JEE Advanced 2021 Paper 1 Q13 Physics Rotational Mechanics Angular Momentum Hard

JEE Advanced 2021 Paper 1 · Q13 · Angular Momentum

A particle of mass M = 0.2 kg is initially at rest in the xy-plane at a point (x = −ℓ, y = −h), where ℓ = 10 m and h = 1 m. The particle is accelerated at time t = 0 with a constant acceleration a = 10 m/s² along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by L⃗ and τ⃗ respectively. î, ĵ and k̂ are unit vectors along the positive x, y and z-directions, respectively. If k̂ = î × ĵ then which of the following statement(s) is(are) correct?

  1. A. The particle arrives at the point (x = ℓ, y = −h) at time t = 2 s
  2. B. τ⃗ = 2k̂ when the particle passes through the point (x = ℓ, y = −h)
  3. C. L⃗ = 4k̂ when the particle passes through the point (x = ℓ, y = −h)
  4. D. τ⃗ = k̂ when the particle passes through the point (x = 0, y = −h)
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

Displacement to reach (ℓ, −h) from (−ℓ, −h) = 2ℓ = 20 m. Using 2ℓ = ½at²: 20 = ½(10)t² ⇒ t = 2 s ⇒ (A) correct. Velocity at this point v = at = 20 m/s along î. Position r⃗ = ℓ î − h ĵ. L⃗ = M r⃗ × v⃗ = 0.2 · (ℓ î − h ĵ) × (20 î) = 0.2 · (−h)·20 · (ĵ×î) = 0.2 · 20 · h · k̂ = 0.2·20·1·k̂ = 4 k̂ ⇒ (C) correct. τ⃗ = r⃗ × F⃗ = (ℓ î − h ĵ) × (Ma î) = −h·Ma·(ĵ×î) = h·M·a·k̂ = 1·0.2·10·k̂ = 2 k̂; this holds at any x because only y-component contributes ⇒ (B) correct. At (0, −h), τ⃗ is also 2k̂ (not k̂) ⇒ (D) incorrect.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →