CBSE Class 12 Physics: 2024 board paper, set 55/1/1

1. Every small element of the uniformly charged ring has an identical element diametrically opposite it, at the same distance R from the centre. Their fields at the centre are equal in magnitude and opposite in direction, so each pair cancels. Summing over the whole ring gives $E=0$. Option (B).

   1 mark

Answer: Zero, option (B)

1. For capacitors in parallel, $C=10\times 1\mu F=10^{-5}$ F. The stored energy is $U=\frac{1}{2}C{V}^2=\frac{1}{2}\times 10^{-5}\times (100{)}^2=5.0\times 10^{-2}$ J. Option (D).

   1 mark

Answer: $5.0\times 10^{-2}$ J, option (D)

1. The two cells are in parallel, so the potential difference between A and B is $V=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$.

2. Substituting: $V=\frac{12/1+6/0.5}{1+2}=\frac{24}{3}=8$ V. Option (B).

   1 mark

Answer: 8 V, option (B)

1. A clockwise current in the x–y plane, seen from +z, has magnetic moment along −z. Taking the field along +z, the force $Id\vec{l}\times \vec{B}$ on each element points towards the centre of the loop.

2. The net force is zero, so the loop does not move. The inward forces tend to make the loop shrink. Option (C).

   1 mark

Answer: Shrink, option (C)

1. Use $\vec{F}=I\vec{l}\times \vec{B}$ with $\vec{l}=0.1\hat{j}$ m and $I=1$ A.

2. Then $\vec{F}=(0.1\hat{j})\times (5\hat{j}-8\hat{k})\times 10^{-3}=-0.8\times 10^{-3}(\hat{j}\times \hat{k})=-(0.8\text{ mN})\hat{i}$, since $\hat{j}\times \hat{j}=0$ and $\hat{j}\times \hat{k}=\hat{i}$. Option (B).

   1 mark

Answer: $-(0.8\text{ mN})\hat{i}$, option (B)

1. With $I_g=0.001I$, the shunt is $S=\frac{I_{g}G}{I-I_g}=\frac{G}{999}$.

2. The ammeter resistance is G in parallel with S: $\frac{GS}{G+S}=\frac{G^2/999}{1000G/999}=\frac{G}{1000}\Omega$. Option (B).

   1 mark

Answer: $G/1000\Omega$, option (B)

1. Capacitive reactance is $X=\frac{1}{\omega C}$. With $C\to 2C$ and $\omega \to 3\omega$: $X^{\prime }=\frac{1}{6\omega C}=\frac{X}{6}$. Option (A).

   1 mark

Answer: X/6, option (A)

1. Displacement current is $i_d={\epsilon }_0\frac{d{\Phi }_E}{dt}$, so it exists only where E changes with time.

2. Only in region I, $E_0\sin (kz-\omega t)$, does the field depend on t. In the other three regions the field is independent of time. Option (A).

   1 mark

Answer: Region I, option (A)

1. All Balmer lines end at $n_f=2$. The first line (Hα) comes from $n_i=3$, so the second line (Hβ) comes from $n_i=4$. Option (C).

   1 mark

Answer: $n_f=2$, $n_i=4$, option (C)

1. As is pentavalent. Four of its electrons form bonds with neighbouring Ge atoms and the fifth becomes a free electron. This makes an n-type semiconductor, so the number of conduction electrons increases. Option (B).

   1 mark

Answer: Option (B)

1. Maximum kinetic energy is $K=h\nu -\varphi$. For beam A: $K_A=3.3-2.3=1$ eV. For beam B: $K_B=11.3-2.3=9$ eV.

2. Since $v\propto \sqrt{K}$, the ratio is $\frac{v_A}{v_B}=\sqrt{1/9}=\frac{1}{3}$. Option (C).

   1 mark

Answer: 1/3, option (C)

1. The de Broglie relation is $\lambda =h/p$. If the wavelengths are equal, then $p=h/\lambda$ is the same for both particles, i.e. they have the same momentum. Their masses differ, so their speeds and energies differ. Option (A).

