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A body is allowed to fall from a height ‘h’ above the ground. Then match the following. List – I List-II (a) P.E. = K.E. (e) at height h2 (b) P.E. = 2K.E (f) constant at any point (c) K.E = 2P. E (g) at height of 2h3 (d) P.E. + K.E (h) at height of h3

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a

a-e, b-g, c-h, d-f

b

a-g b-e, c-f, d-h

c

a-f, b-g c-e d-h

d

a-e, b-h, c-g, d-f

answer is A.

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Detailed Solution

(a) ⇒ P.E. = mg(height from the ground) KE =12mv2 where v2=2g(displacement from the point where it is dropped) m is mass; g is acceleration due to gravity; v is velocity ***For a freely falling body when it falls through a distance x from a height h its vlocity is given by v=2gx and k.E =mv22=mgx its height from the ground is= h-x k.Ep.E=mg xmgh-x

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A body is allowed to fall from a height ‘h’ above the ground. Then match the following. List – I List-II (a) P.E. = K.E. (e) at height h2 (b) P.E. = 2K.E (f) constant at any point (c) K.E = 2P. E (g) at height of 2h3 (d) P.E. + K.E (h) at height of h3