CBSE Class 12 Physics: 2025 board paper, set 55/7/1

1. For the particle to stay suspended, the electric force balances its weight: $qE=mg$ with $E=V/d$, so $V=\frac{mgd}{q}$.

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2. $V=\frac{(5\times 10^{-6})(10)(10^{-2})}{2\times 10^{-9}}=250\text{V}$

   ½ mark

Answer: 250 V, option (A)

1. Reduce the network step by step: combine resistors in series, then combine parallel pairs, and check whether the 7 Ω branch forms a balanced bridge (equal potential at its two ends, so it carries no current).

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2. The reduced network gives an equivalent resistance of $R_{AB}=\frac{16}{3}\Omega$.

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Answer: 16/3 Ω, option (C)

1. Flux is $\Phi =BA\cos \theta$, where $\theta$ is the angle between $\vec{B}$ and the normal to the coil. The normal makes an angle $90^{\circ }-\alpha$ with $\vec{B}$.

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2. So $\Phi =BA\cos (90^{\circ }-\alpha )=BA\sin \alpha$.

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Answer: BA sin α, option (A)

1. $\omega =100\pi$, so $T=\frac{2\pi }{\omega }=\frac{1}{50}$ s.

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2. The current falls from maximum to zero in a quarter cycle: $t=\frac{T}{4}=\frac{1}{200}$ s.

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Answer: 1/200 s, option (A)

1. Below resonance, $X_C=\frac{1}{\omega C}>X_L=\omega L$, so the circuit is capacitive.

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2. Above resonance, $X_L>X_C$, so the circuit is inductive.

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Answer: Capacitive and inductive respectively, option (C)

1. A vertical rod moving east cuts the horizontal (south–north) field, so $e=B_{H}vl$.

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2. $e=(0.4\times 10^{-4})(10)(0.5)=2\times 10^{-4}\text{V}=0.2\text{mV}$

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Answer: 0.2 mV, option (B)

1. $L=\frac{e}{dI/dt}$, and the dimensions of emf are $[M{L}^2{T}^{-3}{A}^{-1}]$.

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2. $[L]=[M{L}^2{T}^{-3}{A}^{-1}]\times [T{A}^{-1}]=[M{L}^2{T}^{-2}{A}^{-2}]$

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Answer: [M L² T⁻² A⁻²], option (D)

1. $\nu =\frac{E}{h}=\frac{1.326\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}$

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2. $\nu =3.2\times 10^{14}$ Hz

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Answer: 3.20 × 10¹⁴ Hz, option (B)

1. Boron is trivalent, so it is an acceptor and the doped crystal is p-type, with holes as majority carriers.

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2. At room temperature, thermal generation also gives a few electrons, which are minority carriers.

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Answer: Holes and few electrons, option (C)

1. $\lambda =\frac{hc}{\Delta E}$, so the maximum wavelength comes from the minimum energy gap.

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2. $\Delta E$ for n = 4 → 3 is $13.6\left(\frac{1}{9}-\frac{1}{16}\right)=0.66$ eV, the smallest of the four options.

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Answer: n = 4 to n = 3, option (A)

1. In forward bias, the applied voltage opposes the barrier potential, so the barrier height falls to $V_0-V$.

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2. Majority carriers are pushed towards the junction, so the depletion layer narrows.

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Answer: Option (A)

1. Isotones have the same number of neutrons (N = A − Z) but different Z, for example ${}_1^3$H and ${}_2^4$He, each with N = 2.

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Answer: Option (C)

1. The velocity component along y is parallel to $\vec{B}$ and stays unchanged, while the x component makes the particle circle. So the helix axis is along y, not x, and A is false.

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2. The magnetic force $q\vec{v}\times \vec{B}$ is perpendicular to $\vec{v}$, not along it, so R is false.

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Answer: Option (D)

1. At normal incidence $r_1=0$, so $r_2=A$. If $A=C$, the emergent ray grazes the second face, so A is true.

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2. The refractive index depends on the material and the wavelength of light, not on the prism angle, so R is false.

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Answer: Option (C)

1. Both statements are true. EM waves travel through vacuum because their oscillating E and B fields sustain each other; being transverse does not explain why no medium is needed.

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Answer: Option (B)

1. The definition of cut-off (stopping) potential is correct, and so is the definition of threshold frequency. However, threshold frequency does not explain the cut-off voltage.

