CBSE Class 12 Physics: sample paper 2025-26

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQElectric Charges and Fields

   If a charged hollow sphere and a solid sphere of aluminum and copper of equal radii are in electrostatic equilibrium, then which of the following statements is true?

   1. (A)Both the spheres are having equal charges.
   2. (B)The hollow sphere will have more charge than solid sphere at its surface.
   3. (C)The aluminum sphere will have more charge on its surface than copper sphere.
   4. (D)If hollow sphere is also made up of aluminum then it will have more charge.

   Try it first, then show our solutionSolution

   1. In electrostatic equilibrium after contact, both conductors are at the same potential $V$.

   2. For an isolated sphere $C=4\pi {\epsilon }_{0}r$. This depends only on the radius, not on whether the sphere is hollow or solid, or on its material.

   3. Equal radii give equal $C$, so $q=CV$ is the same for both. Option (A).

      1 mark

   Answer: Option (A): both spheres carry equal charges.

   Where the marks are

   • Uses equal potential and $C=4\pi {\epsilon }_{0}r$ to conclude the charges are equal (1 mark)

   Open Q1 on its own page

2. Q2 · 1 mark · MCQCurrent Electricity

   A coil contains N turns of insulated copper wire of diameter d and resistivity ρ wound on a cylinder of diameter D. What is the total resistance between the two ends of the coil of copper wire?(given: D>>d)

   1. (A)$\frac{4\rho ND}{d^2}$
   2. (B)$\frac{8\rho ND}{d^2}$
   3. (C)$\frac{2\rho ND}{d^2}$
   4. (D)$\frac{12\rho ND}{d^2}$

   Try it first, then show our solutionSolution

   1. Length of wire: $L=N\pi D$. Cross-sectional area: $A=\pi d^2/4$.

   2. $R=\frac{\rho L}{A}=\frac{\rho N\pi D}{\pi d^2/4}=\frac{4\rho ND}{d^2}$. Option (A).

      1 mark

   Answer: $R=\frac{4\rho ND}{d^2}$, option (A)

   Where the marks are

   • Takes $L=N\pi D$, $A=\pi d^2/4$ and obtains $4\rho ND/d^2$ (1 mark)

   Open Q2 on its own page

3. Q3 · 1 mark · MCQAlternating Current

   If the phasor diagram for a device connected to AC supply is as shown in the fig, then which of the following statements is true?

   1. (A)When the frequency of the AC source is increased than the impedance of the device decreases.
   2. (B)This device behaves as conducting wire when connected across DC source.
   3. (C)When the frequency of the AC source is decreased than the impedance of the device decreases.
   4. (D)This device stores energy in the form of magnetic potential energy.

   Figure in the paper: Phasor diagram with perpendicular phasors: V pointing up-left (blue) and I pointing down-left (grey), with a right-angle mark between them at the origin. See it in the official paper.

   Try it first, then show our solutionSolution

   1. From the phasor diagram, the current leads the voltage by $90^{\circ }$, so the device is a pure capacitor.

   2. Its reactance is $X_C=\frac{1}{2\pi fC}$, so the impedance decreases as the frequency increases. Option (A).

      1 mark

   Answer: Option (A)

   Where the marks are

   • Identifies the device as a capacitor (current leads by 90°) and uses $X_C\propto 1/f$ (1 mark)

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4. Q4 · 1 mark · MCQElectromagnetic Waves

   Which of the following statement is true for the radio waves and the gamma rays?

   1. (A)The energy of gamma rays is lesser than that of the radio waves.
   2. (B)The frequency of the radio waves is higher than that of gamma rays.
   3. (C)The radio waves and the gamma rays have the same energy.
   4. (D)The energy of radio waves is lesser than that of the gamma rays.

   Try it first, then show our solutionSolution

   1. Photon energy is $E=h\nu$.

   2. Radio waves have a much lower frequency than gamma rays, so radio photons carry less energy. Option (D).

      1 mark

   Answer: Option (D)

   Where the marks are

   • Uses $E=h\nu$ with ${\nu }_{radio}<{\nu }_{\gamma }$ (1 mark)

5. Q5 · 1 mark · MCQRay Optics and Optical Instruments

   A glass prism has internal angles of $45^{\circ }$, $45^{\circ }$ and $90^{\circ }$. The glass has a critical angle of $45^{\circ }$. Which of the following ray diagrams depicts the possible path the of light through the prism?

   1. (A)Ray diagram (A): prism ABC with right angle at A on the left; ray enters vertically at the top near B, and a ray emerges downward near C.
   2. (B)Ray diagram (B): prism with right angle at A at top, hypotenuse CB at the bottom; ray enters downward on one side and emerges upward on the other side after two total internal reflections.
   3. (C)Ray diagram (C): prism with right angle at A on the left, B top, C bottom; ray enters horizontally on the face and returns back reversed.
   4. (D)Ray diagram (D): prism with right angle at A at bottom, B and C at top; ray enters horizontally and emerges obliquely out of the face near C.

   Figure in the paper: Four ray diagrams (A)-(D) through a right-angled isosceles prism ABC with right angle at A. See it in the official paper.

   Try it first, then show our solutionSolution

   1. A ray entering normally through a short face meets the hypotenuse at an angle of incidence of $45^{\circ }$, which equals the critical angle.

   2. At or beyond the critical angle the light is totally internally reflected, turned through $90^{\circ }$, and leaves normally through the other face. This path is shown in diagram (A).

      1 mark

   Answer: Option (A): total internal reflection

   Where the marks are

   • Recognises incidence of 45° on the hypotenuse, giving total internal reflection, and chooses (A) (1 mark)

   Open Q5 on its own page

6. Q6 · 1 mark · MCQWave Optics

   The light from a monochromatic source is incident on a single slit and the resulting diffraction pattern is viewed on a screen. The graph shows the variation of the intensity with the distance on the screen. The width of slit is increased keeping the intensity of the source the same. Which of the following graphs is correct? (The original curve is shown with a dashed line.)

   1. (A)Narrower central peak of same height as original
   2. (B)Broader, wider curve of same height
   3. (C)Broader and taller curve
   4. (D)Narrower and taller central peak

   Figure in the paper: Graph of intensity vs distance showing single-slit diffraction pattern with central maximum and small side maxima; four option graphs compare a solid new curve with the dashed original. See it in the official paper.

   Try it first, then show our solutionSolution

   1. The angular half-width of the central maximum is $\lambda /a$, so a wider slit gives a narrower central peak.

   2. A wider slit also lets more light through and concentrates it into a smaller region, so the peak intensity increases. Option (D).

      1 mark

   Answer: Option (D): narrower and taller central peak

   Where the marks are

   • States that the width decreases ($\propto 1/a$) and the intensity increases (1 mark)

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7. Q7 · 1 mark · MCQAtoms

   Which of the following transitions corresponds to the emission of the radiation of the maximum wavelength?

   1. (A)I
   2. (B)III
   3. (C)IV
   4. (D)VI

   Figure in the paper: Energy level diagram with levels 1, 2, 3, 4 and ∞. Transitions: I from 2 to 1 (down), II from 3 to 1... as drawn: I down from 2 to 1; II down from 3 to 2 region; III up from 3 to 4; IV down from 3 to 2; V up from 1 to 3; VI up from 2 to 3. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Transitions III, V and VI go upward, so they are absorptions. Only I, II and IV are emissions.

