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JEE Advanced Physics Oscillations 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Physics Question (2023) — Solution

Question

Two point-like objects of masses 20   gm and 30   gm are fixed at the two ends of a rigid massless rod of length 10   cm . This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is 1 . 2 × 10 − 8   N   m   rad − 1 . The angular frequency of the oscillations in n × 10 − 3   rad   s − 1 . The value of n is _____.

Options

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

Answer

A. A

Step-by-step solution

Given: m 1 = 30   gm  and  m 2 = 20   gm . Moment of inertia about the axis of rotation is I = m 1 r 1 2 + m 2 r 2 2 X-coordinate of the centre of mass can be written as, x C M = m 1 x 1 + m 2 x 2 m 1 + m 2 = 0 + 20 × 10 - 3 × 10 × 10 - 2 30 × 10 - 3 + 20 × 10 - 3 = 4 × 10 - 2   m = 4   cm = r 1 Therefore,  And  r 2 = 10 - 4   cm = 6   cm ∴   I = 30 × 10 - 3 × 16 × 10 - 4 + 20 × 10 - 3 × 36 × 10 - 4 ⇒   I = 1200 × 10 - 7   kg   m 2 If the system is rotated by small angle ‘ θ ’, the restoring torque is  τ R = - k θ And  d 2 θ d t 2 = - k I · θ = - ω 2 θ = - 1 . 2 × 10 - 8 1200 × 10 - 7 · θ ∴   ω 2 = 10 - 4 ⇒   ω = 10 × 10 - 3   rad   s - 1 Therefore,  n = 10 .

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