eg ≈ 5.4 eV) is an insulator; tin (overlapping bands, Eg ≈ 0) is a conductor.

   Where the marks are

   • Correct comparison of band diagrams of the three classes (1 mark)
   • Explains how band gap controls number of carriers and conductivity (1 mark)
   • Diamond large gap → insulator; tin overlapping bands → conductor (1 mark)

   Open Q22 on its own page

2. Q23 · 3 marks · Short answerCompetency-basedCurrent Electricity

   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.

   Try it first, then show our solutionSolution

   1. I. The internal resistance of the car battery (electrolyte) decreases as temperature rises. On a warm day it is lower, so the battery can deliver the large current needed by the starter motor more easily.

      1 mark

   2. II. Even though body resistance is large, $I=V/R$ at 240 V can give a current of several milliamperes or more, and the human body (heart and nerves) is extremely sensitive to such small currents, so the shock is strong or even fatal.

      1 mark

   3. III. As the heater gets hotter than its surroundings, it loses heat at a rate that increases with temperature. When the rate of heat loss equals the rate of heat production ($I^{2}R$), a steady state is reached and the temperature stays constant.

      1 mark

   Answer: I. Battery's internal resistance falls with temperature; II. body is sensitive to even a few mA; III. heat produced = heat lost at steady state.

   Where the marks are

   • Internal resistance decreases on warm day → larger current (1 mark)
   • Body sensitive to currents of a few mA (1 mark)
   • Steady state when rate of heat produced equals rate of heat lost (1 mark)

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3. Q24 · 3 marks · Short answerNuclei

   The atomic mass of ${}_8^{16}\mathrm{O}$ 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/$c^2$)

   Try it first, then show our solutionSolution

   1. For ${}_8^{16}$O: $Z=8$, $N=16-8=8$. Mass of nucleons: $Z{m}_p+N{m}_n=8(1.00727)+8(1.00866)=8.05816+8.06928=16.12744$ u.

      1 mark

   2. Mass defect: $\Delta m=16.12744-15.99493=0.13251$ u.

      1 mark

   3. Binding energy: $B=\Delta m\times 931.5=0.13251\times 931.5\approx 123.4$ MeV. Binding energy per nucleon: $\frac{B}{A}=\frac{123.4}{16}\approx 7.71$ MeV.

      1 mark

   Answer: ≈ 7.71 MeV per nucleon (B ≈ 123.4 MeV)

   Where the marks are

   • Total nucleon mass 16.12744 u (1 mark)
   • Mass defect 0.13251 u (1 mark)
   • B ≈ 123.4 MeV and B/A ≈ 7.71 MeV (1 mark)

   Open Q24 on its own page

4. Q25 · 3 marks · Short answerRay Optics and Optical Instruments

   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.
   II. Explain why diamonds sparkle when cut properly.

   Try it first, then show our solutionSolution

   1. Critical angle for diamond–air: $\sin {\theta }_c=\frac{1}{\mu }=\frac{1}{2.42}=0.413\Rightarrow {\theta }_c\approx 24.4^{\circ }$.

      1 mark

   2. I. The ray travels from denser (diamond) to rarer (air) medium and $i=30^{\circ }>{\theta }_c\approx 24.4^{\circ }$, so it undergoes total internal reflection, not refraction.

      1 mark

   3. II. Because diamond's critical angle is very small, a properly cut diamond has facets angled so that light entering it strikes the inner faces at angles greater than ${\theta }_c$ and is totally internally reflected many times. Light finally leaves through only a few faces, concentrated and dispersed into colours, so the diamond sparkles.

      1 mark

   Answer: θc ≈ 24.4°; I. total internal reflection; II. repeated TIR due to small critical angle and proper cut.

   Where the marks are

   • Calculates critical angle ≈ 24.4° (1 mark)
   • Concludes TIR since 30° > θc (1 mark)
   • Explains sparkle by multiple TIR from cut facets (1 mark)

   Open Q25 on its own page

5. Q26 · 3 marks · Short answerWave Optics

   Derive Snell’s law of refraction using Huygens’ wave theory. What prediction of the wave theory was later experimentally verified?

