Physics (Theory)

A positive charge means the body has LOST electrons, so its mass decreases.
Number of electrons removed: n = q/e = 8.0 × 10⁻¹² / 1.6 × 10⁻¹⁹ = 5 × 10⁷.
Mass lost = n mₑ = 5 × 10⁷ × 9.1 × 10⁻³¹ kg = 4.55 × 10⁻²³ kg.
Common mistake: thinking a charged body gains mass – that happens only when it becomes negatively charged.

Take a thin ring of radius r and width dr: area dA = 2πr dr.
dI = j dA = αr · 2πr dr = 2πα r² dr.
\( I = 2\pi\alpha \int_0^{R/2} r^2\,dr = \frac{2\pi\alpha}{3}\left(\frac{R}{2}\right)^3 = \frac{\pi\alpha R^3}{12} \)
So I ∝ R³.

v is perpendicular to B, so the magnetic force F = q v × B is always perpendicular to v and lies in the XY plane → circular motion in the XY plane.
Direction: v = +ŷ, B = −ẑ ⇒ ŷ × (−ẑ) = −x̂, so the particle curves towards −X.
Since the field exists only on one side (y > 0), the particle completes half a circle and leaves the field moving along −Y. Hence a semicircle in the XY plane.
No velocity component along B, so no helix.

τ = mB sin θ, where θ is the angle between m and B.
Initially θ = 90°, so τ₀ = mB.
After turning through angle φ, θ = 90° − φ and τ = mB cos φ.
mB cos φ = mB/2 ⇒ cos φ = 1/2 ⇒ φ = 60°.
Common mistake: answering 30° – that is the final angle between m and B, not the angle turned.

Sodium chloride has no unpaired electrons (Na⁺ and Cl⁻ have closed shells), so it is diamagnetic.
Aluminium, calcium and copper chloride (Cu²⁺ has an unpaired electron) are paramagnetic.

AC ammeters read the rms value: I_rms = 10 A.
I₀ = √2 × I_rms = 10√2 A ≈ 14.1 A.

B₀ = E₀/c = 1000 / (3 × 10⁸) T = 3.33 × 10⁻⁶ T.

Negative magnification ⇒ image is inverted ⇒ image is real.
Only a concave mirror forms real (and enlarged, |m| = 2) images; a convex mirror always gives virtual, erect, diminished images.

Electrons show diffraction (Davisson–Germer), so (A) is wrong.
Photons carry momentum p = h/λ, so (B) is wrong.
All EM radiation is quantised into photons, so (D) is wrong.
Both light and matter show wave as well as particle behaviour → (C).

Intensity I = (number of photons per second per unit area) × hν.
Red light has lower ν, so each red photon carries less energy (E = hν).
For the same intensity, the red beam must carry more photons.
(C) and (D) are wrong: red has the longer wavelength and lower photon energy.

Tin (Sn) has essentially zero band gap and behaves as a metal (conductor).
Mica and carbon (diamond) are insulators; Si is a semiconductor with Eg ≈ 1.1 eV.

For a straight current, field lines are circles centred on the wire, in planes perpendicular to it. The wire is vertical, so the circles are horizontal; their sense is given by the right-hand thumb rule (anticlockwise seen from above for upward current).

Each donor atom that releases a free electron becomes a fixed positive ion. Number of free electrons ≈ number of positive donor ions (+ equal numbers of thermally generated electrons and holes), so the total charge is zero.
Thus R directly explains why the n-type crystal stays neutral.

Z = √(R² + (X_L − X_C)²). When X_L = X_C, Z = R and the phase angle tan φ = (X_L − X_C)/R = 0, so current and voltage are in phase – purely resistive behaviour.
X_L = X_C ⇒ ωL = 1/(ωC) ⇒ ω = 1/√(LC). R is the reason for A (as per the CBSE key).

A is true: with one slit open, the light from that single slit spreads and gives a single-slit diffraction pattern (wide central maximum with weaker side maxima).
R is taken as false (CBSE key): a diffraction pattern is itself produced by superposition (interference) of secondary wavelets coming from different parts of a single wavefront – it does not need two separate sources/waves.

A is true: electrons at the surface leave with K_max = hν − φ₀, but electrons from deeper inside lose extra energy in collisions, so they emerge with less KE – down to zero.
R is true (φ₀ is a property of the metal), but it only fixes the maximum KE; it does not explain why there is a spread of speeds. Hence (B).

Let each bulb have resistance R and current through S be I.
(i) K closed: P and Q are in parallel, so each carries I/2 (identical bulbs). Power in S = I²R; in P or Q = (I/2)²R = I²R/4. S glows brighter (four times the power); P and Q glow equally. (Equivalently, the p.d. across S is twice that across P or Q.)
(ii) K open: P's branch is broken, so S and Q are simply in series and carry the same current ⇒ same power ⇒ equal brightness.
Key point for marks: brightness ∝ power = I²R, so compare currents.

