CBSE Class 12 Physics: sample paper 2024-25

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQElectrostatic Potential and Capacitance

   A uniform electric field pointing in positive X-direction exists in a region. Let A be the origin, B be the point on the X-axis at x = +1 cm and C be the point on the Y-axis at y = +1 cm. Then the potential at points A, B and C satisfy.

   1. (A)$V_A<V_B$
   2. (B)$V_A>V_B$
   3. (C)$V_A<V_C$
   4. (D)$V_A>V_C$

   Try it first, then show our solutionSolution

   1. The electric potential falls along the direction of the field. B lies 1 cm from A along +X, so $V_B=V_A-E(0.01)$, which gives $V_A>V_B$.

      ½ mark

   2. C lies on the Y-axis, perpendicular to the field, so A and C are on the same equipotential: $V_A=V_C$. Options (C) and (D) are therefore wrong, and the answer is (B).

      ½ mark

   Answer: $V_A>V_B$, option (B)

   Where the marks are

   • Potential decreases along the field, so $V_A>V_B$ (0.5 mark)
   • $V_A=V_C$, giving option (B) (0.5 mark)

   Open Q1 on its own page

2. Q2 · 1 mark · MCQElectrostatic Potential and Capacitance

   A conducting wire connects two charged conducting spheres of radii $r_1$ and $r_2$ such that they attain equilibrium with respect to each other. The distance of separation between the two spheres is very large as compared to either of their radii. The ratio of the magnitudes of the electric fields at the surfaces of the spheres of radii $r_1$ and $r_2$ is

   1. (A)$\frac{r_1}{r_2}$
   2. (B)$\frac{r_2}{r_1}$
   3. (C)$\frac{r_2^2}{r_1^2}$
   4. (D)$\frac{r_1^2}{r_2^2}$

   Try it first, then show our solutionSolution

   1. The connected spheres reach equal potential: $\frac{k{q}_1}{r_1}=\frac{k{q}_2}{r_2}$, so $\frac{q_1}{q_2}=\frac{r_1}{r_2}$.

      ½ mark

   2. The surface field is $E=\frac{kq}{r^2}$, so $\frac{E_1}{E_2}=\frac{q_1}{q_2}\cdot \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}$. The answer is (B).

      ½ mark

   Answer: $r_2/r_1$, option (B)

   Where the marks are

   • Equal potentials give $q_1/q_2=r_1/r_2$ (0.5 mark)
   • $E_1/E_2=r_2/r_1$, option (B) (0.5 mark)

   Open Q2 on its own page

3. Q3 · 1 mark · MCQMoving Charges and Magnetism

   A long straight wire of circular cross section of radius 'a' carries a steady current I. The current is uniformly distributed across its cross section. The ratio of magnitudes of the magnetic field at a point a/2 above the surface of wire to that of a point a/2 below its surface is

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

   Try it first, then show our solutionSolution

   1. Outside, at $r=3a/2$: $B_1=\frac{{\mu }_{0}I}{2\pi (3a/2)}=\frac{{\mu }_{0}I}{3\pi a}$.

      ½ mark

   2. Inside, at $r=a/2$: $B_2=\frac{{\mu }_{0}Ir}{2\pi a^2}=\frac{{\mu }_{0}I}{4\pi a}$. The ratio is $B_1:B_2=4:3$, so the answer is (C).

      ½ mark

   Answer: 4:3, option (C)

   Where the marks are

   • Field outside the wire at $3a/2$ (0.5 mark)
   • Field inside at $a/2$ and ratio 4:3 (0.5 mark)

   Open Q3 on its own page

4. Q4 · 1 mark · MCQWave Optics

   The diffraction effect can be observed in

   1. (A)sound waves only
   2. (B)light waves only
   3. (C)ultrasonic waves only
   4. (D)sound waves as well as light waves

   Try it first, then show our solutionSolution

   1. Diffraction is a property of all waves, so it occurs for both sound and light. It is noticeable when the obstacle or aperture size is comparable to the wavelength. The answer is (D).

      1 mark

   Answer: Option (D)

   Where the marks are

   • Diffraction is common to all waves (1 mark)

5. Q5 · 1 mark · MCQElectromagnetic Waves

   A capacitor consists of two parallel plates, with an area of cross-section of 0.001 m${}^2$, separated by a distance of 0.0001 m. If the voltage across the plates varies at the rate of $10^8$ V/s, then the value of displacement current through the capacitor is

   1. (A)$8.85\times 10^{-3}$ A
   2. (B)$8.85\times 10^{-4}$ A
   3. (C)$7.85\times 10^{-3}$ A
   4. (D)$9.85\times 10^{-3}$ A

   Try it first, then show our solutionSolution

   1. Capacitance: $C=\frac{{\epsilon }_{0}A}{d}=\frac{8.85\times 10^{-12}\times 10^{-3}}{10^{-4}}=8.85\times 10^{-11}$ F.

      ½ mark

   2. Displacement current: $I_d=C\frac{dV}{dt}=8.85\times 10^{-11}\times 10^8=8.85\times 10^{-3}$ A. The answer is (A).

      ½ mark

   Answer: $8.85\times 10^{-3}$ A, option (A)

   Where the marks are

   • Finds $C$ (0.5 mark)
   • $I_d=CdV/dt$ (0.5 mark)

   Open Q5 on its own page

6. Q6 · 1 mark · MCQAlternating Current

   In a series LCR circuit, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited the voltage across the inductance will be

   1. (A)10 V
   2. (B)$10\sqrt{2}$ V
   3. (C)$10/\sqrt{2}$ V
   4. (D)20 V

   Try it first, then show our solutionSolution

   1. Equal voltages mean $R=X_L=X_C$. The circuit is at resonance, so the source voltage is $V=IR=10$ V.

      ½ mark

   2. With C shorted, $Z=\sqrt{R^2+R^2}=R\sqrt{2}$, so $I^{\prime }=\frac{10}{R\sqrt{2}}$. Then $V_L=I^{\prime }R=\frac{10}{\sqrt{2}}=5\sqrt{2}$ V. The answer is (C).

      ½ mark

   Answer: $10/\sqrt{2}$ V, option (C)

   Where the marks are

   • $R=X_L=X_C$ and source voltage 10 V (0.5 mark)
   • $Z=R\sqrt{2}$, $V_L=10/\sqrt{2}$ V (0.5 mark)

   Open Q6 on its own page

7. Q7 · 1 mark · MCQElectromagnetic Waves

   Correct match of column I with column II is

   C-I (waves)	C-II (Production)
   (1) Infra-red	P. Rapid vibration of electrons in aerials
   (2) Radio	Q. Electrons in atoms emit light when they move from higher to lower energy level.
   (3) Light	R. Klystron valve
   (4) Microwave	S. Vibration of atoms and molecules

   1. (A)1-P, 2-R, 3-S, 4-Q
   2. (B)1-S, 2-P, 3-O, 4-R
   3. (C)1-Q, 2-P, 3-S, 4-R
   4. (D)1-S, 2-R, 3-P, 4-Q

   Try it first, then show our solutionSolution

   1. Infrared comes from vibrations of atoms and molecules (S). Radio waves come from rapid vibration of electrons in aerials (P).

      ½ mark

   2. Light comes from electronic transitions in atoms (Q). Microwaves come from a klystron valve (R). The match 1-S, 2-P, 3-Q, 4-R is option (B); its 'O' is a misprint for Q.

      ½ mark

   Answer: Option (B)

   Where the marks are

   • Correct matching 1-S, 2-P, 3-Q, 4-R (1 mark)

   Open Q7 on its own page

8. Q8 · 1 mark · MCQAtoms

   The distance of closest approach of an alpha particle is $d$ when it moves with a speed V towards a nucleus. Another alpha particle is projected with higher energy such that the new distance of the closest approach is $d/2$. What is the speed of projection of the alpha particle in this case?