   Try it first, then show our solutionSolution

   1. Huygens' principle: every point on a wavefront is a source of secondary wavelets; the new wavefront is the common tangent (envelope) to these wavelets. Let a plane wavefront AB be incident at angle $i$ on the surface separating medium 1 (speed $v_1$) and medium 2 (speed $v_2$). (Figure: A touches the surface first; B is still in medium 1 at distance BC from the surface point C.)

      1 mark

   2. In time $t$, B reaches C: $BC=v_{1}t$. Meanwhile the wavelet from A spreads into medium 2 to radius $AE=v_{2}t$; the tangent CE is the refracted wavefront at angle $r$. From triangles ABC and AEC (common hypotenuse AC): $\sin i=\frac{BC}{AC}$, $\sin r=\frac{AE}{AC}$, so $\frac{\sin i}{\sin r}=\frac{v_{1}t}{v_{2}t}=\frac{v_1}{v_2}$. With $n=c/v$, $\frac{v_1}{v_2}=\frac{n_2}{n_1}$, giving $n_1\sin i=n_2\sin r$ — Snell's law.

      1 mark

   3. Prediction verified: wave theory predicts that light travels slower in a denser medium ($v_2<v_1$ when $n_2>n_1$). Foucault's experiment showed the speed of light in water is less than in air, confirming wave theory and contradicting the corpuscular theory.

      1 mark

   Answer: sin i/sin r = v₁/v₂ = n₂/n₁, i.e. n₁ sin i = n₂ sin r; speed of light is less in a denser medium (verified by Foucault).

   Where the marks are

   • Huygens' principle and construction with BC = v₁t, AE = v₂t (1 mark)
   • Geometry giving sin i/sin r = v₁/v₂ and Snell's law (1 mark)
   • Prediction: light slower in denser medium, verified experimentally (1 mark)

   Open Q26 on its own page

6. Answer one of these 2 alternatives.

   Q27(A) · 3 marks · Short answerMoving Charges and Magnetism

   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?

   Try it first, then show our solutionSolution

   1. I. Treat the ammeter (G = 13 Ω, $I_g$ = 100 A) as a galvanometer. In parallel, p.d. is the same: $I_{g}G=(I-I_g)S\Rightarrow S=\frac{I_{g}G}{I-I_g}$.

      1 mark

   2. $S=\frac{100\times 13}{750-100}=\frac{1300}{650}$

      ½ mark

   3. $S=2\Omega$ (connected in parallel).

      ½ mark

   4. II. When the coil oscillates, eddy currents are induced in the metal frame which oppose the motion (Lenz's law). This provides electromagnetic damping, so the coil comes to rest quickly at its deflected position.

      1 mark

   Answer: I. S = 2 Ω; II. for electromagnetic damping (dead-beat galvanometer).

   Where the marks are

   • Uses equal p.d. across meter and shunt: S = IgG/(I − Ig) (1 mark)
   • Correct substitution (½ mark)
   • S = 2 Ω (½ mark)
   • Metal frame gives eddy-current (electromagnetic) damping (1 mark)

   Open Q27(A) on its own page

   OR

   Q27(B) · 3 marks · Short answerMoving Charges and Magnetism

   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?

   Try it first, then show our solutionSolution

   1. I. The magnetic dipole moment $\vec{m}=NI\vec{A}$ determines both the strength (magnitude $NIA$) and orientation (along the area vector, given by the right-hand rule) of the loop's dipole.

      ½ mark

   2. SI unit: A m${}^2$ (equivalently J T${}^{-1}$).

      ½ mark

   3. II. Pole strength of the straight wire: $q_m=\frac{m}{l}$; it stays the same on bending.

      ½ mark

   4. On bending into a semicircle of radius r, the poles lie at the ends of a diameter, so the new separation is $2r$.

      ½ mark

   5. Length of arc: $l=\pi r\Rightarrow r=\frac{l}{\pi }$.

      ½ mark

   6. New moment: $m^{\prime }=q_m\times 2r=\frac{m}{l}\times \frac{2l}{\pi }=\frac{2m}{\pi }$.

      ½ mark

   Answer: I. Magnetic dipole moment, unit A m² (J/T); II. m′ = 2m/π

   Where the marks are

   • Identifies magnetic dipole moment m = NIA (½ mark)
   • SI unit A m² or J/T (½ mark)
   • Pole strength m/l (½ mark)
   • New pole separation 2r (½ mark)
   • r = l/π (½ mark)
   • m′ = 2m/π (½ mark)

   Open Q27(B) on its own page

7. Q28 · 3 marks · Short answerElectromagnetic Induction

   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?

