CBSE Class 12 Physics: 2024 board paper, set 55/5/1

1. $E=I(R+r)$: $0.9(2+r)=0.3(7+r)$, so $1.8+0.9r=2.1+0.3r$.

   ½ mark

2. $0.6r=0.3\Rightarrow r=0.5\Omega$. Option (D).

   ½ mark

Answer: $r=0.5\Omega$, option (D)

1. $R=\frac{mv}{qB}$. With mass $2m$ and charge $q/2$ (same $v$, $B$): $R^{\prime }=\frac{2mv}{(q/2)B}=4R$. Option (D).

   1 mark

Answer: 4R, option (D)

1. Copper and lead are diamagnetic; iron, nickel, cobalt are ferromagnetic. Sodium and calcium (like aluminium) are paramagnetic. Option (B).

   1 mark

Answer: Sodium and Calcium, option (B)

1. $I_g=\frac{2}{50+1000}=\frac{2}{1050}$ A.

   ½ mark

2. $R=\frac{10}{I_g}-50=5250-50=5200\Omega =5.2\mathrm{k\Omega }$. Option (C).

   ½ mark

Answer: $5.2\mathrm{k\Omega }$, option (C)

1. $M=\frac{e}{dI/dt}=\frac{2\times 10^{-3}}{5}=4\times 10^{-4}$ H $=0.4$ mH. Option (A).

   1 mark

Answer: 0.4 mH, option (A)

1. UV radiation kills bacteria and germs, so UV lamps are used in water purifiers. Option (B).

   1 mark

Answer: Ultraviolet rays, option (B)

1. Relaxed eye (image at infinity): $m=\frac{L}{f_o}\cdot \frac{D}{f_e}=\frac{10}{1}\times \frac{25}{2}=125$. Option (D).

   1 mark

Answer: 125, option (D)

1. $K_{max}=h\nu -h{\nu }_0$. For the same $\nu$, $K_{max}$ is largest for the smallest threshold frequency ${\nu }_0$.

   ½ mark

2. Metal A has the smallest intercept on the $\nu$-axis, so its photoelectrons have maximum KE. Option (A).

   ½ mark

Answer: Metal A, option (A)

1. First excited state $n=2$: $E_2=-13.6/4=-3.4$ eV.

   ½ mark

2. $K=-E=3.4$ eV, $U=2E=-6.8$ eV. Option (C).

   ½ mark

Answer: 3.4 eV, –6.8 eV, option (C)

1. Position of $n$th maximum $y=n\lambda D/d$; in medium ${\lambda }^{\prime }=\lambda /\mu =3\lambda /4$.

   ½ mark

2. $n\cdot \frac{3\lambda }{4}=6\lambda \Rightarrow n=8$. Option (C).

   ½ mark

Answer: 8th maximum, option (C)

1. The nuclear potential energy curve has its minimum at about $r_0\approx 0.8$ fm; below this the force is repulsive, above it attractive. Option (A).

   1 mark

Answer: 0.8 fm, option (A)

1. $n_e\approx N_D=5\times 10^{28}\times 10^{-6}=5\times 10^{22}{\mathrm{m}}^{-3}$.

   ½ mark

2. $n_i^2=n_e{n}_h=5\times 10^{22}\times 4.5\times 10^9=2.25\times 10^{32}$, so $n_i=1.5\times 10^{16}{\mathrm{m}}^{-3}$. Option (B).

   ½ mark

Answer: $1.5\times 10^{16}{\mathrm{m}}^{-3}$, option (B)

1. Field of each element points away from + charges and towards − charges; horizontal components cancel by symmetry, vertical components add downward, so the net field at O is along OC. Assertion true.

   ½ mark

2. Net potential at O = $k(Q-Q)/R=0$, so Reason is false. Option (C).

   ½ mark

Answer: Option (C)

1. Magnetic force $q\vec{v}\times \vec{B}$ is always perpendicular to velocity, so it does no work; hence kinetic energy stays constant. Both true and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. $\beta =\lambda D/d$; as $d\to 0$, fringe width becomes very large, so a single fringe fills the screen and no pattern is seen. Assertion true.

   ½ mark

2. Fringe width is inversely (not directly) proportional to $d$, so Reason is false. Option (C).

   ½ mark

Answer: Option (C)

1. $b\propto Z\cot (\theta /2)$; at $\theta =180^{\circ }$, $b=0$ (head-on), the minimum, so Assertion is false.

   ½ mark

2. $b$ depends on $Z$ of the target, so Reason is also false. Option (D).

