CBSE Class 12 Physics: 2026 board paper, set 55/5/1

1. Initially the force is $F=\frac{k(q)(2q)}{r^2}=\frac{2k{q}^2}{r^2}$, and it is attractive.

2. On contact the total charge $-q$ is shared equally, so each ball has $-q/2$. The new force is $F^{\prime }=\frac{k(q/2{)}^2}{(r/2{)}^2}=\frac{k{q}^2}{r^2}=\frac{F}{2}$. Both charges are now negative, so the balls repel.

   1 mark

Answer: Repel with force F/2, option (B)

1. The field is $E=-\frac{dV}{dx}$, so the force is largest where the V–x graph is steepest.

2. The graph is steepest at S, so the force is maximum there.

   1 mark

Answer: S, option (D)

1. Each wire is at distance $a/\sqrt{2}$ from O, so each field has magnitude $\frac{{\mu }_{0}I}{2\pi (a/\sqrt{2})}=\frac{{\mu }_{0}I}{\sqrt{2}\pi a}$.

2. B and D are opposite corners with currents in the same direction, so their fields cancel. A and C are opposite corners with opposite currents, so their fields add. By the right-hand rule both point along OB.

3. The net field is $B=2\cdot \frac{{\mu }_{0}I}{\sqrt{2}\pi a}=\frac{\sqrt{2}{\mu }_{0}I}{\pi a}$, directed along OB.

   1 mark

Answer: $\frac{\sqrt{2}{\mu }_{0}I}{\pi a}$ along OB, option (C)

1. Flux is $\Phi =BA\cos \theta$. Changing B, A or θ, or any combination of them, changes the flux.

   1 mark

Answer: Option (D)

1. The peak value of ac is $\sqrt{2}$ times its rms value, so ac is more dangerous than dc, not less. Statement (D) is therefore false.

   1 mark

Answer: Option (D)

1. Speed in glass: $v=c/n=3\times 10^8/1.5=2\times 10^8$ m/s.

2. Wave number: $\alpha =\omega /v=1.5\times 10^{11}/(2\times 10^8)=7.5\times 10^2{\text{m}}^{-1}$.

   1 mark

Answer: $7.5\times 10^2$ m⁻¹, option (C)

1. Photon energy is $E=hc/\lambda$. Blue light has the shortest wavelength of the four colours, so its photons carry the most energy.

   1 mark

Answer: Blue light, option (D)

1. Nuclides with the same neutron number N are called isotones. Isotopes share Z and isobars share A.

   1 mark

Answer: Isotones, option (B)

1. $R=R_0{A}^{1/3}=1.2\times (125{)}^{1/3}=1.2\times 5=6.0$ fm.

   1 mark

Answer: 6.0 fm, option (A)

1. From $E_n=-13.6/n^2$ eV, an energy of $-3.4$ eV gives $n=2$.

2. Angular momentum: $L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }$.

   1 mark

Answer: $h/\pi$, option (C)

1. A diode conducts in forward bias, so it shows low resistance, and blocks current in reverse bias, so it shows high resistance.

   1 mark

Answer: Option (A)

1. The rms value is $V_0/\sqrt{2}$. Over a full cycle the positive and negative halves cancel, so the average value is 0.

   1 mark

Answer: $V_0/\sqrt{2}$, 0, option (C)

1. By Faraday's law, $e=-\frac{d\Phi }{dt}$. The induced emf depends on the rate of change of flux, not on the flux itself.

2. A large flux that is not changing produces no emf, so both A and R are false.

   1 mark

Answer: Option (D)

1. Bright and dark fringes are both equally spaced with $\beta =\lambda D/d$. This formula is the reason the two widths are the same, so R correctly explains A.

   1 mark

Answer: Option (A)

1. Fission of heavy nuclei and fusion of light nuclei both increase the binding energy per nucleon, so energy is released. A is true.

2. R has the trend reversed: binding energy per nucleon decreases with mass number for heavy nuclei and increases for light nuclei. So R is false.

