Work Energy And Power Question 224

Question: A rod of mass m and length (is made to stand at an angle of $ 60{}^\circ $ with the vertical. Potential energy of the rod in this position is

Options:

A) $ mg\ell $

B) $ \frac{mg\ell }{2} $

C) $ \frac{mg\ell }{3} $

D) $ \frac{mg\ell }{4} $

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Answer:

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=( 1/2 )\sin {30^{o}} $

$ \therefore $ Potential energy of the rod $ =m\times g\times \frac{\ell }{2}\sin {30^{o}} $

$ =m\times g\times \frac{\ell }{2}\times \frac{1}{2}=\frac{mg\ell }{4} $



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