   ½ mark

Answer: Option (D)

1. Each corner is at the same distance from the centre O: $r=\frac{30\sqrt{2}}{2}\text{ cm}=15\sqrt{2}\text{ cm}$, so $r^2=0.045{\text{m}}^2$.

   Diagram (in words): draw square ABCD with O at the centre.

   • ${\vec{F}}_{OA}$ and ${\vec{F}}_{OC}$ are repulsions, since $+1\mu$C acts on $+4\mu$C. They point along AO and CO, which are opposite directions.
   • ${\vec{F}}_{OB}$ and ${\vec{F}}_{OD}$ are attractions, since $-2\mu$C acts on $+4\mu$C. They point from O towards B and from O towards D, which are also opposite directions.
   1 mark

2. $F_{OA}=F_{OC}=\frac{9\times 10^9\times 4\times 10^{-6}\times 1\times 10^{-6}}{0.045}=0.8\text{N}$.

   These two forces are equal and opposite, so ${\vec{F}}_{OA}+{\vec{F}}_{OC}=0$.

   ½ mark

3. $F_{OB}=F_{OD}=\frac{9\times 10^9\times 4\times 10^{-6}\times 2\times 10^{-6}}{0.045}=1.6\text{N}$.

   These two forces are also equal and opposite, so ${\vec{F}}_{OB}+{\vec{F}}_{OD}=0$.

   Hence the net force on the $4\mu$C charge is $\vec{F}=0$.

   ½ mark

Answer: Net force on the 4 μC charge = 0

1. The centroid O is equidistant from all three vertices: $r=\frac{a}{\sqrt{3}}=\frac{0.1}{\sqrt{3}}$ m, so $r^2=\frac{0.01}{3}{\text{m}}^2$.

   Each field has magnitude
   $E=\frac{kq}{r^2}=\frac{9\times 10^9\times 10^{-12}\times 3}{0.01}=2.7{\text{N C}}^{-1}$.

   So $E_{OA}=E_{OB}=E_{OC}=2.7{\text{N C}}^{-1}$.

   1 mark

2. The three fields point away from their vertices and are at $120^{\circ }$ to each other.

   Resultant of $E_{OB}$ and $E_{OC}$:
   $E_{BC}=\sqrt{E^2+E^2+2E^2\cos 120^{\circ }}=E=2.7{\text{N C}}^{-1}$.

   This resultant is directed opposite to ${\vec{E}}_{OA}$. Therefore ${\vec{E}}_O={\vec{E}}_{OA}+{\vec{E}}_{BC}=0$.

   (Equivalently, three equal vectors at $120^{\circ }$ form a closed triangle, so their sum is zero.)

   1 mark

Answer: Net electric field at the centroid = 0

1. Let the conductor have length $L$, cross-section $A$ and $n$ mobile carriers per unit volume, each of charge $q$.

   Total number of mobile carriers $=nAL$.

   Force on one carrier moving with drift velocity ${\vec{v}}_d$ is $q({\vec{v}}_d\times \vec{B})$. So the total force is
   $\vec{F}=nALq({\vec{v}}_d\times \vec{B})$ ...(1)

   ½ mark

2. The current through the conductor is $I=nqA{v}_d$ ...(2)

   ½ mark

3. Define $\vec{L}$ as a vector of magnitude $L$ along the direction of the current. From (1) and (2), $nqA{v}_d\vec{L}=I\vec{L}$, so
   $$\vec{F}=I(\vec{L}\times \vec{B})$$

   ½ mark

4. Zig-zag conductor: Yes, the result still holds.

   Divide the conductor into small straight elements $d\vec{l}$ and add their forces vectorially: $\vec{F}=I\int d\vec{l}\times \vec{B}$.

   For a uniform $\vec{B}$ this becomes $I{\vec{L}}_{\text{eff}}\times \vec{B}$, where ${\vec{L}}_{\text{eff}}$ is the vector joining the two ends of the conductor.

   ½ mark

Answer: $\vec{F}=I(\vec{L}\times \vec{B})$; yes, it is valid for a zig-zag conductor (sum over small straight elements $d\vec{l}$)

1. In normal adjustment, magnifying power
   $m=\frac{f_o}{f_e}=\frac{150}{5}=30$.

   1 mark

2. The object (a distant one) is effectively at infinity, $u=-\infty$.

   Lens formula for the objective: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_o}$, so $\frac{1}{v}=\frac{1}{150}$.

   Hence $v=150$ cm. The image is formed at the focus of the objective, 150 cm behind it.

   1 mark

Answer: Magnifying power = 30; image formed by the objective at 150 cm from it

1. (a) The photon energy equals the energy gap: $E_2-E_1=\frac{hc}{\lambda }$.

   $E_2-E_1=2.55\times 1.6\times 10^{-19}$ J.

   ½ mark

2. $\lambda =\frac{6.63\times 10^{-34}\times 3\times 10^8}{2.55\times 1.6\times 10^{-19}}=4.875\times 10^{-7}\text{m}=487.5\text{nm}$

   1 mark

3. (b) 487.5 nm lies in the visible region. It corresponds to the transition $n=4\to n=2$, since $-0.85-(-3.40)=2.55$ eV.

   So the line belongs to the Balmer series.

   ½ mark

Answer: (a) λ ≈ 487.5 nm; (b) Balmer series

1. Bohr's quantisation condition: $mvr=\frac{nh}{2\pi }$, so $n=\frac{2\pi mvr}{h}$.

   1 mark

2. Substituting: $n=\frac{2\times 3.14\times 6.0\times 10^{24}\times 3\times 10^4\times 1.5\times 10^{11}}{6.63\times 10^{-34}}$

   ½ mark

3. $n\approx 2.56\times 10^{74}$

   This is enormously large, so quantisation of the earth's orbit is not noticeable.

   ½ mark

Answer: n ≈ 2.56 × 10⁷⁴

1. (a) Einstein's photoelectric equation:
   $$K_{max}=h\nu -{\varphi }_0=h\nu -h{\nu }_0,\text{with }{K}_{max}=e{V}_0$$

   ½ mark

2. Millikan's test: Millikan measured the stopping potential $V_0$ for different frequencies $\nu$ using metals such as sodium.

   The equation predicts $V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$. Millikan's $V_0$ versus $\nu$ graph was indeed a straight line.

   Its slope $h/e$ gave a value of $h$ that agreed closely with Planck's constant. This verified the equation.

   1 mark

3. (b) $K_{max}$ cannot be negative, so $h\nu -{\varphi }_0\ge 0$, which gives $\nu \ge \frac{{\varphi }_0}{h}={\nu }_0$.

   A photon gives all its energy to a single electron. If $h\nu <{\varphi }_0$, that energy is not enough to remove the electron, however intense the light.

   Hence emission occurs only above a minimum frequency ${\nu }_0$, which is the threshold frequency of the surface.

   1½ marks

Answer: (a) $K_{max}=h\nu -{\varphi }_0$ (= eV₀), verified by Millikan's linear V₀–ν plot whose slope h/e gave Planck's constant; (b) K_max ≥ 0 requires ν ≥ ν₀ = φ₀/h

1. (a) Electric flux through a surface is a measure of the number of electric field lines crossing that surface normally.

   Mathematically it is the surface integral of the field, ${\varphi }_E=\oint \vec{E}\cdot d\vec{S}$ (for a uniform field, $\vec{E}\cdot \vec{A}$).

   1 mark

2. Dimensions: $[{\varphi }_E]=[E][A]=[ML{T}^{-3}{A}^{-1}][L^2]=[M{L}^3{T}^{-3}{A}^{-1}]$

   ½ mark

3. (b) Area: $A=(1\text{cm}{)}^2=10^{-4}{\text{m}}^2$, so $\vec{A}=10^{-4}(0.8\hat{i}+0.6\hat{k}){\text{m}}^2$.

   ½ mark

4. ${\varphi }_E=\vec{E}\cdot \vec{A}=(100\hat{i})\cdot (0.8\hat{i}+0.6\hat{k})\times 10^{-4}$

   ½ mark

5. ${\varphi }_E=80\times 10^{-4}=8\times 10^{-3}{\text{N m}}^2{\text{C}}^{-1}$

   ½ mark

Answer: (a) Dimensions [ML³T⁻³A⁻¹]; (b) φ = 8 × 10⁻³ N m² C⁻¹

1. (i) Lenz's law: The induced emf has a polarity such that the current it drives opposes the change in magnetic flux that produced it.

   1 mark

2. Justification: Suppose the induced current aided the change in flux. The flux would then increase further, which would induce a still larger current, and so on without limit. Current and energy would grow without any work being done by an external agent.

   This violates the law of conservation of energy. So the induced current must oppose the change, and the work done against this opposition appears as electrical energy.

   ½ mark

3. (ii) The emf between the centre and one end equals that of a rod of length $l$ rotating about one end: $\epsilon =\frac{1}{2}B\omega l^2$.