   1. (A)V/2
   2. (B)$\sqrt{2}$ V
   3. (C)2 V
   4. (D)4 V

   Try it first, then show our solutionSolution

   1. At closest approach, kinetic energy equals potential energy: $\frac{1}{2}m{V}^2=\frac{kZe(2e)}{d}$, so $d\propto \frac{1}{V^2}$.

      ½ mark

   2. Halving d doubles $V^2$, so $V^{\prime }=\sqrt{2}V$. The answer is (B).

      ½ mark

   Answer: $\sqrt{2}V$, option (B)

   Where the marks are

   • $d\propto 1/V^2$ (0.5 mark)
   • $V^{\prime }=\sqrt{2}V$ (0.5 mark)

   Open Q8 on its own page

9. Q9 · 1 mark · MCQRay Optics and Optical Instruments

   A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface of the sphere is

   1. (A)2 cm
   2. (B)4 cm
   3. (C)6 cm
   4. (D)12 cm

   Try it first, then show our solutionSolution

   1. Apply $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ with $n_1=1.5$, $n_2=1$, $u=-6$ cm and $R=-6$ cm: $\frac{1}{v}+\frac{1.5}{6}=\frac{-0.5}{-6}$.

      ½ mark

   2. This gives $\frac{1}{v}=\frac{1}{12}-\frac{3}{12}=-\frac{1}{6}$, so $v=-6$ cm. The image is at the centre, 6 cm from the surface. The answer is (C).

      ½ mark

   Answer: 6 cm, option (C)

   Where the marks are

   • Correct refraction formula with sign convention (0.5 mark)
   • $v=-6$ cm (0.5 mark)

   Open Q9 on its own page

10. Q10 · 1 mark · MCQWave Optics

    Colours observed on a CD (Compact Disk) is due to

    1. (A)Reflection
    2. (B)Diffraction
    3. (C)Dispersion
    4. (D)Absorption

    Try it first, then show our solutionSolution

    1. The closely spaced tracks on a CD act as a diffraction grating. Different wavelengths are diffracted at different angles, producing colours. The answer is (B).

       1 mark

    Answer: Diffraction, option (B)

    Where the marks are

    • CD tracks act as a grating (1 mark)

11. Q11 · 1 mark · MCQSemiconductor Electronics: Materials, Devices and Simple Circuits

    The number of electrons made available for conduction by dopant atoms depends strongly upon

    1. (A)doping level
    2. (B)increase in ambient temperature
    3. (C)energy gap
    4. (D)options (A) and (B) both

    Try it first, then show our solutionSolution

    1. Each dopant atom contributes about one carrier, and at room temperature nearly all dopants are ionised. The number of carriers from dopants is therefore fixed mainly by the doping level, not by ambient temperature. The answer is (A).

       1 mark

    Answer: Doping level, option (A)

    Where the marks are

    • Dopant carriers depend on doping level (1 mark)

12. Q12 · 1 mark · MCQCurrent Electricity

    If copper wire is stretched to make its radius decrease by 0.1%, then the percentage change in its resistance is approximately

    1. (A)–0.4%
    2. (B)+0.8%
    3. (C)+0.4%
    4. (D)+0.2%

    Try it first, then show our solutionSolution

    1. Stretching keeps the volume constant, so $l\propto 1/r^2$. Then $R=\frac{\rho l}{\pi r^2}\propto r^{-4}$.

       ½ mark

    2. $\frac{\Delta R}{R}=-4\frac{\Delta r}{r}=-4(-0.1\%)=+0.4\%$. The answer is (C).

       ½ mark

    Answer: +0.4%, option (C)

    Where the marks are

    • $R\propto r^{-4}$ (0.5 mark)
    • +0.4% (0.5 mark)

    Open Q12 on its own page

13. Q13 · 1 mark · Assertion-reasonCompetency-basedMoving Charges and Magnetism

    Assertion (A): On increasing the current sensitivity of a galvanometer by increasing the number of turns may not necessarily increase its voltage sensitivity.
    Reason (R): The resistance of the coil of the galvanometer increases on increasing the number of turns.

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

    Try it first, then show our solutionSolution

    1. Current sensitivity is $\frac{NBA}{k}$ and voltage sensitivity is $\frac{NBA}{kR}$. More turns raise both N and the coil resistance R, so voltage sensitivity may stay the same.

       ½ mark

    2. Both statements are true, and R explains A. The answer is (A).

       ½ mark

    Answer: Option (A)

    Where the marks are

    • Voltage sensitivity $=$ current sensitivity$/R$, and R increases with N (1 mark)

    Open Q13 on its own page

14. Q14 · 1 mark · Assertion-reasonCompetency-basedAtoms

    Assertion (A): In a hydrogen atom there is only one electron but its emission spectrum shows many lines.
    Reason (R): In a given sample of hydrogen there are many atoms each containing one electron; hence many electrons in different atoms may be in different orbits so many transitions from higher to lower orbits are possible.

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

    Try it first, then show our solutionSolution

    1. A sample contains a huge number of atoms whose electrons are excited to different levels. Their many different downward transitions produce many spectral lines. Both statements are true and R explains A, so the answer is (A).

       1 mark

    Answer: Option (A)

    Where the marks are

    • Many atoms undergo different transitions (1 mark)

15. Q15 · 1 mark · Assertion-reasonCompetency-basedNuclei

    Assertion (A): Nuclei having mass number about 60 are least stable.
    Reason (R): When two or more light nuclei are combined into a heavier nucleus then the binding energy per nucleon will decrease.

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

    Try it first, then show our solutionSolution

    1. Nuclei with $A\approx 60$ (such as Fe) have the maximum binding energy per nucleon, so they are the most stable. The Assertion is false.

       ½ mark

    2. In fusion of light nuclei, binding energy per nucleon increases, so the Reason is false too. The answer is (D).

       ½ mark

    Answer: Option (D)

    Where the marks are

    • A is false: A≈60 is most stable (0.5 mark)
    • R is false: fusion increases BE/nucleon (0.5 mark)

    Open Q15 on its own page

16. Q16 · 1 mark · Assertion-reasonCompetency-basedDual Nature of Radiation and Matter

    Assertion (A): de Broglie's wavelength of a freely falling body keeps decreasing with time.
    Reason (R): The momentum of the freely falling body increases with time.

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

    Try it first, then show our solutionSolution

    1. For a falling body $v=gt$, so its momentum $p=mgt$ increases with time. Since $\lambda =h/p$, the wavelength decreases.

       ½ mark

    2. Both statements are true and R explains A. The answer is (A).

       ½ mark

    Answer: Option (A)

    Where the marks are

    • $\lambda =h/p$ with p increasing (1 mark)

    Open Q16 on its own page

Section B

1. Q17 · 2 marks · Very short answerDual Nature of Radiation and Matter

   A platinum surface having work function 5.63 eV is illuminated by a monochromatic source of $1.6\times 10^{15}$ Hz. What will be the minimum wavelength associated with the ejected electron.

   Try it first, then show our solutionSolution

   1. Energy of the incident photon: $E=h\nu =6.63\times 10^{-34}\times 1.6\times 10^{15}=1.06\times 10^{-18}\text{J}=\frac{1.06\times 10^{-18}}{1.6\times 10^{-19}}=6.63\text{eV}$

      ½ mark

   2. By Einstein's photoelectric equation, the maximum kinetic energy of the ejected electrons is $K_{max}=h\nu -{\varphi }_0=6.63-5.63=1.0\text{eV}=1.6\times 10^{-19}\text{J}$

      ½ mark

   3. The electron with the largest kinetic energy has the largest momentum, so it has the minimum de Broglie wavelength: ${\lambda }_{min}=\frac{h}{p_{max}}=\frac{h}{\sqrt{2m{K}_{max}}}$

      ½ mark

   4. ${\lambda }_{min}=\frac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}}}=\frac{6.63\times 10^{-34}}{5.40\times 10^{-25}}\approx 1.23\times 10^{-9}\text{m}=1.23\text{nm}$

      ½ mark

   Answer: $K_{max}=1$ eV; minimum de Broglie wavelength $\approx 1.23\times 10^{-9}$ m (1.23 nm)

   Where the marks are

   • Calculates photon energy $h\nu \approx 6.63$ eV (½ mark)
   • Finds $K_{max}=h\nu -{\varphi }_0=1$ eV (½ mark)
   • Uses $\lambda =h/\sqrt{2mK}$ for the electron (½ mark)
   • Correct value $\approx 1.23\times 10^{-9}$ m (½ mark)

   Open Q17 on its own page

2. Answer one of these 2 alternatives.

   Q18(I) · 2 marks · Very short answerWave Optics

   A beam of light consisting of two wavelengths, 4000 Å and 6000 Å, is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from the central maximum where the dark fringe is obtained?

