Physics · Simple Harmonic Motion

JEE Advanced 2023 — Paper 1 — Question 29

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 $1.2 \times 10^{-8} \mathrm{Nm} \mathrm{rad}^{-1}$ . The angular frequency of the oscillations in $\mathrm{n} \times 10^{-3} \mathrm{rad}$ $\mathrm{s}^{-1}$ . The value of n is $\qquad$

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

Time period of oscillation

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{K}}}$

I = Moment of inertia

$\mathrm{K}=$ Torsional constant

moment of inertia $\mathrm{I}=30 \times 16+20 \times 36$

\mathrm{I}=12 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^{2}

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{K}}}$

$=2 \pi \sqrt{\frac{12 \times 10^{-5}}{1.2 \times 10^{-8}}}=200 \pi \mathrm{sec}$

$\omega=\frac{2 \pi}{\mathrm{~T}}=10 \times 10^{-3} \mathrm{rad} / \mathrm{s}$

$\mathrm{n}=10$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Physical Pendulum and Torsional Pendulum