   Try it first, then show our solutionSolution

   1. I. As the magnet falls towards the ring, the flux linked with the ring changes, inducing a current. By Lenz's law this current opposes the cause (the magnet's motion), exerting an upward force on the magnet; hence its acceleration is less than $g$.

      1 mark

   2. II. Self-inductance of the coil: $L\propto \frac{N^{2}A}{l}$ (equivalently $L={\mu }_0{n}^{2}Al$). With turns per unit length fixed, making each side 5 times gives $N\to 5N$ and $l\to 5l$.

      ½ mark

   3. Area of an equilateral triangle ∝ (side)${}^2$, so $A\to 25A$.

      ½ mark

   4. $\frac{L^{\prime }}{L}=\frac{(5N{)}^2(25A)/(5l)}{N^{2}A/l}=\frac{25\times 25}{5}$

      ½ mark

   5. $=125$, so $L^{\prime }=125L$ — the self-inductance becomes 125 times.

      ½ mark

   Answer: I. Induced current opposes fall (Lenz's law); II. L′ = 125 L

   Where the marks are

   • Lenz's law: induced current opposes motion of magnet (1 mark)
   • L ∝ N²A/l with N → 5N, l → 5l (½ mark)
   • A → 25A (½ mark)
   • Correct ratio expression (½ mark)
   • L′ = 125L (½ mark)

   Open Q28 on its own page

Section D

case study

1. Q29 · 4 marks · Case-basedCompetency-basedSemiconductor Electronics: Materials, Devices and Simple Circuits

   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.

   1. I.

      Which advantage of semiconductors makes them ideal for wearable devices like smart shirts?

      1. (A)High voltage operation (~100 V)
      2. (B)Bulky design and short life
      3. (C)Low power consumption and reliability
      4. (D)Requirement of external heating

      1 mark
   2. 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. (A)Diamond has too many free electrons.
      2. (B)Diamond’s band gap is too large for thermal excitation at room temperature.
      3. (C)Silicon has weaker covalent bonds than diamond.
      4. (D)Diamond cannot form covalent bonds.

      1 mark
   3. III.

      In an intrinsic semiconductor at room temperature

      1. (A)only electrons contribute to conduction.
      2. (B)only holes contribute to conduction.
      3. (C)number of electrons equals number of holes.
      4. (D)no conduction occurs at all.

      1 mark
   4. IV.

      How will you use the concept of majority and minority carriers to explain the efficiency of polymer semiconductors in wearable technology?

      1 mark

   Try it first, then show our solutionSolution

   1. (I) Wearables run continuously on tiny batteries, so they need devices with low power consumption and high reliability — option (C).

      1 mark

   2. (II) Thermal energy at room temperature (~0.025 eV) can excite electrons across 1.1 eV in Si but practically none across 5.4 eV in diamond, so diamond behaves as an insulator — option (B).

      1 mark

   3. (III) In a pure semiconductor each broken bond gives one free electron and one hole, so $n_e=n_h=n_i$ — option (C).

      1 mark

   4. (IV) Doped polymer semiconductors carry current mainly by majority carriers (electrons in n-type, holes in p-type) whose concentration and mobility can be controlled, so they conduct well even when bent; minority carriers are few and govern recombination. Hence a useful current flows at low voltage and low power, making them efficient for flexible wearable devices.

      1 mark

   Answer: I. (C); II. (B); III. (C); IV. Conduction by controllable majority carriers gives efficient low-power conduction in flexible films.

