CBSE Class 12 Physics: 2023 board paper, set 55/1/1

1. On the axis of a short dipole, $E=\frac{2kp}{r^3}$, so the force on the charge is $F=qE\propto \frac{1}{r^3}$.

   ½ mark

2. When the distance becomes $2r$: $F^{\prime }=\frac{F}{2^3}=\frac{\vec{F}}{8}$. This is option (B).

   ½ mark

Answer: $\vec{F}/8$, option (B)

1. For a metal, $\rho ={\rho }_0[1+\alpha (T-T_0)]$ holds only over a limited range. Over a wide range the resistance rises non-linearly, more and more steeply, as collisions of electrons with the vibrating ions increase.

   ½ mark

2. A curve rising with increasing steepness matches this behaviour, so the answer is option (D).

   ½ mark

Answer: Option (D)

1. In open circuit the terminal voltage equals the emf, so $\epsilon =8$ V. With current drawn, $V=\epsilon -Ir$.

   ½ mark

2. $4=8-4r\Rightarrow r=1\Omega$. This is option (D).

   ½ mark

Answer: $1\Omega$, option (D)

1. For a steady current, $I=neA{v}_d$ is the same at every cross-section, so $v_d=\frac{I}{neA}\propto \frac{1}{A}$.

   ½ mark

2. A $1/A$ dependence gives a hyperbola-like decreasing curve, which is option (A).

   ½ mark

Answer: Option (A)

1. A diamagnetic substance is weakly magnetised opposite to the applied field, whichever pole is near.

   ½ mark

2. It moves from a stronger field towards a weaker field, so it is repelled by both poles. This is option (A).

   ½ mark

Answer: Repelled by both poles, option (A)

1. The maximum emf is $e_0=NBA\omega =40\times 3\times 10^{-2}\times \pi (0.08{)}^2\times \frac{25}{\pi }$.

   ½ mark

2. $e_0=40\times 0.03\times 0.0064\times 25=0.192\approx 0.19$ V. This is option (C).

   ½ mark

Answer: 0.19 V, option (C)

1. The motional emf is $e=Blv$, which depends only on $B$, $l$ and $v$.

   ½ mark

2. The resistance $r$ affects only the current, not the emf. So the answer is option (C).

   ½ mark

Answer: Resistance r, option (C)

1. Between the plates there is no conduction current. Maxwell's displacement current there is $i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$, which equals the charging current. This is option (C).

   1 mark

Answer: ${\epsilon }_{0}d{\varphi }_E/dt$, option (C)

1. When the object is placed at the centre of curvature (2f), the real image is formed at the same point.

   ½ mark

2. So the minimum distance between the object and its real image is zero. This is option (A).

   ½ mark

Answer: Zero, option (A)

1. In the Bohr model, $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}$, so $r_n\propto n^2$. This is option (C).

   1 mark

Answer: $n^2$, option (C)

1. The energy absorbed is $\Delta E=13.6\left(1-\frac{1}{25}\right)=13.06$ eV.

   ½ mark

2. $\lambda =\frac{hc}{\Delta E}=\frac{1240\text{eV nm}}{13.06\text{eV}}\approx 95$ nm. This is option (D).

   ½ mark

Answer: 95 nm, option (D)

1. The nuclear radius is $R=R_0{A}^{1/3}$, so the volume is proportional to $A$. The mass is also proportional to $A$.

   ½ mark

2. Density = mass/volume, so $A$ cancels and the density is independent of $A$. This is option (D).

   ½ mark

Answer: Independent of A, option (D)

1. A single diode conducts only in forward bias, that is, during one half-cycle of the ac input.

   ½ mark

2. The output across the load is therefore a half-wave rectified wave (positive humps with gaps). This is option (C).

   ½ mark

Answer: Half-wave rectified output, option (C)

1. A trivalent impurity (acceptor) creates holes, so holes become the majority carriers and the semiconductor becomes p-type. This is option (B).

   1 mark

Answer: It becomes p-type, option (B)

1. The donor level lies just below the bottom of the conduction band; the gap $E_C-E_D$ is about 0.01 eV for Si (roughly 0.01–0.05 eV). This is option (D).

   1 mark

Answer: 0.01 eV, option (D)

1. Copper is diamagnetic, so field lines are expelled from (thinned inside) the bar, not concentrated in it. Both Assertion (A) and Reason (R) are false, so the answer is option (D).

