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Q.

A body of mass m kg is rotating in a vertical circle at the end of a string of length r metre. The difference in the kinetic energy at the top and the bottom of the circle is

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a

mg / r

b

2 mg / r

c

2 mgr

d

mgr

answer is C.

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Detailed Solution

K.E. at top  = 12mv2 K.E. at the bottom = 12mu2Conserving mechanical  energy , 12mu2 = 12mv2 +mg2r ⇒12mu2 - 12mv2 = 2mgr
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