A) \[m\,(g+\pi \sqrt{2g\,h})\]
B) \[m\,(g+\sqrt{{{\pi }^{2}}g\,h})\]
C) \[m\,\left( g+\sqrt{\frac{{{\pi }^{2}}}{2}g\,h} \right)\]
D) \[m\,\left( g+\sqrt{\frac{{{\pi }^{2}}}{3}g\,h} \right)\]
Correct Answer: A
Solution :
Tension in the string when bob passes through lowest point \[T=mg+\frac{m{{v}^{2}}}{r}=mg+mv\omega \] (\[\because \] v = rw) putting \[v=\sqrt{2gh}\]and \[\omega ==\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \] we get \[T=m\ (g+\pi \sqrt{2gh})\]You need to login to perform this action.
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