   2. Since $\lambda =hc/\Delta E$, the longest wavelength comes from the smallest energy gap. Among the emissions, IV ($3\to 2$) has the smallest gap, so the answer is option (C).

      1 mark

   Answer: Transition IV, option (C)

   Where the marks are

   • Excludes the absorption transitions and picks the smallest emission gap (1 mark)

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8. Q8 · 1 mark · MCQMoving Charges and Magnetism

   A charged particle is projected along the axis of a current carrying loop. Which of the following statements is true?

   1. (A)The acceleration of the charged particle will depend on the velocity with which it is projected.
   2. (B)The acceleration of the charged particle will depend on the magnitude of the current passing through the coil.
   3. (C)The acceleration of the charged particle will depend on the radius of the coil.
   4. (D)The charged particle will move with constant velocity.

   Try it first, then show our solutionSolution

   1. On the axis of a current loop, the magnetic field $\vec{B}$ points along the axis.

   2. The velocity is parallel to $\vec{B}$, so $\vec{F}=q(\vec{v}\times \vec{B})=0$. There is no acceleration and the particle moves with constant velocity. Option (D).

      1 mark

   Answer: Option (D)

   Where the marks are

   • Notes that $\vec{v}\parallel \vec{B}$, so the force is zero and the velocity stays constant (1 mark)

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9. Q9 · 1 mark · MCQElectromagnetic Induction

   Two small identical magnets are allowed to fall freely one through a vertical solenoid of 20 m made up of copper and another in air through the same vertical distance. The time taken by the two magnets to fall will be

   1. (A)same in both the cases.
   2. (B)more for the magnet falling in air.
   3. (C)more for the magnet falling through the solenoid.
   4. (D)infinite.

   Try it first, then show our solutionSolution

   1. The falling magnet changes the flux through the copper solenoid, which induces currents in it.

   2. By Lenz's law these induced currents oppose the magnet's motion, so the magnet falls with acceleration less than $g$ and takes longer. Option (C).

      1 mark

   Answer: Option (C)

   Where the marks are

   • Uses electromagnetic induction and Lenz's law opposition (1 mark)

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10. Q10 · 1 mark · MCQAlternating Current

    The emf generated by an AC generator is given by $V=V_o\sin \omega t$, where ω is angular frequency of armature of generator. What will be the emf if the angular frequency is doubled

    1. (A)$V=V_o\sin 2\omega t$
    2. (B)$V=2V_o\sin \omega t$
    3. (C)$V=2V_o\sin 2\omega t$
    4. (D)$V=V_o\sin \omega t$

    Try it first, then show our solutionSolution

    1. The generator emf is $V=NBA\omega \sin \omega t$, so the peak value is $V_0=NBA\omega$.

    2. Doubling $\omega$ doubles both the peak and the argument: $V=2V_0\sin 2\omega t$. Option (C).

       1 mark

    Answer: $V=2V_0\sin 2\omega t$, option (C)

    Where the marks are

    • Recognises that the amplitude is proportional to $\omega$ (1 mark)

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11. Q11 · 1 mark · MCQNuclei

    The ratio of the nuclear densities of two nuclei having the mass numbers 8 and 27 is

    1. (A)8:27
    2. (B)3:2
    3. (C)2:3
    4. (D)1:1

    Try it first, then show our solutionSolution

    1. Nuclear radius is $R=R_0{A}^{1/3}$, so the volume is proportional to $A$, the same as the mass.

    2. Density = mass/volume is therefore independent of $A$, so the ratio is 1 : 1. Option (D).

       1 mark

    Answer: 1:1, option (D)

    Where the marks are

    • Uses $R\propto A^{1/3}$ to show the density is independent of $A$ (1 mark)

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12. Q12 · 1 mark · MCQMagnetism and Matter

    When we move magnetic compass from point P to Q then which of the following statement is true

    1. (A)The deflection of the magnetic needle at P and Q will be in the same direction.
    2. (B)The deflection of the magnetic needle at P and Q will be in the opposite directions.
    3. (C)The deflection of the magnetic needle at P and Q will be perpendicular to each other.
    4. (D)The deflection of the magnetic needle at P and Q will be inclined at $45^{\circ }$ with respect to each other.

    Figure in the paper: A horizontal solenoid carrying current I (marked at its ends); point P lies on the solenoid's axis to the right, and point Q lies on a vertical line above the middle of the solenoid. See it in the official paper.

    Try it first, then show our solutionSolution

    1. A solenoid behaves like a bar magnet.

    2. On the axis (P) the field is along the axis. At an equatorial point (Q) the field is antiparallel to the axial field, so the needle points in opposite directions at P and Q. Option (B).

       1 mark

    Answer: Option (B)

    Where the marks are

    • Uses the bar-magnet analogy: equatorial field is antiparallel to axial field (1 mark)

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

    Assertion (A): Total energy of an electron in hydrogen atom is negative.
    Reason (R): The centripetal force is provided by electrostatic force.

    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. The assertion is true: $E=-13.6/n^2$ eV is negative because the electron is bound to the nucleus.

    2. The reason is true, since the Coulomb attraction provides the centripetal force. However, the negative sign comes from the energy being a bound state, not from that fact alone, so the reason is not the correct explanation. Option (B).

       1 mark

    Answer: Option (B)

    Where the marks are

    • Judges both statements true but R not the explanation (1 mark)

    Open Q13 on its own page

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

    Assertion (A): The critical angle of light passing from glass to air is minimum for violet colour.
    Reason (R): The wavelength of blue light is greater than the light of other colours.

    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. Since $\sin C=1/\mu$ and $\mu$ is largest for violet, the critical angle is least for violet, so the assertion is true.

    2. Red light, not blue, has the longest wavelength in the visible range, so the reason is false. Option (C).

       1 mark

    Answer: Option (C)

    Where the marks are

    • Uses $\sin C=1/\mu$ with ${\mu }_{violet}$ maximum, and rejects R (1 mark)

    Open Q14 on its own page

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

    Assertion (A): Two light sources emitting waves of similar wavelengths are coherent.
    Reason (R): Two light sources emitting waves having zero or constant phase difference are known as coherent sources.

    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. Similar wavelengths alone do not make two sources coherent, because their phase difference must also stay constant. The assertion is false.

    2. The scheme marks both statements false and gives option (D).

       1 mark

    Answer: Option (D)

    Where the marks are

    • Recognises that the assertion is false and selects (D) (1 mark)

    Open Q15 on its own page

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

    Assertion (A): For three point charges to be in equilibrium, they must be collinear.
    Reason(R): One of the three charges must have different polarity than rest of the two.

    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. The forces on each charge must cancel, which is possible only if they act along one line. So the charges must be collinear and the assertion is true.

    2. The middle charge must have opposite polarity to the other two, so the reason is true. However, it does not explain why the charges must be collinear. Option (B).

       1 mark

    Answer: Option (B)

    Where the marks are

    • Judges both statements true but R not explaining collinearity (1 mark)

    Open Q16 on its own page

Section B

1. Q17 · 2 marks · Very short answerElectromagnetic Waves

   The amplitude of the magnetic field of a plane electromagnetic wave propagating along positive X axis in vacuum is $510\mathrm{nT}\hat{k}$ and its angular frequency is $60\times 10^6$ rad/sec. Write the expression for the electric field $(\vec{E})$.

   Try it first, then show our solutionSolution

   1. For an EM wave, $\vec{E}$, $\vec{B}$ and the direction of propagation are mutually perpendicular, with $\vec{E}\times \vec{B}$ along the direction of travel. Here $\vec{B}$ is along $\hat{k}$ and the wave travels along $+\hat{i}$. Since $\hat{j}\times \hat{k}=\hat{i}$, $\vec{E}$ must be along $\hat{j}$.