   1 mark

Answer: Option (C)

1. Photoemission begins immediately (within about $10^{-9}$ s), so A is taken as true.

2. R correctly states the wave-theory prediction of a long delay. That prediction contradicts what is observed, so it does not explain A. Hence both are true but R is not the correct explanation.

   1 mark

Answer: Option (B)

1. (i) The resistance is fixed by the rating: $R=\frac{V^2}{P}=\frac{(220{)}^2}{2200}=22\Omega$

   1 mark

2. (ii) At 110 V the resistance is still 22 Ω, so $P^{\prime }=\frac{V^{\prime 2}}{R}=\frac{(110{)}^2}{22}=550$ W.

   Heat in $t=10$ min $=600$ s: $H=P^{\prime }t=550\times 600=3.3\times 10^5$ J

   1 mark

Answer: (i) 22 Ω; (ii) 3.3 × 10^5 J

1. Resistance of the wire: $R=\frac{V}{I}=\frac{2}{4.0}=0.5\Omega$

   1 mark

2. From $R=\frac{\rho l}{A}$: $\rho =\frac{RA}{l}=\frac{0.5\times 1.0\times 10^{-6}}{1}=5\times 10^{-7}\Omega \text{m}$

   1 mark

Answer: ρ = 5 × 10^-7 Ω m

1. At $t=0$ the plane of the coil is parallel to $\vec{B}$, so its normal is perpendicular to $\vec{B}$ and the flux is zero. After time t the normal has turned through $\omega t$ from this position, so the angle between normal and $\vec{B}$ is $(90^{\circ }-\omega t)$:
   $\varphi =NBA\cos (90^{\circ }-\omega t)=NBA\sin \omega t$

2. (a) Plot of $\varphi$ vs $\omega t$: a sine curve starting at zero, rising to maximum $+NBA$ at $\omega t=\pi /2$, zero at $\pi$, minimum $-NBA$ at $3\pi /2$, zero at $2\pi$.

   1 mark

3. (b) $\epsilon =-\frac{d\varphi }{dt}=-NBA\omega \cos \omega t$.
   Plot of $\epsilon$ vs $\omega t$: a (negative) cosine curve: $-NBA\omega$ at $\omega t=0$, zero at $\pi /2$, $+NBA\omega$ at $\pi$, zero at $3\pi /2$, $-NBA\omega$ at $2\pi$. The emf is maximum in magnitude when the flux is zero, and zero when the flux is maximum (90° phase difference).

   1 mark

Answer: φ = NBA sin ωt (sine curve from zero); ε = −NBAω cos ωt (cosine curve, maximum magnitude at ωt = 0)

1. Refractive index of the liquid: $n=\frac{\text{real depth}}{\text{apparent depth}}=\frac{12.5}{9.0}\approx 1.39$

   1 mark

2. $v=\frac{c}{n}=\frac{3\times 10^8\times 9.0}{12.5}=2.16\times 10^8{\text{m s}}^{-1}$

   1 mark

Answer: v ≈ 2.16 × 10^8 m/s

1. Let an object O be at distance u from the combination (lenses thin and in contact, so distances are measured from a common optical centre). Lens 1 alone forms image $I_1$ at $v_1$:
   $\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$ ...(1)

   ½ mark

2. $I_1$ acts as a virtual object for lens 2, which forms the final image I at v:
   $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$ ...(2)

   ½ mark

3. Adding (1) and (2): $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$

   ½ mark

4. If the combination is replaced by a single lens of focal length f giving the same image, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Hence $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1{f}_2}$, i.e. $f=\frac{f_1{f}_2}{f_1+f_2}$.
   (Ray diagram: object O, image $I_1$ by first lens beyond the lenses, final image I by second lens.)