   1 mark

Answer: Momentum, option (A)

1. The maximum kinetic energy of photoelectrons depends on the frequency of the light, not its intensity, so A is false.

2. The photoelectric current (above threshold) depends on intensity, not on wavelength, so R is also false. Option (D).

   1 mark

Answer: Option (D)

1. Mutual inductance is $M=N_2{\Phi }_2/I_1$. When the coils are wound on each other, the flux linking them is maximum, so M is maximum. R is true and it explains A. Option (A).

   1 mark

Answer: Option (A)

1. When two parallel wires are connected in series to a battery, the current flows through them in opposite directions. Antiparallel currents repel, so the wires move apart. A is true and R explains it. Option (A).

   1 mark

Answer: Option (A)

1. A plane or convex mirror can form a real image when the object is virtual, i.e. rays converging towards a point behind the mirror. So A is false.

2. A virtual image can act as an object for another lens or mirror and produce a real image, so R is also false. Option (D).

   1 mark

Answer: Option (D)

1. The resistance varies with temperature as $R_T=R_0[1+\alpha (T-T_0)]$, where $T_0=20^{\circ }$C is the reference temperature. We need the temperature at which $R_T=2R_0$.

   ½ mark

2. Substituting: $2R_0=R_0[1+4.0\times 10^{-3}(T-20)]$, so $4.0\times 10^{-3}(T-20)=1$.

   ½ mark

3. Hence $T-20=\frac{1}{4.0\times 10^{-3}}=250$, which gives $T=270^{\circ }\text{C}$ (about 543 K).

   1 mark

Answer: $T=270^{\circ }$C (≈ 543 K)

1. (i) The reflected light stays in air, so its speed is $c$ and the frequency is unchanged. Using $c=\nu \lambda$: $\lambda =\frac{3\times 10^8}{5.0\times 10^{14}}=6\times 10^{-7}$ m = 600 nm.

   1 mark

2. (ii) On refraction the frequency stays the same but the speed falls to $c/n$, so ${\lambda }_{medium}=\frac{{\lambda }_{air}}{n}=\frac{600}{1.5}=400$ nm $=4\times 10^{-7}$ m.

   1 mark

Answer: (i) 600 nm; (ii) 400 nm

1. Lens maker's formula: $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. For a plano-convex lens, $R_1=R$ and $R_2=\infty$, so $\frac{1}{f}=(n-1)\frac{1}{R}$.

   1 mark

2. Substituting: $\frac{1}{16}=(1.4-1)\frac{1}{R}$, so $R=16\times 0.4=6.4$ cm.

   1 mark

Answer: R = 6.4 cm

1. (i) By the sign convention, $u=-30$ cm and $f=-R/2=-20$ cm. Mirror formula: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $\frac{1}{v}+\frac{1}{-30}=\frac{1}{-20}$.

   ½ mark

2. $\frac{1}{v}=-\frac{1}{20}+\frac{1}{30}=-\frac{1}{60}$, so $v=-60$ cm. The image is real and lies 60 cm in front of the mirror.

   ½ mark

3. (ii) Magnification: $m=-\frac{v}{u}=-\frac{(-60)}{(-30)}$.

   ½ mark

4. $m=-2$. The image is inverted and twice the size of the object.

   ½ mark

Answer: (i) v = −60 cm (real, 60 cm in front of mirror); (ii) m = −2

1. For the photon, $E=\frac{hc}{{\lambda }_p}$, so ${\lambda }_p=\frac{hc}{E}$.

   ½ mark

2. For the neutron, $E=\frac{p^2}{2m}$ gives $p=\sqrt{2mE}$, so ${\lambda }_n=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$.

   ½ mark

3. Ratio: $\frac{{\lambda }_n}{{\lambda }_p}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}$.

   ½ mark

4. $\frac{{\lambda }_n}{{\lambda }_p}=\sqrt{\frac{E}{2m{c}^2}}$

   ½ mark

Answer: $\frac{{\lambda }_n}{{\lambda }_p}=\sqrt{\frac{E}{2m{c}^2}}$

1. Graph (described): Current I is on the y-axis and voltage V on the x-axis. Starting from the origin, I rises linearly with V up to a peak. Beyond the peak, I decreases as V increases, falling to a minimum. After the minimum, I rises again with V.

