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Question

When a body is projected vertically up from the ground with certain velocity, its potential energy and kinetic energy at a point $$A$$ are in the ratio $$2 : 3$$. If the same body is projected with double the previous velocity, then at the same point $$A$$ the ratio of its potential energy to kinetic energy is


A
9:1
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B
2:9
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C
1:9
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D
9:2
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E
3:2
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Solution

The correct option is C $$1 : 9$$
Let in the first case, the total energy at the throwing point be $$5E$$.
$$KE$$ at a particular height $$= 3E$$, potential energy at the height $$= 2E$$
As velocity of thrown is doubled, kinetic energy becomes $$4$$ times.
$$\therefore$$ Total $$KE$$ at the throwing point $$20\ E$$
$$PE$$ at the same height $$= 2E$$ (will remain same as the previous value) and $$KE$$ at the same height $$= 18 E$$
$$\dfrac {PE}{KE} = \dfrac {2E}{18E} = \dfrac {1}{9} = 1 : 9$$.

Physics

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