   Here $l$ is the centre-to-end length $=\frac{2}{2}=1$ m, and $\omega =2\pi \nu =2\pi \times 60=120\pi$ rad/s.

   ½ mark

4. $\epsilon =\frac{1}{2}\times 2\times 120\pi \times (1{)}^2=120\pi \text{V}$

   ½ mark

5. $\epsilon \approx 377\text{V}$

   ½ mark

Answer: (ii) emf between centre and end = ½Bωl² = 120π V ≈ 377 V (with l = 1 m)

1. (i) Ampere's circuital law: The line integral of the magnetic field around any closed loop in vacuum equals ${\mu }_0$ times the net current threading the loop:
   $$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$$

   The integral is taken along the closed loop (Amperian loop) that bounds the surface. $I$ is the net current passing through that surface, with its sign given by the right-hand rule.

   1 mark

2. (ii) Field of a long straight wire: $B=\frac{{\mu }_{0}I}{2\pi r}$. The midpoint is at $r=0.1$ m from each wire.

   $B_1=\frac{4\pi \times 10^{-7}\times 5}{2\pi \times 0.1}=1\times 10^{-5}$ T (due to the 5 A wire)

   ½ mark

3. $B_2=\frac{4\pi \times 10^{-7}\times 10}{2\pi \times 0.1}=2\times 10^{-5}$ T (due to the 10 A wire)

   ½ mark

4. For currents in the same direction, the two fields at the midpoint are opposite to each other. So
   $B=B_2-B_1=1\times 10^{-5}$ T

   ½ mark

5. The net field is $10^{-5}$ T. It is perpendicular to the plane of the wires, in the direction of the field due to the 10 A wire (given by the right-hand rule for that wire).

   ½ mark

Answer: (ii) B = 1 × 10⁻⁵ T, perpendicular to the plane of the wires, along the field of the 10 A wire

1. $\vec{B}$ is along $\hat{j}$. So the velocity component perpendicular to $\vec{B}$ is $v_x=1.0\times 10^7$ m/s, and the component parallel to $\vec{B}$ is $v_y=0.5\times 10^7$ m/s.

   Radius: $r=\frac{m{v}_x}{eB}$

   ½ mark

2. $r=\frac{9.1\times 10^{-31}\times 1.0\times 10^7}{1.6\times 10^{-19}\times 0.5\times 10^{-3}}=0.114\text{m}(\approx 11.4\text{cm})$

   ½ mark

3. Yes. The parallel component $v_y$ is unaffected by the field, so the electron also moves uniformly along $\vec{B}$. It therefore follows a helical path.

   ½ mark

4. Linear distance covered in one revolution (the pitch): $p=v_{y}T=v_y\frac{2\pi m}{eB}$

   ½ mark

5. $p=\frac{2\times 3.14\times 9.1\times 10^{-31}\times 0.5\times 10^7}{1.6\times 10^{-19}\times 0.5\times 10^{-3}}$

   ½ mark

6. $p\approx 0.357\text{m}\approx 0.36\text{m}$

   ½ mark

Answer: r ≈ 0.114 m; yes (helical path); linear distance per revolution ≈ 0.36 m

1. (a)(i) Heat waves are infrared waves.

   ½ mark

2. (a)(ii) Radiation absorbed by the ozone layer is ultraviolet rays.

   ½ mark

3. (b) Production of infrared: emitted by hot bodies, through vibrations of atoms and molecules (e.g. a heated filament or a room heater).

   ½ mark

4. Production of ultraviolet: emitted by very hot bodies such as the sun, and by special UV lamps (e.g. mercury vapour lamps).

   ½ mark

5. Detection of infrared: thermopile, bolometer or infrared photographic film.

   ½ mark

6. Detection of ultraviolet: photocells or photographic film.

   ½ mark

Answer: (i) Infrared; (ii) Ultraviolet. IR: produced by hot bodies, detected by thermopile/bolometer. UV: produced by the sun or UV lamps, detected by photocells/photographic film.

1. (a) A p-n junction diode conducts easily only when forward biased. Under reverse bias it allows almost no current, only a tiny reverse saturation current.

   This unidirectional conduction lets it pass only one polarity of an ac input, which converts ac into dc.