   1 mark

2. (a) The portion between the peak and the minimum, where I decreases while V increases ($dV/dI<0$), is marked as the negative resistance region.

   ½ mark

3. (b) The initial straight-line portion through the origin, where $I\propto V$, is marked as the region where Ohm's law is obeyed.

   ½ mark

Answer: Graph rises linearly, falls after a peak, then rises again. Falling part = negative resistance; initial linear part = Ohm's law obeyed.

1. (a) $\vec{E}=500x\hat{i}$ is along x, so flux passes only through the two faces perpendicular to x. Each has area $A=(0.1{)}^2=0.01{\text{m}}^2$. The left face is at $x=0.1$ m and the right face at $x=0.2$ m.

2. Left face (outward normal along $-\hat{i}$): $E_L=500\times 0.1=50$ N/C, so ${\varphi }_L=-50\times 0.01=-0.5{\text{N m}}^2{\text{C}}^{-1}$.

   ½ mark

3. Right face (outward normal along $+\hat{i}$): $E_R=500\times 0.2=100$ N/C, so ${\varphi }_R=100\times 0.01=1.0{\text{N m}}^2{\text{C}}^{-1}$.

   ½ mark

4. The other four faces are parallel to $\vec{E}$ and carry zero flux. Net flux: $\varphi ={\varphi }_L+{\varphi }_R=0.5{\text{N m}}^2{\text{C}}^{-1}$.

   1 mark

5. (b) By Gauss's law, $\varphi =\frac{q}{{\epsilon }_0}$, so $q={\epsilon }_0\varphi =8.85\times 10^{-12}\times 0.5$.

   ½ mark

6. $q\approx 4.4\times 10^{-12}$ C

   ½ mark

Answer: (a) 0.5 N m² C⁻¹; (b) ≈ 4.4 × 10⁻¹² C

1. Current density at a point is the current flowing per unit area of cross-section held normal to the direction of flow: $j=I/A$. Its SI unit is A m${}^{-2}$.

   ½ mark

2. It is a vector quantity, directed along the flow of (positive) charge.

   ½ mark

3. An electron accelerates as $\vec{a}=-e\vec{E}/m$. Averaging over the relaxation time τ gives the drift velocity ${\vec{v}}_d=-\frac{e\vec{E}}{m}\tau$. The charge crossing area A in time Δt is $I\Delta t=neA|{\vec{v}}_d|\Delta t$.

   ½ mark

4. Substituting $|{\vec{v}}_d|=\frac{eE\tau }{m}$: $I\Delta t=\frac{e^{2}A}{m}\tau n\Delta t|\vec{E}|$, so $I=\frac{n{e}^2\tau }{m}A|\vec{E}|$.

   ½ mark

5. Since $I=|\vec{j}|A$, we get $|\vec{j}|=\frac{n{e}^2}{m}\tau |\vec{E}|$.

   ½ mark

6. The electrons drift opposite to $\vec{E}$ and carry negative charge, so the current density points along $\vec{E}$. Hence $\vec{j}=\left(\frac{n{e}^2}{m}\tau \right)\vec{E}=\alpha \vec{E}$.

   ½ mark

Answer: Current density = current per unit normal area (vector); $\vec{j}=\frac{n{e}^2\tau }{m}\vec{E}$

1. A Wheatstone bridge is an arrangement of four resistances forming a closed quadrilateral ABCD. A cell is connected across one diagonal (A and C) and a galvanometer of resistance G across the other (B and D). It is used to find an unknown resistance. Circuit (described): $R_1$ in arm AB, $R_2$ in BC, $R_3$ in AD and $R_4$ in DC.

   1 mark

2. Let currents $I_1,I_2,I_3,I_4$ flow in $R_1,R_2,R_3,R_4$ and $I_g$ through the galvanometer from B to D. Kirchhoff's loop rule for loop ABDA: $-I_1{R}_1-I_{g}G+I_3{R}_3=0$ ...(1)

   ½ mark

3. Loop rule for loop BCDB: $-I_2{R}_2+I_4{R}_4+I_{g}G=0$ ...(2)

   ½ mark

4. The bridge is balanced when $I_g=0$. Then the junction rule at B and D gives $I_2=I_1$ and $I_4=I_3$.

   ½ mark

5. From (1): $I_1{R}_1=I_3{R}_3$. From (2): $I_1{R}_2=I_3{R}_4$. Dividing, the balance condition is $\frac{R_2}{R_1}=\frac{R_4}{R_3}$.

   ½ mark

Answer: Balanced when no current flows through the galvanometer, i.e. $\frac{R_2}{R_1}=\frac{R_4}{R_3}$

1. (a) From $K=\frac{1}{2}m{v}^2$: $v=\sqrt{\frac{2K}{m}}$.

   ½ mark

2. With $m=1.673\times 10^{-27}$ kg: $v=\sqrt{\frac{2\times 1.3384\times 10^{-14}}{1.673\times 10^{-27}}}=\sqrt{1.6\times 10^{13}}=4\times 10^6$ m/s.

   ½ mark

3. (b) $\vec{v}$ (southward) is perpendicular to $\vec{B}$ (eastward), so the magnetic force is $F=qvB$ and $a=\frac{qvB}{m}=\left(\frac{q}{m}\right)vB$.

   ½ mark

4. $a=1.0\times 10^8\times 4\times 10^6\times 2.0\times 10^{-3}=8\times 10^{11}{\text{m/s}}^2$

   ½ mark

5. (c) The magnetic force supplies the centripetal force, $qvB=\frac{m{v}^2}{r}$, so $r=\frac{mv}{qB}=\frac{v}{(q/m)B}$.

   ½ mark

6. $r=\frac{4\times 10^6}{1.0\times 10^8\times 2.0\times 10^{-3}}=20$ m

   ½ mark

Answer: (a) 4 × 10⁶ m/s; (b) 8 × 10¹¹ m/s²; (c) 20 m

1. In a series LCR circuit with $v=v_m\sin \omega t$, the current is $i=i_m\sin (\omega t+\varphi )$, where $\tan \varphi =\frac{X_C-X_L}{R}$.

   ½ mark

2. Instantaneous power: $p=vi=v_m{i}_m\sin \omega t\sin (\omega t+\varphi )=\frac{v_m{i}_m}{2}[\cos \varphi -\cos (2\omega t+\varphi )]$.

   ½ mark

3. Over a complete cycle, $\cos \varphi$ is constant. The time-dependent term $\cos (2\omega t+\varphi )$ averages to zero.

   ½ mark

4. Hence $\bar{P}=\frac{v_m{i}_m}{2}\cos \varphi =\frac{v_m}{\sqrt{2}}\frac{i_m}{\sqrt{2}}\cos \varphi =VI\cos \varphi =I^{2}Z\cos \varphi$, with power factor $\cos \varphi =R/Z$.

   ½ mark

5. At resonance the impedance is minimum and the current is maximum, which happens when $X_L=X_C$, i.e. $\omega L=\frac{1}{\omega C}$.

   ½ mark

6. ${\omega }_0=\frac{1}{\sqrt{LC}}$, so the resonant frequency is ${\nu }_0=\frac{1}{2\pi \sqrt{LC}}$.

   ½ mark

Answer: $\bar{P}=V_{rms}{I}_{rms}\cos \varphi$; ${\nu }_0=\frac{1}{2\pi \sqrt{LC}}$

1. Example 1 – Gamma rays: they have wavelengths of about $10^{-14}$–$10^{-15}$ m. This is comparable to nuclear size (≈ 10⁻¹⁵ m), and gamma rays are indeed emitted by atomic nuclei.

   1½ marks

2. Example 2 – Radio waves: they are produced by accelerating electrons in circuits and antennas. An antenna radiates most efficiently at wavelengths comparable to its own size (metres), so radio wavelengths match antenna dimensions. (Also, X-rays of wavelength ≈ 10⁻¹⁰ m come from inner-shell transitions in heavy atoms, whose size is of the same order.)

   1½ marks

Answer: Gamma rays (~10⁻¹⁵ m) come from nuclei of similar size; radio waves (metres) come from antennas of comparable size (also X-rays from heavy atoms).

