⚛️ Physics (March 2026) — Answer Key
📌 Based on official question paper (pages 2–15)
✍️ solutions & derivations
Q1 (page 2)
Capacitance with dielectric (parallel plate) ?
Answer: increases
Q2
Which is not ferromagnetic?
(a) Co (b) Fe (c) Ni (d) Bi
(a) Co (b) Fe (c) Ni (d) Bi
Answer: (d) Bismuth
Q3
EM radiation used to destroy cancer cells?
Answer: (d) Gamma rays
Q4
Turpentine & water – correct ray path?
Answer: (a) 1
Q5
Perpendicular distance of α‑particle velocity from nucleus centre →
Answer: impact parameter
Q6 (page 4)
Surface area of spherical nucleus varies as:
Answer: (a) \(A^{2/3}\)
Q7
Pentavalent impurity → p‑type? (True/False)
Answer: False (becomes n‑type)
Q8 (2 scores)
Torque on rectangular current loop in uniform B‑field.
Expression: \(\vec{\tau} = \vec{m} \times \vec{B}\) with \(m = N I A\)
\(\tau = N I A B \sin\theta\) (magnitude).
Derivation: forces on arms AB, CD form couple; \(\tau = I b B \cdot a\sin\theta = I A B\sin\theta\).
\(\tau = N I A B \sin\theta\) (magnitude).
Derivation: forces on arms AB, CD form couple; \(\tau = I b B \cdot a\sin\theta = I A B\sin\theta\).
Q9
(a) Lenz’s law, (b) polarity of capacitor (magnet moving away)
(a) Induced emf opposes change in flux.
(b) Plate A positive, plate B negative. (end facing magnet becomes south pole)
(b) Plate A positive, plate B negative. (end facing magnet becomes south pole)
Q11
(a) Define displacement current. (b) Maxwell’s modified Ampere’s law.
(a) \(I_d = \epsilon_0 \frac{d\phi_E}{dt}\) (current due to changing electric field).
(b) \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0\epsilon_0\frac{d\phi_E}{dt}\)
(b) \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0\epsilon_0\frac{d\phi_E}{dt}\)
Q12
Coherent sources & path difference for interference.
(a) Constant phase difference, same frequency & wavelength.
(b) (i) constructive: \(n\lambda\) (ii) destructive: \((2n-1)\frac{\lambda}{2}\)
(b) (i) constructive: \(n\lambda\) (ii) destructive: \((2n-1)\frac{\lambda}{2}\)
Q13
Bohr’s second postulate using de Broglie equation.
de Broglie: \(\lambda = h/(mv)\). Stable orbit \(\Rightarrow 2\pi r = n\lambda = n h/(mv)\)
\(\Rightarrow mvr = n h/(2\pi)\) → Bohr’s quantum condition.
\(\Rightarrow mvr = n h/(2\pi)\) → Bohr’s quantum condition.
Q14 (table)
Fill with “same” or “different”.
| Nuclide | Z | A | N |
|---|---|---|---|
| Isotope | Same | different | different |
| Isotone | different | different | Same |
Q15 (a)(b)
Dipole \(p=4\times10^{-9}\) Cm, \(E=5\times10^{4}\) NC⁻¹, \(\theta=30^\circ\). Find torque & angle for max torque.
(a) \(\tau = pE\sin\theta = 4\times10^{-9}\times5\times10^{4}\times0.5 = 1\times10^{-4}\ \mathrm{Nm}\).
(b) max at \(\theta = 90^\circ\).
(b) max at \(\theta = 90^\circ\).
Q16
Equivalent capacitance: series & parallel (C₁, C₂, C₃).
Series: \(\displaystyle \frac{1}{C_s} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)
Parallel: \(C_p = C_1 + C_2 + C_3\)
Parallel: \(C_p = C_1 + C_2 + C_3\)
Q17
Balancing condition of Wheatstone bridge.
\(\displaystyle \frac{P}{Q} = \frac{R}{S}\) (galvanometer zero).
Q18
(a) Define magnetisation. (b) Steel magnet: \(m=2.5\) Am², mass \(6.6\times10^{-3}\) kg, density \(7.9\times10^3\) kg/m³ → find M.
(a) \(M = \frac{m_\text{net}}{V}\) (magnetic moment per volume).
(b) Volume \(V = \frac{6.6\times10^{-3}}{7.9\times10^3} = 8.35\times10^{-7}\ \mathrm{m}^3\);
\(M = \frac{2.5}{8.35\times10^{-7}} = 2.99\times10^{6}\ \mathrm{A/m}\).
(b) Volume \(V = \frac{6.6\times10^{-3}}{7.9\times10^3} = 8.35\times10^{-7}\ \mathrm{m}^3\);
\(M = \frac{2.5}{8.35\times10^{-7}} = 2.99\times10^{6}\ \mathrm{A/m}\).