   2. $E_0=c{B}_0=3\times 10^8\times 510\times 10^{-9}=153N/C$ and $k=\frac{\omega }{c}=\frac{60\times 10^6}{3\times 10^8}=0.2rad/m$.

      1 mark

   3. $\vec{E}=153\sin \left(60\times 10^{6}t-0.2x\right)\hat{j}N/C$

      1 mark

   Answer: $\vec{E}=153\sin (6\times 10^{7}t-0.2x)\hat{j}$ N/C

   Where the marks are

   • Calculates $E_0=c{B}_0=153$ N/C and $k=\omega /c=0.2$ rad/m (1 mark)
   • Writes the correct expression $E=153\sin (60\times 10^{6}t-0.2x)$ N/C along $\hat{j}$ (1 mark)

   Open Q17 on its own page

2. Q18 · 2 marks · Very short answerCompetency-basedCurrent Electricity

   The following graph shows the potential difference across the terminals of a cell against its load current.

   Find,
   (I) the emf of the cell and
   (II) the internal resistance of the cell.

   Figure in the paper: Graph of potential difference (V, vertical, 0-6) against load current (A, horizontal, 0-12): a straight line falling from 6 V at 0 A to 0 V at 12 A. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) The terminal voltage is $V=\epsilon -Ir$. When the load current $I=0$, $V=\epsilon$. From the graph, $V=6$ V at $I=0$, so emf $\epsilon =6$ V.

      1 mark

   2. (II) The slope of the $V$–$I$ line is $-r$: slope $=\frac{0-6}{12-0}=-0.5\Omega$. Hence internal resistance $r=0.5\Omega$. (Check: at $V=0$, $I=\epsilon /r=6/0.5=12$ A.)

      1 mark

   Answer: (I) emf = 6 V; (II) r = 0.5 Ω

   Where the marks are

   • Reads emf as 6 V from the intercept at zero current, with reason (1 mark)
   • Finds internal resistance from the slope, $r=0.5\Omega$ (1 mark)

   Open Q18 on its own page

3. Q19 · 2 marks · Very short answerElectric Charges and Fields

   A charge q is placed inside a sphere of radius ‘a’ filled with water and another charge 2q is placed inside cube of side ‘2a’ which is vacuumed inside. Find the ratio of the flux linked with the sphere to that linked with the cube. (Take relative permittivity of water as 80)

   Try it first, then show our solutionSolution

   1. By Gauss's law in a medium, $\varphi =\frac{q_{enclosed}}{{\epsilon }_r{\epsilon }_0}$ (independent of the size or shape of the closed surface).

   2. For the water-filled sphere: ${\varphi }_{sphere}=\frac{q}{80{\epsilon }_0}$

      ½ mark

   3. For the vacuum cube: ${\varphi }_{cube}=\frac{2q}{{\epsilon }_0}$

      ½ mark

   4. $\frac{{\varphi }_{sphere}}{{\varphi }_{cube}}=\frac{q/80{\epsilon }_0}{2q/{\epsilon }_0}=\frac{1}{160}$, i.e. ratio $1:160$.

      1 mark

   Answer: 1 : 160

   Where the marks are

   • Flux through sphere $q/(80{\epsilon }_0)$ (½ mark)
   • Flux through cube $2q/{\epsilon }_0$ (½ mark)
   • Correct ratio 1:160 (1 mark)

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4. Answer one of these 2 alternatives.

   Q20(I) · 2 marks · Very short answerMoving Charges and Magnetism

   Write an expression for the magnetic force per unit length between two parallel thin current carrying wires. Hence define one ampere.

   Try it first, then show our solutionSolution

   1. Force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$: $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$ (attractive for like currents, repulsive for unlike).

      1 mark

   2. Putting $I_1=I_2=1$ A, $r=1$ m: $\frac{F}{L}=\frac{4\pi \times 10^{-7}}{2\pi }=2\times 10^{-7}$ N/m. Hence one ampere is that steady current which, flowing in each of two very long, straight, parallel conductors placed 1 m apart in vacuum, produces a force of $2\times 10^{-7}$ N per metre of length on each conductor.

      1 mark

   Answer: $F/L={\mu }_0{I}_1{I}_2/(2\pi r)$; 1 A gives $2\times 10^{-7}$ N/m between wires 1 m apart in vacuum

   Where the marks are

   • Correct expression $F/L={\mu }_0{I}_1{I}_2/2\pi r$ (1 mark)
   • Correct definition of one ampere with $2\times 10^{-7}$ N/m and 1 m separation in vacuum (1 mark)

   Open Q20(I) on its own page

   OR

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

   Draw a diagram representing the behaviour of magnetic field lines for a
   (A) diamagnetic &
   (B) paramagnetic substance.

   Try it first, then show our solutionSolution

   1. (A) Diamagnetic substance: draw a specimen placed in a uniform external field with the field lines bending away (repelled) around it; fewer lines pass through the specimen, so the line density inside is less than outside (field inside is weakened).

      1 mark

   2. (B) Paramagnetic substance: draw the field lines bending towards and crowding into the specimen; the line density inside is more than outside (field inside is slightly enhanced).

      1 mark

   Answer: Diamagnetic: lines expelled/fewer inside; Paramagnetic: lines drawn in/concentrated inside

   Where the marks are

   • Diagram showing field lines repelled from a diamagnetic specimen (1 mark)
   • Diagram showing field lines concentrated inside a paramagnetic specimen (1 mark)

   Open Q20(II) on its own page

5. Answer one of these 2 alternatives.

   Q21(I) · 2 marks · Very short answerAtoms

   How does the impact parameter affect the trajectory of a α – particles scattered by a heavy nucleus? What is the value of impact parameter for head on collision of α – particles with the nucleus?

   Try it first, then show our solutionSolution

   1. The smaller the impact parameter $b$, the closer the α-particle passes to the nucleus, the stronger the repulsion and the larger the scattering angle; as $b\to$ large, the α-particle is hardly deflected and nearly keeps its original straight path.

      1 mark

   2. For a head-on collision, the impact parameter is zero ($b=0$) and the α-particle rebounds back (scattering angle ≈ 180°).

      1 mark

   Answer: Smaller b → larger deflection; larger b → small deflection; b = 0 for head-on collision

   Where the marks are

   • Relation between impact parameter and scattering angle (1 mark)
   • Impact parameter zero for head-on collision (1 mark)

   Open Q21(I) on its own page

   OR

   Q21(II) · 2 marks · Very short answerDual Nature of Radiation and Matter

   Plot a graph showing variation of de-Broglie wavelength λ versus $\frac{1}{\sqrt{V}}$, where V is accelerating potential for a particle of mass m and charge q. Obtain the slope of this graph.

   Try it first, then show our solutionSolution

   1. Kinetic energy gained: $\frac{1}{2}m{v}^2=qV\Rightarrow p=\sqrt{2mqV}$, so $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mq}}\cdot \frac{1}{\sqrt{V}}$. The graph of $\lambda$ (y-axis) against $1/\sqrt{V}$ (x-axis) is a straight line passing through the origin.