   Solution being checked. Our step-by-step solution is written and is waiting for an independent check before we show it.

   OR

   Q18(II) · 2 marks · Very short answerWave Optics

   In Young's double-slit experiment using monochromatic light of wavelength λ, the intensities of two sources are I. What is the intensity of light at a point where path difference between wavefronts is λ/4?

   Try it first, then show our solutionSolution

   1. Resultant intensity of two interfering waves: $I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

      ½ mark

   2. With $I_1=I_2=I$: $I_R=2I(1+\cos \varphi )=4I{\cos }^2\frac{\varphi }{2}$

      ½ mark

   3. Phase difference for a path difference of $\lambda /4$: $\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

      ½ mark

   4. $I_R=4I{\cos }^2\frac{\pi }{4}=4I\times \frac{1}{2}=2I$

      ½ mark

   Answer: $I_R=2I$

   Where the marks are

   • Writes the resultant-intensity formula (½ mark)
   • Simplifies it to $4I{\cos }^2(\varphi /2)$ (½ mark)
   • Converts the path difference to phase difference $\pi /2$ (½ mark)
   • Gets the result $2I$ (½ mark)

   Open Q18(II) on its own page

3. Q19 · 2 marks · Very short answerMoving Charges and Magnetism

   P and Q are two identical charged particles each of mass $4\times 10^{-26}$ kg and charge $4.8\times 10^{-19}$ C, each moving with the same speed of $2.4\times 10^5$ m/s as shown in the figure. The two particles are equidistant (0.5 m) from the vertical Y-axis. At some instant, a magnetic field B is switched on so that the two particles undergo head-on collision.

   Find –
   (I) the direction of the magnetic field and
   (II) the magnitude of the magnetic field applied in the region.

   Figure in the paper: X-Y axes. Particle P (positive) at upper left of a 1 m vertical marker, 1 m above the X-axis on the right side of Y-axis, moving left (toward Y-axis). Particle Q (positive) on the X-axis left of the Y-axis, moving right. Each is 0.5 m from the Y-axis; vertical separation 1 m. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) Set up axes: +x to the right, +y up, +z out of the page. P (upper, moving in −x) must curve downward and Q (lower, moving in +x) must curve upward, so that they meet head-on on the Y-axis.

   2. For P: $\vec{F}=q(\vec{v}\times \vec{B})$ must point along $-\hat{y}$. Since $(-\hat{i})\times (-\hat{k})=-\hat{j}$, B must be along $-\hat{k}$. Check for Q: $\hat{i}\times (-\hat{k})=+\hat{j}$, upward, as required. So the field is perpendicular to the plane of the paper, directed inwards.

      ½ mark

   3. (II) Each particle travels a quarter circle and both reach the point (0, 0.5 m) at the same instant, moving in opposite directions. This requires a radius equal to the distance from the Y-axis: $r=0.5$ m.

      ½ mark

   4. $r=\frac{mv}{qB}\Rightarrow B=\frac{mv}{qr}$

      ½ mark

   5. $B=\frac{4\times 10^{-26}\times 2.4\times 10^5}{4.8\times 10^{-19}\times 0.5}=\frac{9.6\times 10^{-21}}{2.4\times 10^{-19}}=0.04\text{T}$

      ½ mark

   Answer: (I) B perpendicular to the plane of the paper, into it; (II) B = 0.04 T

   Where the marks are

   • Correct direction of B: into the plane of the paper (½ mark)
   • Recognises that the radius must be 0.5 m (½ mark)
   • Uses $r=mv/qB$ (½ mark)
   • Gets B = 0.04 T (½ mark)

   Open Q19 on its own page

4. Q20 · 2 marks · Very short answerCompetency-basedNuclei

   Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z are four nuclei indicated on the curve. Identify which of the following nuclei is most likely to undergo
   (i) Nuclear Fission
   (ii) Nuclear Fusion.
   Justify your answer.

   Figure in the paper: Graph of B.E./A (MeV) vs mass number A. Curve rises steeply, peaks near A=60 and then slowly falls. Points: Z at A=30 with 5.0 MeV; Y at A=60 with 8.5 MeV (peak); X at A=90 with 8.0 MeV; W at A=190 with 7.5 MeV. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (i) Nuclear fission: nucleus W (A = 190).

      ½ mark

   2. W is a heavy nucleus with a lower binding energy per nucleon (7.5 MeV) than medium nuclei such as X and Y. Splitting it into two medium-mass nuclei raises the B.E./A, so energy is released.

      ½ mark

   3. (ii) Nuclear fusion: nucleus Z (A = 30).

      ½ mark

   4. Z is a light nucleus with a low B.E./A (5.0 MeV). Two such nuclei fusing give a nucleus of A ≈ 60 near the peak (8.5 MeV per nucleon), so the B.E./A increases and energy is released.

      ½ mark

   Answer: (i) Fission: W; (ii) Fusion: Z

   Where the marks are

   • Identifies W for fission (½ mark)
   • Justifies it: heavy nucleus, lower B.E./A, products have higher B.E./A (½ mark)
   • Identifies Z for fusion (½ mark)
   • Justifies it: light nucleus, low B.E./A, fused product has higher B.E./A (½ mark)

   Open Q20 on its own page

5. Q21 · 2 marks · Very short answerCurrent Electricity

   A cylindrical conductor of length l and cross-section area A is connected to a DC source. Under the influence of electric field set up due to source, the free electrons begin to drift in the opposite direction of the electric field.
   (I) Draw the curve showing the dependency of drift velocity on relaxation time.
   (II) If the DC source is replaced by a source whose current changes its magnitude with time such that $I=I_o\sin 2\pi \nu t$, where $\nu$ is the frequency of variation of current, then determine the average drift velocity of the free electrons over one complete cycle.

   Try it first, then show our solutionSolution

   1. (I) Drift speed: $v_d=\frac{eE}{m}\tau$. For a given field E, $v_d\propto \tau$.

      ½ mark

   2. So the graph of $v_d$ (y-axis) against $\tau$ (x-axis) is a straight line through the origin with slope $eE/m$.

      ½ mark

   3. (II) $I=neA{v}_d$, so $v_d=\frac{I}{neA}=\frac{I_0}{neA}\sin 2\pi \nu t$. The drift velocity varies sinusoidally and reverses direction every half cycle.

      ½ mark

   4. Average over one cycle: $⟨v_d⟩=\frac{I_0}{neA}\cdot \frac{1}{T}{\int }_0^T\sin 2\pi \nu tdt=0$. The average drift velocity over a complete cycle is zero.

      ½ mark

   Answer: (I) Straight line through the origin ($v_d\propto \tau$); (II) average drift velocity = 0

   Where the marks are

   • States $v_d=eE\tau /m$ (½ mark)
   • Graph: straight line through the origin (½ mark)
   • Relates $v_d$ to I: $v_d=I/neA$, sinusoidal (½ mark)
   • Average over a cycle = 0 (½ mark)

   Open Q21 on its own page

Section C

1. Answer one of these 2 alternatives.

   Q22(I) · 3 marks · Short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

   Identify the circuit elements X and Y as shown in the given block diagram and draw the output waveforms of X and Y.