   1 mark

Answer: Option (D)

1. A wavefront is the locus of points oscillating in the same phase, so the assertion is true.

   ½ mark

2. For a point source, all points on the wavefront are equidistant from it, which explains the equal phase. Both are true and R explains A, so the answer is option (A).

   ½ mark

Answer: Option (A)

1. The photoelectric effect is explained only by photons (quanta of $h\nu$), so the assertion is true. Photocurrent is proportional to intensity above threshold, so the reason is also true.

   ½ mark

2. However, the particle nature is shown by the threshold frequency and instantaneous emission, not by the proportionality of current to intensity. So R does not explain A: option (B).

   ½ mark

Answer: Option (B)

1. Force on a moving charge: $\vec{F}=q(\vec{v}\times \vec{B})=q\left[(3.0\times 10^5)\hat{i}\times (0.4\hat{i}+0.3\hat{j})\right]$

   ½ mark

2. Using $\hat{i}\times \hat{i}=0$ and $\hat{i}\times \hat{j}=\hat{k}$: $\vec{F}=q(0.9\times 10^5)\hat{k}$ N

   ½ mark

3. By Newton's second law, $\vec{a}=\frac{\vec{F}}{m}=\frac{q}{m}(0.9\times 10^5)\hat{k}$

   ½ mark

4. $\vec{a}=4.8\times 10^7\times 0.9\times 10^5\hat{k}=4.32\times 10^{12}\hat{k}{\text{m s}}^{-2}$, i.e. along the +z axis.

   ½ mark

Answer: $\vec{a}=4.32\times 10^{12}\hat{k}$ m s${}^{-2}$ (along +z axis)

1. The induced electric field due to a changing magnetic field (electromagnetic induction) is more easily observed.

   1 mark

2. Justification: an induced emf/current is readily produced and detected in everyday ways — moving a bar magnet towards a coil, rotating or moving a coil in a magnetic field, changing the area/shape of a coil in a field (as in generators and transformers). In contrast, the magnetic field due to a changing electric field (displacement current ${\epsilon }_{0}d{\Phi }_E/dt$) is extremely small because ${\epsilon }_0$ is very small, so it needs very rapidly varying fields to be noticed.

   1 mark

Answer: Induced electric field due to changing magnetic field is more easily observed.

1. Ray diagram (described): a plane surface PP′ separates medium 1 (speed $v_1$, upper) and medium 2 (speed $v_2$, lower, $v_2<v_1$). Plane incident wavefront AB meets the surface at A, making angle $i$ with it. In time $\tau$, the disturbance from B travels BC $=v_1\tau$ to reach C on the surface. From A, a secondary wavelet of radius AE $=v_2\tau$ is drawn in medium 2. The tangent CE from C to this sphere is the refracted wavefront, making angle $r$ with the surface. Rays are drawn perpendicular to AB and CE.

   1 mark

2. In right triangle ABC: $\sin i=\frac{BC}{AC}=\frac{v_1\tau }{AC}$; in right triangle AEC: $\sin r=\frac{AE}{AC}=\frac{v_2\tau }{AC}$

   ½ mark

3. Dividing: $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{c/n_1}{c/n_2}=\frac{n_2}{n_1}=n_{21}$, a constant — this is Snell's law ($n_1\sin i=n_2\sin r$).

   ½ mark

Answer: $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21}$ (Snell's law verified)

1. Reason 1: No chromatic aberration — a mirror forms the image by reflection, not refraction, so there is no dispersion of colours.

   1 mark

2. Reason 2: Easier mechanical support and larger aperture — a mirror weighs much less than a lens of equal optical quality and can be supported over its entire back surface, so very large diameters are possible, giving higher resolving power and brighter images (more light gathered). (Also: a parabolic mirror removes spherical aberration.)

   1 mark

Answer: No chromatic aberration; mirror lighter and supported over whole back, allowing large aperture (high resolving power, brighter image); parabolic mirror removes spherical aberration.

