A typical rain drop formed in a cloud at an altitude of 800 m reaches the earths surface with a speed of 8 m/s. The ratio of the mechanical energy of the rain drop before and after the trip is

500 : 1

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250 : 1

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125 : 1

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1000 : 1

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Solution

The correct option is B $250 : 1$
M.E before the trip $= P E = m g (800) = M E_{I}$
M.E after the trip $= K . E = \frac{1}{2} m (8)^{2} = M E_{I I}$
$M E_{I} : M E_{I I} = 800 \times 10 : 8 \times 4$ $= 250 : 1$

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