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Answer: Option (B)

1. Current in the circuit: $I=\frac{E}{R+r}$. Terminal voltage: $V=E-Ir=\frac{ER}{R+r}=\frac{E}{1+r/R}$.

2. (a) V versus R graph: at $R=0$, $V=0$. As $R$ increases, $V$ rises quickly at first and then more slowly. The curve is concave downward and approaches $V=E$ asymptotically as $R\to \infty$. It never crosses the line $V=E$, and $V=E/2$ when $R=r$.

   1 mark

3. (b) I versus R graph: at $R=0$, $I=E/r$ (the maximum, short-circuit current). As $R$ increases, $I$ falls along a hyperbola-like curve, concave upward, and tends to zero as $R\to \infty$.

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Answer: (a) V rises from 0 and tends asymptotically to E; (b) I falls from E/r and tends to 0.

1. For a convex mirror, take $f>0$ (focus behind the mirror). For a real object, $u=-x$ with $x>0$. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{x}$.

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2. Both terms on the right are positive, so $v>0$: the image is always behind the mirror, i.e. virtual. Also $\frac{1}{v}>\frac{1}{f}\Rightarrow v<f$. Hence $0<v<f$: the image lies between the pole and the focus.

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3. Solving, $v=\frac{fx}{x+f}$. Magnification: $m=-\frac{v}{u}=\frac{v}{x}=\frac{f}{x+f}$.

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4. Since $f<x+f$, we get $0<m<1$. The image is therefore always erect and diminished.

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Answer: 0 < v < f (between pole and focus, virtual) and 0 < m = f/(x+f) < 1 (diminished, erect).

1. For thin lenses in contact: $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$, with $f_1=+10$ cm, $f_2=-15$ cm and $F=+12$ cm.

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2. Substituting: $\frac{1}{f_3}=\frac{1}{12}-\frac{1}{10}+\frac{1}{15}=\frac{5-6+4}{60}=\frac{3}{60}$.

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3. $f_3=+20$ cm. The third lens is convex (converging).

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4. Yes, the answer would change. The formula $1/F=\sum 1/f_i$ holds only for thin lenses in contact; for thick lenses, the separations between their optical centres cannot be neglected.

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Answer: Third lens is convex with f = +20 cm; yes, the result would change for thick lenses.

1. Difference 1: The double-slit interference pattern has many equally spaced bright and dark fringes of equal width. In the single-slit diffraction pattern, the central bright maximum is twice as wide as the other maxima.

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2. Difference 2: In interference, all bright fringes have (nearly) the same intensity. In diffraction, the intensity of successive maxima falls rapidly on either side of the centre. (Another valid difference: interference results from superposing two waves from two slits, while diffraction results from superposing a continuous family of wavelets from different points of a single slit.)

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3. Two separate sodium lamps are independent, incoherent sources. Each emits light in random bursts, so the phase difference between the waves from the two pinholes changes rapidly and randomly with time.

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4. The positions of maxima and minima therefore shift too quickly for the eye to follow. Only uniform average illumination is seen, and no sustained interference pattern forms.

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Answer: Interference: equally spaced, equal-width, equal-intensity fringes. Diffraction: central maximum twice as wide, with secondary maxima of decreasing intensity. The sodium lamps are incoherent (no constant phase difference), so no steady pattern forms.

1. n-type diagram: draw the conduction band (top) and valence band (bottom), separated by the band gap $E_g\approx 1$ eV. A few electrons are shown in the conduction band and a few holes in the valence band at T > 0 K.

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2. Mark the **donor energy level $E_D$** as a dashed line just below the bottom of the conduction band, at $E_C-E_D\approx 0.01$ eV (Ge) to $0.05$ eV (Si). Electrons are shown leaving this level for the conduction band.

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3. p-type diagram: draw the same two bands with gap $E_g\approx 1$ eV. Show many holes in the valence band and a few electrons in the conduction band.

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4. Mark the **acceptor energy level $E_A$** as a dashed line just above the top of the valence band, at $E_A-E_V\approx 0.01$–$0.05$ eV. Electrons are shown jumping from the valence band to this level, leaving holes behind.

