Q.1 (1) — Dimensions of sin θ
Ans: (d) sin θ is a pure ratio — it is dimensionless → [M⁰L⁰T⁰]
Q.1 (2) — Distance of planet from Earth
Ans: (d) Parallax method — used for astronomical distances by measuring the parallax angle from two positions on Earth.
Q.1 (3) — Unit vector along î + ĵ
Ans: (c) Magnitude of (î + ĵ) = √(1²+1²) = √2 → Unit vector = (î + ĵ)/√2
Q.1 (4) — Kepler's law of equal areas
Ans: (c) Law of equal areas → Conservation of angular momentum. The torque of gravitational force about the Sun is zero, so angular momentum is conserved.
Q.1 (5) — Potential difference between two points 1 cm apart
E = F/q = 3000/3 = 1000 V/m
V = E × d = 1000 × 0.01 = 10 V
Ans: (a) 10 V
Q.1 (6) — Pole strengths of broken magnet (2:1 length ratio)
Ans: (d) Pole strength depends on the cross-sectional area, not length. Both pieces have the same cross-section → pole strength ratio = 1:1
Q.1 (7) — Magnetic dip when horizontal = vertical component
tan(δ) = B_V/B_H = 1 → δ = 45°
Ans: (b) 45°
Q.1 (8) — Potential barrier in p-n diode
Ans: (d) Positive donor ions accumulate on n-side and negative acceptor ions on p-side near the junction, creating the built-in potential barrier.
Q.1 (9) — n-type and p-type silicon doping
Ans: (c) n-type → doped with Arsenic (pentavalent); p-type → doped with Boron (trivalent).
Q.1 (10) — Pure semiconductor
Ans: (b) A pure (undoped) semiconductor is an intrinsic semiconductor. Equal numbers of electrons and holes are generated by thermal excitation.
Q.2 (1) — Heavy and light objects released from same height
Both experience the same acceleration — acceleration due to gravity g is independent of mass (g ≈ 9.9 m/s²). Both reach the ground simultaneously.
Q.2 (2) — Deformation
Change in the shape or size of a body produced by an applied force (stress). The body may or may not recover its original shape depending on whether the force is within elastic limits.
Q.2 (3) — Static friction: self-adjusting force
Static friction automatically adjusts its magnitude to match the applied force (up to its maximum limiting value f_s = μN). It never exceeds what is needed to prevent motion — hence "self-adjusting."
Q.2 (4) — Freely suspended bar magnet direction
It aligns in the geographic North-South direction, with its North pole pointing toward geographic North (magnetic south pole of Earth).
Q.2 (5) — Critical frequency
The maximum frequency of a radio wave that can be reflected back to Earth by the ionosphere when transmitted vertically upward. Formula: f_c = 9√N_max Hz, where N_max is the maximum electron density.
Q.2 (6) — Wave
A disturbance (periodic) that propagates through a medium (or vacuum), transferring energy from one point to another without any net transport of matter.
Q.2 (7) — p-n junction
A junction formed by bringing p-type and n-type semiconductors into intimate contact. A depletion layer forms at the interface due to diffusion of majority carriers, creating a built-in electric field.
Q.2 (8) — Charge on p-type and n-type semiconductor
Both are electrically neutral (charge = 0). Although p-type has more holes and n-type has more electrons, the total positive and negative charges in each are equal (dopant atoms are neutral overall).
Q.3 — Vector representation
Graphical: A vector is represented by a directed line segment (arrow). The length (to scale) represents its magnitude; the arrowhead shows its direction.
Symbolic: Written as Ā or →A. Magnitude is |Ā|. In component form: Ā = Aₓî + Aᵧĵ + Aᵤk̂
Q.4 — Derivative of e²ˣ − tan x
d/dx(e²ˣ) = 2e²ˣ (chain rule)
d/dx(tan x) = sec²x
Answer: 2e²ˣ − sec²x
Q.5 — Seat belt safety
During sudden braking, passengers continue forward due to inertia (Newton's 1st law). The seat belt: (1) applies a retarding force to stop the passenger with the car, (2) spreads the stopping force over a longer time interval, reducing the peak force (Impulse = F×Δt = Δp), and (3) prevents the passenger from hitting the dashboard or windshield.
Q.6 — Torque of F̄ = 3î + ĵ − 4k̂ at (3,4,−2) about (−1,2,4)
r̄ = (3−(−1))î + (4−2)ĵ + (−2−4)k̂ = 4î + 2ĵ − 6k̂
τ = r̄ × F̄ = |î ĵ k̂|
|4 2 −6|
|3 1 −4|
î: (2×(−4) − (−6)×1) = −8+6 = −2
ĵ: −(4×(−4) − (−6)×3) = −(−16+18) = −2
k̂: (4×1 − 2×3) = 4−6 = −2
τ̄ = −2î − 2ĵ − 2k̂ N·m
Q.7 — Work done moving 4 m along z-axis, F̄ = (−î + 2ĵ + 3k̂) N
Displacement s̄ = 4k̂ m (only z-direction)
W = F̄ · s̄ = (−1)(0) + (2)(0) + (3)(4)
W = 12 J
Q.8 — Weight zero at Earth's centre
At the centre of Earth, the gravitational pull from all surrounding mass acts equally in all directions and cancels out, giving g = 0. Since W = mg, weight = 0. (Also: g = g₀(1 − d/R), at d = R, g = 0.)
Q.9 — Cooking faster in pressure cooker
In a pressure cooker, the steam cannot escape, so pressure inside rises above atmospheric. Higher pressure raises the boiling point of water (above 100°C). Food cooks at this higher temperature, significantly speeding up cooking.
Q.10 — Dispersive power
Definition: Ability of a prism to split white light into constituent colours (dispersion). It is the ratio of angular dispersion to mean deviation.
ω = (μᵥ − μᵣ)/(μᵧ − 1)
In terms of angles: ω = (δᵥ − δᵣ)/δᵧ
where μᵥ, μᵣ, μᵧ = refractive indices for violet, red, yellow; δ = deviation angles.
Q.11 — Current density in copper wire (r = 0.6 mm, I = 1 A)
r = 0.6 × 10⁻³ m
A = πr² = π × (0.6 × 10⁻³)² = π × 3.6 × 10⁻⁷ = 1.131 × 10⁻⁶ m²
J = I/A = 1 / (1.131 × 10⁻⁶)
J ≈ 8.84 × 10⁵ A/m²
Q.12 — SI units of pole strength & magnetic dipole moment
Pole strength: A·m (ampere-metre)
Magnetic dipole moment: A·m² (ampere-metre²)
Q.13 — Skip distance
The minimum distance from the transmitter at which a sky wave (of given frequency) first returns to Earth after reflection from the ionosphere. Inside the skip zone, sky waves are not received.
Q.14 — Forward current in zero biased p-n junction
At zero bias, diffusion current (majority carriers) and drift current (minority carriers) are equal and opposite. Net current = 0. Therefore, forward current at zero bias is zero.
Q.15 — Five spheres on ruler (ruler: 1 cm to 11 cm → total 10 cm)
Diameter of each sphere = 10 cm / 5 = 2 cm → radius = 1 cm
(i) Area of central sphere (cross-section) = πr² = π × (1)² ≈ 3.14 cm²
(ii) Absolute error in reading = Least count of ruler = 1 mm = 0.1 cm
Absolute error = ±0.1 cm
Q.16 — Dimensional analysis: correct ≠ physically correct
Dimensionally correct but wrong: The equation may lack a dimensionless constant. E.g., s = ut + at² is dimensionally correct but physically wrong (missing ½).
Dimensionally incorrect = necessarily wrong: If dimensions of LHS ≠ RHS, the equation violates the principle of homogeneity — it cannot be physically valid under any circumstances.
Q.17 — Ā = 3î + 4ĵ, B̄ = 7î + 24ĵ; find vector with magnitude |B̄| parallel to Ā
|B̄| = √(7² + 24²) = √(49 + 576) = √625 = 25
|Ā| = √(3² + 4²) = √(9 + 16) = √25 = 5
Unit vector along Ā: â = (3î + 4ĵ)/5
Required vector = 25 × (3î + 4ĵ)/5 = 5(3î + 4ĵ)
= 15î + 20ĵ
Q.18 — Partially inelastic head-on collision (identical marbles)
Let mass = m, initial velocity = u (one at rest).
Conservation of momentum: u = v₁ + v₂ ... (1)
Coefficient of restitution: e = (v₂ − v₁)/u → v₂ − v₁ = eu ... (2)
Solving (1) & (2):
v₁ = u(1−e)/2 v₂ = u(1+e)/2
Ratio v₁ : v₂ = (1−e) : (1+e)
Q.19 — Mass of Earth (g = 9.81 m/s², Rₑ = 6.37×10⁶ m, G = 6.67×10⁻¹¹)
From g = GM/R² → M = gR²/G
M = (9.81 × (6.37×10⁶)²) / (6.67×10⁻¹¹)
= (9.81 × 40.577×10¹²) / (6.67×10⁻¹¹)
= 398.06×10¹² / 6.67×10⁻¹¹
M ≈ 5.97 × 10²⁴ kg
Q.20 — Why balloon bursts when filled with air
As air is blown in, pressure inside increases. The rubber stretches elastically to accommodate the increased volume. Beyond the elastic limit, the rubber cannot withstand the excess pressure and ruptures (bursts). The stress exceeds the tensile strength of rubber.