   Figure in the paper: Block diagram: AC source connected to block X, then block Y, then a D.C. Appliance. See it in the official paper.

   Try it first, then show our solutionSolution

   1. X is a rectifier (p-n junction diode circuit). It converts the alternating input into unidirectional (pulsating) current.

      ½ mark

   2. Y is a filter circuit (e.g. a capacitor across the load). It removes the ac ripple and gives a nearly steady dc for the appliance.

      ½ mark

   3. Output waveform of X (full-wave rectifier): a series of positive half-sine humps, one per half cycle, all above the time axis. The voltage rises from zero to the peak and falls back to zero each half cycle. (For a half-wave rectifier, alternate humps are missing.)

      1 mark

   4. Output waveform of Y: a nearly flat line at about the peak value. The capacitor charges up to the peak and discharges slightly between humps, giving a small saw-tooth ripple on a steady dc level.

      1 mark

   Answer: X = rectifier (output: pulsating unidirectional dc); Y = filter (output: nearly steady dc with small ripple)

   Where the marks are

   • Identifies X as a rectifier (½ mark)
   • Identifies Y as a filter (½ mark)
   • Correct pulsating-dc waveform for X (1 mark)
   • Correct smoothed dc waveform with small ripple for Y (1 mark)

   Open Q22(I) on its own page

   OR

   Q22(II) · 3 marks · Short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

   If the centre tapping is shifted towards Diode D1 as shown in the diagram, draw the output waveform of the given circuit.

   Figure in the paper: Full-wave rectifier with centre-tap transformer; secondary ends A and B connect through diodes D1 and D2 to a common point X of load resistor R_L; centre tap connected to ground and the other end Y of R_L; output across R_L. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Shifting the centre tap towards D1 gives the half of the secondary feeding D1 (A to centre tap) fewer turns than the half feeding D2 (centre tap to B). So the emf across the D1 half is smaller than the emf across the D2 half.

      1 mark

   2. In one half cycle D1 is forward biased and conducts. The current through $R_L$ is from X to Y, and the output is a half-sine hump of smaller peak. In the next half cycle D2 conducts, the current through $R_L$ is in the same direction, and the hump has a larger peak.

      1 mark

   3. Output waveform: still full-wave (both humps on the same side of the time axis, no gaps), but alternate humps have unequal heights: small, large, small, large, …

      1 mark

   Answer: Full-wave rectified output with alternate half-cycle humps of unequal amplitude (smaller when D1 conducts, larger when D2 conducts)

   Where the marks are

   • Explains unequal emf in the two halves of the secondary (1 mark)
   • Both diodes conduct in alternate half cycles, with current in the same direction through $R_L$ (1 mark)
   • Draws/describes the output with alternately smaller and larger humps (1 mark)

   Open Q22(II) on its own page

2. Q23 · 3 marks · Short answerElectrostatic Potential and Capacitance

   Find the expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d when (I) a dielectric slab of thickness t and (II) a metallic slab of thickness t, where (t < d) are introduced one by one between the plates of the capacitor. In which case would the capacitance be more and why?

   Try it first, then show our solutionSolution

   1. (I) Let the plates carry charges $\pm q$, with surface density $\sigma =q/A$. In the air gaps the field is $E_0=\frac{\sigma }{{\epsilon }_0}=\frac{q}{{\epsilon }_{0}A}$. Inside the slab, polarisation reduces the field to $E=\frac{E_0}{K}$.

      ½ mark

   2. Potential difference between the plates, with air thickness $(d-t)$ and slab thickness $t$: $V=E_0(d-t)+\frac{E_0}{K}t=\frac{q}{{\epsilon }_{0}A}\left(d-t+\frac{t}{K}\right)$

      ½ mark

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

      ½ mark

   4. (II) Metallic slab: the field inside a conductor is zero, so $V=E_0(d-t)=\frac{q}{{\epsilon }_{0}A}(d-t)$ and $C=\frac{{\epsilon }_{0}A}{d-t}$. This is the same as putting $K\to \infty$ in (I).

      ½ mark

   5. The capacitance is greater with the metallic slab. For the dielectric, the term $t/K$ is still present in the denominator. For the metal it is zero, so the denominator $(d-t)$ is smaller and C is larger.

      1 mark

   Answer: (I) $C=\frac{{\epsilon }_{0}A}{d-t+t/K}$; (II) $C=\frac{{\epsilon }_{0}A}{d-t}$; capacitance is more with the metallic slab (field inside metal is zero, $K\to \infty$)

   Where the marks are

   • Fields: $E_0=q/{\epsilon }_{0}A$ in air, $E_0/K$ in the slab (½ mark)
   • Potential difference $V=E_0(d-t)+E_{0}t/K$ (½ mark)
   • Expression for C with the dielectric slab (½ mark)
   • Expression for C with the metallic slab (½ mark)
   • Concludes the metallic slab gives more capacitance, with reason (1 mark)

   Open Q23 on its own page

3. Q24 · 3 marks · Short answerRay Optics and Optical Instruments

   (I) Draw a ray diagram for the formation of image by a Cassegrain telescope.
   (II) Why these types of telescopes are preferred over refracting type telescopes. (Write 2 points)

   Try it first, then show our solutionSolution

   1. (I) Ray diagram (described): parallel rays from a distant object fall on a large concave (paraboloidal) primary mirror with a small hole at its centre. The reflected rays converge towards the primary's focus.

      1 mark

   2. Before reaching that focus, the rays are intercepted by a small convex (hyperbolic) secondary mirror placed coaxially. It reflects them back through the central hole of the primary, forming the image just behind the primary, where it is viewed through an eyepiece. Arrows show the direction of light; primary, secondary, hole, image and eyepiece are labelled.

      1 mark

   3. (II) Advantage 1: the objective is a mirror, so there is no chromatic aberration, and a paraboloidal mirror also removes spherical aberration.

      ½ mark

   4. Advantage 2: a mirror is much lighter and easier to support than a lens of the same size, so very large apertures can be made. This gathers more light, giving brighter images and higher resolving power. (The folded path also makes the telescope compact.)

      ½ mark

   Answer: (I) Concave primary → convex secondary → through hole in primary → image viewed by eyepiece; (II) free from chromatic/spherical aberration, large aperture gives brighter image and better resolution

   Where the marks are

   • Correct ray diagram: primary concave mirror, secondary convex mirror, image behind the primary through the hole, with labels (2 marks)
   • Any two valid advantages over refracting telescopes (½ + ½ mark)

   Open Q24 on its own page

4. Q25(I) · 3 marks · Short answerSemiconductor Electronics: Materials, Devices and Simple Circuits

   Draw the energy band diagram for P-type semiconductor at (i) T=0K and (ii) room temperature.
   (II) In the given diagram considering an ideal diode, in which condition will the bulb glow
   (a) when the switch is open
   (b) when the switch is closed
   Justify your answer.

   Figure in the paper: Circuit: a cell in series with a bulb; a branch consisting of a switch in series with a diode is connected in parallel with the bulb. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I)(i) T = 0 K: the valence band is completely filled and the conduction band is completely empty, separated by the energy gap $E_g$. Acceptor energy level $E_A$ is a short dashed line just above the top of the valence band (≈ 0.01–0.05 eV); it is empty.

      1 mark

   2. (ii) Room temperature: electrons from the valence band have jumped into the acceptor level, so $E_A$ is shown filled and many holes appear in the valence band (majority carriers). A few electrons have crossed into the conduction band by thermal excitation (minority carriers).

      1 mark

   3. (II) The bulb glows in case (a), when the switch is open. The diode branch is then broken, so the full current of the cell passes through the bulb.

      ½ mark

   4. When the switch is closed, the ideal diode is forward biased and has zero resistance. It short-circuits the bulb, so the current by-passes the bulb and it does not glow.

      ½ mark

   Answer: (I) At 0 K: empty acceptor level just above a full valence band, empty conduction band; at room temperature: acceptor level filled, holes in valence band, few electrons in conduction band. (II) Bulb glows when the switch is open (a).

