A) \[mg\,\ell \]
B) \[\frac{mg\,\ell }{2}\]
C) \[\frac{mg\,\ell }{3}\]
D) \[\frac{mg\,\ell }{4}\]
Correct Answer: D
Solution :
[d] For any uniform rod, the mass is supposed to be concentrated at its centre. \[\therefore \] height of the mass from ground is \[h=\left( 1/2 \right)\sin {{30}^{\operatorname{o}}}\] \[\therefore \] Potential energy of the rod \[=m\times g\times \frac{\ell }{2}\sin {{30}^{\operatorname{o}}}\] \[=m\times g\times \frac{\ell }{2}\times \frac{1}{2}=\frac{mg\ell }{4}\]You need to login to perform this action.
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