Take the potential of the left middle junction (between the cells) as 0 and the right junction as x.
Top of upper cell = +10 V, bottom of lower cell = −10 V (cells aiding).
KCL at the right junction (currents in = currents out):
from 20 Ω: (10 − x)/20, from bulb: (0 − x)/10, out through 10 Ω: (x − (−10))/10.
(10 − x)/20 − x/10 = (x + 10)/10
⇒ 10 − x − 2x = 2x + 20 ⇒ x = −2 V.
Current through bulb = (0 − (−2))/10 = 0.2 A (from the cells' junction towards the right junction).
Check with loop rule (Kirchhoff): the same result I = 1/5 A is obtained by writing loop equations for the upper and lower meshes.

Secondary maxima: a sin θ ≈ aθ = (n + ½)λ; first secondary maximum n = 1 ⇒ θ = 3λ/(2a).
θ = 3 × 550 × 10⁻⁹ / (2 × 0.55 × 10⁻³) = 1.5 × 10⁻³ rad.
In degrees: 1.5 × 10⁻³ × (180/π) ≈ 0.086°.
Common mistake: using θ = λ/a – that gives the first minimum, not the first secondary maximum.

Lens maker's formula: \( \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).
Equiconvex: R₁ = +R, R₂ = −R ⇒ 1/f = (n − 1)(2/R).
R = 2(n − 1)f = 2 × 0.55 × 15.0 = 16.5 cm.

1 u = (mass of C-12 atom)/12 = 1.9926 × 10⁻²⁶ / 12 = 1.66 × 10⁻²⁷ kg.
Mass of 2 electrons in u = 2 × 9.1 × 10⁻³¹ / 1.66 × 10⁻²⁷ ≈ 0.001096 u.
An α-particle is a helium nucleus = helium atom minus its 2 electrons:
m_α = 4.002603 − 0.001096 ≈ 4.001507 u.

(a) Holes increased far above nᵢ ⇒ acceptor (trivalent, e.g. B, Al, In) impurity ⇒ p-type semiconductor.
(b) Mass-action law: nₑ n_h = nᵢ².
nₑ = (5 × 10⁸)² / (8 × 10¹²) = 25 × 10¹⁶ / 8 × 10¹² = 3.125 × 10⁴ m⁻³.

(i) In field E, each electron accelerates a = −eE/m; average over relaxation time τ gives drift speed v_d = eEτ/m.
Current I = n e A v_d = n e A (eEτ/m) ⇒ current density j = I/A = (n e² τ/m) E.
Comparing with Ohm's law j = σE = E/ρ:
\( \rho = \frac{m}{n e^2 \tau} \)
(ii) Charge = area under the I–t graph.
Q₁ = 2.0 A × 1 s = 2 C (rectangle).
Q₂ = ½ × 2 s × 1.5 A = 1.5 C (triangle).
Q₁/Q₂ = 2/1.5 = 4/3.

(i)(I) Total resistance R + r ⇒ I = E/(R + r). I is largest when R = 0 (short circuit): I_max = E/r.
(II) V = E − Ir = ER/(R + r). V is largest when I = 0 (R → ∞, open circuit): V_max = E.
(ii) E = I₁(R₁ + r) = I₂(R₂ + r)
⇒ I₁R₁ + I₁r = I₂R₂ + I₂r
⇒ r(I₁ − I₂) = I₂R₂ − I₁R₁
⇒ r = (I₂R₂ − I₁R₁)/(I₁ − I₂).

(a) \( d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\vec{r}}{r^3} \)
(b) Field of a long wire: B = μ₀I/(2πd) = 2 × 10⁻⁷ I/d.
Wire along x: distance of P = 5 m ⇒ B₁ = 2 × 10⁻⁷ × 2/5 = 8 × 10⁻⁸ T; direction x̂ × ŷ = +ẑ.
Wire along y: distance = 4 m ⇒ B₂ = 2 × 10⁻⁷ × 2/4 = 1 × 10⁻⁷ T; direction ŷ × x̂ = −ẑ.
Net: B = (8 − 10) × 10⁻⁸ ẑ = −2 × 10⁻⁸ ẑ T.
Magnitude 2 × 10⁻⁸ T, directed along −z.

Lenz's law: the induced current opposes the change of flux.
(a) A south pole approaches coil 1 ⇒ coil 1's near face becomes a south pole to repel it ⇒ current appears clockwise when viewed from coil 2's side.
(b) The south pole moves away ⇒ coil 1's near face becomes a north pole to attract it ⇒ anticlockwise as seen from coil 2.
(c) Increasing R₂ lowers the current in coil 2, so the flux through coil 1 decreases ⇒ coil 1 tries to keep the flux, making its near face a north pole ⇒ anticlockwise current, only while R₂ is changing; it stops once R₂ is steady.

