A body of mass 500 g is attached to a string of length 1m and whirled in vertical circle about the free end of the string with critical velocity. Net force on the body when the string is in horizontal position is (g=10m/s²)

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$\sqrt{10} N$

5 N

$5 \sqrt{10} N$

10 N

answer is C.

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Detailed Solution

$While moving with critcal velocity at horizontal position its velocity becomes \sqrt{3 gl} and centripetal force F = m \frac{v^{2}}{l} = 3 mg F = \sqrt{T^{2} + W^{2}} = (\sqrt{10}) m g$

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