Water drops fall at regular intervals from a hole at the bottom of a vessel placed at a high level. The ninth drop is about to fall when the first drop just falls on the floor after being in the air for 2 second. The distance between the 3rd and the 5th drop at this instant is: (taking g = 10 m/s²)

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$\frac{40}{9} m$

$\frac{25}{4} m$

$\frac{5}{16} m$

answer is B.

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Detailed Solution

Let time interval between two drops be T, then
$8 T = 2 \Rightarrow T = \frac{1}{4} s$
Distance between 3rd and 5th drops $= \frac{1}{2} g {(6 T)}^{2} - \frac{1}{2} g {(4 T)}^{2}$
$= \frac{1}{2} g [36 T^{2} - 16 T^{2}]$
$= \frac{1}{2} g \times 20 T^{2}$
$= \frac{1}{2} \times 10 \times 20 \times {(\frac{1}{2})}^{2} = \frac{25}{4} m$