   Where the marks are

   • I: option C (1 mark)
   • II: option B (1 mark)
   • III: option C (1 mark)
   • IV: majority carriers carry current, controllable concentration/mobility → efficient low-power conduction (1 mark)

   Open Q29 on its own page

2. Q30 · 4 marks · Case-basedCompetency-basedDual Nature of Radiation and Matter

   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 $K_{max}=h\nu -{\varphi }_0$. 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.

   1. I.

      Define the term work function.

      1 mark
   2. II.

      Why the increase in the intensity of light does not increase the maximum kinetic energy of photoelectrons?

      1 mark
   3. III.

      A metal surface has a work function of 2 eV. The light of frequency $1.2\times 10^{15}$ Hz falls on it. Which statement is correct?

      1. (A)No electron is emitted because the intensity is low.
      2. (B)Electrons are emitted with maximum kinetic energy less than 2 eV.
      3. (C)Electrons are emitted with maximum kinetic energy greater than 0 eV.
      4. (D)Electron emission depends only on the intensity of light and not on the frequency.

      1 mark
   4. IV.

      Which observation/s directly contradicts the classical wave theory of light?

      1. (A)Photoelectric current increases with intensity
      2. (B)The maximum kinetic energy depends only on the frequency and not on the intensity.
      3. (C)The electrons are emitted instantaneously.
      4. (D)Both B and C.

      1 mark

   Try it first, then show our solutionSolution

   1. (I) Work function ${\varphi }_0$ is the minimum energy that must be given to an electron to just remove it from the surface of a metal.

      1 mark

   2. (II) Intensity only increases the number of photons per second, not the energy $h\nu$ of each photon. Since one photon is absorbed by one electron, $K_{max}=h\nu -{\varphi }_0$ depends only on frequency; greater intensity only raises the photoelectric current.

      1 mark

   3. (III) $E=h\nu =4.14\times 10^{-15}\text{eV s}\times 1.2\times 10^{15}\text{Hz}\approx 5$ eV; $K_{max}=5-2=3$ eV $>0$. Option (C).

      1 mark

   4. (IV) Wave theory predicts $K_{max}$ should rise with intensity and that emission should need a time lag; observations B (K_max depends only on frequency) and C (instantaneous emission) contradict it. Option (D).

      1 mark

   Answer: I. Minimum energy to free an electron from the metal surface; II. photon energy depends only on ν; III. (C), K_max ≈ 3 eV; IV. (D)

   Where the marks are

   • Correct definition of work function (1 mark)
   • Intensity changes number of photons, not energy per photon (1 mark)
   • III: hν ≈ 5 eV, Kmax ≈ 3 eV, option C (1 mark)
   • IV: option D (1 mark)

   Open Q30 on its own page

Section E

1. Answer one of these 2 alternatives.

   Q31(A) · 5 marks · Long answerElectrostatic Potential and Capacitance

   I. An isolated conductor cannot have a large capacitance. Why?
   II. How would you connect 6 μF, 9 μF and 18 μF capacitors to obtain-
   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?

   Try it first, then show our solutionSolution

   1. I. Capacitance of an isolated spherical conductor is $C=4\pi {\epsilon }_{0}R$; since ${\epsilon }_0$ is fixed, C depends only on size.

      ½ mark

   2. Even for the Earth ($R\approx 6.4\times 10^6$ m), $C\approx 711\mu$F, so a large capacitance would need an impractically huge conductor; hence an isolated conductor has small capacitance.

      ½ mark

   3. II(i). Minimum capacitance: connect all three in series.

      ½ mark

   4. $\frac{1}{C_{min}}=\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=\frac{3+2+1}{18}=\frac{1}{3}$

      ½ mark

   5. $C_{min}=3\mu$F

      ½ mark

   6. II(ii). Maximum capacitance: connect all three in parallel.

      ½ mark

   7. $C_{max}=6+9+18$

      ½ mark

   8. $C_{max}=33\mu$F

      ½ mark

   9. III. From $C=\frac{{\epsilon }_{0}KA}{d}$: (i) separation between the plates, $C\propto \frac{1}{d}$; (ii) permittivity (dielectric constant) of the medium between the plates, $C\propto \epsilon$.

      1 mark

   Answer: I. C = 4πε₀R is tiny for practical sizes; II. series: 3 μF (min), parallel: 33 μF (max); III. plate separation and permittivity of medium.

   Where the marks are

   • C = 4πε₀R depends only on size (½ mark)
   • Large C needs impractically large conductor (½ mark)
   • Series for minimum (½ mark), correct reciprocal sum (½ mark), 3 μF (½ mark)
   • Parallel for maximum (½ mark), sum (½ mark), 33 μF (½ mark)
   • Distance between plates and permittivity of medium (½ × 2 marks)

   Open Q31(A) on its own page

   OR

   Q31(B) · 5 marks · Long answerElectrostatic Potential and Capacitance

   I. Show that the electric field at any point is equal to the negative of the potential gradient at that point.
   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 $2.5\times 10^6$ V m${}^{-1}$. If the sphere is allowed to attain a maximum potential of $5\times 10^5$ V without causing any discharge, determine the minimum radius the insulated sphere must have.

   Try it first, then show our solutionSolution

   1. I. Let a small test charge $+q$ be in field $\vec{E}$. Force on it: $\vec{F}=q\vec{E}$.

      ½ mark

   2. Work done by the field in a small displacement $d\vec{r}$: $dW=\vec{F}\cdot d\vec{r}=q\vec{E}\cdot d\vec{r}$.

      ½ mark

   3. Work done by the field equals decrease in potential energy: $dW=-qdV$, so $dV=-\frac{dW}{q}$.

      ½ mark

   4. Substituting: $dV=-\frac{q\vec{E}\cdot d\vec{r}}{q}=-\vec{E}\cdot d\vec{r}$.

      ½ mark

   5. Taking $d\vec{r}$ along $\vec{E}$: $dV=-Edr$.

      ½ mark

   6. $\Rightarrow E=-\frac{dV}{dr}$: the electric field equals the negative of the potential gradient, and points in the direction of steepest decrease of potential.

      ½ mark

   7. II. For a charged sphere of radius r: $V=\frac{kQ}{r}$ and surface field $E=\frac{kQ}{r^2}$, so $E=\frac{V}{r}$.

      ½ mark

   8. No breakdown requires $E\le 2.5\times 10^6$ V m${}^{-1}$ when $V=5\times 10^5$ V, so $r\ge \frac{V}{E_{max}}$.

      ½ mark

   9. $r_{min}=\frac{5\times 10^5}{2.5\times 10^6}$

      ½ mark

   10. $r_{min}=0.2$ m (20 cm).

       ½ mark

   Answer: I. E = −dV/dr; II. minimum radius = 0.2 m

   Where the marks are

   • F = qE on test charge (½ mark)
   • dW = qE·dr (½ mark)
   • dV = −dW/q (½ mark)
   • Substitution dV = −E·dr (½ mark)
   • Along field: dV = −E dr (½ mark)
   • E = −dV/dr with statement (½ mark)
   • Relation E = V/r for sphere (½ mark)
   • r = V/E_max condition (½ mark)
   • Substitution (½ mark)
   • r = 0.2 m (½ mark)

   Open Q31(B) on its own page

2. Answer one of these 2 alternatives.

   Q32(A) · 5 marks · Long answerRay Optics and Optical Instruments

   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.
   II. The eyepiece is so placed that the final image is formed at infinity. Find the magnifying power of the microscope.
   III. Explain briefly why the compound microscope provides much higher magnification compared to a simple magnifier.

   Try it first, then show our solutionSolution

   1. I. Image by the objective. Use the lens formula with the sign convention: $f_o=+2$ cm and $u_o=-2.2$ cm (the object is on the incident side).

      $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}\Rightarrow \frac{1}{2}=\frac{1}{v_o}+\frac{1}{2.2}$

      1 mark

   2. $\frac{1}{v_o}=\frac{1}{2}-\frac{1}{2.2}=\frac{2.2-2}{4.4}=\frac{0.2}{4.4}=\frac{1}{22}$

      So $v_o=+22$ cm. A real, inverted image forms 22 cm from the objective, on the side opposite to the object.

      1 mark

   3. II. Magnifying power, final image at infinity. For normal adjustment of a compound microscope:

      $M=m_o\times m_e=\frac{v_o}{|u_o|}\cdot \frac{D}{f_e}$, with $D=25$ cm.

