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Q.

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-h,   c-g,   d-f

b

a-g    b-e,    c-f,   d-h

c

a-f,    b-g   c-e    d-h

d

a-e,    b-g,  c-h,  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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