Work Energy And Power Question 329

Question: The particle is released from a height h. At a certain height, its KE is two times its potential energy Height and speed of the particle at that instant are

Options:

A) $ \frac{h}{3},\frac{\sqrt{2gh}}{3} $

B) $ \frac{h}{3},\frac{\sqrt{gh}}{3} $

C) $ \frac{2h}{3},\frac{\sqrt{2gh}}{3} $

D) $ \frac{h}{3},\sqrt{2gh} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Total mechanical energy = mgh

As, $ \frac{KE}{PE}=\frac{2}{1} $

$ \frac{KE}{PE}=\frac{2}{1} $ mgh and $ PE=\frac{1}{3}mgh $

Height from the ground at this instant,

$ h’=\frac{h}{3}, $ and speed of particle at this instant,

$ v=\sqrt{2g(h-h’)}=\sqrt{2g( \frac{2h}{3} )}=2\sqrt{\frac{gh}{3}} $



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