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 1.2×10−8Nmrad−1. The angular frequency of the oscillations is n×10−3rads−1. The value of n is ____.
Answer: 10
Solution
Center of mass from the 20 g mass: xcm=5030×10=6 cm
Moment of inertia about the center of mass: I=m1r12+m2r22=20×62+30×42=720+480=1200 g·cm² =1.2×10−4 kg·m²
Angular frequency of torsional pendulum: ω=IC=1.2×10−41.2×10−8=10−4=10−2 rad/s
ω=n×10−3 rad/s
n=10.
JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer
