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JEE Advanced 2023 · Paper 1, Physics question 13

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−11.2\times10^{-8}\ \mathrm{N\,m\,rad^{-1}}. The angular frequency of the oscillations in n×10−3 rad s−1n\times10^{-3}\ \mathrm{rad\,s^{-1}}. The value of nn is _____.

A vertical wire hangs from a rigid ceiling (curved double arrow on the wire indicates twisting). Its lower end is attached to a horizontal massless rod at the system's centre of mass; a 30 gm mass is at the left end of the rod and a 20 gm mass at the right end. Gravity g points down.
Answer and worked solution

Official answer: 10

Distances from the centre of mass. The rod is 10 cm long:

  • 20 g mass: 3050×10=6\frac{30}{50}\times10=6 cm;
  • 30 g mass: 44 cm.

Moment of inertia.
I=0.02(0.06)2+0.03(0.04)2=7.2×10−5+4.8×10−5=1.2×10−4 kg m2.I=0.02(0.06)^2+0.03(0.04)^2=7.2\times10^{-5}+4.8\times10^{-5}=1.2\times10^{-4}\ \text{kg m}^2.

Frequency.
ω=κI=1.2×10−81.2×10−4=10−2=10×10−3 rad/s.\omega=\sqrt{\frac{\kappa}{I}}=\sqrt{\frac{1.2\times10^{-8}}{1.2\times10^{-4}}}=10^{-2}=10\times10^{-3}\ \text{rad/s}.

Topics:Simple harmonic motionMoment of inertiaCentre of mass
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