   Where the marks are

   • Band diagram at 0 K with empty acceptor level near the valence band (1 mark)
   • Band diagram at room temperature with filled acceptor level, holes in VB and few electrons in CB (1 mark)
   • Bulb glows when the switch is open (½ mark)
   • Justification: closed switch puts a forward-biased ideal diode (zero resistance) across the bulb, shorting it (½ mark)

   Open Q25(I) on its own page

5. Q26 · 3 marks · Short answerCompetency-basedMoving Charges and Magnetism

   A boy is holding a smooth, hollow and non-conducting pipe vertically with charged spherical ball of mass 10 g carrying a charge of +10 mC inside it which is free to move along the axis of the pipe. The boy is moving the pipe from East to West direction in the presence of magnetic field of 2T. With what minimum velocity, should the boy move the pipe such that the ball does not move along the axis. Also determine the direction of the magnetic field.

   Try it first, then show our solutionSolution

   1. For the ball not to move along the vertical pipe, the upward magnetic force must balance its weight: $qvB\sin \theta =mg\Rightarrow v=\frac{mg}{qB\sin \theta }$

      ½ mark

   2. v is minimum when $\sin \theta =1$, i.e. B is horizontal and perpendicular to the velocity: $v_{min}=\frac{mg}{qB}$

      ½ mark

   3. With $m=10\text{ g}=10^{-2}$ kg and $q=10\text{ mC}=10^{-2}$ C: $v_{min}=\frac{10^{-2}\times 9.8}{10^{-2}\times 2}$

      ½ mark

   4. $v_{min}=4.9{\text{m s}}^{-1}$

      ½ mark

   5. Direction: $\vec{F}=q\vec{v}\times \vec{B}$ must be vertically upward, with $\vec{v}$ towards West. Take East = $\hat{i}$, North = $\hat{j}$, Up = $\hat{k}$. Then $(-\hat{i})\times (-\hat{j})=+\hat{k}$, so $\vec{B}$ is along $-\hat{j}$. This is consistent with Fleming's left-hand rule: the magnetic field is horizontal, from North to South.

      1 mark

   Answer: $v_{min}=4.9$ m/s; magnetic field horizontal, from North to South

   Where the marks are

   • Equates magnetic force to weight, $qvB\sin \theta =mg$ (½ mark)
   • Uses $\sin \theta =1$ for minimum v (½ mark)
   • Correct substitution in SI units (½ mark)
   • Result 4.9 m/s (½ mark)
   • Direction of B: North to South, with reasoning (1 mark)

   Open Q26 on its own page

6. Q27 · 3 marks · Short answerRay Optics and Optical Instruments

   A light ray entering a right-angled prism undergoes refraction at the face AC as shown in Fig. 1.
   (I) What is the refractive index of the material of the prism in Fig. 1?
   (II) (a) If the side AC of the above prism is now surrounded by a liquid of refractive index $\frac{2}{\sqrt{3}}$, as shown in Fig. 2, determine if the light ray continues to graze along the interface AC or undergoes total internal reflection or undergoes refraction into the liquid.
   (b) Draw the ray diagram to represent the path followed by the incident ray with the corresponding angle values.
   (Given, ${\sin }^{-1}(\frac{\sqrt{2}}{\sqrt{3}})=54.6^{\circ }$)

   Figure in the paper: Right-angled prism ABC with right angle at B, angle 45° at C, vertical side AB, horizontal base BC, hypotenuse AC. A horizontal ray enters face AB normally and strikes AC; in Fig. 1 it grazes along AC after refraction. In Fig. 2 the region on the AC side is filled with a liquid of refractive index 2/√3. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) The ray enters face AB normally and goes undeviated. From the geometry of the prism (∠C = 45°), it meets face AC at an angle of incidence of 45°.

      ½ mark

   2. Since the ray grazes along AC, 45° is the critical angle: $\sin C=\frac{1}{n}\Rightarrow \sin 45^{\circ }=\frac{1}{n}\Rightarrow n=\sqrt{2}$

      ½ mark

   3. (II)(a) Refractive index of glass relative to the liquid: ${}^l{n}_g=\frac{\sqrt{2}}{2/\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{2}}$, so $\sin C^{\prime }=\frac{\sqrt{2}}{\sqrt{3}}$, giving $C^{\prime }=54.6^{\circ }$. The angle of incidence (45°) is less than C', so the ray neither grazes nor suffers total internal reflection; it is refracted into the liquid.

      1 mark

   4. Snell's law at AC: $\sqrt{2}\sin 45^{\circ }=\frac{2}{\sqrt{3}}\sin r\Rightarrow 1=\frac{2}{\sqrt{3}}\sin r\Rightarrow \sin r=\frac{\sqrt{3}}{2}\Rightarrow r=60^{\circ }$

      ½ mark

   5. (b) Ray diagram (described): the ray enters AB normally (0°), goes straight to AC and meets it at 45° to the normal. It then enters the liquid bent away from the normal, making 60° with the normal to AC.

      ½ mark

   Answer: (I) $n=\sqrt{2}$; (II) critical angle 54.6° > 45°, so the ray refracts into the liquid at $r=60^{\circ }$

   Where the marks are

   • Angle of incidence on AC is 45° (½ mark)
   • n = √2 from the critical-angle condition (½ mark)
   • Relative critical angle 54.6° compared with 45°: refraction occurs (1 mark)
   • Angle of refraction 60° and ray diagram (1 mark)

   Open Q27 on its own page

7. Answer one of these 2 alternatives.

   Q28(I) · 3 marks · Short answerElectric Charges and Fields

   State Gauss's theorem in electrostatics. Using this theorem, derive an expression for the electric field due to an infinitely long straight wire of linear charge density $\lambda$.

   Try it first, then show our solutionSolution

   1. Gauss's theorem: the total electric flux through any closed surface equals $\frac{1}{{\epsilon }_0}$ times the net charge enclosed by it: $\oint \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{{\epsilon }_0}$

      1 mark

   2. Take an infinitely long straight wire with uniform linear charge density λ. To find E at a point P at distance r, choose a cylindrical Gaussian surface of radius r and length l, coaxial with the wire. By symmetry, E is radial and has the same magnitude at all points of the curved surface.

      ½ mark

   3. $\oint \vec{E}\cdot d\vec{S}={\int }_{\text{curved}}EdS\cos 0^{\circ }+{\int }_{\text{two end faces}}EdS\cos 90^{\circ }$

      ½ mark

   4. The flat ends contribute zero [E is parallel to the end faces], so $\varphi =E\times 2\pi rl$

      ½ mark

   5. Charge enclosed $=\lambda l$. By Gauss's theorem $E\cdot 2\pi rl=\frac{\lambda l}{{\epsilon }_0}\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$, directed radially outward for λ > 0.

      ½ mark

   Answer: $\oint \vec{E}\cdot d\vec{S}=q/{\epsilon }_0$; $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ (radial)

   Where the marks are

   • Correct statement of Gauss's theorem with formula (1 mark)
   • Chooses a coaxial cylindrical Gaussian surface, with symmetry argument (½ mark)
   • Splits the flux over the curved surface and end faces (½ mark)
   • Flux = $E\cdot 2\pi rl$ (½ mark)
   • Charge enclosed λl and final result $\lambda /2\pi {\epsilon }_{0}r$ (½ mark)

   Open Q28(I) on its own page

   OR

   Q28(II) · 3 marks · Short answerElectric Charges and Fields

   (a) Define electric flux and write its SI unit.
   (b) Use Gauss's law to obtain the expression for the electric field due to a uniformly charged infinite plane sheet of charge.

   Try it first, then show our solutionSolution

   1. (a) Electric flux through a surface is the total number of electric field lines crossing it normally, i.e. the surface integral of the field: $\varphi =\int \vec{E}\cdot d\vec{S}$. It is a scalar. SI unit: N m² C⁻¹ (equivalently V m).

