A) \[\frac{1}{2\pi }\frac{\sqrt{l}}{g}\]
B) \[\upsilon \,\sin \frac{\theta }{2}\]
C) \[\sqrt{mg}\]
D) zero
Correct Answer: D
Solution :
From the formula the work done is the product of force and displacement. As displacement after one complete oscillation of simple pendulum is zero. Hence, work done will also be zero.
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