1. (i) Short-wave radio waves (about 3–30 MHz) are reflected back to the Earth by the ionosphere (sky-wave propagation). Repeated reflection between the ionosphere and the ground lets the signal travel very long distances, beyond the curvature of the Earth.

   1½ marks

2. (ii) The Earth's atmosphere strongly absorbs X-rays, so they cannot reach ground-based telescopes. Visible light and radio waves pass through the atmospheric 'windows'. Hence optical and radio telescopes work on the ground, but X-ray astronomy needs satellites above the atmosphere.

   1½ marks

Answer: (i) Ionosphere reflects short waves (sky waves); (ii) atmosphere absorbs X-rays but transmits visible and radio waves.

1. Drawbacks of Rutherford's model: (1) An electron revolving round the nucleus is accelerated, so by classical electromagnetic theory it must radiate energy continuously. It would spiral into the nucleus, so the atom could not be stable. (2) As it spirals in, its frequency of revolution changes continuously, so the atom should emit a continuous spectrum. Atoms actually show line spectra.

   1 mark

2. Bohr's remedy: Bohr postulated that an electron revolves only in certain stable (stationary) orbits, for which its angular momentum $mvr=\frac{nh}{2\pi }$. In these orbits it does not radiate. Radiation is emitted only when the electron jumps from a higher orbit to a lower one, giving a photon of energy $h\nu =E_i-E_f$. This explains both the stability of the atom and the line spectrum.

   1 mark

3. In Bohr's model the radius of the nth orbit is $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi mZ{e}^2}$, so $r_n=k{n}^2$ and $r_1:r_2:r_3=1:4:9$. The spacing between successive orbits is $r_{n+1}-r_n=k[(n+1{)}^2-n^2]=k(2n+1)$. This depends on n (3k, 5k, 7k, …), so the orbits are not equally spaced.

   1 mark

Answer: Rutherford: atom unstable and spectrum should be continuous; Bohr: stationary non-radiating orbits with $L=nh/2\pi$ and quantised transitions; $r_n\propto n^2$ so spacing $k(2n+1)$ is not constant.

1. (a) Any two properties: (1) The nucleus is positively charged, with charge $+Ze$. (2) It consists of protons and neutrons (nucleons). Others: its radius is $R=R_0{A}^{1/3}$; nuclear density is independent of mass number.

   1 mark

2. (b) Almost the entire mass of an atom is concentrated in the nucleus, whose radius (~10⁻¹⁵ m) is about 10⁻⁵ times that of the atom (~10⁻¹⁰ m). Most of the atom's volume is empty space, so the nucleus has a far larger mass per unit volume.

   1 mark

3. (c) Let m be the mass of a nucleon. For mass number A, the mass is $mA$ and the radius is $R=R_0{A}^{1/3}$. Density $\rho =\frac{mA}{\frac{4}{3}\pi R^3}=\frac{mA}{\frac{4}{3}\pi R_0^{3}A}=\frac{3m}{4\pi R_0^3}$. This is independent of A, so nuclear density is the same for all nuclei (≈ $2.3\times 10^{17}$ kg m⁻³).

   1 mark

Answer: (a) Positively charged; made of protons and neutrons (R = R₀A^{1/3}). (b) Mass concentrated in tiny volume, atom mostly empty. (c) ρ = 3m/(4πR₀³), independent of A.

1. (i) $P=\frac{1}{f}=(n-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(n-1)}{R}$. Option (A).

   1 mark

2. (ii) Cutting perpendicular to the axis gives two plano-convex halves, each with $P^{\prime }=\frac{(n-1)}{R}=\frac{P}{2}$. Option (D).

   1 mark

3. (iii) The two halves in contact give $P_{net}=\frac{P}{2}+\frac{P}{2}=P$. Option (B).

   1 mark

4. (iv)(a) A cut along the principal axis leaves both curved faces on each half, so each half keeps power P. Placed one after the other along the axis, as in the figure, they give $P+P=2P$. Option (C).

   1 mark

5. (iv)(b) $P=\frac{1}{0.60}+\frac{1}{0.20}=1.67+5=6.67\text{D}\approx 6.6$ D. Option (A).