1. Intensity $\propto$ (amplitude)${}^2$: $\frac{a^2}{b^2}=\frac{4I}{I}=4\Rightarrow \frac{a}{b}=2$

   ½ mark

2. $I_{max}\propto (a+b{)}^2$ and $I_{min}\propto (a-b{)}^2$

   ½ mark

3. $\frac{I_{max}}{I_{min}}=\frac{(a/b+1{)}^2}{(a/b-1{)}^2}=\frac{(2+1{)}^2}{(2-1{)}^2}$

   ½ mark

4. $=\frac{9}{1}$, i.e. 9 : 1

   ½ mark

Answer: $I_{max}:I_{min}=9:1$

1. Total energy $E_n=-\frac{13.6}{n^2}$ eV; third excited state means $n=4$.

   ½ mark

2. $E_4=-\frac{13.6}{16}=-0.85$ eV

   ½ mark

3. Potential energy $U=2E_4=2(-0.85)=-1.70$ eV

   ½ mark

4. Kinetic energy $K=-E_4=0.85$ eV

   ½ mark

Answer: PE = $-1.70$ eV, KE = $0.85$ eV

1. Intrinsic semiconductor	Extrinsic semiconductor
   Pure semiconductor (e.g. pure Si, Ge) with no added impurity.	Semiconductor doped with a suitable impurity (pentavalent or trivalent).

   1 mark

2. Intrinsic	Extrinsic
   Number of free electrons equals number of holes, $n_e=n_h$; conductivity is low at room temperature.	$n_e\ne n_h$ ($n_e\gg n_h$ in n-type, $n_h\gg n_e$ in p-type); conductivity is much higher at room temperature.

   1 mark

Answer: Intrinsic: pure, $n_e=n_h$, low conductivity; Extrinsic: doped, $n_e\ne n_h$, high conductivity.

1. Forward bias circuit (described): battery through a rheostat/potential divider; p-side of diode connected to positive terminal and n-side to negative; a milliammeter in series with the diode and a voltmeter across it.

   ½ mark

2. Reverse bias circuit (described): p-side connected to negative terminal and n-side to positive terminal through a potential divider; a microammeter in series and a voltmeter across the diode.

   ½ mark

3. V–I plot, forward region (described): in the first quadrant, current (mA) is almost zero until the threshold (knee) voltage ≈ 0.7 V for silicon, after which it rises steeply (exponentially) with voltage.

   ½ mark

4. V–I plot, reverse region (described): in the third quadrant, a very small, nearly constant reverse current (μA) flows with increasing reverse voltage, until at breakdown voltage the reverse current suddenly increases sharply.

   ½ mark

Answer: Forward bias: p to +, n to −, mA meter; reverse bias: p to −, n to +, μA meter. Si characteristic: knee ≈ 0.7 V in forward, tiny μA current then breakdown in reverse.

1. Due to the concentration gradient, holes diffuse from p to n side and electrons from n to p side, giving a diffusion current. They leave behind immobile ions (negative on p-side, positive on n-side) near the junction, forming a depletion (space charge) region.

   1 mark

2. This space charge sets up an electric field from n to p, which drives minority carriers across the junction, giving a drift current opposite to the diffusion current. As diffusion continues, the field grows until drift current equals diffusion current (equilibrium, no net current); the potential difference then existing across the junction is the potential barrier.

   1 mark

Answer: Diffusion creates the space charge region; its field causes opposing drift; equilibrium (diffusion = drift) fixes the barrier potential.

1. (a) Potential due to a point charge is $V=\frac{kq}{R}$; each of the 12 charges is $-q$ at distance R from C, and potential is a scalar, so $V=\frac{k(-12q)}{R}=-\frac{12q}{4\pi {\epsilon }_{0}R}=-\frac{3q}{\pi {\epsilon }_{0}R}$

   1 mark

2. (a) Field: the charges are symmetrically placed, so field of each charge is cancelled by that of the diametrically opposite charge (equal magnitude, opposite direction). Hence $E=0$ at C.

   1 mark

3. (b) Potential is a scalar and depends only on the distance of each charge from C, not on their arrangement. All 12 charges are still at distance R, so $V=-\frac{12q}{4\pi {\epsilon }_{0}R}$ (same as in (a)).

   1 mark

Answer: (a) $V=-\frac{12q}{4\pi {\epsilon }_{0}R}$, $E=0$; (b) $V=-\frac{12q}{4\pi {\epsilon }_{0}R}$

1. Difference: resistance is the opposition offered by a resistor to both AC and DC and is independent of frequency, whereas impedance is the combined opposition of R, L and C to AC only and depends on the frequency of the source.