(a) Any three:
• They are transverse; E, B and direction of propagation are mutually perpendicular.
• They need no material medium and travel in vacuum at c = 1/√(μ₀ε₀) = 3 × 10⁸ m/s.
• They carry energy and momentum (and exert radiation pressure).
• They consist of oscillating E and B fields in phase, with E₀/B₀ = c.
(b) While charging, the charge on the plates and hence the electric field/flux between them changes with time. This changing flux acts as a current, the displacement current i_d = ε₀ dΦ_E/dt, which exists in the gap between the plates, where no conduction current flows. It equals the conduction current in the wires, keeping current continuous.

(a) v = c/n: larger n ⇒ speed decreases.
Frequency is set by the source ⇒ unchanged.
λ' = λ₀/n ⇒ wavelength decreases.
(b) β = λ'D/d ⇒ fringe width decreases.
The fringes remain straight bright and dark bands (shape unchanged).
The whole set-up is in the same medium, so the central maximum stays at the centre (no shift).

Einstein's equation: hν = φ₀ + K_max, with φ₀ = hν₀ ⇒ K_max = h(ν − ν₀).
(a) If ν < ν₀, K_max would be negative, which is impossible; one photon cannot give an electron enough energy to escape. So no emission below threshold frequency, however intense the light.
(b) K_max = h(ν − ν₀) contains only ν (and the metal's ν₀). Greater intensity means more photons, so more electrons (larger current), but each photon still has energy hν, so K_max is unchanged.
(c) K_max = eV₀ ⇒ V₀ = (h/e)ν − (h/e)ν₀. This is of the form y = mx + c, so V₀ vs ν is a straight line of slope h/e (same for all metals) and intercept depending on ν₀.

(a) p-type: majority = holes, minority = electrons.
n-type: majority = electrons, minority = holes.
(b) Diffusion current: when the junction forms, holes are far more concentrated on the p-side and electrons on the n-side. Because of this concentration difference, holes diffuse p → n and electrons diffuse n → p, giving the diffusion current. This leaves charged ions behind, forming the depletion layer and a junction electric field (n → p).
Drift current: this field sweeps minority carriers across – electrons from the p-side to the n-side and holes from the n-side to the p-side. This field-driven motion is the drift current, opposite to the diffusion current; in equilibrium the two are equal.

P₂ forms one capacitor with P₁ and another with P₃; both share the p.d. V (A to B), so they are in parallel.
Each: C₁ = Kε₀L²/d.
(i) C = 2Kε₀L²/d → (C).
(ii) P₁ belongs to one capacitor: |Q₁| = C₁V = ε₀KVL²/d (it is negative, since B is at lower potential) → (B).
(iii) Uniform field E = V/d (p.d. V across gap d) → (A).
(iv) ε₀L²/x = 2Kε₀L²/d ⇒ x = d/(2K) → (C).

(i) mv²/r = Ke²/r² ⇒ v = √(Ke²/(mr)) → (C).
(ii) KE = ½mv² = Ke²/(2r); PE = −Ke²/r; total E = −Ke²/(2r) → (B).
(iii) Photon energy = 1240 eV·nm / 500 nm = 2.48 eV. Emission ⇒ total energy decreases: ΔE = −2.48 eV. Since KE = −E, ΔKE = +2.48 eV → (C).
(iv) v ∝ 1/n and r ∝ n² ⇒ f = v/(2πr) ∝ 1/n³ → (D).

(i) Distance of (0, y) from each charge = √(y² + a²).
Magnitude due to each: E₊ = E₋ = q/[4πε₀(y² + a²)].
The y-components cancel; the x-components add, both pointing towards −x (away from +q, towards −q).
Each x-component = E cos θ with cos θ = a/√(y² + a²).
\( \vec{E} = -\frac{2qa}{4\pi\varepsilon_0 (y^2+a^2)^{3/2}}\,\hat{i} \), i.e. opposite to the dipole moment p = 2qa x̂.
For y ≫ a: \( \vec{E} = -\frac{2qa}{4\pi\varepsilon_0 y^3}\hat{i} = -\frac{\vec{p}}{4\pi\varepsilon_0 y^3} \).
(ii) U = k Σ qᵢqⱼ/rᵢⱼ.
Initial (side 0.2 m): Σqᵢqⱼ = (−2)(−1) + (−2)(5) + (−1)(5) = 2 − 10 − 5 = −13 (nC)².
U₁ = 9 × 10⁹ × (−13 × 10⁻¹⁸)/0.2 = −5.85 × 10⁻⁷ J.
The midpoints form an equilateral triangle of side 0.1 m, and the same three charges sit at its corners:
U₂ = 9 × 10⁹ × (−13 × 10⁻¹⁸)/0.1 = −11.7 × 10⁻⁷ J.
Work done (by external agent) W = U₂ − U₁ = −5.85 × 10⁻⁷ J.