      1 mark

   4. $M=\frac{22}{2.2}\times \frac{25}{5}=10\times 5=50$

      The magnifying power of the microscope is 50.

      1 mark

   5. III. Why the magnification is higher.

      • In a compound microscope the objective first forms a large, real, magnified image of the tiny object.
      • The eyepiece then magnifies this image again.
      • The total magnification is the product $m_o\times m_e$, so it is far larger than what a single lens (simple magnifier) can give.
      1 mark

   Answer: I. $v_o=22$ cm from the objective (real image); II. $M=50$; III. The total magnification is the product of the objective and eyepiece magnifications.

   Where the marks are

   • Applies the lens formula with correct signs ($u_o=-2.2$ cm, $f_o=2$ cm) (1 mark)
   • Obtains $v_o=22$ cm (1 mark)
   • Writes $M=\frac{v_o}{|u_o|}\cdot \frac{D}{f_e}$ for the final image at infinity (1 mark)
   • Calculates $M=50$ (1 mark)
   • Explains the two-stage magnification: total $M=m_o\times m_e$ (1 mark)

   Open Q32(A) on its own page

   OR

   Q32(B) · 5 marks · Long answerRay Optics and Optical Instruments

   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.
   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.
   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.

   Try it first, then show our solutionSolution

   1. I. Normal adjustment (final image at infinity). Given $f_o=100$ cm and $f_e=5$ cm:

      $M=\frac{f_o}{f_e}$

      1 mark

   2. $M=\frac{100}{5}=20$

      1 mark

   3. **II. Final image at the least distance of distinct vision ($D=25$ cm):**

      $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$

      1 mark

   4. $M=\frac{100}{5}\left(1+\frac{5}{25}\right)=20\times 1.2=24$

      1 mark

   5. III. Design of the lenses.

      • The objective has a large focal length, because the magnifying power is $M\propto f_o/f_e$. It also has a large aperture, which gathers more light from faint, distant objects, so the image is brighter and the resolving power is higher.
      • The eyepiece has a small focal length, so it magnifies the intermediate image strongly. A small $f_e$ also makes $f_o/f_e$ large.
      1 mark

   Answer: I. $M=20$; II. $M=24$; III. A large $f_o$ and small $f_e$ give a high $M=f_o/f_e$; a large aperture gives a brighter image and better resolution.

   Where the marks are

   • Writes $M=f_o/f_e$ for normal adjustment (1 mark)
   • Gets $M=20$ (1 mark)
   • Writes $M=\frac{f_o}{f_e}(1+f_e/D)$ (1 mark)
   • Gets $M=24$ (1 mark)
   • Reasons for a large $f_o$ and aperture (magnification, light gathering, resolution) and a small $f_e$ (1 mark)

   Open Q32(B) on its own page

3. Answer one of these 2 alternatives.

   Q33(A) · 5 marks · Long answerAlternating Current

   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.
   II. State the basic working mechanism of the device and mention the condition required in its input signal for proper functioning.
   III. Give one reason why this device does not operate with 100% efficiency in practical situations.

   Try it first, then show our solutionSolution

   1. I. Device: step-up transformer.

      Diagram (in words): A laminated soft-iron core carries two separate windings.

      • The primary coil $P$ has fewer turns $N_p$ and is connected to the a.c. input $V_p$.
      • The secondary coil $S$ has more turns $N_s$ and is connected to the output/load $V_s$.
      • Label the core, primary, secondary, input a.c. and output.
      1 mark

   2. Principle: The transformer works on mutual induction. A changing current in the primary sets up a changing magnetic flux in the core. This flux links the secondary and induces an emf in it (Faraday's law).

      ½ mark

   3. How the voltage increases: The same flux links every turn, so for an ideal transformer $\frac{V_s}{V_p}=\frac{N_s}{N_p}$. With $N_s>N_p$, we get $V_s>V_p$.

      Energy is conserved because $V_p{I}_p=V_s{I}_s$: as the voltage is stepped up, the current falls in the same ratio.