   ½ mark

Answer: 1/f = 1/f₁ + 1/f₂ ⇒ f = f₁f₂/(f₁ + f₂)

1. Arsenic is pentavalent, so each As atom donates one electron. Since $N_D=5\times 10^{22}\gg n_i$, majority carriers (electrons): $n_e\approx N_D=5\times 10^{22}{\text{m}}^{-3}$

   1 mark

2. Using $n_e{n}_h=n_i^2$: $n_h=\frac{(1.5\times 10^{16}{)}^2}{5\times 10^{22}}=\frac{2.25\times 10^{32}}{5\times 10^{22}}=4.5\times 10^9{\text{m}}^{-3}$ (minority carriers: holes)

   1 mark

Answer: Electrons (majority) ≈ 5 × 10^22 m^-3; holes (minority) = 4.5 × 10^9 m^-3

1. Same A and d, so $C_X=\frac{{\epsilon }_{0}A}{d}=C$ and $C_Y=\frac{K{\epsilon }_{0}A}{d}=4C$.

   ½ mark

2. Series: $C_{eq}=\frac{C\cdot 4C}{C+4C}=\frac{4C}{5}=4\mu \text{F}\Rightarrow C=5\mu \text{F}$.
   So $C_X=5\mu$F, $C_Y=20\mu$F.

   1 mark

3. (b) Charge on each (series): $Q=C_{eq}V=4\mu \text{F}\times 6\text{V}=24\mu$C

   ½ mark

4. $V_X=\frac{Q}{C_X}=\frac{24}{5}=4.8$ V; $V_Y=\frac{Q}{C_Y}=\frac{24}{20}=1.2$ V (check: 4.8 + 1.2 = 6 V)

   1 mark

Answer: (a) C_X = 5 μF, C_Y = 20 μF; (b) V_X = 4.8 V, V_Y = 1.2 V

1. Biot–Savart law: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3}$

   1 mark

2. Here $d\vec{l}=0.01\hat{i}$ m, $\vec{r}=(\hat{i}+\hat{j})$ m, $r=\sqrt{2}$ m, $r^3=2\sqrt{2}$ m³.
   $d\vec{l}\times \vec{r}=0.01(\hat{i}\times \hat{i}+\hat{i}\times \hat{j})=0.01\hat{k}$

   1 mark

3. $\vec{B}=10^{-7}\times \frac{10\times 0.01}{2\sqrt{2}}\hat{k}=\frac{10^{-8}}{2.83}\hat{k}\approx 3.5\times 10^{-9}\hat{k}$ T

   1 mark

Answer: B ≈ 3.5 × 10^-9 T along +z (k̂)

1. Let current $I_1$ flow in the solenoid. Field inside the long solenoid (uniform, along axis): $B_1=\frac{{\mu }_0{N}_1{I}_1}{L}$; outside it is negligible.

   1 mark

2. The coil (radius $r_2>r_1$) links flux only over the solenoid's cross-section $\pi r_1^2$:
   Total flux linkage $N_2{\varphi }_2=N_2{B}_1\pi r_1^2=\frac{{\mu }_0{N}_1{N}_2\pi r_1^2}{L}{I}_1$

   1 mark

3. Since $N_2{\varphi }_2=M_{21}{I}_1$: $M=M_{21}=\frac{{\mu }_0{N}_1{N}_2\pi r_1^2}{L}$

   ½ mark

4. Yes, $M_{12}=M_{21}$ is valid here (mutual inductance is reciprocal for any pair of coils). However, $M_{12}$ cannot be easily calculated directly, since the field of the short coil is non-uniform over the solenoid; so we compute $M_{21}$ and use the equality.

   ½ mark

Answer: M = μ₀N₁N₂πr₁²/L; yes, M₁₂ = M₂₁ holds

1. Displacement current is the current that arises due to a time-varying electric field (changing electric flux) in a region, e.g. in the gap between capacitor plates, where no charges flow; it produces a magnetic field just like conduction current.

   1 mark

2. While a parallel plate capacitor (plate area A) is charging, let the charge be q at time t. Field between plates: $E=\frac{q}{{\epsilon }_{0}A}$, so electric flux ${\varphi }_E=EA=\frac{q}{{\epsilon }_0}$.