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Answer: n-type: donor level ~0.01–0.05 eV below the conduction band. p-type: acceptor level ~0.01–0.05 eV above the valence band. Band gap ~1 eV.

1. Energy in stars comes from thermonuclear fusion. The core temperature is about $10^7$ K, so light nuclei (mainly hydrogen) have enough kinetic energy to overcome Coulomb repulsion and fuse.

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2. The heavier nucleus formed has a higher binding energy per nucleon, so its mass is less than the total mass of the reacting nuclei. This mass defect $\Delta m$ is released as energy $E=\Delta m{c}^2$.

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3. Example 1 (proton–proton cycle): ${}_1^1\text{H}+{}_1^1\text{H}\to {}_1^2\text{H}+e^{+}+\nu +0.42$ MeV.

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4. Example 2: ${}_1^2\text{H}+{}_1^1\text{H}\to {}_2^3\text{He}+\gamma +5.49$ MeV. Further, ${}_2^3\text{He}+{}_2^3\text{He}\to {}_2^4\text{He}+{}_1^1\text{H}+{}_1^1\text{H}+12.86$ MeV. Overall, four protons combine to form one ${}_2^4$He nucleus and about 26.7 MeV is released.

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Answer: By nuclear fusion of hydrogen into helium (proton–proton cycle), e.g. ¹H + ¹H → ²H + e⁺ + ν, and ³He + ³He → ⁴He + 2¹H.

1. All positive terminals are joined at one junction and all negative terminals at the other. Let $I_1$ flow through cell C (5 V, 1 Ω) in the direction of its emf. Let $I_2$ flow through cell B (3 V, 5 Ω) against its emf. By the junction rule, $I_1-I_2$ flows through cell A (2 V, 5 Ω) against its emf.

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2. Loop C–B (Kirchhoff's loop rule): $5-3=1\cdot I_1+5I_2\Rightarrow I_1+5I_2=2$ ...(1)

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3. Loop C–A: $5-2=1\cdot I_1+5(I_1-I_2)\Rightarrow 6I_1-5I_2=3$ ...(2)

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4. Adding (1) and (2): $7I_1=5\Rightarrow I_1=\frac{5}{7}$ A through cell C.

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5. From (1): $5I_2=2-\frac{5}{7}=\frac{9}{7}\Rightarrow I_2=\frac{9}{35}$ A through cell B.

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6. Current through A: $I_1-I_2=\frac{25}{35}-\frac{9}{35}=\frac{16}{35}$ A. Cells A and B are being charged by C. (Check: common terminal voltage $=5-\frac{5}{7}\times 1=\frac{30}{7}$ V.)

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Answer: I_C = 5/7 A ≈ 0.71 A; I_B = 9/35 A ≈ 0.26 A; I_A = 16/35 A ≈ 0.46 A (A and B are being charged).

1. (i) Biot–Savart law: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I(d\vec{l}\times \vec{r})}{r^3}$. Here $d\vec{l}$ is the current element and $\vec{r}$ is the position vector of the field point from that element.

   1 mark

2. (ii) Coil A lies in the XY plane, so its field at the centre is along the z-axis: $B_1=\frac{{\mu }_{0}I}{2R}$. Coil B lies in the YZ plane, so its field is along the x-axis: $B_2=\frac{{\mu }_0\sqrt{3}I}{2R}$.

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3. The two fields are perpendicular, so $B=\sqrt{B_1^2+B_2^2}=\frac{{\mu }_{0}I}{2R}\sqrt{1+3}$.

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4. Net field: $B=\frac{{\mu }_{0}I}{R}$.

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5. Direction: $\tan \theta =\frac{B_1}{B_2}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ }$. The net field lies in the XZ plane at $30^{\circ }$ to ${\vec{B}}_2$ (the x-axis), i.e. $60^{\circ }$ to ${\vec{B}}_1$ (the z-axis).

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Answer: B = μ₀I/R, in the XZ plane at 30° to the x-axis (field of coil B), i.e. 60° to the z-axis.

1. (i) $\vec{m}=I\vec{A}$, with magnitude $m=Ilb$. By the right-hand thumb rule, a clockwise current (as seen by the viewer) gives $\vec{m}$ perpendicular to the plane of the loop, pointing into the page (away from the viewer). In the diagram, draw the rectangle with clockwise arrows and a ⊗ at its centre marking $\vec{m}$.

