Quantrex Academy · Free JEE Advanced PYQ solutions
JEE Advanced Physics Rotational Motion 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Physics Question (2023) — Solution

Question

A thin circular coin of mass 5   gm and radius 4 3   cm  is initially in a horizontal x y -plane. The coin is tossed vertically up ( + z direction) by applying an impulse of  π 2 × 10 - 2   N - s  at a distance 2 3   cm  from its center. The coin spins about its diameter and moves along the + z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is _____. [Given: The acceleration due to gravity g = 10   m   s − 2 ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

By impulse - momentum theorem: J = M V C M ⇒   V C M = J M = π 2 100 × 5 1000 = 2 π   m   s - 1 ⇒  Total time of journey  = 2 V C M g = 2 g × 2 π ⇒   ∆ t = 2 π 5   s Also, by angular impulse - momentum theorem: J × R 2 = I ω ⇒ J × R 2 = M R 2 4 ω ⇒   ω = J × R 2 M R 2 4 = J M R × 2 = π 2 100 × 2 5 1000 × 4 3 × 1 100 = 150 2 π   rad   s - 1 ⇒  Number of rotations  = ω ∆ t 2 π = 150 2 π × 2 π 5 2 π = 30 Therefore,  n = 30 .

Practice more on Quantrex App →

Related: Physics — Rotational Motion · All PYQ Banks