      1 mark

   2. (b) Take an infinite plane sheet with uniform surface charge density σ. By symmetry, E is perpendicular to the sheet and has the same magnitude on both sides. Choose a Gaussian cylinder of cross-section A piercing the sheet, with its flat faces equidistant from it on either side and point P on one face.

      ½ mark

   3. $\varphi ={\int }_{\text{RCF}}\vec{E}\cdot d\vec{A}+{\int }_{\text{LCF}}\vec{E}\cdot d\vec{A}+{\int }_{\text{CS}}\vec{E}\cdot d\vec{A}=\int EdA\cos 0^{\circ }+\int EdA\cos 0^{\circ }+\int EdA\cos 90^{\circ }$

      ½ mark

   4. $\varphi =EA+EA+0=2EA$ [the curved surface is parallel to E]

      ½ mark

   5. Charge enclosed $=\sigma A$. By Gauss's law $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}$, directed normally away from the sheet for σ > 0 and independent of distance.

      ½ mark

   Answer: (a) $\varphi =\oint \vec{E}\cdot d\vec{S}$, unit N m² C⁻¹; (b) $E=\frac{\sigma }{2{\epsilon }_0}$

   Where the marks are

   • Definition of electric flux with SI unit (1 mark)
   • Gaussian cylinder with symmetry argument (½ mark)
   • Flux split into the three surfaces (½ mark)
   • Total flux = 2EA (½ mark)
   • Charge σA and result $E=\sigma /2{\epsilon }_0$ (½ mark)

   Open Q28(II) on its own page

Section D

case study

1. Q29 · 4 marks · Case-basedCompetency-basedMoving Charges and Magnetism

   Case Study Based Question: Motion of Charge in Magnetic Field
   An electron with speed $v_o\ll c$ moves in a circle of radius $r_o$ in a uniform magnetic field. This electron is able to traverse a circular path as the magnetic force acting on the electron is perpendicular to both $v_o$ and B, as shown in the figure. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is $T_o$.

   1. (i)

      If the speed of the electron is now doubled to $2v_o$. The radius of the circle will change to

      1. (A)$4r_o$
      2. (B)$2r_o$
      3. (C)$r_o$
      4. (D)$r_o/2$

      1 mark
   2. (ii)

      If $v=2v_o$, then the time required for one revolution of the electron ($T_o$) will change to

      1. (A)$4T_o$
      2. (B)$2T_o$
      3. (C)$T_o$
      4. (D)$T_o/2$

      1 mark
   3. (iii)

      A charged particle is projected in a magnetic field $\mathbf{B}=(2\mathbf{i}+4\mathbf{j})\times 10^2$ T. The acceleration of the particle is found to be $\mathbf{a}=(x\mathbf{i}+2\mathbf{j})$ m/s${}^2$. Find the value of x.

      1. (A)4 ms${}^{-2}$
      2. (B)-4 ms${}^{-2}$
      3. (C)-2 ms${}^{-2}$
      4. (D)2 ms${}^{-2}$

      1 mark
   4. (iv)

      If the given electron has a velocity not perpendicular to B, then trajectory of the electron is

      1. (A)straight line
      2. (B)circular
      3. (C)helical
      4. (D)zig-zag

      1 mark

   5. OR

      (iv)

      If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is

      1. (A)Bev
      2. (B)Be/v
      3. (C)B/ev
      4. (D)Zero

      1 mark

   Figure in the paper: Electron moving in a circle in a uniform magnetic field B directed into the page (crosses); velocity v_o tangent to circle and magnetic force F directed toward the centre at several positions. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (i) $r=\frac{mv}{eB}\propto v$, so doubling v doubles the radius: $r=2r_o$. Option (B).

      1 mark

   2. (ii) $T=\frac{2\pi r}{v}=\frac{2\pi m}{eB}$, which is independent of speed, so T stays $T_o$. Option (C).

      1 mark

   3. (iii) The magnetic force, and hence the acceleration, is perpendicular to B, so $\vec{a}\cdot \vec{B}=0$: $(x)(2)+(2)(4)=0\Rightarrow x=-4{\text{m s}}^{-2}$. Option (B).

      1 mark

   4. (iv) The component of v perpendicular to B gives circular motion and the component parallel to B gives uniform motion along B, so the path is a helix. Option (C).

      1 mark

   5. (iv) Alternative: $F=evB\sin 0^{\circ }=0$ when v is parallel to B. Option (D).

   Answer: (i) B; (ii) C; (iii) B (x = −4 m s⁻²); (iv) C [alternative (iv): D]

   Where the marks are

   • (i) $2r_o$, option B (1 mark)
   • (ii) $T_o$, option C (1 mark)
   • (iii) x = −4, option B (1 mark)
   • (iv) helical, option C; or alternative: zero, option D (1 mark)

   Open Q29 on its own page

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

   Case Study Based Question: Photoelectric effect
   It is the phenomenon of emission of electrons from a metallic surface when light of a suitable frequency is incident on it. The emitted electrons are called photoelectrons. Nearly all metals exhibit this effect with ultraviolet light but alkali metals like lithium, sodium, potassium, cesium etc. show this effect even with visible light. It is an instantaneous process i.e. photoelectrons are emitted as soon as the light is incident on the metal surface. The number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation. The maximum kinetic energy of the photoelectrons emitted from a given metal surface is independent of the intensity of the incident light and depends only on the frequency of the incident light. For a given metal surface there is a certain minimum value of the frequency of the incident light below which emission of photoelectrons does not occur.

   1. (I)

      In a photoelectric experiment plate current is plotted against anode potential.

      1. (A)A and B will have same intensities while B and C will have different frequencies
      2. (B)B and C will have different intensities while A and B will have different frequencies
      3. (C)A and B will have different intensities while B and C will have equal frequencies
      4. (D)B and C will have equal intensities while A and B will have same frequencies.

      1 mark
   2. (II)

      Photoelectrons are emitted when a zinc plate is

      1. (A)Heated
      2. (B)hammered
      3. (C)Irradiated by ultraviolet light
      4. (D)subjected to a high pressure

      1 mark
   3. (III)

      The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of 500 nm. Its work function is about

      1. (A)$4\times 10^{-19}$ J
      2. (B)1 J
      3. (C)$2\times 10^{-19}$ J
      4. (D)$3\times 10^{-19}$ J

      1 mark
   4. (IV)

      The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is

      1. (A)2 V
      2. (B)4 V
      3. (C)6 V
      4. (D)10 V

      1 mark

   5. OR

      (IV)

      The minimum energy required to remove an electron from a substance is called its

      1. (A)work function
      2. (B)kinetic energy
      3. (C)stopping potential
      4. (D)potential energy

      1 mark

   Figure in the paper: Graph of plate current I vs anode potential V with three curves A, B, C. A and B start rising from the same (stopping potential) point on the V axis; A saturates at a lower current than B; C starts at a more negative potential (larger stopping potential) and saturates at the same current as B. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (I) A and B have the same stopping potential, so the same frequency, but different saturation currents, so different intensities. B and C have the same saturation current, so equal intensity, but different stopping potentials, so different frequencies. Option (D).

      1 mark

   2. (II) Zinc needs ultraviolet light (frequency above its threshold) for photoemission; heating, hammering or pressure do not supply photons. Option (C).

      1 mark

   3. (III) ${\varphi }_0=\frac{hc}{{\lambda }_0}=\frac{6.63\times 10^{-34}\times 3\times 10^8}{500\times 10^{-9}}\approx 3.98\times 10^{-19}\text{J}\approx 4\times 10^{-19}$ J. Option (A).

      1 mark

   4. (IV) $e{V}_0=K_{max}=4$ eV, so $V_0=4$ V. Option (B).

      1 mark

   5. (IV) Alternative: the minimum energy needed to remove an electron from a substance is its work function. Option (A).

