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JEE Advanced Physics Rotational Motion 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

A particle of mass  M = 0 . 2   kg  is initially at rest in the  x y -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   s - 2  along the positive  x -direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by  L →  and   τ → , respectively.  i ^ ,   j ^  and  k ^  are unit vectors along the positive  x , y  and  z -directions, respectively. If  k ^ = i ^ × j ^  then which of the following statement(s) is (are) correct?

Options

  1. A. The particle arrives at the point  x = l ,   y = - h  at time  t = 2   s .
  2. B. τ → = 2 k ^  when the particle passes through the point  x = l ,   y = - h .
  3. C. L → = 4 k ^  when the particle passes through the point  x = l ,   y = - h .
  4. D. τ → = k ^  when the particle passes through the point  x = 0 ,   y = - h .

Answer

C. L → = 4 k ^  when the particle passes through the point  x = l ,   y = - h .

Step-by-step solution

At  t = 2   s : displacement along + X  axis = 1 2 × 10 × t 2 = 1 2 × 10 × 4 = 20   m Hence, particle arrives at  10 , - 1 At  x = l ,   y = − h , τ → = r → × F → = 10 i ^ − j ^ × 0 .2 × 10 i ^ = 2 k ^   N   m L → = r → × p → = 10 i ^ − j ^ × 0 .2 × v i ^ and  v at   t = 2 s = 0 + 10 × 2 = 20   m   s - 1 ∴     L → = 10 i ^ − j ^ × 4 i ^ = 4 k ^   N   m   s At  x = 0 , y = − h : r → = 0 i ^ − j ^ F → = 2 i ^ ∴     τ → = r → × F → = 2 k ^   N   m

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Related: Physics — Rotational Motion · All PYQ Banks