   ½ mark

3. $\frac{d{\varphi }_E}{dt}=\frac{1}{{\epsilon }_0}\frac{dq}{dt}\Rightarrow {\epsilon }_0\frac{d{\varphi }_E}{dt}=\frac{dq}{dt}=i$. This current in the gap (equal to the conduction current in the wires, maintaining continuity) is $i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$.

   1 mark

4. For a conductor with constant voltage, the electric field (and flux) is constant, so $\frac{d{\varphi }_E}{dt}=0$ and $i_d=0$.

   ½ mark

Answer: i_d = ε₀ dφ_E/dt; i_d = 0 for constant voltage

1. (i) Features of nuclear forces (any two):
   • They are very strong — much stronger than Coulomb force between protons.
   • They are short-range — negligible beyond a few femtometres (attractive for separation > ~0.8 fm, strongly repulsive below it).
   • They are charge independent — n–n, p–p and n–p forces are about the same.

   1 mark

2. (ii) Although the total numbers of protons and neutrons are unchanged, the binding energies of the nuclei before and after the reaction are different. The mass of a nucleus is less than the sum of the masses of its nucleons by the mass defect $\Delta m=B/c^2$.

   1 mark

3. Hence the total rest mass of the reactant nuclei differs from that of the product nuclei. If the products are more tightly bound, the difference in mass $\Delta m$ appears as energy $Q=\Delta m{c}^2$ (released); if less tightly bound, energy is absorbed and converted into mass. Thus mass–energy conversion takes place through the difference in binding energies.

   1 mark

Answer: (i) Strong, short-range, charge-independent (any two); (ii) Difference in binding energy (mass defect) of initial and final nuclei appears as energy Q = Δm c²

1. (i) Graph: number of scattered α-particles N(θ) (log scale) on y-axis vs scattering angle θ (0° to 180°) on x-axis — very large at small angles and falling steeply as θ increases ($N\propto 1/{\sin }^4(\theta /2)$), with only a very few particles at angles close to 180°.

   1 mark

2. Conclusions:
   • Most α-particles pass undeflected, so most of the atom is empty space.
   • A very few are deflected through large angles (some > 90°), so the positive charge and nearly all the mass of the atom are concentrated in a tiny central nucleus.

   1 mark

3. (ii) Angular momentum of a planet is enormous: for the Earth, $L=mvr\approx 6\times 10^{24}\times 3\times 10^4\times 1.5\times 10^{11}\approx 2.7\times 10^{40}$ J s, so $n=\frac{2\pi L}{h}\sim 10^{74}$. For such huge n, the difference between successive allowed angular momenta and energies is negligibly small compared to the values themselves, so the allowed orbits appear continuous; quantization is unobservable and is not spoken of.

   1 mark

Answer: (i) N(θ) falls sharply with θ; atom mostly empty, positive charge and mass concentrated in tiny nucleus; (ii) n ~ 10^74 for planets, so quantized levels are practically continuous

1. (a) $E=h\nu =6.63\times 10^{-34}\times 6.4\times 10^{14}=4.24\times 10^{-19}$ J $=\frac{4.24\times 10^{-19}}{1.6\times 10^{-19}}\approx 2.65$ eV

   1 mark

2. (b) Einstein's equation: $K_{max}=h\nu -{\varphi }_0=2.65-1.96=0.69$ eV $(\approx 1.1\times 10^{-19}$ J)

   1 mark

3. (c) $e{V}_0=K_{max}\Rightarrow V_0=\frac{0.69\text{eV}}{e}=0.69$ V

   1 mark

Answer: (a) 4.24 × 10^-19 J ≈ 2.65 eV; (b) 0.69 eV; (c) 0.69 V

1. Circuit: AC input applied to the primary of a centre-tap transformer. The two ends A and B of the secondary are connected to the p-sides of diodes D₁ and D₂; the n-sides of both diodes are joined together and connected through the load resistance $R_L$ to the centre tap C. Output is taken across $R_L$.