   1 mark

2. With current $i=i_m\sin \omega t$ common to all elements: $V_R=i_{m}R$ (in phase with $i$), $V_L=i_m{X}_L$ (leads $i$ by $\pi /2$), $V_C=i_m{X}_C$ (lags $i$ by $\pi /2$). Phasor diagram (described): ${\vec{V}}_R$ along the current phasor, ${\vec{V}}_L$ and ${\vec{V}}_C$ perpendicular to it in opposite directions; their resultant $({\vec{V}}_C-{\vec{V}}_L)$ combines with ${\vec{V}}_R$ to give ${\vec{V}}_m={\vec{V}}_R+{\vec{V}}_L+{\vec{V}}_C$.

   1 mark

3. From the right triangle of phasors: $V_m^2=V_R^2+(V_C-V_L{)}^2=i_m^2[R^2+(X_C-X_L{)}^2]$, so $i_m=\frac{V_m}{\sqrt{R^2+(X_C-X_L{)}^2}}=\frac{V_m}{Z}$, where impedance $Z=\sqrt{R^2+(X_C-X_L{)}^2}$.

   1 mark

Answer: $Z=\sqrt{R^2+(X_C-X_L{)}^2}$

1. $i_m=\frac{V_m}{Z}$ with $Z=\sqrt{R^2+(X_L-X_C{)}^2}$. Current is maximum when Z is minimum, i.e. when $X_L=X_C$: $\omega L=\frac{1}{\omega C}\Rightarrow {\omega }_r=\frac{1}{\sqrt{LC}}$. This is the condition for resonance (then $Z=R$).

   1 mark

2. Resonant frequency ${\nu }_r=\frac{1}{2\pi \sqrt{LC}}$ depends only on the inductance L and capacitance C (not on R).

   1 mark

3. Graph (described): current amplitude $i_m$ on y-axis vs angular frequency $\omega$ on x-axis; current rises from a small value, reaches a sharp peak $V_m/R$ at $\omega ={\omega }_r$ and then decreases. For smaller R the peak is higher and sharper; for larger R it is lower and flatter.

   1 mark

Answer: Resonance when $X_L=X_C$, ${\omega }_r=1/\sqrt{LC}$; depends on L and C; current peaks at ${\omega }_r$.

1. (a) Field inside a long solenoid near the centre: $B={\mu }_{0}nI$

   ½ mark

2. Flux linked with the coil of N turns (cross-section area $\pi r^2$, since wound tightly): $\Phi =NBA=N\pi r^2{\mu }_{0}nI$

   ½ mark

3. Induced emf: $e=-\frac{d\Phi }{dt}=-{\mu }_0\pi r^{2}nN\frac{dI}{dt}$ ...(i)

   ½ mark

4. With $I=I_0\sin \omega t$, $\frac{dI}{dt}=I_0\omega \cos \omega t$, so $e=-{\mu }_0\pi r^{2}nN{I}_0\omega \cos \omega t$

   ½ mark

5. (b) Comparing (i) with $e=-M\frac{dI}{dt}$:

   ½ mark

6. $M={\mu }_0\pi r^{2}nN$

   ½ mark

Answer: (a) $e=-{\mu }_0\pi r^{2}nN{I}_0\omega \cos \omega t$; (b) $M={\mu }_0\pi r^{2}nN$

1. Einstein's equation: $K_{max}=h\nu -{\varphi }_0$. Light consists of photons of energy $h\nu$; an electron absorbs one whole photon. If $h\nu$ exceeds the work function ${\varphi }_0$ (minimum energy to escape), the electron is emitted and the excess energy appears as its maximum kinetic energy. If $h\nu <{\varphi }_0$ (i.e. $\nu <{\nu }_0$), no emission occurs however intense the light; emission is instantaneous.

   1 mark

2. (a) Graph (described): photocurrent (y-axis) vs collector plate potential (x-axis) for intensities $I_3>I_2>I_1$ at fixed frequency. Each curve rises with positive potential and becomes flat at a saturation current; higher intensity gives higher saturation current. On the negative side all curves fall to zero at the same stopping potential $-V_0$.

   1 mark

3. (b) Graph (described): photocurrent vs intensity of incident radiation (frequency fixed above threshold) is a straight line through the origin — photocurrent is directly proportional to intensity.

   1 mark

Answer: $K_{max}=h\nu -{\varphi }_0$; (a) saturation current rises with intensity, same stopping potential; (b) straight line through origin.

1. Energy level diagram (described): horizontal lines for $n=1,2,3,4,5,\dots ,\infty$ with energies $E_n=-13.6/n^2$ eV: $-13.6$ eV ($n=1$), $-3.4$ eV ($n=2$), $-1.51$ eV ($n=3$), $-0.85$ eV ($n=4$), $-0.54$ eV ($n=5$), 0 eV ($n=\infty$); levels crowd together as n increases.