   1 mark

2. (ii) Let the loop ABCD have arms AB and CD of length $b$ and arms BC and DA of length $l$. The rotation axis is perpendicular to $\vec{B}$, and the normal to the loop makes angle $\theta$ with $\vec{B}$. The forces on BC and DA are equal, opposite and act along the same line (the axis), so they cancel.

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3. Arms AB and CD are perpendicular to $\vec{B}$, so each experiences force $F=IbB$. Because their currents are opposite, one force is into and the other out of the plane. The vector sum of all forces is zero, so the net force is zero.

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4. The forces on AB and CD have different lines of action and form a couple. The perpendicular distance between them is $l\sin \theta$, so $\tau =F\cdot l\sin \theta =IbBl\sin \theta$.

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5. $\tau =I(lb)B\sin \theta =mB\sin \theta$. The torque turns $\vec{m}$ towards $\vec{B}$, so in vector form $\vec{\tau }=\vec{m}\times \vec{B}$.

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Answer: m = I l b, perpendicular to the loop (into the plane for a clockwise current); net force = 0; τ = IlbB sin θ, i.e. τ⃗ = m⃗ × B⃗.

1. (a) Faraday's law: the magnitude of the emf induced in a circuit equals the time rate of change of magnetic flux linked with it: $\epsilon =-\frac{d{\Phi }_B}{dt}$ (for N turns, $\epsilon =-N\frac{d{\Phi }_B}{dt}$).

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2. Role of the negative sign (Lenz's law): the induced emf, and hence the induced current, is in a direction that opposes the change in flux that produces it.

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3. (b) Example: push the north pole of a bar magnet towards a closed coil. The induced current makes the near face of the coil a north pole, which repels the magnet. When the magnet is pulled away, that face becomes a south pole and attracts it. In both cases, work must be done against this opposing force. That mechanical work appears as electrical energy (heat in the coil). If the induced current aided the motion, energy would be created from nothing, violating conservation of energy. Hence Lenz's law is consistent with energy conservation.

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Answer: ε = −dΦ/dt; the minus sign shows the induced emf opposes the change in flux. Work done against the opposition (e.g. magnet approaching a coil) is converted into electrical energy, as energy conservation requires.

1. (a) Similarity: both conduction current and displacement current produce a magnetic field (both appear as sources in the Ampere–Maxwell law).

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2. Dissimilarity: conduction current is due to the actual flow of charges in a conductor. Displacement current, $I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}$, arises from a time-varying electric flux, even in vacuum (e.g. between capacitor plates), with no flow of charge.

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3. (b) By Faraday's law, a time-varying magnetic field produces an electric field. By the Ampere–Maxwell law, $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_c+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}$, a time-varying electric field (displacement current) produces a magnetic field. In free space $I_c=0$, so the changing E and B fields keep regenerating each other. The result is a self-sustaining disturbance of mutually perpendicular E and B fields that propagates as an electromagnetic wave with speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$.

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Answer: Both produce magnetic fields. Conduction current is a flow of charges; displacement current comes from changing electric flux. Changing E and B fields generate each other via displacement current, so EM waves propagate in free space.

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

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2. From Einstein's equation, $e{V}_0=h\nu -{\varphi }_0\Rightarrow V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$. The $V_0$–$\nu$ graph is a straight line. Extended back, it cuts the $V_0$ axis at $-{\varphi }_0/e$, so ${\varphi }_0=e\times |\text{intercept on }{V}_0\text{ axis}|$. Equivalently, the intercept on the $\nu$ axis is the threshold frequency ${\nu }_0$, and ${\varphi }_0=h{\nu }_0$.

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3. (b) Energy of the incident photon: $\frac{hc}{\lambda }={\varphi }_0+e{V}_0=2.4+0.6=3.0$ eV.

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4. $\lambda =\frac{hc}{E}=\frac{6.63\times 10^{-34}\times 3\times 10^8}{3.0\times 1.6\times 10^{-19}}$

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5. $\lambda \approx 4.14\times 10^{-7}$ m $\approx 414$ nm.

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Answer: (a) φ₀ = e × |V₀-intercept| (or hν₀ from the ν-intercept); (b) λ ≈ 414 nm (≈ 4.14 × 10⁻⁷ m).