   1 mark

2. Working: During the positive half cycle of input, end A is positive and B negative with respect to C; D₁ is forward biased and conducts, D₂ is reverse biased. Current flows through $R_L$ (from the diode junction towards C).
   During the negative half cycle, B is positive and A negative; D₂ conducts and D₁ is cut off. Current again flows through $R_L$ in the same direction. Thus output is obtained in both half cycles — full-wave rectification.

   1 mark

3. Waveforms: Input — a sinusoidal AC voltage with positive and negative half cycles. Output across $R_L$ — a series of positive half-sine pulses, one for every half cycle (pulsating unidirectional DC), the pulses due to D₁ and D₂ alternating.

   1 mark

Answer: Centre-tap full-wave rectifier: D₁ conducts in positive half cycle, D₂ in negative half cycle; output across R_L is unidirectional for both halves (pulsating DC).

1. (i) $E=\frac{V}{d}=\frac{200}{0.01}=2\times 10^4$ V/m, directed from the positive (upper) plate to the negative (lower) plate, i.e. along $-\hat{k}$. $\vec{E}=-(2\times 10^4\text{V/m})\hat{k}$ — option (D).

   1 mark

2. (ii) Force on electron is opposite to $\vec{E}$, i.e. along $+\hat{k}$. $a=\frac{eE}{m}=\frac{1.6\times 10^{-19}\times 2\times 10^4}{9.1\times 10^{-31}}\approx 3.5\times 10^{15}$ m s⁻². $\vec{a}=(3.5\times 10^{15}{\text{m s}}^{-2})\hat{k}$ — option (B).

   1 mark

3. (iii)(a) Horizontal velocity is unchanged: $t=\frac{l}{v}=\frac{0.05}{3\times 10^7}=1.67\times 10^{-9}$ s — option (C).

   1 mark

4. (iii)(b) [Alternative to (iii)(a), 1 mark] $y=\frac{1}{2}a{t}^2=\frac{1}{2}\times 3.5\times 10^{15}\times (1.67\times 10^{-9}{)}^2\approx 4.9\times 10^{-3}$ m $=4.9$ mm — option (B). (Just less than 5 mm, so the electron just leaves the plates.)

5. (iv) Inside the plates the electron has uniform velocity horizontally and uniform acceleration towards the positive (upper) plate, so it follows a parabola curving upward — path b, option (B).

   1 mark

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

1. (i) Interference (superposition of waves giving bright and dark fringes) shows the wave nature of light; it does not show transverse nature (polarisation does). Option (A).

   1 mark

2. (ii)(a) From the figure, successive bright fringes are 1.2 mm apart, so $\beta =1.2$ mm. $\lambda =\frac{\beta d}{D}=\frac{1.2\times 10^{-3}\times 2\times 10^{-3}}{5.0}=4.8\times 10^{-7}$ m $=480$ nm — option (C).

   1 mark

3. (ii)(b) [Alternative to (ii)(a), 1 mark] Fringe width read from the pattern (distance between consecutive maxima), consistent with $\beta =\lambda D/d$: $\beta =1.2$ mm — option (A).

4. (iii) P is the second minimum, at $y=1.5\beta =1.8$ mm. Path difference $=\frac{yd}{D}=\frac{1.8\times 10^{-3}\times 2\times 10^{-3}}{5}=7.2\times 10^{-7}$ m $(=\frac{3}{2}\lambda )$ — option (B).

   1 mark

5. (iv) In a liquid of refractive index μ, wavelength becomes $\lambda /\mu$, so ${\beta }^{\prime }=\frac{\lambda D}{\mu d}<\beta$: fringes become narrower — the pattern is compressed. Option (D).

   1 mark

Answer: (i) A; (ii)(a) C (480 nm); (ii)(b) A (1.2 mm); (iii) B (7.2 × 10^-7 m); (iv) D

1. (i) Bridge: resistances P (A–B), Q (B–C), R (A–D), S (D–C); battery across A and C; galvanometer G (resistance $R_g$) between B and D. Let $I_1$ flow through P, $I_2$ through R, $I_g$ through G (B to D).