   1½ marks

2. Ultraviolet region: Lyman series — downward arrows from $n=2,3,4,\dots$ to $n=1$.

   ½ mark

3. Visible region: Balmer series — downward arrows from $n=3,4,5,\dots$ to $n=2$.

   ½ mark

4. Infrared region: Paschen series — arrows from $n=4,5,\dots$ to $n=3$ (also Brackett to $n=4$ and Pfund to $n=5$ lie in infrared).

   ½ mark

Answer: UV: Lyman ($\to n=1$); Visible: Balmer ($\to n=2$); IR: Paschen ($\to n=3$), Brackett, Pfund.

1. Graph (described): binding energy per nucleon $E_{bn}$ (MeV) on y-axis vs mass number A on x-axis. It rises steeply for light nuclei (H-2, He-4 peak etc.), reaches a broad maximum about 8.75 MeV near A = 56 (Fe), and then decreases slowly to about 7.6 MeV at A = 238 (U).

   1 mark

2. Feature 1: $E_{bn}$ is nearly constant (about 8 MeV) for nuclei of middle mass numbers 30 < A < 170, showing saturation of nuclear forces.

   ½ mark

3. Feature 2: $E_{bn}$ is lower for both light nuclei (A < 30) and heavy nuclei (A > 170); maximum ≈ 8.75 MeV at A = 56.

   ½ mark

4. Lighter nuclei have low binding energy per nucleon. When two of them fuse, the product nucleus has a higher binding energy per nucleon, so the final system is more tightly bound (more stable) and energy is released. Hence light nuclei tend to undergo fusion.

   1 mark

Answer: Curve peaks ≈ 8.75 MeV at A = 56, nearly flat for 30 < A < 170, lower at both ends; fusion gives higher $E_{bn}$, hence more stable product with energy release.

1. (i) Coulomb's law: the electrostatic force between two stationary point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them; it acts along the line joining them. $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

   1 mark

2. Vector form: force on $q_1$ due to $q_2$: ${\vec{F}}_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}^2}{\hat{r}}_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}^3}{\vec{r}}_{12}$, where ${\vec{r}}_{12}$ is the vector from $q_2$ to $q_1$.

   1 mark

3. (ii) Flux through a sphere of radius r around a charge q is $\Phi =E\times 4\pi r^2=\frac{q}{4\pi {\epsilon }_0{r}^2}\times 4\pi r^2=\frac{q}{{\epsilon }_0}$. The $1/r^2$ of Coulomb's law exactly cancels the $r^2$ of the surface area, making flux independent of r and of the shape of the surface. If the force law were not inverse-square, this cancellation would fail and Gauss's law would not hold.

   1 mark

4. (iii) Position vector of B relative to A: ${\vec{r}}_{AB}=a\hat{i}+a\hat{j}$, $|{\vec{r}}_{AB}|=\sqrt{a^2+a^2}=\sqrt{2}a$; $\vec{F}=\frac{1}{4\pi {\epsilon }_0}\frac{q(2q)}{r^3}{\vec{r}}_{AB}$

   ½ mark

5. $\vec{F}=\frac{1}{4\pi {\epsilon }_0}\frac{2q^2(a\hat{i}+a\hat{j})}{(\sqrt{2}a{)}^3}=\frac{1}{4\pi {\epsilon }_0}\frac{2q^{2}a(\hat{i}+\hat{j})}{2\sqrt{2}{a}^3}$

   ½ mark

6. $\vec{F}=\frac{q^2}{4\sqrt{2}\pi {\epsilon }_0{a}^2}(\hat{i}+\hat{j})$ (magnitude $\frac{q^2}{4\pi {\epsilon }_0{a}^2}$, directed away from A along the line AB).