      1 mark

   2. Comparing with $y=mx$, slope $=\frac{h}{\sqrt{2mq}}$.

      1 mark

   Answer: Straight line through origin; slope $=h/\sqrt{2mq}$

   Where the marks are

   • Derives $\lambda =h/\sqrt{2mqV}$ and plots straight line through origin (1 mark)
   • Slope $h/\sqrt{2mq}$ (1 mark)

   Open Q21(II) on its own page

Section C

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

   With the help of circuit diagram explain working of the full wave rectifier.

   Try it first, then show our solutionSolution

   1. Circuit: AC input is applied to the primary of a centre-tapped transformer. The two ends A and B of the secondary are connected to the p-sides of diodes $D_1$ and $D_2$; their n-sides are joined at point Q. The load resistance $R_L$ is connected between Q and the centre tap P, and output is taken across $R_L$.

      1 mark

   2. During one half cycle, end A is positive with respect to the centre tap: $D_1$ is forward biased and conducts, $D_2$ is reverse biased. Current flows through $R_L$ from Q to P.

      1 mark

   3. During the next half cycle, end B becomes positive: $D_2$ is forward biased and conducts while $D_1$ is reverse biased. Current again flows through $R_L$ in the same direction (Q to P). Thus output is obtained in both half cycles — a pulsating unidirectional (DC) voltage, with output waveform showing a positive hump for every half cycle of the input.

      1 mark

   Answer: Centre-tap full wave rectifier: D1 and D2 conduct alternately, current through load in the same direction in both half cycles, giving pulsating DC

   Where the marks are

   • Correct labelled circuit diagram with centre-tapped transformer, two diodes and load (1 mark)
   • Working in first half cycle (1 mark)
   • Working in second half cycle and unidirectional output waveform (1 mark)

   Open Q22 on its own page

2. Answer one of these 2 alternatives.

   Q23(I) · 3 marks · Short answerCurrent Electricity

   The current $I_1$ in a wire is getting divided in two wires with currents $I_2$ and $I_3$ at a junction in a circuit.The currents in the three wires are related by $I_1=I_2+I_3$.
   (A) State the fundamental law from which this relation is derived.
   (B) Explain the validation of law of conservation of energy in Kirchhoff's voltage law?

   Try it first, then show our solutionSolution

   1. (A) The relation $I_1=I_2+I_3$ is Kirchhoff's junction rule, which follows from the law of conservation of electric charge: charge cannot accumulate at a junction, so charge entering per second equals charge leaving per second.

      1 mark

   2. (B) Kirchhoff's voltage law states that the algebraic sum of changes in potential around any closed loop is zero: $\sum \epsilon =\sum IR$.

      1 mark

   3. A charge taken once round a closed loop returns to its starting point, i.e. to the same potential, so its net change in energy is zero. Hence the energy supplied by the emfs equals the energy dissipated in the resistors — this is the law of conservation of energy.

      1 mark

   Answer: (A) Conservation of charge (junction rule); (B) KVL: net potential change round a closed loop is zero, so energy gained from sources = energy lost in resistors

   Where the marks are

   • Identifies conservation of charge (1 mark)
   • States KVL as algebraic sum of potential changes round a loop = 0 (1 mark)
   • Explains this as conservation of energy for a charge going round the loop (1 mark)

   Open Q23(I) on its own page

   OR

   Q23(II) · 3 marks · Short answerCurrent Electricity

   How the balancing condition gets affected if you are interchanging the galvanometer and the cell in the Wheat stone bridge?

   Try it first, then show our solutionSolution

   1. Let arms be AB = P, BC = Q, AD = R, DC = S, with cell across A–C and galvanometer across B–D. Balance condition: $\frac{P}{Q}=\frac{R}{S}$.

      1 mark

   2. After interchange, cell is across B–D and galvanometer across A–C. Current from B divides along B→A→D (P, R) and B→C→D (Q, S): $I_1=\frac{V}{P+R}$, $I_2=\frac{V}{Q+S}$. No deflection requires $V_A=V_C$, i.e. $I_{1}P=I_{2}Q$: $\frac{P}{P+R}=\frac{Q}{Q+S}$.

      1 mark

   3. This gives $PQ+PS=QP+QR\Rightarrow PS=QR\Rightarrow \frac{P}{Q}=\frac{R}{S}$. So the balance condition is unchanged; the bridge remains balanced on interchanging the cell and galvanometer.

      1 mark

   Answer: No change; the balance condition P/Q = R/S remains the same

   Where the marks are

   • States the balance condition of the bridge (1 mark)
   • Sets up the condition for null deflection after interchange (1 mark)
   • Shows it reduces to the same P/Q = R/S, so no change (1 mark)

   Open Q23(II) on its own page

3. Q24 · 3 marks · Short answerNuclei

   A fast-moving neutron collides with the nucleus of Plutonium (Pu), thereby producing Xenon (Xe) and Zirconium (Zr) along with neutrons.
   (I) Write the nuclear fission reaction.
   (II) Find the energy released in the above nuclear reaction.
   Given atomic masses:
   m(${}_{94}^{239}Pu$) = 239.052157u,
   m(${}_{40}^{103}Zr$) = 102.926597u,
   m(${}_{54}^{134}Xe$) = 133.905040u &
   m(${}_0^{1}n$) = 1.00866u.

   Try it first, then show our solutionSolution

   1. (I) ${}_{94}^{239}\mathrm{Pu}+{}_0^{1}n\to {}_{54}^{134}\mathrm{Xe}+{}_{40}^{103}\mathrm{Zr}+3{}_0^{1}n$ (mass number $240=134+103+3$; charge $94=54+40$).

      1 mark

   2. (II) Mass defect: $\Delta m=[m(\mathrm{Pu})+m(n)]-[m(\mathrm{Xe})+m(\mathrm{Zr})+3m(n)]=[239.052157+1.00866]-[133.905040+102.926597+3.02598]$

   3. $\Delta m=240.060817-239.857617=0.2032\mathrm{u}$

      1 mark

   4. Energy released $Q=\Delta m\times 931.5MeV/u=0.2032\times 931.5\approx 189.3\mathrm{MeV}$

      1 mark

   Answer: (I) Pu-239 + n → Xe-134 + Zr-103 + 3n; (II) Δm = 0.2032 u, Q ≈ 189.3 MeV

   Where the marks are

   • Correct balanced fission equation with 3 neutrons (1 mark)
   • Mass defect 0.2032 u (1 mark)
   • Energy released ≈ 189.3 MeV (1 mark)

   Open Q24 on its own page

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

   A compound microscope consists of an objective lens of focal length 0.82 cm and an eyepiece lens of focal length 2.9 cm. An object is placed 0.91 cm from the objective lens. The image is formed at the near point (25 cm) from the eye.
   (I) Calculate that the angular magnification of the microscope.
   (II) Draw the ray diagram of compound microscope in normal adjustment.

   Try it first, then show our solutionSolution

   1. (I) Objective: $\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$ with $u_o=-0.91$ cm, $f_o=0.82$ cm: $\frac{1}{v_o}=\frac{1}{0.82}-\frac{1}{0.91}=1.2195-1.0989=0.1206\Rightarrow v_o\approx 8.3$ cm.

      1 mark

   2. Image at near point: $M=m_o{m}_e=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)=\frac{8.3}{-0.91}\left(1+\frac{25}{2.9}\right)=(-9.12)(9.62)\approx -87.7$ (magnitude ≈ 88, image inverted).