1. Postulate 1: the centripetal force is provided by the Coulomb attraction, and the electron moves in certain stable (stationary) orbits without radiating: $\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$ ...(1)

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2. Postulate 2 (quantisation of angular momentum): $L=mvr=\frac{nh}{2\pi }$, n = 1, 2, 3, ... ...(2)

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3. Postulate 3 (frequency condition): $h\nu =E_i-E_f$ when the electron jumps from a higher orbit to a lower one.

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4. (a) From (2), $v=\frac{nh}{2\pi mr}$. Substituting into (1), $m{v}^{2}r=\frac{e^2}{4\pi {\epsilon }_0}$ gives $m\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}r=\frac{e^2}{4\pi {\epsilon }_0}$.

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5. $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}$. Hence $r_n\propto n^2$.

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6. (b) KE $=\frac{1}{2}m{v}^2=\frac{e^2}{8\pi {\epsilon }_{0}r}$ and PE $=-\frac{e^2}{4\pi {\epsilon }_{0}r}$, so $E_n=-\frac{e^2}{8\pi {\epsilon }_0{r}_n}$. Substituting $r_n$: $E_n=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$. Hence $E_n\propto \frac{1}{n^2}$.

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Answer: r_n = n²h²ε₀/(πme²) ∝ n²; E_n = −me⁴/(8ε₀²h²n²) ∝ 1/n².

1. Depletion layer: when the p-n junction forms, electrons diffuse from the n-side (high concentration) to the p-side and holes diffuse from the p-side to the n-side. Each electron leaving the n-side leaves behind an immobile positive donor ion. Each hole leaving the p-side leaves behind an immobile negative acceptor ion. This thin region of immobile ions near the junction, with no free charge carriers, is the depletion layer. Diagram: a p-region and an n-region with a narrow central strip showing ⊖ ions on the p-side and ⊕ ions on the n-side, and an electric field pointing from n to p.

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2. Potential barrier: the space charge sets up an electric field from n to p, which drives a drift current opposite to the diffusion current. As diffusion continues, the layer widens and the field grows until drift current equals diffusion current, giving equilibrium. The potential difference across the layer then opposes further movement of electrons from n to p (and holes from p to n). This is called the potential barrier ($V_B\approx 0.7$ V for Si, 0.3 V for Ge).

   1 mark

3. Rectifier feature: a junction diode conducts easily (low resistance) only when forward biased and offers very high resistance when reverse biased. Because it allows current in one direction only (unidirectional conduction), it converts AC into DC.

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Answer: Diffusion of carriers leaves immobile ions forming the depletion layer; their field creates the potential barrier opposing further diffusion. The diode's unidirectional conduction (low forward, high reverse resistance) makes it a rectifier.

1. (i) Mobility $\mu =\frac{e\tau }{m}$ depends only on the material (and temperature), not on the dimensions of the wire. Since A and B are of the same metal, ${\mu }_A/{\mu }_B=1$. Option (D)

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2. (ii) $\sigma =\frac{1}{\rho }=\frac{l}{RA}=\frac{Il}{VA}=\frac{1.5\times 0.5}{2\times 1.0\times 10^{-7}}=3.75\times 10^6{\Omega }^{-1}{\text{m}}^{-1}$. Option (C)

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3. (iii) $\frac{\Delta R}{R_0}=\alpha \Delta T\Rightarrow 0.085=1.70\times 10^{-4}\Delta T\Rightarrow \Delta T=500^{\circ }$C. Option (B)

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4. (iv)(a) $\rho =\frac{m}{n{e}^2\tau }$. In a metal, n hardly changes with temperature, but more frequent collisions reduce $\tau$. So I decreases and II is almost constant. Option (D)

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5. (iv)(b) I is the same everywhere, so $j=I/A$ changes with A. Then $v_d=j/(ne)$ changes, and $E=\rho j$ also changes. All three change. Option (D)

Answer: (i) D; (ii) C, 3.75 × 10⁶ Ω⁻¹ m⁻¹; (iii) B, 500 °C; (iv)(a) D; (iv)(b) D

1. (i) Light emerges only within the critical angle: $\sin i_c=\frac{1}{n}$. Then $\frac{R}{H}=\tan i_c=\frac{1/n}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{n^2-1}}$. Option (A)

