JEE Advanced 2023 Paper 1 · Q13 · Moment of Inertia
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 $1.2 \times 10^{-8}$ N·m·rad$^{-1}$. The angular frequency of the oscillations is $n \times 10^{-3}$ rad·s$^{-1}$. The value of $n$ is ___.
Reveal answer + step-by-step solution
Correct answer:10
Solution
Position of CM: from $20$ g mass: $\bar{x} = (30\cdot 10)/50 = 6$ cm from $20$g end, so distances to masses are $6$ cm and $4$ cm. Moment of inertia about CM-axis: $I = 20(6)^2 + 30(4)^2 = 720 + 480 = 1200$ g·cm$^2 = 1.2\times 10^{-4}$ kg·m$^2$. Angular frequency: $\omega = \sqrt{c/I} = \sqrt{1.2\times 10^{-8}/1.2\times 10^{-4}} = \sqrt{10^{-4}} = 10^{-2}$ rad/s $= 10 \times 10^{-3}$ rad/s. So $n = 10$.
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