   1 mark

Answer: ${\vec{F}}_{BA}=\frac{q^2}{4\sqrt{2}\pi {\epsilon }_0{a}^2}(\hat{i}+\hat{j})$

1. (i) Dipole: $-q$ at A and $+q$ at B, AB = 2a, $\vec{p}$ from $-q$ to $+q$. Point P on the perpendicular bisector at distance r from the centre O; each charge is at distance $\sqrt{r^2+a^2}$ from P. Magnitudes: $E_{+q}=E_{-q}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2+a^2}$

   ½ mark

2. ${\vec{E}}_{+q}$ is directed away from +q and ${\vec{E}}_{-q}$ towards −q. Their components normal to the dipole axis cancel; components parallel to the axis add, both pointing opposite to $\vec{p}$. With $\cos \theta =\frac{a}{\sqrt{r^2+a^2}}$: $E=2E_{+q}\cos \theta =\frac{1}{4\pi {\epsilon }_0}\frac{2qa}{(r^2+a^2{)}^{3/2}}$

   1 mark

3. In vector form, with $\vec{p}=q(2a)\hat{p}$: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$

   ½ mark

4. (ii) For a far point ($r\gg a$): $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$, so $E\propto 1/r^3$.

   ½ mark

5. When r is halved: $E^{\prime }=\frac{p}{4\pi {\epsilon }_0(r/2{)}^3}=\frac{8p}{4\pi {\epsilon }_0{r}^3}=8E$. The field becomes 8 times (direction unchanged).

   ½ mark

6. (iii) Diagonal = $\sqrt{2}\times \sqrt{2}=2$ m, so each charge is 1 m from O. Pair A(+q), C(−q): field of A at O points away from A (towards C), field of C points towards C; each $kq/1^2$, so $E_1=2kq$ along OC.

   ½ mark

7. Pair B(−q), D(+q): field of B points towards B, field of D points away from D (towards B); so $E_2=2kq$ along OB.

   ½ mark

8. OB ⟂ OC, so $E_{net}=\sqrt{E_1^2+E_2^2}=2\sqrt{2}kq=\frac{2\sqrt{2}q}{4\pi {\epsilon }_0}=\frac{q}{\sqrt{2}\pi {\epsilon }_0}$ N/C

   ½ mark

9. Direction: along the bisector of OB and OC, i.e. parallel to DC (from D towards C, horizontally to the right).

   ½ mark

Answer: (i) $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$; (ii) becomes 8 times; (iii) $E=2\sqrt{2}kq=\frac{q}{\sqrt{2}\pi {\epsilon }_0}$, along DC

1. (i) Biot–Savart law: the magnetic field $d\vec{B}$ at a point due to a current element $Id\vec{l}$ is proportional to I, to the length dl and to $\sin \theta$ (θ between $d\vec{l}$ and $\vec{r}$), and inversely proportional to the square of the distance r: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3}$, $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$.

   1 mark

2. For a circular loop of radius a, every element $d\vec{l}$ is perpendicular to the radius joining it to the centre, so $\theta =90^{\circ }$, $\sin \theta =1$ and $r=a$: $dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{a^2}$, and all $d\vec{B}$ point in the same direction (perpendicular to the plane of the loop).

   ½ mark

3. $B=\int dB=\frac{{\mu }_{0}I}{4\pi a^2}{\int }_0^{2\pi a}dl=\frac{{\mu }_{0}I}{4\pi a^2}(2\pi a)$

   1 mark

4. $B=\frac{{\mu }_{0}I}{2a}$, directed along the axis of the loop (given by the right-hand thumb rule).

   ½ mark

5. Field lines (described): closed loops encircling the wire on each side; near the centre they become nearly straight and parallel to the axis, passing through the loop. For anticlockwise current (seen from the front) the field at centre points out of the loop towards the viewer; the loop behaves like a magnetic dipole with one face N and other S.

   ½ mark

6. (ii) Current due to revolving electron: $i=\frac{e}{T}=\frac{ev}{2\pi r}$

   ½ mark

7. $i=\frac{1.6\times 10^{-19}\times 10^7}{2\pi \times 10^{-10}}=\frac{0.8\times 10^{-2}}{\pi }$ A

   ½ mark

8. $i\approx 2.55\times 10^{-3}$ A $=2.55$ mA

   ½ mark

Answer: (i) $B=\frac{{\mu }_{0}I}{2a}$ along the axis; (ii) $i\approx 2.55$ mA

1. (i) Consider a straight rod of length l, cross-section A, with n mobile charge carriers (charge q) per unit volume; total carriers $=nlA$.