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2. (ii)(a) $n=\frac{\sin \frac{A+D_m}{2}}{\sin \frac{A}{2}}=\frac{\sin 45^{\circ }}{\sin 30^{\circ }}=\sqrt{2}\approx 1.41$. Option (B)

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3. (ii)(b) For a thin prism, $D_m=(n-1)A$, so $\frac{D^{\prime }}{D}=\frac{0.62}{0.52}\approx 1.19$: an increase of about 19%. Option (D)

4. (iii) Refraction formula: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$. With $u=-x$: $\frac{1.5}{v}=\frac{0.5}{R}-\frac{1}{x}$. So $v>0$ (real) when $x>2R$ and $v<0$ (virtual) when $x<2R$. Starting at 4R and moving inwards, the image is first real, then virtual. Option (C)

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5. (iv) Lens maker's formula: $\frac{1}{f}=(n-1)\left(\frac{1}{R}+\frac{1}{R}\right)=0.5\times \frac{2}{R}=\frac{1}{R}$, so $R=f=10$ cm. Option (A)

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Answer: (i) A; (ii)(a) B, ≈1.4; (ii)(b) D, increase by 19%; (iii) C; (iv) A, 10 cm

1. (i) Without the slab, $C_0=\frac{{\epsilon }_{0}A}{d}$. With the slab, $E_0=\frac{\sigma }{{\epsilon }_0}$ in the air gap (thickness $d-t$) and $E=\frac{E_0}{K}$ inside the dielectric (thickness $t$).

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2. $V=E_0(d-t)+\frac{E_0}{K}t=\frac{Q}{{\epsilon }_{0}A}\left(d-t+\frac{t}{K}\right)$, so $C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{K}}$.

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3. With $t=\frac{d}{4}$: $C=\frac{{\epsilon }_{0}A}{\frac{3d}{4}+\frac{d}{4K}}=\frac{{\epsilon }_{0}A}{d}\cdot \frac{4K}{3K+1}$.

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4. $C=\frac{4K}{3K+1}{C}_0$

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5. (ii) Each capacitor can take at most 200 V, so each series row must have $n_s=\frac{1200}{200}=6$ capacitors. Capacitance of one row $=\frac{1}{6}\mu$F.

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6. Let m such rows be connected in parallel: $C_{eq}=\frac{m}{6}=2\Rightarrow m=12$.

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7. Minimum number $=6\times 12=72$ capacitors: 12 parallel rows, each with 6 capacitors in series.

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Answer: (i) C = 4K C₀/(3K+1); (ii) 72 capacitors: 12 rows in parallel, each with 6 capacitors in series.

1. (i)(I) Axial point P at distance x from the centre, on the side of +q. Distance from +q is $(x-a)$ and from −q is $(x+a)$. So $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{x-a}-\frac{q}{x+a}\right]$.

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2. $V=\frac{1}{4\pi {\epsilon }_0}\frac{q(2a)}{x^2-a^2}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{x^2-a^2}$, where $p=q\cdot 2a$.

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3. For $x\gg a$: $V=\frac{1}{4\pi {\epsilon }_0}\frac{p}{x^2}$.

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4. (II) Equatorial (bisector) point at distance x from the centre. Both charges are at the same distance $\sqrt{x^2+a^2}$.

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5. $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{\sqrt{x^2+a^2}}-\frac{q}{\sqrt{x^2+a^2}}\right]$

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6. $V=0$ at every point on the bisector.

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7. (ii) $\vec{\tau }=\vec{p}\times \vec{E}=(0.8\hat{i}+0.6\hat{j})\times 10^{-29}\times (1.0\times 10^7\hat{k})$. Using $\hat{i}\times \hat{k}=-\hat{j}$ and $\hat{j}\times \hat{k}=\hat{i}$: $\vec{\tau }=(0.6\hat{i}-0.8\hat{j})\times 10^{-22}$ N m.

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8. $|\vec{\tau }|=\sqrt{0.6^2+0.8^2}\times 10^{-22}$

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9. $|\vec{\tau }|=1.0\times 10^{-22}$ N m.

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10. Angle of $\vec{\tau }$ with the x-axis: $\tan \theta =\frac{0.8}{0.6}=\frac{4}{3}\Rightarrow \theta \approx 53^{\circ }$, measured below the x-axis towards −y. (The dipole itself makes ${\tan }^{-1}(0.6/0.8)\approx 37^{\circ }$ with the x-axis.)

    ½ mark

Answer: (I) V = p/(4πε₀x²) on the axis; (II) V = 0 on the bisector; (ii) |τ| = 1.0 × 10⁻²² N m, at about 53° to the x-axis (towards −y).

1. Labelled diagram: a rectangular coil of many turns wound on a soft iron cylindrical core, placed between concave pole pieces N and S of a permanent magnet. The coil is suspended or pivoted, with a spring for restoring torque and a pointer moving over a scale.

   1 mark

2. Principle and working: a current-carrying coil in a magnetic field experiences a torque $\tau =NIAB$. This torque rotates the coil until it is balanced by the restoring torque of the spring, $k\varphi$: $NIAB=k\varphi \Rightarrow \varphi =\left(\frac{NAB}{k}\right)I$. So the deflection is proportional to the current.

   1 mark

3. (i) Radial magnetic field: the plane of the coil is always parallel to the field, so the torque $NIAB$ is maximum and independent of the coil's position. This makes $\varphi \propto I$ and the scale linear.

   ½ mark

4. (ii) Soft iron core: it has high permeability, so it concentrates and strengthens the magnetic field (and helps make it radial). This increases the sensitivity of the galvanometer.

   ½ mark

5. (ii) Current sensitivity is the deflection per unit current: $I_s=\frac{\varphi }{I}=\frac{NAB}{k}$.

   1 mark

6. Voltage sensitivity is $V_s=\frac{\varphi }{V}=\frac{\varphi }{IR}=\frac{NAB}{kR}$.

   ½ mark

7. If N is doubled to increase $I_s$, the length of wire, and hence the coil resistance R, also roughly doubles. $V_s=\frac{NAB}{kR}$ therefore stays essentially unchanged. So increasing current sensitivity does not necessarily increase voltage sensitivity.

   ½ mark

Answer: φ = (NAB/k) I. A radial field gives a linear scale; the soft iron core strengthens B and raises sensitivity. Current sensitivity = φ/I = NAB/k. Increasing N raises I_s, but R rises in the same proportion, so V_s = NAB/(kR) is unchanged.

1. (i)(I) Ampere's circuital law: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_e$. Here $\vec{B}$ is the magnetic field at the small element $d\vec{l}$ of a closed (Amperian) loop, $d\vec{l}$ is a small length element along the loop, $I_e$ is the net current enclosed by (passing through the surface bounded by) the loop, and ${\mu }_0$ is the permeability of free space.

   1 mark

2. (II) As the solenoid is made longer, its closely wound turns behave like a long continuous cylindrical current sheet. The field lines spread out over a very large region outside, so the field density outside becomes negligible. Hence the field outside approaches zero, while the field inside stays uniform and strong.

   1 mark

3. (III) In a uniform magnetic field, a current loop experiences forces that make it spread out so that the magnetic flux through it (and its magnetic moment $m=IA$) becomes maximum. For a given perimeter, a circle encloses the greatest area of any shape. So the flexible loop takes a circular shape, which gives maximum area and maximum flux.

   1 mark

4. (ii) Let $I_g$ be the full-scale current of the galvanometer. For a voltmeter of range $V$: $V=I_g(G+R)$.

   With $R_1$: $\frac{V}{I_g}=R_1+G$ ...(1)

   With $R_2$ (range $V/2$): $\frac{V}{2I_g}=R_2+G$ ...(2)

   Putting (1) in (2): $R_1+G=2(R_2+G)$, so $G=R_1-2R_2$.

   1 mark

5. For range $2V$: $R_3=\frac{2V}{I_g}-G=2(R_1+G)-G=2R_1+G$.

   Substituting $G$: $R_3=2R_1+R_1-2R_2=3R_1-2R_2$.