Answer: (i) A; (ii) D; (iii) B; (iv)(a) C; (iv)(b) A

1. (i) Full-wave rectification only flips the negative half cycles to positive. The square of the voltage, and hence the rms value, is unchanged: $V_{rms}=\frac{V_0}{\sqrt{2}}$. Option (A).

   1 mark

2. (ii) Each diode conducts only in the half cycle in which it is forward biased. Option (B).

   1 mark

3. (iii) In each half cycle one diode is forward biased while the other is reverse biased. Option (C).

   1 mark

4. (iv)(a) A half-wave rectifier gives one output pulse per input cycle, so the ripple frequency is 50 Hz. Option (B).

   1 mark

5. (iv)(b) The diode conducts only for the +5 V part of the input and blocks the −5 V part. The output across $R_L$ is a set of pulses rising to +5 V, and zero otherwise. Option (D).

Answer: (i) A; (ii) B; (iii) C; (iv)(a) B; (iv)(b) D

1. (i) A dipole in a uniform field $\vec{E}$ feels a torque of magnitude $\tau =pE\sin \theta$. To rotate it slowly from ${\theta }_0$ to ${\theta }_1$, an external torque equal and opposite to this must act. The work it does is $W={\int }_{{\theta }_0}^{{\theta }_1}{\tau }_{ext}d\theta ={\int }_{{\theta }_0}^{{\theta }_1}pE\sin \theta d\theta$.

   ½ mark

2. $W=pE(\cos {\theta }_0-\cos {\theta }_1)$. This work is stored as potential energy: $U({\theta }_1)-U({\theta }_0)=pE(\cos {\theta }_0-\cos {\theta }_1)$.

   ½ mark

3. Take the zero of potential energy at ${\theta }_0=\pi /2$ (dipole perpendicular to the field) and write ${\theta }_1=\theta$: $U(\theta )=pE\left(\cos \frac{\pi }{2}-\cos \theta \right)$.

   ½ mark

4. $U(\theta )=-pE\cos \theta =-\vec{p}\cdot \vec{E}$

   ½ mark

5. (1) U is maximum ($+pE$) when $\theta =180^{\circ }$, i.e. $\vec{p}$ is antiparallel to $\vec{E}$.

   ½ mark

6. (2) U is minimum ($-pE$) when $\theta =0^{\circ }$, i.e. $\vec{p}$ is along $\vec{E}$ (stable equilibrium).

   ½ mark

7. (ii) The dipole moment points from −q to +q: $\vec{p}=q\vec{d}=1.0\times 10^{-12}(3\times 10^{-3}\hat{i}+4\times 10^{-3}\hat{j})$.

   ½ mark

8. $\vec{p}=(3\hat{i}+4\hat{j})\times 10^{-15}$ C m. The torque is $\vec{\tau }=\vec{p}\times \vec{E}$.

   ½ mark

9. $\vec{\tau }=(3\hat{i}+4\hat{j})\times 10^{-15}\times 10^3\hat{i}=4\times 10^{-12}(\hat{j}\times \hat{i})$, using $\hat{i}\times \hat{i}=0$.

   ½ mark

10. Since $\hat{j}\times \hat{i}=-\hat{k}$: $\vec{\tau }=-4\times 10^{-12}\hat{k}$ N m, i.e. magnitude $4\times 10^{-12}$ N m along the negative z-direction.

    ½ mark

Answer: (i) $U=-\vec{p}\cdot \vec{E}$; maximum at θ = 180°, minimum at θ = 0°. (ii) $\vec{\tau }=-4\times 10^{-12}\hat{k}$ N m

1. (i) Figure (described): $-q$ at $x=-a$ and $+q$ at $x=+a$ on the x-axis, centre O at the origin, and point P at distance x on the axis beyond $+q$. Then $\vec{p}=2qa\hat{i}$.

   ½ mark

2. Potential of a point charge is $\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$. By superposition, $V=V_{+q}+V_{-q}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{x-a}-\frac{q}{x+a}\right]$.