Q.21 — Electric dipole: 2l = 2 cm, θ = 30°, E = 10⁵ N/C, τ = 10√3 N·m → find q
τ = pE sin θ
10√3 = p × 10⁵ × sin 30° = p × 10⁵ × 0.5
p = 10√3 / (0.5 × 10⁵) = 2√3 × 10⁻⁴ C·m
p = q × 2l = q × 2 × 10⁻² m
q = (2√3 × 10⁻⁴) / (2 × 10⁻²)
q = √3 × 10⁻² ≈ 1.73 × 10⁻² C
Q.22 — Bar magnet cut transverse / along its length
Cut transverse (across width/breadth): Two shorter magnets each of half the original length but same pole strength m. Magnetic moment M' = m × (L/2) = M/2 for each.
Cut along its length: Two thinner magnets, each with half the original pole strength (m/2) but same length L. Magnetic moment M' = (m/2) × L = M/2 for each. Each piece is a complete magnet with both N and S poles.
Q.23 — Physicist–Work matching
(i) H. Hertz → (a) Existence of EM waves, (b) Properties of EM waves
(ii) J. Maxwell → (b) Properties of EM waves, (d) Displacement current
(iii) G. Marconi → (c) Wireless communication
Q.24 — Frequency of red light (λ = 6.5×10⁻⁷ m, c = 3×10⁸ m/s)
f = c/λ = (3×10⁸) / (6.5×10⁻⁷)
f ≈ 4.6 × 10¹⁴ Hz
Q.25 — Carrier wave
A high-frequency sinusoidal electromagnetic wave generated by an oscillator, used to carry information/signals over long distances. The signal (audio/video) is superimposed on it by modulation (AM or FM). The carrier itself carries no useful information — it only serves as a vehicle.
Q.26 — Light travels in vacuum; sound cannot
Light is an electromagnetic wave — it consists of oscillating electric and magnetic fields that sustain each other and require no medium to propagate. Sound is a mechanical wave — it needs a material medium (particles) to transfer compression and rarefaction. In vacuum there are no particles, so sound cannot propagate.
Q.27 — Impulse: quantity related to change in momentum
Impulse (J̄): The product of force and the time interval for which it acts. It equals the change in momentum of a body.
For constant force: J = F × t
For variable force: J = ∫F dt = Δp = mv − mu
When used instead of force: When force acts for a very short duration (collision, explosion, a bat hitting a ball), it is practically impossible to measure the force and time separately. It is easier and more meaningful to measure the net change in momentum — hence we use impulse.
Q.28 — Block on rough table: min and max weights in pan
Max static friction = 10% of block's weight = 0.1 × 35 × 9.9 = 34.65 N
Left: 20 kg load → pulls left with 20g N
Right: (2 + n) kg pan → pulls right with (2+n)g N
For block NOT to slide (net force ≤ friction):
Minimum n: 20g − (2+n)g ≤ 34.65 → (18−n)×9.9 ≤ 34.65 → n ≥ 14.5 → n_min = 15
Maximum n: (2+n)g − 20g ≤ 34.65 → (n−18)×9.9 ≤ 34.65 → n ≤ 21.5 → n_max = 21
Weights in pan: minimum 15, maximum 21 ✓
Q.29 — Net force on each object
(1) Rain drop at constant speed → Net force = 0 (gravity = air drag + buoyancy)
(2) Cork floating on water → Net force = 0 (weight = buoyant force)
(3) Kite stationary in sky → Net force = 0 (gravity = wind lift + string tension)
(4) Car at constant 30 km/h → Net force = 0 (engine force = friction)
(5) High-speed electron far from all fields → Net force = 0 (no gravitational or electromagnetic force acts on it)
Q.30 — EM wave: f = 50 MHz along +x, Ē = 9.6 ĵ V/m → find B̄
B₀ = E₀/c = 9.6 / (3×10⁸) = 3.2 × 10⁻⁸ T
Direction: Wave travels along +x̂, Ē along +ŷ
B̄ must be ⊥ to both → along +ẑ (using right-hand rule: x̂ × ŷ = ẑ)
B̄ = 3.2 × 10⁻⁸ k̂ T (along +z direction)
Q.31 — Intrinsic vs Extrinsic semiconductors
| Property | Intrinsic | Extrinsic |
|---|
| Definition | Pure semiconductor (no dopant) | Doped with impurity atoms |
| Conductivity | Very low at room temp | Much higher |
| Charge carriers | nₑ = nₕ (electrons = holes) | nₑ ≠ nₕ (depends on dopant) |
| Types | Only one type | p-type and n-type |
| Example | Pure Si, pure Ge | Si+B (p-type), Si+As (n-type) |
| Temp dependence | Strongly increases with temp | Less sensitive at room temp |
| Fermi level | Exactly at middle of band gap | Shifts toward valence or conduction band |