   ½ mark

2. Kirchhoff's loop rule for loop ABDA: $I_{1}P+I_g{R}_g-I_{2}R=0$; for loop BCDB: $(I_1-I_g)Q-(I_2+I_g)S-I_g{R}_g=0$.

   1 mark

3. For balance $I_g=0$: $I_{1}P=I_{2}R$ and $I_{1}Q=I_{2}S$. Dividing: $\frac{P}{Q}=\frac{R}{S}$ — the balance condition.

   ½ mark

4. (ii) Here $\frac{20}{2}=10\ne \frac{12}{1}=12$, so the bridge is not balanced and current flows through 3 Ω. Let $I_1$ flow in 20 Ω, $I_2$ in 12 Ω, $I_3$ in 3 Ω (top to bottom). Then 2 Ω carries $(I_1-I_3)$ and 1 Ω carries $(I_2+I_3)$.

   ½ mark

5. Loop (left–top–bottom–left): $20I_1+3I_3-12I_2=0$ ...(1)
   Loop (top–right–bottom–top): $2(I_1-I_3)-1(I_2+I_3)-3I_3=0\Rightarrow 2I_1-I_2-6I_3=0$ ...(2)
   Loop with battery (left–top–right): $20I_1+2(I_1-I_3)=6\Rightarrow 22I_1-2I_3=6$ ...(3)

   1½ marks

6. From (2): $I_2=2I_1-6I_3$. Put in (1): $20I_1+3I_3-24I_1+72I_3=0\Rightarrow I_1=\frac{75}{4}{I}_3$.
   Put in (3): $22\cdot \frac{75}{4}{I}_3-2I_3=6\Rightarrow 821I_3=12\Rightarrow I_3=\frac{12}{821}\approx 0.0146$ A $\approx 14.6$ mA (from the top junction to the bottom junction).

   1 mark

Answer: (i) P/Q = R/S; (ii) I(3 Ω) = 12/821 A ≈ 0.0146 A (≈ 14.6 mA)

1. (i) In time Δt, electrons within distance $v_d\Delta t$ cross a section: charge $=neA{v}_d\Delta t$, so $I=neA{v}_d$.

   1 mark

2. Substituting $v_d=\frac{eE\tau }{m}$ and $E=\frac{V}{l}$: $I=neA\cdot \frac{eV\tau }{ml}=\frac{n{e}^2\tau A}{ml}V$

   1 mark

3. So $R=\frac{V}{I}=\frac{m}{n{e}^2\tau }\cdot \frac{l}{A}$. Comparing with $R=\rho \frac{l}{A}$: $\rho =\frac{m}{n{e}^2\tau }$, hence $\sigma =\frac{1}{\rho }=\frac{n{e}^2\tau }{m}$.

   1 mark

4. (ii) $R_T=R_0[1+\alpha (T-T_0)]$ with $T_0=20^{\circ }$C: $\alpha =\frac{R_{100}-R_{20}}{R_{20}(100-20)}$

   1 mark

5. $\alpha =\frac{1.38-1.05}{1.05\times 80}=\frac{0.33}{84}\approx 3.9\times 10^{-3}{}^{\circ }{\text{C}}^{-1}$

   1 mark

Answer: (i) σ = ne²τ/m; (ii) α ≈ 3.9 × 10^-3 °C^-1

1. (i) Diagram: rectangular loop PQRS with sides PQ = RS = b (perpendicular to $\vec{B}$, along the rotation axis direction) and QR = SP = a, in uniform $\vec{B}$; the normal $\hat{n}$ (area vector) makes angle θ with $\vec{B}$.
   Forces on sides QR and SP (length a) are equal, opposite and collinear (along the axis), so they cancel and give no torque.

   1 mark

2. Forces on PQ and RS: each of magnitude $F=IbB$ (since these sides are ⟂ $\vec{B}$), equal and opposite but along different lines — they form a couple. Perpendicular distance between their lines of action $=a\sin \theta$.