      1 mark

   3. (II) Ray diagram (normal adjustment): object AB placed just beyond $F_o$ of the objective; objective forms a real, inverted, magnified image A′B′ exactly at the focus $F_e$ of the eyepiece; rays from A′B′ emerge parallel from the eyepiece, so the final, magnified image is formed at infinity (inverted with respect to the object). Arrows on rays, both lenses and foci labelled.

      1 mark

   Answer: (I) $v_o\approx 8.3$ cm, M ≈ −87.7; (II) ray diagram with final image at infinity

   Where the marks are

   • Finds image distance of objective ≈ 8.3 cm (1 mark)
   • Magnification $M=(v_o/u_o)(1+D/f_e)\approx -87.7$ (1 mark)
   • Correct labelled ray diagram in normal adjustment (1 mark)

   Open Q25 on its own page

5. Q26 · 3 marks · Short answerWave Optics

   Draw the reflected wave front for a plane wave front incident on a plane reflecting surface. Hence verify the laws of reflection using Huygen’s principle.

   Try it first, then show our solutionSolution

   1. Diagram: plane reflecting surface MN; incident plane wavefront AB meets it at A at angle $i$; point B travels to C on MN. With A as centre, a secondary wavelet (hemisphere) of radius $AE=BC$ is drawn; the tangent CE from C is the reflected wavefront, making angle $r$ with MN. Rays ⟂ to the wavefronts are shown.

      1 mark

   2. Construction: if $v$ is the speed of light and $\tau$ the time for the disturbance at B to reach C, then $BC=v\tau$. In the same time the wavelet from A spreads to radius $AE=v\tau$ (same medium). By Huygens' principle CE, the common tangent, is the reflected wavefront.

      1 mark

   3. In right triangles ABC and AEC: $\angle ABC=\angle AEC=90^{\circ }$ [wavefront ⟂ ray], AC is common, $BC=AE=v\tau$. So the triangles are congruent (RHS), giving $\angle BAC=\angle ECA$, i.e. $i=r$. Also incident ray, reflected ray and normal lie in the same plane (plane of the figure). Hence the laws of reflection are verified.

      1 mark

   Answer: Reflected wavefront CE drawn; congruent triangles ABC and AEC give i = r

   Where the marks are

   • Correct diagram of incident and reflected wavefronts (1 mark)
   • Huygens construction with $BC=AE=v\tau$ (1 mark)
   • Proof of congruence giving $i=r$ (1 mark)

   Open Q26 on its own page

6. Answer one of these 2 alternatives.

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

   If a point sized object having charge 1C and mass 1g is projected with velocity of $2\hat{i}$ m/s from a point (0,2cm,0) in the region of magnetic field $-0.1\hat{k}$ T which spreads in the first quadrant.
   (A) What will be the shape of the path followed by the given charged particle?
   (B) At what point it will cross the X-axis?
   (C) What will be the kinetic energy of particle when it will enter in the fourth quadrant?

   Figure in the paper: First quadrant with X and Y axes, filled with magnetic field into the page (crosses). Charge q at (0, 2 cm) on the Y-axis with velocity V directed along +X. See it in the official paper.

   Try it first, then show our solutionSolution

   1. $\vec{v}\perp \vec{B}$, so the magnetic force provides centripetal force and the path is circular with radius $r=\frac{mv}{qB}=\frac{10^{-3}\times 2}{1\times 0.1}=0.02\mathrm{m}=2$ cm. Initial force: $\vec{F}=q\vec{v}\times \vec{B}=1(2\hat{i})\times (-0.1\hat{k})=+0.2\hat{j}$ N.

      1 mark

   2. (A), (B) With $\vec{B}$ along $-\hat{k}$ (into the page) the force at (0, 2 cm) is along $+y$, so the charge turns anticlockwise about centre (0, 4 cm): it describes a semicircle of radius 2 cm and leaves the field at (0, 6 cm) on the Y-axis; it does not reach the X-axis. (If the field were along $+\hat{k}$, the motion would be clockwise about the origin: a quarter circle meeting the X-axis at (2 cm, 0) and entering the fourth quadrant — the answer given in the scheme.)

      1 mark

   3. (C) Magnetic force is always ⟂ velocity, so it does no work and speed is unchanged: $K=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10^{-3}\times 2^2=2\times 10^{-3}$ J when it leaves the field.

      1 mark

   Answer: r = 2 cm; with B = −0.1k̂ the path is a semicircle leaving at (0, 6 cm) (scheme, assuming clockwise motion: quarter circle crossing X-axis at (2 cm, 0)); KE = 2×10⁻³ J

   Where the marks are

   • Circular path with radius $mv/qB=2$ cm (1 mark)
   • Correct shape/exit point of the path (1 mark)
   • KE unchanged, $2\times 10^{-3}$ J, since magnetic force does no work (1 mark)

   Open Q27(I) on its own page

   OR

   Q27(II) · 3 marks · Short answerMagnetism and Matter

   A solenoid has a core of material with relative permeability 200. The windings of the solenoid are insulated from the core and carry a current of 1A. If the number of turns is 2000 per metre, calculate
   (A) magnetic intensity,
   (B) magnetic field &
   (C) magnetisation

   Try it first, then show our solutionSolution

   1. (A) Magnetic intensity $H=nI=2000\times 1=2\times 10^3$ A/m

      1 mark

   2. (B) $B={\mu }_0{\mu }_{r}H=4\pi \times 10^{-7}\times 200\times 2\times 10^3=0.16\pi \approx 0.50$ T

      1 mark

   3. (C) Magnetisation $M=\chi H=({\mu }_r-1)H=199\times 2\times 10^3=3.98\times 10^5$ A/m

      1 mark

   Answer: (A) 2×10³ A/m; (B) ≈ 0.50 T; (C) 3.98×10⁵ A/m

   Where the marks are

   • H = nI = 2×10³ A/m (1 mark)
   • B = μ₀μᵣH ≈ 0.50 T (1 mark)
   • M = (μᵣ − 1)H = 3.98×10⁵ A/m (1 mark)

   Open Q27(II) on its own page

7. Q28 · 3 marks · Short answerElectromagnetic Induction

   A conducting coil of 50 turns and area $\frac{5}{\pi }$ cm${}^2$ is rotating along the axis of solenoid of length 50cm and 2000 turns, carrying current of 5 A. What will be the value of maximum emf generated?

   Try it first, then show our solutionSolution

   1. Solenoid: $n=\frac{2000}{0.5}=4000$ turns/m; $B={\mu }_{0}nI=4\pi \times 10^{-7}\times 4000\times 5=8\pi \times 10^{-3}$ T

      1 mark

   2. When the coil rotates with angular speed $\omega$ (about a diameter ⟂ the solenoid axis), flux $\varphi =N_{c}BA\cos \omega t$, so $\epsilon =-\frac{d\varphi }{dt}=N_{c}BA\omega \sin \omega t$ and ${\epsilon }_{max}=N_{c}BA\omega$.

      1 mark

   3. ${\epsilon }_{max}=50\times 8\pi \times 10^{-3}\times \frac{5}{\pi }\times 10^{-4}\times \omega =2\times 10^{-4}\omega$ V. Since $\omega$ is not given, ${\epsilon }_{max}=2\times 10^{-4}\omega$ volt (e.g. $2\times 10^{-4}$ V for $\omega =1$ rad/s).