   Answer: (I) D; (II) C; (III) A; (IV) B [alternative (IV): A]

   Where the marks are

   • (I) option D (1 mark)
   • (II) option C (1 mark)
   • (III) $\approx 4\times 10^{-19}$ J, option A (1 mark)
   • (IV) 4 V, option B; or alternative: work function, option A (1 mark)

   Open Q30 on its own page

Section E

1. Answer one of these 2 alternatives.

   Q31(I) · 5 marks · Long answerCurrent Electricity

   a) Write two limitations of ohm's law. Plot their I-V characteristics.
   b) A heating element connected across a battery of 100 V having an internal resistance of 1 Ω draws an initial current of 10 A at room temperature 20.0 °C which settles after a few seconds to a steady value. What is the power consumed by battery itself after the steady temperature of 320.0 °C is attained? Temperature coefficient of resistance averaged over the temperature range involved is $3.70\times 10^{-4}$ °C${}^{-1}$.

   Try it first, then show our solutionSolution

   1. (a) Limitation 1: V is not always proportional to I; the V-I graph is non-linear. Example: a p-n junction diode or a filament lamp. Graph: I against V is a curve, not a straight line through the origin.

      1 mark

   2. Limitation 2: the relation between V and I depends on the sign of V. Reversing V with the same magnitude does not give the same magnitude of current, e.g. a diode. Graph: a large forward current for small positive V, but almost no current for negative V. (Another limitation: the relation may not be unique, with more than one V for the same I, as in GaAs.)

      1 mark

   3. (b) At room temperature (20 °C): $I=\frac{\epsilon }{R_0+r}\Rightarrow R_0=\frac{\epsilon }{I}-r=\frac{100}{10}-1$

      ½ mark

   4. $R_0=9\Omega$

      ½ mark

   5. At 320 °C: $R=R_0(1+\alpha \Delta T)=9(1+3.70\times 10^{-4}\times 300)$

      ½ mark

   6. $R=9\times 1.111\approx 10\Omega$

      ½ mark

   7. Steady current: $I^{\prime }=\frac{100}{10+1}=\frac{100}{11}=9.09$ A. Power consumed by the battery itself: $P=I^{\prime 2}r={\left(\frac{100}{11}\right)}^2\times 1$

      ½ mark

   8. $P\approx 82.6\text{W}$

      ½ mark

   Answer: (a) Non-linear V-I relation; dependence on the sign of V (or non-unique relation). (b) R₀ = 9 Ω, R = 10 Ω, power in battery ≈ 82.6 W

   Where the marks are

   • First limitation with its V-I graph (1 mark)
   • Second limitation with its V-I graph (1 mark)
   • Finds $R_0=9\Omega$ from $I=\epsilon /(R_0+r)$ (1 mark)
   • Finds R ≈ 10 Ω at 320 °C (1 mark)
   • Power in internal resistance $I^{2}r\approx 82.6$ W (1 mark)

   Open Q31(I) on its own page

   OR

   Q31(II) · 5 marks · Long answerCurrent Electricity

   a) Using Kirchhoff's laws obtain the equation of the balanced state in Wheatstone bridge.
   b) A wire of uniform cross-section and resistance of 12 ohm is bent in the shape of circle as shown in the figure. A resistance of 10 ohms is connected to diametrically opposite ends C and D. A battery of emf 8V is connected between A and B. Determine the current flowing through arm AD.

   Figure in the paper: Circle of wire with centre O; C at top and D at bottom, joined by a vertical diameter containing a 10 Ω resistor. A and B are points on the circle on left and right, each making 60° with the vertical diameter at O (angle AOD = 60°, BOD = 60°). An 8 V battery is connected between A and B. See it in the official paper.

   Try it first, then show our solutionSolution

   1. (a) Circuit: four resistances P (AB), Q (BC), R (AD) and S (DC) form a quadrilateral. A cell is connected across A and C, and a galvanometer of resistance G across B and D. The current $I_1$ flows through P, $I_2$ through R, and $I_g$ through G from B to D.

      1 mark

   2. Kirchhoff's loop law for mesh ABDA: $I_{1}P+I_{g}G-I_{2}R=0$ …(1)

      ½ mark

   3. For mesh BCDB: $(I_1-I_g)Q-(I_2+I_g)S-I_{g}G=0$ …(2)

      ½ mark

   4. At balance $I_g=0$. Then (1) gives $I_{1}P=I_{2}R$ and (2) gives $I_{1}Q=I_{2}S$. Dividing: $\frac{P}{Q}=\frac{R}{S}$, the balance condition.

      1 mark

   5. (b) The wire is uniform, so resistance ∝ arc length; 12 Ω for 360°. Arc AD = arc DB = 60°, so each is $12\times \frac{60}{360}=2\Omega$. Arc AC = arc CB = 120°, so each is $4\Omega$. Since $\frac{R_{AC}}{R_{CB}}=\frac{4}{4}=\frac{R_{AD}}{R_{DB}}=\frac{2}{2}$, the bridge is balanced and no current flows through the 10 Ω resistor in CD.

      1 mark

   6. The circuit reduces to two parallel branches between A and B: ACB = 8 Ω and ADB = 4 Ω, each with 8 V across it. Current through arm AD: $I=\frac{8}{4}=2$ A.

      1 mark

   Answer: (a) $P/Q=R/S$; (b) current through AD = 2 A (no current in the 10 Ω arm)

   Where the marks are

   • Labelled Wheatstone bridge circuit (1 mark)
   • Loop equation for ABDA (½ mark)
   • Loop equation for BCDB (½ mark)
   • Puts $I_g=0$ and derives $P/Q=R/S$ (1 mark)
   • Arc resistances 2, 2, 4, 4 Ω and recognises a balanced bridge (1 mark)
   • Current in AD = 8/4 = 2 A (1 mark)

   Open Q31(II) on its own page

2. Answer one of these 2 alternatives.

   Q32(I) · 5 marks · Long answerAlternating Current

   Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c. generator. In an a.c. generator, coil of N turns and area A is rotated at an angular velocity ω in a uniform magnetic field B. Derive an expression for the instantaneous value of the emf induced in coil. What is the source of energy generation in this device?

   Try it first, then show our solutionSolution

   1. Principle: An a.c. generator works on electromagnetic induction. When a coil is rotated in a uniform magnetic field, the magnetic flux linked with it changes continuously. This induces an emf (and a current in a closed circuit) that keeps reversing direction.

      1 mark

   2. Labelled diagram (described):

      • A rectangular armature coil ABCD of many turns of insulated copper wire is wound on a soft-iron core.
      • It is placed between the concave pole pieces N and S of a strong field magnet.
      • It rotates about an axis perpendicular to $\vec{B}$.
      • The coil ends are joined to two slip rings $R_1,R_2$ that rotate with the coil.
      • Two stationary carbon brushes $B_1,B_2$ press on the slip rings and connect to the external load.
      1 mark

   3. Flux at time t: Let the normal to the coil be along $\vec{B}$ at $t=0$, and let the coil rotate with angular velocity $\omega$, so $\theta =\omega t$.

      Flux linkage: $\varphi =NBA\cos \omega t$

      1 mark

   4. Instantaneous emf: By Faraday's law,
      $e=-\frac{d\varphi }{dt}=-NBA\frac{d}{dt}(\cos \omega t)=NBA\omega \sin \omega t$

      Hence $e=e_0\sin \omega t$, where $e_0=NBA\omega$ is the peak emf.

      The current through a load $R$ is $i=\frac{e_0}{R}\sin \omega t=i_0\sin \omega t$, which varies sinusoidally.

      1 mark

   5. Source of energy: The electrical energy comes from the mechanical energy spent in rotating the armature coil against the opposing (Lenz's law) torque, for example by a turbine.

      1 mark

   Answer: $e=NBA\omega \sin \omega t$ (peak $e_0=NBA\omega$); source of energy is the mechanical work done in rotating the coil.