   1 mark

2. (b) Circuit (described):

   • The ac input is applied to the primary of a centre-tapped transformer.
   • The ends A and B of the secondary are connected to the p-sides of diodes D₁ and D₂ respectively.
   • The n-sides of both diodes are joined together.
   • The load resistance $R_L$ is connected between this common point and the centre tap of the secondary.
   • The output is taken across $R_L$.
   1 mark

3. Working:

   • When A is positive with respect to the centre tap, B is negative, because it is out of phase. D₁ is then forward biased and conducts, while D₂ is reverse biased and does not. Current flows through $R_L$.
   • In the next half cycle A is negative and B is positive. D₂ conducts and D₁ does not, and current flows through $R_L$ in the same direction.

   So a unidirectional (pulsating dc) output is obtained across $R_L$ in both halves of the ac cycle.

   1 mark

Answer: Diode conducts only in forward bias; in a centre-tap full wave rectifier D₁ and D₂ conduct in alternate half cycles, giving output in both halves.

1. (a) Every extra free charge carrier contributed by a dopant atom leaves behind an ionised core of equal and opposite charge. Electrons are left behind by donor ions, and holes come with acceptor ions.

   The total positive charge therefore equals the total negative charge, so the doped semiconductor stays electrically neutral.

   1 mark

2. (b) At equilibrium the diffusion current of majority carriers across the junction is exactly balanced by the oppositely directed drift current of minority carriers, which is driven by the barrier field.

   The two cancel, so the net current is zero.

   1 mark

3. (c) The reverse current is carried by minority carriers, which drift across the junction under the barrier field.

   Its size is limited by the concentration of minority carriers on either side. This concentration depends on temperature, not on the applied voltage. Even a small reverse voltage sweeps all available minority carriers across, so the current saturates and hardly changes with voltage (until breakdown).

   1 mark

Answer: (a) Carrier charge balanced by ionised cores; (b) diffusion current = drift current; (c) reverse current limited by the minority carrier concentration

1. (i) HCl has an unsymmetrical charge distribution (Cl is more electronegative than H), so it has a permanent dipole moment. O₂, H₂ and N₂ are symmetric.

   Option (D).

   1 mark

2. (ii) In an external field the induced dipoles line up along the field, so their net dipole moment is along $\vec{E}$.

   Option (A) is wrong because without a field the net dipole moment is zero (random orientation). Option (C) is wrong because dielectrics have no free charges. Option (D) is wrong because the induced field opposes the external field.

   Option (B).

   1 mark

3. (iii) The capacitor is isolated, so $Q$ stays constant while $C$ becomes $KC$.

   $U=\frac{Q^2}{2C}$ decreases, and $E=\frac{E_0}{K}$ also decreases.

   Option (B).

   1 mark

4. (iv)(a) $C=\frac{{\epsilon }_{0}A}{d-t+t/K}$ with $t=\frac{d}{5}$:

   $C=\frac{{\epsilon }_{0}A}{d\left(\frac{4}{5}+\frac{1}{5K}\right)}=\left[\frac{5K}{4K+1}\right]C_0$

   Option (C).

   (iv)(b) $C_s=\frac{2C_0\times 6C_0}{8C_0}=1.5C_0$ and $C_p=8C_0$.

   At the same voltage, $\frac{U_s}{U_p}=\frac{C_s}{C_p}=\frac{1.5}{8}=\frac{3}{16}$

   Option (D).

   1 mark

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

1. (i) Glass–water: $\sin c=\frac{1.33}{1.5}=0.887$, so $c\approx 62.5^{\circ }$.

   For comparison, ${\theta }_1={\sin }^{-1}\frac{1}{1.5}\approx 41.8^{\circ }$ and ${\theta }_2={\sin }^{-1}\frac{1}{1.33}\approx 48.8^{\circ }$.

   So $c>{\theta }_2$. Option (C).

   1 mark

2. (ii) Frequency is set by the source and does not change on refraction. Speed decreases in the denser medium, so $\lambda =v/\nu$ decreases.

   Option (C).

   1 mark

3. (iii)(a) $\sin c=\frac{{\mu }_w}{{\mu }_g}$. Violet has the largest refractive index in glass relative to water, so its critical angle is the smallest.

   Option (D).

   (iii)(b) At minimum deviation the emergent angle equals $i$, and ${\delta }_m=2i-A$. Deviation is greatest for violet and least for red, so the angle of emergence (angle of refraction at the second surface) is greatest for violet.

   Thus $r_R<r_Y<r_V$. Option (C).

   1 mark

4. (iv) The ray enters face AB normally and travels undeviated to face AC. With $\angle A=60^{\circ }$, the angle of incidence on AC is $60^{\circ }$.

   Critical angle: $c={\sin }^{-1}\frac{1}{\sqrt{2}}=45^{\circ }$.

   Since $60^{\circ }>45^{\circ }$, the ray undergoes total internal reflection. Option (D).