   ½ mark

2. Each carrier moving with drift velocity ${\vec{v}}_d$ in field $\vec{B}$ experiences $q({\vec{v}}_d\times \vec{B})$, so total force $\vec{F}=nlAq({\vec{v}}_d\times \vec{B})$

   ½ mark

3. Current density $\vec{j}=nq{\vec{v}}_d$, so $\vec{F}=(\vec{j}\times \vec{B})Al$

   ½ mark

4. Since $jA=I$ and taking $\vec{l}$ along the current: $\vec{F}=I(\vec{l}\times \vec{B})$, magnitude $F=IlB\sin \theta$

   ½ mark

5. Fleming's left-hand rule: stretch the forefinger, middle finger and thumb of the left hand mutually perpendicular; if the forefinger points along the magnetic field and the middle finger along the current, the thumb gives the direction of the force on the conductor.

   ½ mark

6. (1) Maximum force when $\theta =90^{\circ }$ (conductor perpendicular to $\vec{B}$): $F_{max}=IlB$.

   ½ mark

7. (2) Minimum (zero) force when $\theta =0^{\circ }$ or $180^{\circ }$ (conductor parallel or antiparallel to $\vec{B}$): $F=0$.

   ½ mark

8. (ii) Force on length l of wire B: $F=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi d}$

   ½ mark

9. $F=\frac{2\times 10^{-7}\times 5.0\times 2.5\times 0.10}{2.5\times 10^{-2}}$

   ½ mark

10. $F=1.0\times 10^{-5}$ N (repulsive, since currents are opposite).

    ½ mark

Answer: (i) $\vec{F}=I(\vec{l}\times \vec{B})$; max at θ = 90° ($IlB$), zero at θ = 0° or 180°; (ii) $F=10^{-5}$ N (repulsive)

1. (i)(1) Difference 1: An interference pattern has many equally spaced bright and dark fringes. A diffraction pattern has a central bright maximum twice as wide as the other maxima, and the secondary maxima get narrower and fainter.

   1 mark

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.

   (Also acceptable: interference comes from superposing two waves from two coherent narrow slits, while diffraction comes from superposing a continuous family of wavelets from every point of a single slit.)

   1 mark

3. (i)(2) Fringe width is $\beta =\frac{\lambda D}{d}$. So it depends on any two of:

   • the wavelength $\lambda$ of the light;
   • the distance $D$ of the screen from the slits;
   • the slit separation $d$.
   1 mark

4. (ii)(1) For the $n$th maximum, $d\sin \theta =n\lambda$.

   For the maximum next to the central one, $n=1$, so $\sin \theta =\frac{\lambda }{d}$.

   With $d=100\lambda$ and the small-angle approximation $\sin \theta \approx \theta$:

   $\theta =\frac{\lambda }{100\lambda }=\frac{1}{100}=0.01$ rad

   1 mark

5. (ii)(2) The distance between the two maxima on the screen is

   $\beta =D\theta =50\times 10^{-2}\text{m}\times \frac{1}{100}=5\times 10^{-3}\text{m}=5\text{mm}$

   1 mark

Answer: (ii)(1) 0.01 rad; (ii)(2) 5 mm

1. (i) Figure and assumptions. A convex spherical surface has pole M and centre of curvature C, with medium $n_1$ on the object side and $n_2$ on the other side. The object O lies on the principal axis. A ray ON strikes the surface at N, where NC is the normal. It refracts and meets the axis at the image I. The aperture is small, so all angles are small and $\tan \theta \approx \sin \theta \approx \theta$. The distance of N from the axis is taken as MN.

   1 mark

2. In $△NOC$, the angle of incidence $i$ is an exterior angle:

   $i=\angle NOM+\angle NCM\approx \frac{MN}{OM}+\frac{MN}{MC}$ ...(1)

   ½ mark

3. In $△NCI$, $\angle NCM$ is an exterior angle, so $\angle NCM=r+\angle NIM$. Hence

   $r=\angle NCM-\angle NIM\approx \frac{MN}{MC}-\frac{MN}{MI}$ ...(2)

   ½ mark

4. By Snell's law, $n_1\sin i=n_2\sin r$. For small angles this becomes $n_{1}i=n_{2}r$.

   Substituting (1) and (2) and cancelling MN:

   $\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}$

   ½ mark

5. Apply the Cartesian sign convention: $OM=-u$, $MI=+v$, $MC=+R$. This gives

   $\boxed{\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}}$

   ½ mark

6. (ii) Data. The bubble is seen through the nearer surface (the one facing E). Light goes from the bubble (glass) into air, so $n_1=1.5$ and $n_2=1$.

   The bubble is 3.0 cm inside that surface, and the centre lies on the same side as the bubble. Hence $u=-3$ cm and $R=-6$ cm.

