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QJEEAdv23Ph0909: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph0909 Solution: Two point-like objects of masses and are fixed at the two ends…

Two point-like objects of masses 20 gm20 ~\mathrm{gm} and 30 gm30 ~\mathrm{gm} are fixed at the two ends of a rigid massless rod of length 10 cm10 \mathrm{~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 \times 10^{-8} \mathrm{~N} \mathrm{~m} ~\mathrm{rad}^{-1} . The angular frequency of the oscillations in n×10−3 rad s−1n \times 10^{-3} ~\mathrm{rad} ~\mathrm{s}^{-1} . The value of nn is _________ .

JEE Advanced 2023 Paper 1 Online Physics - Rotational Motion Question 6 English

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JEE Advanced 2023 Paper 1 Online Physics - Rotational Motion Question 6 English Explanation

Moment of inertia about the axis of rotation is

I=m1r12+m2r22

Clearly r1=4 cmr_1=4 \mathrm{~cm}

And r2=6 cmr_2=6 \mathrm{~cm}

∴I=(30×10−3×16×10−4)+(20×10−3×36×10−4)⇒I=1200×10−7 kg m2

If the system is rotated by small angle ' θ\theta ', the restoring torque is τ(R)=−kθ\tau_{(R)}=-k \theta

And d2θdt2=−kl⋅θ=−ω2θ=−1.2×10−81200×10−7⋅θ\frac{d^2 \theta}{d t^2}=\frac{-k}{l} \cdot \theta=-\omega^2 \theta=\frac{-1.2 \times 10^{-8}}{1200 \times 10^{-7}} \cdot \theta

∴ω2=10−4

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