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Physics · WORK, ENERGY AND POWER
हिन्दीA bob of mass \( m \) is tied to a light string of length \( L \) and whirled in a vertical circle with the minimum required speed to complete the loop. The ratio of its kinetic energy at the horizontal position B to that at the topmost point C, \( \frac{K_B}{K_C} \), is:
Correct answer · Option 1
\( 3 : 1 \)
Explanation
At position B, \( v_B = \sqrt{3gL} \implies K_B = \frac{3}{2}mgL \). At position C, \( v_C = \sqrt{gL} \implies K_C = \frac{1}{2}mgL \). Thus, \( \frac{K_B}{K_C} = \frac{3}{1} \).
