Question:
Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{\text {th }}$ second after its fall to the next droplet is $34.3 \mathrm{~m}$. At what rate the droplets are coming from the tap ? (Take $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ )
Correct Option: , 3
Solution:
In 4 sec. $1^{\text {st }}$ drop will travel
$\Rightarrow \frac{1}{2} \times(9.8) \times(4)^{2}=78.4 \mathrm{~m}$
$\therefore 2^{\text {nd }}$ drop would have travelled
$\Rightarrow 78.4-34.3=44.1 \mathrm{~m}$
Time for $2^{\text {nd }}$ drop
$\frac{1}{2}(9.8) \mathrm{t}^{2}=44.1$
$t=3 \mathrm{sec}$
$\therefore$ each drop have time gap of $1 \mathrm{sec}$
$\therefore 1$ drop per sec