(i) Consistency: take a sphere of radius r around a point charge q. By symmetry E is radial and the same everywhere on it: ∮E·dS = E(4πr²) = q/ε₀ ⇒ E = q/(4πε₀r²). Force on q₀ = q q₀/(4πε₀r²), which is Coulomb's law. (Gauss's law relies on the inverse-square dependence.)
Shell of charge q (σ = q/4πr²):
(I) y > r: Gaussian sphere of radius y encloses q. E(4πy²) = q/ε₀ ⇒ \( E = \frac{q}{4\pi\varepsilon_0 y^2} \), radially outward – as if all charge were at the centre.
(II) y < r: Gaussian sphere encloses no charge ⇒ E(4πy²) = 0 ⇒ E = 0 inside.
(ii) Distance of (0, 0, 2) from origin = 2 m; from (0, 0, −6) = 8 m.
V = k[2 × 10⁻⁹/2 + Q/8] = 0 ⇒ Q = −8 × 10⁻⁹ C.
A negative charge of 8 nC.

(i) Lenz's law: the induced emf has a polarity that drives a current opposing the change in flux that causes it.
Energy: e.g. pushing a magnet towards a coil makes the near face of the coil the same pole, which repels the magnet; we must do mechanical work against this, and that work appears as electrical energy (heat) in the circuit. If the induced current aided the change, energy would be created from nothing – impossible.
(ii)(I) Entering: inward flux increases ⇒ induced current tries to create outward field ⇒ anticlockwise. Leaving: inward flux decreases ⇒ current makes inward field ⇒ clockwise.
(II) Time to start entering: t₁ = l/v. Fully inside at t₂ = (l + l/2)/v = 3l/(2v). Front edge reaches far side at t₃ = 2l/v. Fully out at t₄ = 5l/(2v).
φ–t: 0 for t < l/v; rises linearly to B(l/2)² = Bl²/4 at 3l/(2v); constant Bl²/4 till 2l/v; falls linearly to 0 at 5l/(2v); zero after.
E–t: E = −dφ/dt. Magnitude = B(l/2)v = Blv/2 during entry (one sign, e.g. −Blv/2) and during exit (opposite sign, +Blv/2); zero when the loop is fully outside or fully inside.

(i) Peak value I₀: the maximum instantaneous value of the current in a cycle.
rms value: the steady (dc) current that would produce the same heating in the same resistor in the same time.
I_rms = I₀/√2 ≈ 0.707 I₀.
(ii)(I) Phasor for X: the V phasor is 90° ahead (anticlockwise) of the I phasor ⇒ X is a pure inductor L.
For Y: the V phasor is 90° behind the I phasor ⇒ Y is a pure capacitor C.
(II) Series LC: Z = |X_L − X_C| = |ωL − 1/(ωC)|.
Resonance: X_L = X_C ⇒ ωL = 1/(ωC) ⇒ ω₀ = 1/√(LC), i.e. \( \nu_0 = \frac{1}{2\pi\sqrt{LC}} \).
At resonance Z = 0 (ideal L and C with no resistance), so the current would be very large in theory.

(i) Convex lens alone: 1/v − 1/u = 1/f ⇒ 1/v = 1/10 − 1/30 = 2/30 ⇒ v = 15 cm.
With the concave lens the image moves to 15 + 45 = 60 cm.
Combination: 1/F = 1/60 + 1/30 = 3/60 ⇒ F = 20 cm.
1/F = 1/f₁ + 1/f₂ ⇒ 1/20 = 1/10 + 1/f₂ ⇒ 1/f₂ = −1/20 ⇒ f₂ = −20 cm.
(ii) \( n = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} \), A = 60°.
√3 = sin((60° + δₘ)/2)/sin 30° ⇒ sin((60° + δₘ)/2) = √3/2 ⇒ (60° + δₘ)/2 = 60° ⇒ δₘ = 60°.
At minimum deviation i = (A + δₘ)/2 = 60°.

(i)(I) β = λD/d ⇒ d = λD/β = 633 × 10⁻⁹ × 5.0 / (5 × 10⁻³) = 6.33 × 10⁻⁴ m = 633 μm.
(II) Dark fringe n: xₙ = (2n − 1)λD/(2d); n = 1 ⇒ x₁ = λD/(2d) = β/2 = 2.5 mm.
(ii) Width of central maximum = 2λD/a = 2 × 650 × 10⁻⁹ × 0.60 / (0.6 × 10⁻³) = 1.3 × 10⁻³ m = 1.3 mm.
Common mistake: forgetting the factor 2 – each first minimum is λD/a from the centre.