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} $
Show Answer
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} $
BETA