      1 mark

   Answer: ${\epsilon }_{max}=N_{c}BA\omega =2\times 10^{-4}\omega$ V (B = 8π×10⁻³ T)

   Where the marks are

   • Field of solenoid $B={\mu }_{0}nI=8\pi \times 10^{-3}$ T (1 mark)
   • Flux and emf relation ${\epsilon }_{max}=NBA\omega$ (1 mark)
   • Correct substitution and value (1 mark)

   Open Q28 on its own page

Section D

case study

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

   When an external voltage is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. When an external voltage is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased. The applied voltage mostly drops across the depletion region.

   1. (I)

      Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of $V_0$ changes by (assume that the Ge diode has large breakdown voltage)

      1. (A)0.2 V
      2. (B)0.4 V
      3. (C)0.6 V
      4. (D)0.8 V

      1 mark
   2. (II)

      The value of $V_0$ and $I_d$ for the network are :

      1. (A)13 V, 2.32mA
      2. (B)11.7 V, 2.08mA
      3. (C)11.3V, 2.01mA
      4. (D)11V, 1.96mA

      1 mark
   3. (III)

      The circuit shown below contains two ideal diodes, each with a forward resistance of $50\Omega$. If the battery voltage is 6 V, the current through the $100\Omega$ resistance (in amperes) is

      1. (A)0.036
      2. (B)0.020
      3. (C)0.030
      4. (D)0.027

      1 mark
   4. (IV)

      In the figure, potential difference between A and B is

      1. (A)Zero
      2. (B)5 V
      3. (C)10 V
      4. (D)15 V

      1 mark

   Figure in the paper: (I) 12 V battery in series with parallel Ge and Si diodes (opposite diodes shown parallel, same direction) then 5 kΩ resistor to ground; V0 at the output. (II) 12 V source, Si diode then Ge diode in series, V0 across 5.6 kΩ resistor to ground. (III) Branches: D1 with 150 Ω, D2 (reversed) with 75 Ω in parallel, in series with 100 Ω and a 6 V battery. (IV) 30 V battery, 10 kΩ in series, then 10 kΩ across A-B, and a diode from A to a 10 kΩ resistor connected to B. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) With both diodes forward, the Ge diode turns on first and clamps the drop at 0.3 V: $V_0=12-0.3=11.7$ V. With Ge reversed, only Si conducts: $V_0=12-0.7=11.3$ V. Change $=0.4$ V → option (B).

      1 mark

   2. (II) Si and Ge in series both conduct: $V_0=12-0.7-0.3=11$ V; $I_d=\frac{11}{5.6\times 10^3}\approx 1.96$ mA → option (D).

      1 mark

   3. (III) $D_2$ is reverse biased, so its 75 Ω branch carries no current. Current flows through $D_1$ (50 Ω), 150 Ω and 100 Ω: $I=\frac{6}{50+150+100}=0.02$ A → option (B).

      1 mark

   4. (IV) The diode is forward biased (acts as a wire), so the two 10 kΩ resistors across AB are in parallel: 5 kΩ. Total resistance 15 kΩ, $I=30/15\text{ k}\Omega =2$ mA, $V_{AB}=2\text{ mA}\times 5\text{ k}\Omega =10$ V → option (C).

      1 mark

   Answer: (I) B – 0.4 V; (II) D – 11 V, 1.96 mA; (III) B – 0.020 A; (IV) C – 10 V

   Where the marks are

   • (I) 0.4 V, option B (1 mark)
   • (II) 11 V, 1.96 mA, option D (1 mark)
   • (III) 0.020 A, option B (1 mark)
   • (IV) 10 V, option C (1 mark)

   Open Q29 on its own page

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

   Photoelectric effect is phenomenon of the ejection of electrons when the radiation of suitable frequency is made to fall on the surface of a metal. When light of suitable wavelength falls on the emitter C given in the diagram, the photoelectrons are emitted. These photoelectrons are drawn to the collector A. The photoelectric current of the order of a few microamperes can be normally obtained from the device given in figure. The device given converts a change in intensity of illumination into a change in photocurrent. This current can be used to operate control systems and in light measuring devices. The devices are made up of metals with low ionization enthalpies, for example platinum whose work function is 6.35 eV.

   1. (I)

      If infrared radiation of $3\times 10^{11}$ Hz is used as incident radiation, determine the reading of microammeter? Justify mathematically.

      2 marks
   2. (II)

      In the given diagram, if terminal B is shifted towards the left then how will it affect the reading of the microammeter?

      1 mark

   3. OR

      (II) (V.I.)

      If the supplied voltage is decreased, then what will be effect on the reading of the microammeter?

      1 mark
   4. (III)

      Plot a graph showing this variation in reading of micrometre on shifting the terminal B towards the right.

      1 mark

   5. OR

      (III) (V.I.)

      If the intensity of incident radiation is doubled, by what factor will the kinetic energy change?

      1 mark

   Figure in the paper: Photocell: evacuated tube with emitter C and collector A, incident light falling on C; external circuit has a battery with sliding terminal B and a microammeter (μA). See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) Threshold frequency ${\nu }_0=\frac{W_0}{h}=\frac{6.35\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}\approx 1.53\times 10^{15}$ Hz.

      1 mark

   2. Incident frequency $3\times 10^{11}$ Hz $<{\nu }_0$, so no photoelectrons are emitted, whatever the intensity. The microammeter reading is zero.

      1 mark

   3. (II) Shifting B to the left lowers the potential of collector A (accelerating potential decreases and finally becomes retarding). The photocurrent decreases and becomes zero when the retarding potential equals the stopping potential. (Assuming light of frequency above threshold.)

      1 mark

   4. (II) (V.I.) On decreasing the supplied (accelerating) voltage below the saturation value, the photocurrent decreases; with a retarding voltage it falls to zero at the stopping potential.

   5. (III) Shifting B to the right increases the accelerating potential of A. Graph of photocurrent (y-axis) vs collector potential (x-axis): current starts from zero at the stopping potential $-V_0$, rises with potential and then becomes constant (saturation current) — a curve flattening into a horizontal line.

      1 mark

   6. (III) (V.I.) Kinetic energy of photoelectrons depends only on frequency ($K_{max}=h\nu -{\varphi }_0$), not on intensity; so it does not change (factor 1). Only the photocurrent doubles.

   Answer: (I) ν₀ ≈ 1.53×10¹⁵ Hz > 3×10¹¹ Hz, reading = 0; (II) current decreases, becomes zero at stopping potential; (III) current rises then saturates; (V.I.) KE unchanged

   Where the marks are

   • Calculates threshold frequency ≈ 1.5×10¹⁵ Hz (1 mark)
   • Concludes no emission, reading zero (1 mark)
   • (II) current decreases to zero as collector potential is reduced (1 mark)
   • (III) graph showing current rising and saturating / (V.I.) KE unchanged (1 mark)

   Open Q30 on its own page

Section E

1. Answer one of these 2 alternatives.

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

   (A) A dielectric slab of thickness t, is introduced between the plates of parallel plate capacitor of area A and separation d (where t<d). Find an expression for the capacitance with the dielectric slab.
   (B) A copper sphere of capacitor C is dropped in ocean. Will the capacitance of the sphere increase, decrease or remain same? Justify.
   (C) A capacitor is connected across a source of potential difference V and then the separation ‘d’ between the plates is increased using insulating stick. Plot ‘V’ vs ‘d’ graph for the given capacitor.

   Try it first, then show our solutionSolution

   1. (A) Charge $\pm q$ on plates, $\sigma =q/A$. Field in the air gaps $E_0=\sigma /{\epsilon }_0$; inside the slab, due to polarisation, $E=E_0/K$.

