A) \[\pi {{R}^{2}}/T\]
B) \[\pi {{R}^{2}}T\]
C) \[2\pi RT\]
D) \[2RT\]
Correct Answer: D
Solution :
Suppose tension in thread is F, then for small part \[\Delta l\] of thread \[\Delta l=R\theta \] and \[2F\sin \theta /2=2T\Delta l=2TR\theta \] \[\Rightarrow F=\frac{TR\theta }{\sin \theta /2}=\frac{TR\theta }{\theta /2}=2TR(\sin \theta /2\approx \theta /2)\]You need to login to perform this action.
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