   ½ mark

3. $V=\frac{q}{4\pi {\epsilon }_0}\frac{(x+a)-(x-a)}{x^2-a^2}=\frac{1}{4\pi {\epsilon }_0}\frac{2qa}{x^2-a^2}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{x^2-a^2}$

   ½ mark

4. Since $\vec{p}$ is along $\hat{i}$, $p=\vec{p}\cdot \hat{i}$, so $V=\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}\cdot \hat{i}}{x^2-a^2}$.

   ½ mark

5. For $x\gg a$, $a^2$ can be neglected compared with $x^2$: $V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\vec{p}\cdot \hat{i}}{x^2}$. Hence proved.

   ½ mark

6. (ii) Initial charges are $Q=\sigma \cdot 4\pi r^2$. $Q_1=\frac{2}{\pi }\times 10^{-9}\times 4\pi (1\times 10^{-2}{)}^2=8\times 10^{-13}$ C.

   ½ mark

7. $Q_2=\frac{2}{\pi }\times 10^{-9}\times 4\pi (3\times 10^{-2}{)}^2=72\times 10^{-13}$ C

   ½ mark

8. When joined by the wire, charge is conserved: $Q_1^{\prime }+Q_2^{\prime }=Q_1+Q_2=80\times 10^{-13}$ C.

   ½ mark

9. The spheres reach a common potential: $\frac{Q_1^{\prime }}{4\pi {\epsilon }_0{r}_1}=\frac{Q_2^{\prime }}{4\pi {\epsilon }_0{r}_2}$, so $\frac{Q_2^{\prime }}{Q_1^{\prime }}=\frac{r_2}{r_1}=3$ and $Q_2^{\prime }=3Q_1^{\prime }$.

   ½ mark

10. $4Q_1^{\prime }=80\times 10^{-13}$, so $Q_1^{\prime }=2\times 10^{-12}$ C. This agrees with $C_{1}V$, where $C_1=\frac{1}{9}\times 10^{-11}$ F and $V=18$ V.

    ½ mark

Answer: (i) $V=\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}\cdot \hat{i}}{x^2}$ for x ≫ a; (ii) new charge on S₁ = 2 × 10⁻¹² C

1. (i) Let the current be $i=i_m\sin \omega t$. Across the resistor, $v_R$ is in phase with $i$ and its peak value is $v_{Rm}=i_{m}R$. Across the capacitor, $v_C$ lags $i$ by $\pi /2$ and its peak value is $v_{Cm}=i_m{X}_C$, where $X_C=\frac{1}{\omega C}$.

   Phasor diagram (in words): the phasor $i_m$ and $V_{Rm}$ are drawn along the same direction. $V_{Cm}$ is drawn at $90^{\circ }$ behind them (clockwise). The source phasor $v_m$ is the resultant of these two perpendicular phasors.

   1 mark

2. Since the two phasors are perpendicular:

   $v_m^2=v_{Rm}^2+v_{Cm}^2=(i_{m}R{)}^2+(i_m{X}_C{)}^2$

   So $i_m=\frac{v_m}{\sqrt{R^2+X_C^2}}$.

   Impedance $Z=\frac{v_m}{i_m}=\sqrt{R^2+X_C^2}=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$.

   The phase angle is given by $\tan \varphi =X_C/R$, with the current leading the voltage.

   1 mark

3. (ii) An inductor acts as a conductor (only its small wire resistance remains) in a dc circuit.

   Reason: $X_L=\omega L=2\pi \nu L$. For dc, $\nu =0$, so $X_L=0$.

   Equivalently, a steady current has $dI/dt=0$, so the induced emf $\epsilon =-LdI/dt=0$ and nothing opposes the current.

   1 mark

4. (iii) Resistance of the lamp:

   $R=\frac{V}{I}=\frac{110}{11}=10\Omega$

   On ac, the same current of 11 A must flow. So the impedance is:

   $Z=\frac{V_{rms}}{I_{rms}}=\frac{220}{11}=20\Omega$

   1 mark

5. Using $Z^2=R^2+X_L^2$:

   $400=100+X_L^2\Rightarrow X_L=\sqrt{300}=10\sqrt{3}\Omega$

   Then $X_L=2\pi fL$ gives:

   $L=\frac{10\sqrt{3}}{2\pi \times 50}=\frac{\sqrt{3}}{10\pi }\text{H}\approx 0.055\text{H}$

   1 mark

Answer: (i) $Z=\sqrt{R^2+(1/\omega C{)}^2}$; (ii) in a dc circuit, since $X_L=2\pi \nu L=0$ for $\nu =0$; (iii) $L=\frac{\sqrt{3}}{10\pi }$ H ≈ 0.055 H

1. (i) Diagram (described in words): A laminated soft-iron core carries two coils, either on separate limbs or wound one over the other.

