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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA non-uniform rod of length \( L = 1.0\,\text{m} \) and weight \( W = 40\,\text{N} \) is balanced horizontally on a knife edge placed at a distance of \( 0.4\,\text{m} \) from end A when a weight of \( 20\,\text{N} \) is suspended from end A. The centre of gravity of the rod is located at a distance from end A equal to:
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Correct answer · Option A
0.60 m
Explanation
Taking torque about the knife edge at \( 0.4\,\text{m} \): clockwise torque = anticlockwise torque. \( 20 \times 0.4 = 40 \times (x_{\text{cg}} - 0.4) \implies 8 = 40(x_{\text{cg}} - 0.4) \implies x_{\text{cg}} - 0.4 = 0.2 \implies x_{\text{cg}} = 0.60\,\text{m} \).