   1 mark

Answer: (i)(I) $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_e$; (II) a long solenoid acts like a cylindrical current sheet, so the outside field tends to zero; (III) a circle encloses maximum area, so the flux is maximum; (ii) $R_3=3R_1-2R_2$

1. (i) Ray diagram (normal adjustment):

   • Draw the objective lens $O$ (large aperture, long focal length $f_o$) and the eyepiece $E$ (short focal length $f_e$) on a common principal axis.
   • Parallel rays from a distant object enter the objective inclined at angle $\alpha$ to the axis.
   • After refraction they meet at the focal plane of the objective and form a real, inverted, diminished image $A^{\prime }{B}^{\prime }$ at distance $f_o$ behind $O$.
   • $A^{\prime }{B}^{\prime }$ lies at the first focus of the eyepiece, so the rays leave $E$ parallel, at angle $\beta$ to the axis.
   • Arrows are shown on all rays.
   1½ marks

2. Working: Light from the distant object forms a real, inverted image $A^{\prime }{B}^{\prime }$ at the focus of the objective. This image acts as the object for the eyepiece. Since it lies at the focus of the eyepiece, the eyepiece forms the final, highly magnified, inverted image at infinity.

   ½ mark

3. Magnifying power: $m=\frac{\beta }{\alpha }=-\frac{f_o}{f_e}$, with magnitude $\frac{f_o}{f_e}$. Tube length $L=f_o+f_e$.

   1 mark

4. (ii) Total magnification $m=m_o{m}_e$, so $20=m_o\times 5$, giving $|m_o|=4$.

   Final image at $D$: $m_e=1+\frac{D}{f_e}$, so $5=1+\frac{25}{f_e}$ and $f_e=\frac{25}{4}=6.25$ cm.

   Object distance for the eyepiece (with $v_e=-25$ cm): $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$, so $-\frac{1}{25}-\frac{1}{u_e}=\frac{4}{25}$, which gives $u_e=-5$ cm.

   1 mark

5. Separation $L=v_o+|u_e|$, so $14=v_o+5$ and $v_o=9$ cm.

   $|m_o|=\frac{v_o}{|u_o|}=4$, so $|u_o|=\frac{9}{4}$ cm, i.e. $u_o=-\frac{9}{4}$ cm.

   Lens formula: $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{9}+\frac{4}{9}=\frac{5}{9}$.

   So $f_o=\frac{9}{5}=1.8$ cm.

   1 mark

Answer: (i) $m=f_o/f_e$ (in magnitude) for final image at infinity; (ii) $f_o=1.8$ cm, $f_e=6.25$ cm

1. (i) Let the two coherent waves at a point be $y_1=a\cos \omega t$ and $y_2=a\cos (\omega t+\varphi )$.

   By the principle of superposition:
   $y=y_1+y_2=a[\cos \omega t+\cos (\omega t+\varphi )]=2a\cos \frac{\varphi }{2}\cos \left(\omega t+\frac{\varphi }{2}\right)$.

   1 mark

2. So the resultant amplitude is $A=2a\cos \frac{\varphi }{2}$.

   Since $I\propto A^2$: $I=k4a^2{\cos }^2\frac{\varphi }{2}$. With $I_0=k{a}^2$:

   $$I=4I_0{\cos }^2\frac{\varphi }{2}$$

   1 mark

3. Maximum: $I_{max}=4I_0$, when ${\cos }^2\frac{\varphi }{2}=1$, i.e. $\varphi =0,2\pi ,4\pi ,\dots$

   Minimum: $I_{min}=0$, when ${\cos }^2\frac{\varphi }{2}=0$, i.e. $\varphi =\pi ,3\pi ,\dots$

   1 mark

4. (ii)(I) Data: $a=3\times 10^{-3}$ m, $D=1.5$ m, $\lambda =600\times 10^{-9}$ m.

   The $n$th minimum is at $y_n=\frac{n\lambda D}{a}$. For $n=1$:

   $y_1=\frac{600\times 10^{-9}\times 1.5}{3\times 10^{-3}}=3\times 10^{-4}$ m $=0.3$ mm.

   1 mark

5. (II) Secondary maxima are at $y=\frac{(2n+1)\lambda D}{2a}$. For the second order maximum, $n=2$:

   $y_2=\frac{5\lambda D}{2a}=\frac{5\times 600\times 10^{-9}\times 1.5}{2\times 3\times 10^{-3}}=7.5\times 10^{-4}$ m $=0.75$ mm.

   1 mark

Answer: (i) $I=4I_0{\cos }^2(\varphi /2)$; $I_{max}=4I_0$, $I_{min}=0$; (ii)(I) $3\times 10^{-4}$ m; (II) $7.5\times 10^{-4}$ m