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Law of Conservation of Mechanical Energy

A stone of ma...

Question

A stone of mass $6 k g$ is revolved in a vertical circle of diameter $6 m$ , such that its speed at a point is minimum . If the K.E at the same point is $250 J$ , then the P.E. at this point is

A

200 J

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B

150 J

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C

100 J

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D

450 J

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Solution

The correct option is A $200 J$
Given: $K . E = 250 J$
But, $T . E . = \frac{5}{2} m g r = 450 J$
$\Rightarrow P . E . = 450 - 250 = 200 J$

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