   1 mark

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

1. (i) Equipotential surfaces of a dipole (figure in words):

   • Draw $+q$ and $-q$ a short distance apart.
   • Around each charge draw closed, distorted oval surfaces that are crowded near the charge.
   • The surfaces around $+q$ are at positive potential and those around $-q$ at negative potential.
   • The perpendicular bisector plane of the dipole is the zero-potential surface.
   • Electric field lines, from $+q$ to $-q$, cut every surface at right angles.
   1 mark

2. (ii) Bring $q_1$ from infinity to ${\vec{r}}_1$ in the external field. Work done:
   $W_1=q_{1}V({\vec{r}}_1)$ ...(1)

   ½ mark

3. Bring $q_2$ from infinity to ${\vec{r}}_2$. Work done against the external field:
   $W_2=q_{2}V({\vec{r}}_2)$ ...(2)

   ½ mark

4. Work done on $q_2$ against the field of $q_1$:
   $W_{12}=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}_{12}}$ ...(3), where $r_{12}=|{\vec{r}}_1-{\vec{r}}_2|$

   ½ mark

5. Potential energy of the system = total work done:
   $$U=q_{1}V({\vec{r}}_1)+q_{2}V({\vec{r}}_2)+\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}_{12}}$$

   ½ mark

6. (iii) Potential energy of a dipole: $U=-pE\cos \theta$.

   The change as $\theta$ goes from $0^{\circ }$ to $60^{\circ }$ is
   $\Delta U=pE(\cos {\theta }_0-\cos {\theta }_1)=pE(\cos 0^{\circ }-\cos 60^{\circ })$

   1 mark

7. Substituting: $\Delta U=10^{-30}\times 10^5\times (1-0.5)$

   ½ mark

8. $\Delta U=5.0\times 10^{-26}$ J (an increase in potential energy)

   ½ mark

Answer: (ii) $U=q_{1}V({\vec{r}}_1)+q_{2}V({\vec{r}}_2)+\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}_{12}}$; (iii) ΔU = 5.0 × 10⁻²⁶ J

1. (i) **Outside ($r>R$):** take a concentric spherical Gaussian surface of radius $r$. By symmetry $\vec{E}$ is radial and has the same magnitude everywhere on it.

   Flux through the surface: $\varphi =E\cdot 4\pi r^2$.

   ½ mark

2. Charge enclosed: $Q=\sigma \cdot 4\pi R^2$.

   ½ mark

3. Gauss's law: $\oint \vec{E}\cdot d\vec{S}=\frac{Q}{{\epsilon }_0}$, so $E\cdot 4\pi r^2=\frac{\sigma 4\pi R^2}{{\epsilon }_0}$.

   ½ mark

4. $$\vec{E}=\frac{\sigma R^2}{{\epsilon }_0{r}^2}\hat{r}(r>R)$$

   ½ mark

5. **Inside ($r<R$):** take a Gaussian sphere of radius $r<R$. Flux $=E\cdot 4\pi r^2$.

   The charge enclosed is $Q=0$, since all the charge lies on the shell.

   ½ mark

6. So $E\cdot 4\pi r^2=0$, which gives $E=0$ everywhere inside the shell.

   ½ mark

7. (ii) Field of a long charged wire: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$. The midpoint is at $r=0.5$ m from each wire.

   ½ mark

8. Directions at the midpoint:

   • The field of AB ($+10\mu$C/m) points away from AB, i.e. towards CD.
   • The field of CD ($-20\mu$C/m) points towards CD.

   Both are along the same direction, so they add:
   $E_{net}=\frac{1}{2\pi {\epsilon }_{0}r}({\lambda }_1+|{\lambda }_2|)$

   ½ mark

9. $E_{net}=\frac{2\times 9\times 10^9\times (10+20)\times 10^{-6}}{0.5}$

   ½ mark

10. $E_{net}=1.08\times 10^6{\text{N C}}^{-1}$, directed towards CD (perpendicular to the wires).

    ½ mark

Answer: (i) E = σR²/(ε₀r²) outside (radial), E = 0 inside; (ii) 1.08 × 10⁶ N/C directed towards CD

1. (i) X: V and I are in phase, so X is a pure resistor.

   ½ mark

2. Y: V leads I by $\pi /4$, so Y is an inductor. A lead of exactly $\pi /4$ means a real inductor with $X_L=R$, since $\tan \varphi =X_L/R=1$.

   ½ mark

3. Z: I leads V by $\pi /4$, so Z is a capacitor. Again this is a real capacitor (in series with resistance) with $X_C=R$.

