A) \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\]
B) \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\]
C) \[[M{{L}^{2}}{{T}^{-2}}{{\theta }^{-1}}]\]
D) \[Mgd\]
Correct Answer: C
Solution :
When the block moves vertically downward with acceleration \[\sqrt{\frac{2h}{g}}\] then tension in the cord \[\sqrt{\frac{h}{g}}\] Work done by the cord \[\sqrt{3}\] \[\frac{R}{\sqrt{3}}\] \[\sqrt{\frac{3}{2}}R\]
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