   1 mark

3. Torque: $\tau =Fa\sin \theta =IbBa\sin \theta =IAB\sin \theta =mB\sin \theta$ (with $A=ab$, $m=IA$). The torque tends to align $\vec{m}$ with $\vec{B}$, and its direction is that of $\vec{m}\times \vec{B}$, so $\vec{\tau }=\vec{m}\times \vec{B}$.

   1 mark

4. (ii)(I) $A=\pi r^2=\pi \times \frac{100}{\pi }\times 10^{-4}=1.0\times 10^{-2}$ m².
   $m=NIA=100\times 5.0\times 1.0\times 10^{-2}=5.0$ A m²

   1 mark

5. (II) $\tau =mB\sin \theta =5.0\times 2.0\times \sin 30^{\circ }=5.0$ N m. A counter torque of 5.0 N m must be applied.

   1 mark

Answer: (i) τ = IAB sin θ ⇒ τ⃗ = m⃗ × B⃗; (ii) (I) 5.0 A m²; (II) 5.0 N m

1. (i) Let the conductor have $n$ free electrons per unit volume, each of charge $-e$, moving with drift velocity ${\vec{v}}_d$. The Lorentz force on one electron is $\vec{f}=-e({\vec{v}}_d\times \vec{B})$.

   1 mark

2. The number of free electrons in the conductor is $nAL$. The total force is $\vec{F}=-nALe({\vec{v}}_d\times \vec{B})$.

   1 mark

3. The current density is $\vec{j}=-ne{\vec{v}}_d$, so $\vec{F}=(\vec{j}AL)\times \vec{B}$. Since $jA=I$, define $\vec{L}$ as a vector of length $L$ along the current. Then $jAL$ in the direction of current $=I\vec{L}$, giving
   $$\vec{F}=I(\vec{L}\times \vec{B}),|\vec{F}|=ILB\sin \theta$$
   where $\theta$ is the angle between $\vec{L}$ and $\vec{B}$.

   1 mark

4. (ii) The magnetic field is uniform. The net force on the bent wire therefore equals the force on a straight wire joining its entry and exit points in the field. The horizontal parts together span the field width $0.50\text{m}$ along $+\hat{i}$. The vertical part is $0.20\text{m}$ along $-\hat{j}$. So:
   $${\vec{L}}_{\text{eff}}=(0.50\hat{i}-0.20\hat{j})\text{m}$$

   1 mark

5. Use $\vec{F}=I{\vec{L}}_{\text{eff}}\times \vec{B}$:
   $$\vec{F}=2.0[(0.50\hat{i}-0.20\hat{j})\times (-0.50\hat{k})]$$
   With $\hat{i}\times \hat{k}=-\hat{j}$ and $\hat{j}\times \hat{k}=\hat{i}$:
   $$\vec{F}=2.0(0.25\hat{j}+0.10\hat{i})=(0.20\hat{i}+0.50\hat{j})\text{N}$$
   The magnitude is:
   $$|\vec{F}|=\sqrt{0.20^2+0.50^2}=\sqrt{0.29}\approx 0.54\text{N}$$
   (Equivalently, horizontal parts: $F=2\times 0.5\times 0.5=0.5$ N along $+\hat{j}$; vertical part: $F=2\times 0.2\times 0.5=0.2$ N along $+\hat{i}$.)

   1 mark

Answer: (i) $\vec{F}=I(\vec{L}\times \vec{B})$; (ii) $\vec{F}=(0.20\hat{i}+0.50\hat{j})$ N, $|\vec{F}|\approx 0.54$ N

1. (i)(I) Central maximum. At $\theta =0$, all secondary wavelets from the slit travel equal paths to the centre of the screen. They arrive in phase and add up, giving the bright central maximum.