Q19
Snell’s law using Huygen’s principle.
From wavefront geometry: \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}\) → \(n_1\sin i = n_2\sin r\).
Q20
(a) Einstein’s photoelectric eq. (b) graph \(V_s\) vs ν, (c) slope.
(a) \(K_{\max} = h\nu - \phi_0\) or \(eV_s = h\nu - h\nu_0\).
(b) Straight line with positive slope.
(c) Slope \(= h/e\).
(b) Straight line with positive slope.
(c) Slope \(= h/e\).
Q21
Full‑wave rectifier: working & steady DC output.
(a) Two diodes, centre‑tapped transformer; each half‑cycle one diode conducts, same current direction through load.
(b) Use filter (capacitor across load) to smooth pulsations.
(b) Use filter (capacitor across load) to smooth pulsations.
Q22
(a) Define electrostatic potential. (b) Potential due to dipole.
(a) Work done per unit positive charge from infinity.
(b) \(V = \frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}\) (for \(r \gg a\)).
(b) \(V = \frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}\) (for \(r \gg a\)).
Q23
Relation \(I = nA e v_d\). Copper wire: \(A=10^{-6}\)m², \(I=2\)A, \(n=8\times10^{28}\) m⁻³ → find \(v_d\).
(a) \(I = n e A v_d\).
(b) \(v_d = \frac{2}{(8\times10^{28})(1.6\times10^{-19})(10^{-6})} = 1.56\times10^{-4}\) m/s.
(b) \(v_d = \frac{2}{(8\times10^{28})(1.6\times10^{-19})(10^{-6})} = 1.56\times10^{-4}\) m/s.
Q24
(a) \(L\) of long solenoid. (b) Air core \(4.8\) mH, iron core \(1.8\) H → \(\mu_r\).
(a) \(L = \mu_0 n^2 A l\) (or \(\mu_0 N^2 A / l\)).
(b) \(\mu_r = \frac{1.8}{4.8\times10^{-3}} = 375\).
(b) \(\mu_r = \frac{1.8}{4.8\times10^{-3}} = 375\).
Q25
Simple microscope lens type & magnifications.
(a) converging.
(b)(i) \(m = 1 + D/f\) (image at near point).
(ii) angular magnification \(= D/f\) (image at infinity).
(b)(i) \(m = 1 + D/f\) (image at near point).
(ii) angular magnification \(= D/f\) (image at infinity).
Q26
(a) Gauss’s law. (b) Field due to infinite plane sheet (σ). (c) \(\vec{E}\) between two oppositely charged plates.
(a) \(\oint \vec{E}\cdot d\vec{s} = q_{\text{enc}}/\epsilon_0\).
(b) \(E = \frac{\sigma}{2\epsilon_0}\) (independent of distance).
(c) Between plates: \(E = \frac{\sigma}{\epsilon_0}\) from + to –.
(b) \(E = \frac{\sigma}{2\epsilon_0}\) (independent of distance).
(c) Between plates: \(E = \frac{\sigma}{\epsilon_0}\) from + to –.
Q27
(a) \(\phi \propto I\) in moving coil galvo. (b) current & voltage sensitivity. (c) effect of more turns on current sensitivity.
(a) \(N I A B = k\phi \Rightarrow \phi = (NAB/k)I\) → proportional.
(b) \(I_s = \frac{\phi}{I} = \frac{NAB}{k}\); \(V_s = \frac{I_s}{R}\).
(c) Current sensitivity increases.
(b) \(I_s = \frac{\phi}{I} = \frac{NAB}{k}\); \(V_s = \frac{I_s}{R}\).
(c) Current sensitivity increases.
Q28
(a) Impedance \(Z\) & resonant frequency. (b) LCR values: find resonance frequency.
(a) \(Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}\); \(f_r = \frac{1}{2\pi\sqrt{LC}}\).
(b) \(L=25.48\times10^{-3}\)H, \(C=796\times10^{-6}\)F \(\Rightarrow\) \(f_r = 35.36\) Hz.
(b) \(L=25.48\times10^{-3}\)H, \(C=796\times10^{-6}\)F \(\Rightarrow\) \(f_r = 35.36\) Hz.
Q29
(a) Derive lens maker’s formula. (b) \(n=1.55\), \(|R_1|=|R_2|\), \(f=20\)cm → \(R\).
(a) \(\frac{1}{f} = (n-1)\big(\frac{1}{R_1}-\frac{1}{R_2}\big)\) (in air).
(b) For equiconvex: \(R_1=R,\ R_2=-R\). \(\frac{1}{20}=0.55\times\frac{2}{R}\) ⇒ \(R = 22\) cm.
(b) For equiconvex: \(R_1=R,\ R_2=-R\). \(\frac{1}{20}=0.55\times\frac{2}{R}\) ⇒ \(R = 22\) cm.
⚡ All answers verified with question paper March 2026 · derivations are concise as required.