   ½ mark

4. (ii) Phasor diagram (in words): In a series R, L, C circuit the same current $i_m$ flows through all three, so take the current phasor as reference.

   • $V_{Rm}=i_{m}R$ is in phase with the current.
   • $V_{Lm}=i_m{X}_L$ leads the current by $90^{\circ }$.
   • $V_{Cm}=i_m{X}_C$ lags the current by $90^{\circ }$.
   ½ mark

5. $V_{Lm}$ and $V_{Cm}$ are opposite in direction, so their resultant is $(V_{Cm}-V_{Lm})$, at right angles to $V_{Rm}$.

   Therefore $V_m^2=V_{Rm}^2+(V_{Cm}-V_{Lm}{)}^2$.

   ½ mark

6. Substituting each voltage: $V_m^2=(i_{m}R{)}^2+(i_m{X}_C-i_m{X}_L{)}^2$.

   ½ mark

7. Impedance:
   $$Z=\frac{V_m}{i_m}=\sqrt{R^2+(X_C-X_L{)}^2}$$
   where $X_L=\omega L$ and $X_C=\frac{1}{\omega C}$.

   ½ mark

8. Variation of current with frequency (graph in words): $i_m=\frac{V_m}{Z}$.

   • At low frequency the current is small, because $X_C$ is large.
   • The current rises to a sharp maximum $\frac{V_m}{R}$ at the resonant frequency ${\omega }_0=\frac{1}{\sqrt{LC}}$.
   • At higher frequency it falls again, because $X_L$ becomes large.
   ½ mark

9. (iii)(i) $Z=\sqrt{R^2+(X_C-X_L{)}^2}$ is minimum (equal to $R$) when $X_C=X_L$, i.e. at resonance $\omega =\frac{1}{\sqrt{LC}}$.

   ½ mark

10. (iii)(ii) Average power is $P=V_{rms}{I}_{rms}\cos \varphi$. A wattless current means $P=0$ even though $V\ne 0$ and $I\ne 0$.

    So $\cos \varphi =0$, i.e. $\varphi =\pi /2$. This happens when the circuit is purely reactive, with $R=0$ (pure L, pure C, or an LC combination).

    ½ mark

Answer: (i) X: resistor, Y: inductor (X_L = R), Z: capacitor (X_C = R); (ii) $Z=\sqrt{R^2+(X_C-X_L{)}^2}$, current peaks at ω₀ = 1/√(LC); (iii) minimum Z when X_L = X_C; wattless current when φ = π/2 (cos φ = 0)

1. (i) Construction: A transformer has two coils, the primary ($N_p$ turns) and the secondary ($N_s$ turns), insulated from each other. Both are wound on a common laminated soft-iron core, either one over the other or on separate limbs of the core. *(Diagram: a rectangular laminated soft-iron core with the primary on one limb connected to the AC input $v_p$ and the secondary on the other limb connected to the output/load $v_s$.)*

   1 mark

2. Working: An alternating voltage applied to the primary drives an alternating current in it. This current sets up an alternating magnetic flux in the core. The core carries this flux to the secondary, so the flux linked with the secondary changes continuously and an alternating emf is induced in the secondary by mutual induction.

   1 mark

3. Relation: For an ideal transformer there is no flux leakage and the coil resistances are negligible. The flux $\varphi$ through each turn is then the same in both coils. Induced emf in the secondary: ${\epsilon }_s=-N_s\frac{d\varphi }{dt}$, and $v_s={\epsilon }_s$. Back emf in the primary: ${\epsilon }_p=-N_p\frac{d\varphi }{dt}$, and $v_p={\epsilon }_p$ (since primary resistance $\approx 0$).

   ½ mark

4. Dividing the two equations: $\frac{v_s}{v_p}=\frac{N_s}{N_p}$. If $N_s>N_p$ it is a step-up transformer; if $N_s<N_p$ it is a step-down transformer.

   ½ mark

5. (ii) Cause 1 – Flux leakage: Not all the flux produced by the primary passes through the secondary, because of poor core design or air gaps. It is reduced by winding the primary and secondary one over the other.

   ½ mark

6. Cause 2 – Resistance of windings (copper loss): The windings have resistance, so energy is lost as heat ($I^{2}R$). It is reduced by using thick copper wires.

   ½ mark

7. Cause 3 – Eddy currents (iron loss): The alternating flux induces eddy currents in the iron core, which heat it. It is reduced by using a laminated core.