      ½ mark

   2. Potential difference: $V=E_0(d-t)+\frac{E_0}{K}t=\frac{q}{A{\epsilon }_0}\left(d-t+\frac{t}{K}\right)$

      1 mark

   3. $C=\frac{q}{V}=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$

      ½ mark

   4. (B) The capacitance increases. Justification: $C=q/V$; in water the polar molecules get polarised and reduce the field around the sphere, so for the same charge its potential is lower. Since $C=4\pi {\epsilon }_0{\epsilon }_{r}R$, it becomes about ${\epsilon }_r$ (≈80) times larger.

      2 marks

   5. (C) The capacitor stays connected to the source, so V across it stays equal to the source voltage whatever d is (only C and q decrease). The V–d graph is a straight line parallel to the d-axis at height V.

      1 mark

   Answer: (A) $C={\epsilon }_{0}A/(d-t+t/K)$; (B) increases (≈80 times); (C) horizontal straight line (V constant)

   Where the marks are

   • Fields in air and dielectric (½ mark)
   • Expression for potential difference (1 mark)
   • Capacitance ${\epsilon }_{0}A/(d-t+t/K)$ (½ mark)
   • Capacitance increases, justified by reduced potential due to polarisation (2 marks)
   • V–d graph horizontal, V constant as battery stays connected (1 mark)

   Open Q31(I) on its own page

   OR

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

   (A) If a charge of 1µC is placed at the origin and another charge of 3 µC placed at the point (20m,0m,0m) in an external uniform electric field of $40V/m\hat{i}$ with the electric potential at origin to be zero. Find the electrical potential energy of system.
   (B) If one charge particle is moved from A to C To B and another charge particle of equal magnitude is moved from A to D to B, In uniform external magnetic field. Than for which charge particle more work will be needed? (use fig for reference)
   (C) Electrostatic potential is constant throughout the volume of conductor has the same value on its surface why?

   Figure in the paper: Uniform field E drawn as horizontal parallel arrows pointing right. Points A (bottom left) and D (bottom right) on the lower line, B at upper right; a curved path from A through C to B, and a straight path A to D then D to B. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (A) For a uniform field $\vec{E}=40\hat{i}$ V/m with $V(0)=0$: $V(x)=-Ex=-40x$. So $V(r_1)=0$ and $V(r_2)=-40\times 20=-800$ V.

      1 mark

   2. $U=\frac{k{q}_1{q}_2}{r_{12}}+q_{1}V(r_1)+q_{2}V(r_2)=\frac{9\times 10^9\times 10^{-6}\times 3\times 10^{-6}}{20}+0+3\times 10^{-6}(-800)$

      1 mark

   3. $U=1.35\times 10^{-3}-2.4\times 10^{-3}=-1.05\times 10^{-3}$ J

      1 mark

   4. (B) Equal work is needed for both charges. The electrostatic field is conservative, so work done depends only on the end points A and B, not on the path (A–C–B or A–D–B). (Even if the field were magnetic, it would do no work on either charge.)

      1 mark

   5. (C) Inside a conductor in electrostatic equilibrium $E=0$, so $dV=-\vec{E}\cdot d\vec{l}=0$: no work is needed to move a test charge inside or on the surface. Hence potential is constant throughout the volume and equals its value at the surface.

      1 mark

   Answer: (A) U = −1.05×10⁻³ J; (B) same work for both paths; (C) E = 0 inside, so V is constant and equal to the surface value

   Where the marks are

   • Potentials at the two charge positions using $V=-Ex$ (1 mark)
   • Correct expression for potential energy in external field (1 mark)
   • Correct value $-1.05\times 10^{-3}$ J (1 mark)
   • Work same as field is conservative (1 mark)
   • E = 0 inside conductor ⇒ potential constant (1 mark)

   Open Q31(II) on its own page

2. Answer one of these 2 alternatives.

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

   (A) Derive lens maker’s formula.
   (B) Equi-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 10cm?

   Try it first, then show our solutionSolution

   1. (A) Set-up. Consider a thin lens of refractive index $n_2$ placed in a medium of refractive index $n_1$, with poles B and D of its two surfaces and centres of curvature $C_1$, $C_2$. A point object O on the principal axis forms an image I. Since the lens is thin, BI${}_1\approx$ DI${}_1$, and so on.

   2. Refraction at the first surface (ABC). The first surface alone would form an image I${}_1$ inside the glass. Using the single-surface formula:
      $$\frac{n_1}{OB}+\frac{n_2}{B{I}_1}=\frac{n_2-n_1}{B{C}_1}\dots (1)$$

      1 mark

   3. Refraction at the second surface (ADC). I${}_1$ acts as a virtual object for the second surface, which forms the final image I in the rarer medium:
      $$-\frac{n_2}{D{I}_1}+\frac{n_1}{DI}=\frac{n_2-n_1}{D{C}_2}\dots (2)$$

      1 mark

   4. Adding (1) and (2) with $B{I}_1\approx D{I}_1$ (thin lens):
      $$\frac{n_1}{OB}+\frac{n_1}{DI}=(n_2-n_1)\left(\frac{1}{B{C}_1}+\frac{1}{D{C}_2}\right)$$
      Now apply the Cartesian sign convention: $OB=-u$, $DI=+v$, $B{C}_1=+R_1$ and $D{C}_2=-R_2$. This gives
      $$\frac{1}{v}-\frac{1}{u}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
      For an object at infinity ($u=-\infty$), the image forms at the focus ($v=f$). With $n_{21}=\mu$:
      $$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
      This is the lens maker's formula.

      1 mark

   5. (B) For an equi-convex lens, $R_1=+R$ and $R_2=-R$, with $\mu =1.55$ and $f=10$ cm. Then
      $$\frac{1}{10}=(1.55-1)\left(\frac{1}{R}+\frac{1}{R}\right)=0.55\times \frac{2}{R}$$

      1 mark

   6. Solving: $R=2\times 0.55\times 10=11$ cm. Each face needs a radius of curvature of 11 cm.

      1 mark

   Answer: (A) $\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$; (B) R = 11 cm

   Where the marks are

   • Writes the refraction equation at the first surface (1 mark)
   • Writes the refraction equation at the second surface, treating I${}_1$ as the object (1 mark)
   • Adds the equations, applies the sign convention and puts $u=-\infty$, $v=f$ to get the lens maker's formula (1 mark)
   • Substitutes $R_1=R$, $R_2=-R$, $\mu =1.55$, $f=10$ cm (1 mark)
   • Obtains R = 11 cm (1 mark)

   Open Q32(I) on its own page

   OR

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

   (A) Define angle of deviation in a prism?
   (B) Obtain the relation $A+\delta =i+e$ for a prism where A is the angle of prism, δ is the angle of deviation, i is the angle of incidence and e is the angle of emergence. Write this relation for the minimum deviation?
   (C) Write the condition for minimum deviation.

   Try it first, then show our solutionSolution

   1. (A) The angle of deviation ($\delta$) is the angle between the direction of the incident ray and the direction of the emergent ray after it passes through the prism.

      1 mark

   2. (B) Set-up. Let ray PQ strike face AB at Q with angle of incidence $i$ and refract at angle $r_1$. It strikes face AC at R with internal angle $r_2$ and emerges at angle $e$. The normals at Q and R meet at N.

      • In quadrilateral AQNR, the angles at Q and R are $90^{\circ }$ each. So $\angle A+\angle QNR=180^{\circ }$.
      • In triangle QNR, $r_1+r_2+\angle QNR=180^{\circ }$.