   ½ mark

7. Substituting:

   $\frac{1}{v}-\frac{1.5}{-3}=\frac{1-1.5}{-6}$

   ½ mark

8. $\frac{1}{v}=\frac{0.5}{6}-\frac{1.5}{3}=\frac{0.5-3}{6}=-\frac{2.5}{6}$

   ½ mark

9. $v=-2.4$ cm.

   The bubble appears 2.4 cm from the surface, inside the sphere, on the same side as the bubble. It therefore looks closer to the surface than it really is.

   ½ mark

Answer: (i) $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$; (ii) v = −2.4 cm, i.e. the bubble appears 2.4 cm from the surface, inside the sphere

1. (a) Points joined only by connecting wires of negligible resistance are at the same potential. The groups are:

   • a, b, c (and i, the top end of $R_1$);
   • d and e;
   • f, g, h and j (the bottom end of $R_1$).
   1 mark

2. (b) Let $I_1$ be the current through the battery and $I_2$ the current through arm bg ($R_1$). Then $(I_1-I_2)$ flows through arm cf ($R_2$ and $R_3$ in series, 10 Ω).

   Loop abgha: $-6+10I_2+1\cdot I_1=0\Rightarrow I_1+10I_2=6$ ...(i)

   Loop acfha: $-6+10(I_1-I_2)+1\cdot I_1=0\Rightarrow 11I_1-10I_2=6$ ...(ii)

   ½ mark

3. Adding (i) and (ii): $12I_1=12$, so $I_1=1$ A.

   Then from (i), $I_2=\frac{6-1}{10}=0.5$ A.

   Current through arm bg = 0.5 A.

   ½ mark

4. (c) [first alternative] The current through $R_3$ is $I_1-I_2=1-0.5=0.5$ A.

   $V_{R_3}=(I_1-I_2)R_3=0.5\times 5.0$

   1 mark

5. $V_{R_3}=2.5$ V

   1 mark

6. (c) [second alternative] The current through $R_2$ is also 0.5 A.

   $P=(I_1-I_2{)}^2{R}_2=(0.5{)}^2\times 5.0=1.25$ W

   (This alternative carries the same 2 marks as the first; only one is answered in the exam.)

Answer: (a) a, b, c (with i); d, e; f, g, h, j; (b) 0.5 A; (c) 2.5 V across R₃ / 1.25 W in R₂

1. (a) The ray falls normally on face AB, so it enters the prism undeviated. Geometry of the equilateral prism gives an angle of incidence of $60^{\circ }$ on the next face it meets (BC). Since $60^{\circ }>i_c=24.5^{\circ }$, the ray is totally internally reflected there. The reflected ray then strikes the third face AC normally and emerges into air without deviation.

   Diagram: the ray goes straight through AB, is totally reflected at BC at $60^{\circ }$ to the normal, and leaves perpendicular to AC.

   1 mark

2. (b) $n=\frac{c}{v}\Rightarrow v=\frac{c}{n}=\frac{3\times 10^8}{2.41}\approx 1.24\times 10^8{\text{m s}}^{-1}$

   1 mark

3. (c) [first alternative] Application 1 – Optical fibres: Light enters one end of a thin glass or quartz fibre whose core is optically denser than its cladding. It strikes the core–cladding boundary at angles greater than the critical angle. It therefore undergoes repeated total internal reflections and travels along the fibre with very little loss. Fibres are used in telecommunication and endoscopy.

   1 mark

4. Application 2 – Brilliance of diamond: Diamond has a very small critical angle (about $24.4^{\circ }$). Light entering a suitably cut diamond suffers multiple total internal reflections at its faces and comes out only through a few faces. This makes the diamond sparkle.

   (Mirage or totally reflecting prisms are also acceptable.)

   1 mark

5. (c) [second alternative] Definition: When light travelling in an optically denser medium meets the boundary with a rarer medium at an angle of incidence greater than the critical angle, it is completely reflected back into the denser medium. This phenomenon is called total internal reflection.

6. Conditions:
   1. Light must travel from an optically denser medium towards an optically rarer medium.
   2. The angle of incidence at the interface must be greater than the critical angle for that pair of media.

   (This alternative carries the same 2 marks as the first: definition 1 mark, conditions ½ + ½.)

Answer: (a) Enters undeviated, TIR at the second face (i = 60° > 24.5°), emerges normally from the third face; (b) v ≈ 1.24 × 10⁸ m/s; (c) optical fibres and brilliance of diamond / definition of TIR with its two conditions