   ½ mark

8. Cause 4 – Hysteresis: The core is magnetised and demagnetised repeatedly by the alternating field, and energy is lost as heat in each cycle. It is reduced by using a material of low hysteresis loss, such as soft iron.

   ½ mark

Answer: $\frac{v_s}{v_p}=\frac{N_s}{N_p}$; energy losses: flux leakage, copper ($I^{2}R$) loss, eddy currents, hysteresis (also magnetostriction/humming).

1. (i) Diagram: Let the surface PP′ separate medium 1 (rarer, speed $v_1$) from medium 2 (denser, speed $v_2<v_1$). The incident plane wavefront AB meets the surface at A at angle $i$. In time $\tau$, the disturbance at B travels to C, so $BC=v_1\tau$. With A as centre, draw a sphere (arc) of radius $AE=v_2\tau$ in medium 2. Draw the tangent CE from C to this arc. CE is the refracted wavefront, and it makes angle $r$ with the surface. Since $v_2<v_1$, it bends towards the normal.

   1 mark

2. In right triangle ABC (right angle at B): $\sin i=\frac{BC}{AC}=\frac{v_1\tau }{AC}$ ...(1)

   ½ mark

3. In right triangle AEC (right angle at E): $\sin r=\frac{AE}{AC}=\frac{v_2\tau }{AC}$ ...(2)

   ½ mark

4. Dividing (1) by (2): $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$ ...(3)

   ½ mark

5. Using $n_1=\frac{c}{v_1}$ and $n_2=\frac{c}{v_2}$: $\frac{\sin i}{\sin r}=\frac{n_2}{n_1}$, i.e. $n_1\sin i=n_2\sin r$. This is Snell's law of refraction.

   ½ mark

6. (ii) The $n$th dark fringe is at $x_n=(2n-1)\frac{\lambda D}{2d}$. Here $n=4$, $\lambda =600\times 10^{-9}$ m, $D=1.5$ m and $d=0.30\times 10^{-3}$ m.

   1 mark

7. Substituting: $x_4=\frac{7\times 600\times 10^{-9}\times 1.5}{2\times 0.30\times 10^{-3}}=7\times 1.5\times 10^{-3}=1.05\times 10^{-2}$ m $=10.5$ mm.

   1 mark

Answer: (i) $n_1\sin i=n_2\sin r$ (Snell's law verified); (ii) $1.05\times 10^{-2}$ m (10.5 mm)

1. (i) Diffraction: When a beam of monochromatic light falls normally on a narrow single slit of width $a$ (comparable to $\lambda$), it bends round the edges of the slit into the region of geometrical shadow. This bending is called diffraction. Each point of the wavefront at the slit acts as a source of secondary wavelets (Huygens). These wavelets superpose on the screen to give a pattern of bright and dark bands.

   1 mark

2. At the centre O all the wavelets arrive in phase, so the central maximum is bright, with maximum intensity. At angle $\theta$ with $a\sin \theta =n\lambda$ ($n=\pm 1,\pm 2,\dots$), the slit can be divided into pairs of halves whose contributions cancel, giving minima. Secondary maxima occur near $a\sin \theta =(n+\frac{1}{2})\lambda$. Their intensity falls rapidly with increasing order. The central maximum is twice as wide as the others, with angular width $2\lambda /a$.

   1 mark

3. Shape of pattern: Draw an intensity ($I$) versus $\theta$ graph. It has a tall, broad central peak at $\theta =0$, zeros at $\theta =\pm \lambda /a,\pm 2\lambda /a,\dots$, and much smaller secondary peaks of decreasing height on either side, symmetric about the centre.

   1 mark

4. (ii) Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\Rightarrow v=\frac{uf}{u-f}$. By the Cartesian sign convention, for a concave mirror $u=-x$ and $f=-F$, with $0<x<F$ (object between P and F). Then $v=\frac{(-x)(-F)}{-x+F}=\frac{xF}{F-x}$. Since $F>x$, $v$ is positive, so the image is formed behind the mirror. Hence the image is virtual (and erect).

   1 mark

5. Magnification: $m=-\frac{v}{u}=-\frac{xF/(F-x)}{-x}=\frac{F}{F-x}$. Since $0<F-x<F$, $m>1$ (and positive). Hence the image is enlarged and erect.

   1 mark

Answer: (i) Central bright maximum with minima at $a\sin \theta =n\lambda$ and fainter secondary maxima; (ii) $v=\frac{xF}{F-x}>0$ ⇒ virtual; $m=\frac{F}{F-x}>1$ ⇒ enlarged.