   Where the marks are

   • States the principle of electromagnetic induction with changing flux through a rotating coil (1 mark)
   • Labelled diagram/construction: armature, field magnet, slip rings, brushes (1 mark)
   • Writes flux $\varphi =NBA\cos \omega t$ (1 mark)
   • Derives $e=NBA\omega \sin \omega t$ with $e_0=NBA\omega$ (1 mark)
   • Identifies mechanical energy as the source (1 mark)

   Open Q32(I) on its own page

   OR

   Q32(II) · 5 marks · Long answerAlternating Current

   a) With the help of a diagram, explain the principle of a device which changes a low ac voltage into a high voltage. Deduce the expression for the ratio of secondary voltage to the primary voltage in terms of the ratio of the number of turns of primary and secondary winding. For an ideal transformer, obtain the ratio of primary and secondary currents in terms of the ratio of the voltages in the secondary and primary coils.
   b) Write any two sources of the energy losses which occur in actual transformers.
   c) A step-up transformer converts a low input voltage into a high output voltage. Does it violate law of conservation of energy? Explain.

   Try it first, then show our solutionSolution

   1. (a) Device and principle: The device is a step-up transformer. It works on mutual induction: an alternating current in one coil produces a changing flux that induces an emf in a neighbouring coil.

      Diagram (described):

      • Primary coil P (few turns) and secondary coil S (many turns) are wound on a closed, laminated soft-iron core.
      • P is connected to the a.c. input; S is connected to the load.
      1 mark

   2. Working: The a.c. in the primary sets up an alternating flux $\varphi$ in the core. Assuming no flux leakage, the same flux per turn links both coils.

      Flux linkages: ${\varphi }_P=N_P\varphi$ and ${\varphi }_S=N_S\varphi$

      ½ mark

   3. Voltage ratio:

      • $e_S=N_S\frac{d\varphi }{dt}$
      • $e_P=N_P\frac{d\varphi }{dt}$ (equal to the applied voltage when coil resistance is negligible)

      Dividing: $\frac{e_S}{e_P}=\frac{N_S}{N_P}$

      For step-up, $N_S>N_P$, so $e_S>e_P$.

      ½ mark

   4. Current ratio (ideal transformer, no losses):

      Input power = output power, so $e_P{i}_P=e_S{i}_S$.

      $\Rightarrow \frac{i_P}{i_S}=\frac{e_S}{e_P}=\frac{N_S}{N_P}$

      So in a step-up transformer the current is stepped down.

      ½ mark

   5. Use: Step-up transformers raise the voltage at generating stations for long-distance transmission. Step-down transformers lower it at the receiving substations.

      ½ mark

   6. (b) Energy losses (any two):
      1. Eddy current loss in the core, reduced by using a laminated core.
      2. Flux leakage, since not all of the primary flux links the secondary.

      Other losses: hysteresis loss in the core, and copper ($I^{2}R$) loss in the windings.

      1 mark

   7. (c) No, it does not violate the law of conservation of energy. When the voltage is stepped up, the current is stepped down in the same ratio. So the output power $e_S{i}_S$ is at most equal to the input power $e_P{i}_P$; no extra energy is created.

      1 mark

   Answer: (a) $\frac{e_S}{e_P}=\frac{N_S}{N_P}$; $\frac{i_P}{i_S}=\frac{e_S}{e_P}$. (b) Eddy current loss and flux leakage (also hysteresis, copper loss). (c) No; the current falls as the voltage rises, so power is not increased.

   Where the marks are

   • Diagram and principle of mutual induction, laminated soft-iron core (1 mark)
   • Flux linkages $N_P\varphi$ and $N_S\varphi$ (½ mark)
   • Derives $e_S/e_P=N_S/N_P$ (½ mark)
   • Uses $e_P{i}_P=e_S{i}_S$ to get $i_P/i_S=e_S/e_P$ (½ mark)
   • Step-up/step-down use statement (½ mark)
   • Any two energy losses (1 mark)
   • Explains no violation: current decreases, power same (1 mark)

   Open Q32(II) on its own page

3. Answer one of these 2 alternatives.

   Q33(I) · 5 marks · Long answerRay Optics and Optical Instruments

   a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is angular magnification of the telescope in normal adjustment?
   b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48\times 10^6$ m, and the radius of lunar orbit is $3.8\times 10^8$ m.

   Try it first, then show our solutionSolution

   1. (a) Given $f_o=15$ m and $f_e=1.0$ cm $=0.01$ m.

      In normal adjustment: $M=\frac{f_o}{f_e}=\frac{15}{0.01}=1500$

      1 mark

   2. (b) Angle subtended by the moon at the objective:

      $\alpha =\frac{\text{diameter of moon}}{\text{radius of lunar orbit}}=\frac{3.48\times 10^6}{3.8\times 10^8}$ rad

      1½ marks

   3. The moon is effectively at infinity, so the objective forms its image in the focal plane. If $d$ is the image diameter, it subtends the same angle at the objective:

      $\alpha =\frac{d}{f_o}=\frac{d}{15}$

      1½ marks

   4. Equating the two expressions:

      $d=\frac{3.48\times 10^6}{3.8\times 10^8}\times 15\approx 0.137$ m $=13.7$ cm

      1 mark

   Answer: (a) 1500; (b) diameter of moon's image ≈ 0.137 m = 13.7 cm

   Where the marks are

   • Uses $M=f_o/f_e$ to get 1500 (1 mark)
   • Angle subtended by moon = diameter/orbit radius (1.5 marks)
   • Image angle = $d/f_o$ with image in focal plane (1.5 marks)
   • Correct result $d\approx 13.7$ cm (1 mark)

   Open Q33(I) on its own page

   OR

   Q33(II) · 5 marks · Long answerRay Optics and Optical Instruments

   A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at
   a) the least distance of distinct vision (25 cm) and
   b) infinity? What is the magnifying power of the microscope in each case?

   Try it first, then show our solutionSolution

   1. (a) Final image at D = 25 cm

      For the eyepiece: $v_e=-25$ cm, $f_e=6.25$ cm. Using the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$:

      $\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25}=-\frac{1}{5}$

      ½ mark

   2. $u_e=-5$ cm. So the image formed by the objective lies 5 cm from the eyepiece.

      Its distance from the objective is $v_o=15-5=10$ cm.

      ½ mark

   3. For the objective: $v_o=+10$ cm, $f_o=2$ cm.

      $\frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{10}-\frac{1}{2}=-\frac{4}{10}$

      ½ mark

   4. $u_o=-2.5$ cm. The object must be placed 2.5 cm in front of the objective.

      ½ mark

   5. Magnifying power:

      $M=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4\times 5=20$

      ½ mark

   6. (b) Final image at infinity

      The objective's image must lie at the focus of the eyepiece.

      $v_o=15-6.25=8.75$ cm

      ½ mark

   7. $\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2}=\frac{2-8.75}{17.5}=-\frac{6.75}{17.5}$

      ½ mark

   8. $u_o=-\frac{17.5}{6.75}\approx -2.59$ cm. The object must be placed about 2.6 cm from the objective.

      ½ mark

   9. Magnifying power:

      $M=\frac{v_o}{|u_o|}\cdot \frac{D}{f_e}=\frac{8.75}{2.59}\times \frac{25}{6.25}\approx 3.375\times 4=13.5$

      1 mark

   Answer: (a) Object 2.5 cm from objective, M = 20; (b) object ≈ 2.59 cm (≈2.6 cm) from objective, M ≈ 13.5

   Where the marks are

   • Eyepiece lens formula with $v_e=-25$ cm (½ mark)
   • $u_e=-5$ cm, hence $v_o=10$ cm (½ mark)
   • Objective lens formula substitution (½ mark)
   • $u_o=-2.5$ cm (½ mark)
   • $M=20$ (½ mark)
   • For infinity, $v_o=15-6.25=8.75$ cm (½ mark)
   • Objective lens formula substitution (½ mark)
   • $u_o\approx -2.6$ cm (½ mark)
   • $M\approx 13.5$ (1 mark)

   Open Q33(II) on its own page