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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA rigid system consists of a light rod of length \( l \) with two masses \( M/2 \) attached at each end. If the radius of gyration of this system about an axis perpendicular to the rod passing through its centre of mass is \( k_1 \), and the radius of gyration of a thin uniform ring of radius \( R \) about its diameter is \( k_2 \), then the ratio \( k_1 / k_2 \) when \( l = 2R \) is:
View Answer & Solution
Correct answer · Option C
\( \sqrt{2} \)
Explanation
For the dumbbell: \( I_1 = 2\times (M/2)(l/2)^2 = M l^2/4 \implies k_1 = l/2 = R \). For the thin ring about diameter: \( I_2 = \frac{1}{2}MR^2 \implies k_2 = R/\sqrt{2} \). Hence, \( k_1/k_2 = R / (R/\sqrt{2}) = \sqrt{2} \).