   • The primary coil has few turns ($N_p$) and is connected to the ac input $V_p$.
   • The secondary coil has many turns ($N_s>N_p$) and is connected across the load, giving output $V_s$.
   • Label: core, primary, secondary, input ac, output.
   1 mark

2. Working principle – mutual induction. The alternating voltage applied to the primary drives an alternating current in it. This produces an alternating magnetic flux in the core, which links the secondary coil and induces an emf in it.

   For an ideal transformer, $\frac{V_s}{V_p}=\frac{N_s}{N_p}$. Since $N_s>N_p$, the output voltage is stepped up.

   ½ mark

3. Causes of energy loss (any three):
   1. Flux leakage – not all of the primary's flux links the secondary.
   2. Resistance of the windings – $I^{2}R$ heating in the copper coils.
   3. Eddy currents – currents induced in the iron core heat it. This is reduced by using a laminated core.
   4. Hysteresis – energy is lost in repeatedly magnetising and demagnetising the core. This is reduced by using soft iron.

   1½ marks

4. (ii) No. When the voltage is stepped up, the current is stepped down in the same ratio: $I_s/I_p=N_p/N_s$.

   So the output power $V_s{I}_s$ equals the input power $V_p{I}_p$ (ideal case). In practice it is slightly less, because of losses. No extra energy is created, so conservation of energy is not violated.

   1 mark

5. (iii)(1) $V_s=V_p\frac{N_s}{N_p}=90\times \frac{3000}{200}=1350\text{V}$

   ½ mark

6. (2) $V_p{I}_p=V_s{I}_s\Rightarrow I_p=I_s\frac{N_s}{N_p}=2.0\times \frac{3000}{200}=30\text{A}$

   ½ mark

Answer: (i) works by mutual induction; losses: flux leakage, winding resistance, eddy currents, hysteresis; (ii) No – current is reduced in the same ratio, so power is conserved; (iii)(1) 1350 V, (2) 30 A

1. (i) Graph (described in words): Plot the angle of deviation $\delta$ (y-axis) against the angle of incidence $i$ (x-axis).

   The curve is U-shaped. As $i$ increases, $\delta$ first decreases to a minimum value ${\delta }_m$, and then increases again. The lowest point of the curve is marked ${\delta }_m$.

   1 mark

2. Angle of minimum deviation ${\delta }_m$ is the smallest deviation produced by the prism. At this position, the angle of incidence equals the angle of emergence ($i=e$), and the refracted ray inside the prism travels parallel to the base.

   1 mark

3. (ii) Normal incidence on the first face means $i=0$ and $r_1=0$. Since $r_1+r_2=A$, the ray meets the second face at an angle $r_2=A$.

   The total deviation is $\delta =(i-r_1)+(e-r_2)=e-A$, so $e=A+\delta$.

   Applying Snell's law at the second face (glass to air):

   $n\sin A=1\cdot \sin (A+\delta )\Rightarrow n=\frac{\sin (A+\delta )}{\sin A}$

   1 mark

4. (iii)(1) Use $n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$:

   $\sqrt{2}=\frac{\sin \frac{60^{\circ }+{\delta }_m}{2}}{\sin 30^{\circ }}$

   $\Rightarrow \sin \frac{60^{\circ }+{\delta }_m}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=\sin 45^{\circ }$

   $\frac{60^{\circ }+{\delta }_m}{2}=45^{\circ }\Rightarrow {\delta }_m=30^{\circ }$

   1 mark

5. (2) At minimum deviation, $i=\frac{A+{\delta }_m}{2}=\frac{60^{\circ }+30^{\circ }}{2}=45^{\circ }$

   1 mark