A) \[2\pi \sqrt{\frac{L}{g}}\]
B) \[2\pi \sqrt{\frac{(L-x)}{g}}\]
C) \[2\pi \sqrt{\frac{(L+x)}{\mu g}}\]
D) \[2\pi \sqrt{\frac{L}{\mu g}}\]
Correct Answer: D
Solution :
[d] \[{{R}_{Q}}-{{R}_{P}}=\frac{mgx}{L}\] Frictional force at Q will be more than frictional force at P, and it will bring back the rod. Restoring force. \[F=-({{f}_{Q}}-{{f}_{p}})=-\mu ({{R}_{Q}}-{{R}_{P}})=-\frac{\mu mgx}{L}\] So \[K=\frac{\mu mg}{L};T=2\pi {{\left( \frac{m}{K} \right)}^{1/2}}=2\pi \sqrt{\frac{L}{\mu g}}\]
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