      ½ mark

   4. II. Working mechanism:

      • The alternating voltage across the primary drives an alternating current in it.
      • This current sets up an alternating magnetic flux in the iron core.
      • The changing flux linked with the secondary induces an alternating emf in it.
      • Energy thus passes from the primary to the secondary through the magnetic field, with no electrical contact between the windings.
      1 mark

   5. Condition on the input: The input must be alternating (a.c.). A steady d.c. produces constant flux, so $d\varphi /dt=0$ and no emf is induced in the secondary.

      1 mark

   6. III. Why the efficiency is below 100% (any one):

      • Copper ($I^{2}R$) loss in the windings
      • Eddy-current loss in the core
      • Hysteresis loss in the core
      • Flux leakage

      Because of these losses the output power is less than the input power.

      1 mark

   Answer: Step-up transformer: works on mutual induction; $V_s/V_p=N_s/N_p$ with $N_s>N_p$; the input must be a.c.; efficiency is below 100% due to copper, eddy-current or hysteresis losses, or flux leakage.

   Where the marks are

   • Labelled diagram of a transformer with the core, primary (fewer turns) and secondary (more turns) (1 mark)
   • States the principle of mutual induction (½ mark)
   • $V_s/V_p=N_s/N_p$ with $N_s>N_p$, and $V_p{I}_p=V_s{I}_s$ for energy conservation (½ mark)
   • Describes the working: changing flux in the core induces emf in the secondary (1 mark)
   • Input must be a.c., because d.c. gives no change of flux (1 mark)
   • Any one loss, e.g. eddy currents, hysteresis, copper loss or flux leakage (1 mark)

   Open Q33(A) on its own page

   OR

   Q33(B) · 5 marks · Long answerAlternating Current

   I. What do you mean by the resonance condition of a series LCR–circuit?
   II. Write an expression for the resonant frequency.
   Describe the use of a series resonant circuit in the tuning of a radio receiver.

   Try it first, then show our solutionSolution

   1. I. Resonance condition. A series LCR circuit driven by an a.c. source is at resonance when the inductive reactance equals the capacitive reactance:

      $X_L=X_C\Rightarrow \omega L=\frac{1}{\omega C}$

      1 mark

   2. At resonance the net reactance is $X_L-X_C=0$, so the impedance $Z=\sqrt{R^2+(X_L-X_C{)}^2}=R$, which is its minimum value. The circuit then behaves as purely resistive, and the voltage and current are in phase.

      ½ mark

   3. The current $I=V/Z=V/R$ is therefore maximum at resonance.

      ½ mark

   4. II. Resonant frequency:

      ${\omega }_0=\frac{1}{\sqrt{LC}}$, or $f_0=\frac{1}{2\pi \sqrt{LC}}$

      1 mark

   5. Use in tuning a radio. Many stations broadcast at different frequencies, and all of their signals reach the antenna.

      The receiver's input stage has a resonant circuit with a fixed inductor $L$ and a variable capacitor $C$. Turning the tuning knob changes $C$, and hence $f_0=\frac{1}{2\pi \sqrt{LC}}$.

      1 mark

   6. When $f_0$ is made equal to the frequency of the desired station:

      • the impedance is minimum for that frequency;
      • its current, and hence the signal, is maximum.

      Other stations are off resonance, so they give very small currents and are rejected. In this way one station is selected.

      1 mark

   Answer: Resonance: $X_L=X_C$, so $Z=R$ (minimum) and the current is maximum; $f_0=\frac{1}{2\pi \sqrt{LC}}$; tuning is done by varying $C$ until $f_0$ equals the station frequency.

   Where the marks are

   • States the resonance condition $X_L=X_C$, i.e. $\omega L=1/(\omega C)$ (1 mark)
   • At resonance $Z=R$, the minimum value (½ mark)
   • At resonance the current is maximum (½ mark)
   • Resonant frequency $f_0=\frac{1}{2\pi \sqrt{LC}}$ (1 mark)
   • Variable capacitor changes $f_0$ in the receiver (1 mark)
   • Matching $f_0$ to the station gives maximum current for that station and rejects others (1 mark)

   Open Q33(B) on its own page