   ½ mark

2. Minima. At angle $\theta$, the path difference between waves from the two edges of the slit is $a\sin \theta$. Suppose $a\sin \theta =\lambda$. Divide the slit into two equal halves. Each point in the upper half has a partner in the lower half with path difference $\lambda /2$, so every pair cancels and the intensity is zero. In general, minima occur at $a\sin \theta =n\lambda$ ($n=\pm 1,\pm 2,\dots$), i.e. at positions $y_n=n\lambda D/a$.

   1 mark

3. Secondary maxima. Suppose $a\sin \theta =\frac{3}{2}\lambda$. Divide the slit into three equal parts. The wavelets from two of the parts cancel, and the third part contributes, giving a bright fringe. In general, secondary maxima occur at $a\sin \theta =(n+\frac{1}{2})\lambda$.

   ½ mark

4. (i)(II) For the $n$th secondary maximum, the slit is divided into $(2n+1)$ equal parts. All but one part cancel in pairs, so only $\frac{1}{2n+1}$ of the slit contributes. As $n$ increases, this fraction shrinks, so successive maxima become weaker and weaker.

   1 mark

5. (ii) Differences (any two):
   1. Interference results from superposition of waves from two coherent sources (two slits). Diffraction results from superposition of secondary wavelets from different parts of the same wavefront.
   2. In interference, all bright fringes are of (nearly) equal intensity. In diffraction, the intensity of successive maxima falls rapidly.
   3. Interference fringes are of equal width. In diffraction, the central maximum is twice as wide as the other maxima.
   4. In double-slit interference, a maximum occurs at $\theta =\lambda /d$. In single-slit diffraction, a minimum occurs at $\theta =\lambda /a$.

   2 marks

Answer: Minima at $a\sin \theta =n\lambda$; secondary maxima at $a\sin \theta =(n+\frac{1}{2})\lambda$; the $n$th maximum gets only a $1/(2n+1)$ contribution of the slit, so it is weaker; differences: source of superposition, intensity distribution, fringe width

1. (i) Ray diagram (described). Two coaxial convex lenses are drawn: a small objective O and a larger eyepiece E.

   • Object AB is placed just beyond the focus $F_o$ of the objective.
   • Rays from B through the objective form a real, inverted, magnified image A′B′. This image lies between E and its focus $F_e$.
   • Rays from B′ through the eyepiece diverge.
   • When produced backwards, these rays meet at B″, forming the final virtual, inverted, highly magnified image A″B″ at the near point (distance D).
   1 mark

2. Construction. The microscope has two convex lenses mounted coaxially in a tube:

   • The objective has a small focal length $f_o$ and a small aperture.
   • The eyepiece has a somewhat larger focal length $f_e$ and a larger aperture.
   • The separation between them can be adjusted.
   1 mark

3. Working.

   • The objective forms a real, inverted, magnified image of the object, with $m_o=\frac{v_o}{u_o}$.
   • The eyepiece then acts as a simple magnifier, giving $m_e=1+\frac{D}{f_e}$ when the final image is at the near point.
   • The total magnification is $m=m_o{m}_e=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)$.
   • For the final image at infinity, $m\approx \frac{L}{f_o}\cdot \frac{D}{f_e}$, where $L$ is the tube length.
   1 mark

4. (ii)(I) Yes, the image is still there. A real image is the point where the refracted rays actually meet. These rays keep travelling and diverge beyond that point. An eye placed beyond the image position, in the path of these diverging rays, sees the image in space. The screen only scatters the light so that the image is visible from all directions.

   1 mark

5. (ii)(II) Yes, provided the light falling on the mirror is converging, i.e. the object is virtual (behind the mirror).

   • Plane mirror: rays converging towards a point behind it reflect and converge to a point in front of it, forming a real image.
   • Convex mirror: a real image forms if the virtual object lies between its pole and focus.
   1 mark

Answer: (i) Objective forms a real, magnified image within the focus of the eyepiece; eyepiece forms the final virtual magnified image; $m=\frac{v_o}{|u_o|}(1+\frac{D}{f_e})$. (ii)(I) Yes, the rays still converge there and can be seen by an eye beyond it. (II) Yes, when a converging beam (virtual object) falls on them.