      Comparing the two gives $A=r_1+r_2$.

      1 mark

   3. Deviation. The deviation at the first face is $(i-r_1)$ and at the second face is $(e-r_2)$. So
      $$\delta =(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)=i+e-A$$
      Hence $A+\delta =i+e$.

      1 mark

   4. At minimum deviation, $i=e$ and $r_1=r_2=r=A/2$. The relation becomes $A+{\delta }_m=2i$, i.e. $i=\frac{A+{\delta }_m}{2}$.

      This gives the refractive index of the prism:
      $$\mu =\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$$

      1 mark

   5. (C) Condition for minimum deviation:

      • The angle of incidence equals the angle of emergence ($i=e$), so $r_1=r_2$.
      • Equivalently, the refracted ray inside the prism is parallel to its base (symmetric passage of the ray).
      1 mark

   Answer: (A) Angle between incident and emergent rays; (B) $A+\delta =i+e$, at minimum deviation $A+{\delta }_m=2i$; (C) $i=e$ ($r_1=r_2$), ray inside the prism parallel to the base

   Where the marks are

   • Correct definition of the angle of deviation (1 mark)
   • Shows $A=r_1+r_2$ from the geometry (1 mark)
   • Shows $\delta =i+e-A$, hence $A+\delta =i+e$ (1 mark)
   • Writes $A+{\delta }_m=2i$ for minimum deviation (1 mark)
   • States the condition $i=e$ / ray parallel to the base inside the prism (1 mark)

   Open Q32(II) on its own page

3. Answer one of these 2 alternatives.

   Q33(I) · 5 marks · Long answerMoving Charges and Magnetism

   (A) State the working principle of a moving coil galvanometer? What modification is required in the galvanometer to make its scale linear?
   (B) If a galvanometer of resistance 49.5Ω has range of 0.05A. What will be the value of resistance needed to convert it in ammeter of range 5A?
   (C) How these two resistors should be connected to galvanometer in both cases?

   Try it first, then show our solutionSolution

   1. (A) Principle. A current-carrying coil placed in a magnetic field experiences a torque, $\tau =NIAB\sin \theta$. A spring provides a restoring torque $k\varphi$. At equilibrium the deflection $\varphi$ is proportional to the current: $\varphi =\frac{NAB}{k}I$.

      1 mark

   2. Modification for a linear scale. The magnetic field is made radial. To do this:

      • the pole pieces of the magnet are made concave (cylindrical), and
      • a soft-iron cylindrical core is placed inside the coil.

      The plane of the coil then always stays parallel to B ($\sin \theta =1$), so $\varphi \propto I$ and the scale is linear.

      1 mark

   3. (B) Given $G=49.5\Omega$, $I_g=0.05$ A and $I=5$ A. The shunt carries $I-I_g$, and the potential difference across the shunt equals that across the galvanometer:
      $$I_{g}G=(I-I_g)S$$

      1 mark

   4. Solving:
      $$S=\frac{I_{g}G}{I-I_g}=\frac{0.05\times 49.5}{5-0.05}=\frac{2.475}{4.95}=0.5\Omega$$

      1 mark

   5. (C)

      • To make an ammeter, the low resistance (shunt, here 0.5 Ω) is connected in parallel with the galvanometer.
      • To make a voltmeter, a high resistance is connected in series with the galvanometer.
      1 mark

   Answer: (A) Torque on a current loop in a magnetic field; use a radial field (concave pole pieces + soft-iron core); (B) 0.5 Ω shunt; (C) shunt in parallel for ammeter, high resistance in series for voltmeter

   Where the marks are

   • States the principle: torque on a current-carrying coil in a magnetic field, giving $\varphi \propto I$ (1 mark)
   • Radial field using concave pole pieces and a soft-iron core (1 mark)
   • Uses $I_{g}G=(I-I_g)S$ (1 mark)
   • Gets S = 0.5 Ω (1 mark)
   • Ammeter: shunt in parallel; voltmeter: high resistance in series (1 mark)

   Open Q33(I) on its own page

   OR

   Q33(II) · 5 marks · Long answerAlternating Current

   (A) An input potential $V_{in}=200\sin 100\pi t$ V is provided to an ideal transformer having 1000 turns in primary coil and 100 turns in secondary coil as shown in figure. The load circuit has a resistance of 4Ω, a capacitive reactance of 2Ω and an inductive reactance of 6Ω.
   Find:
   (i) the output voltage across the load circuit
   (ii) the current flowing through the load circuit
   (iii) the power supplied to the load circuit by the transformer
   (B) State the working principle of a transformer and explain how it is a key component in the transfer of electrical power over long distances.

   Figure in the paper: AC source connected to primary coil of a transformer (many turns); secondary coil (fewer turns) connected to a series load of R, L and C. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (A)(i) For an ideal transformer:
      $$\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{100}{1000}=0.1$$
      So $V_{out}=0.1\times 200\sin 100\pi t=20\sin 100\pi t$ V, i.e. peak value 20 V.

      1 mark

   2. (ii) Impedance:
      $$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{4^2+(6-2{)}^2}=4\sqrt{2}\Omega$$
      Peak current:
      $$I_m=\frac{V_m}{Z}=\frac{20}{4\sqrt{2}}=\frac{5}{\sqrt{2}}\approx 3.54\text{ A}$$
      Phase: $\tan \varphi =\frac{X_L-X_C}{R}=1$, so $\varphi =\pi /4$. Since $X_L>X_C$, the circuit is inductive and the current lags the voltage.

      $$i=\frac{5}{\sqrt{2}}\sin \left(100\pi t-\frac{\pi }{4}\right)\text{ A}$$

      1 mark

   3. (iii) Average power:
      $$P=V_{rms}{I}_{rms}\cos \varphi =\frac{V_m{I}_m}{2}\cos \varphi =\frac{20\times \frac{5}{\sqrt{2}}}{2}\times \frac{1}{\sqrt{2}}=25\text{ W}$$
      Check: $P=I_{rms}^{2}R={\left(\frac{5}{2}\right)}^2\times 4=25$ W.

      1 mark

   4. (B) Principle. A transformer works on mutual induction. An alternating current in the primary produces a changing magnetic flux in the core. This changing flux links the secondary and induces an emf in it, with $\frac{V_s}{V_p}=\frac{N_s}{N_p}$.

      1 mark

   5. Role in power transmission. At the power station, a step-up transformer raises the voltage to a very high value. For the same power $P=VI$, the current in the lines becomes very small. The line loss $I^{2}R$ therefore drops greatly. Near the cities, step-down transformers reduce the voltage to safe values for use.

      1 mark

   Answer: (i) $V_{out}=20\sin 100\pi t$ V; (ii) $i=\frac{5}{\sqrt{2}}\sin (100\pi t-\pi /4)$ A ($Z=4\sqrt{2}\Omega$, current lags); (iii) P = 25 W; (B) mutual induction; step-up gives high V, low I, so low $I^{2}R$ loss

   Where the marks are

   • Uses the turns ratio to get $V_{out}=20\sin 100\pi t$ V (1 mark)
   • Finds $Z=4\sqrt{2}\Omega$, $I_m=5/\sqrt{2}$ A and phase $\pi /4$ to write i(t) (1 mark)
   • Computes average power = 25 W (1 mark)
   • States the principle of mutual induction (1 mark)
   • Explains that step-up gives high voltage and low current, reducing $I^{2}R$ losses, with step-down at the user end (1 mark)

   Open Q33(II) on its own page