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 $34.3\, m$. At what rate the droplets are coming from the tap ? (Take $g=9.8\, m / s ^{2}$ )

• A. 3 drops / 2 seconds
• B. 2 drops / second
• C. 1 drop / second
• D. 1 drop / 7 seconds

Answer 1

Answer: (C)

In $4\, \sec .1^{\text {st }}$ drop will travel
$\Rightarrow \frac{1}{2} \times(9.8) \times(4)^{2}=78.4\, m$
$\therefore 2^{\text {nd }}$ drop would have travelled
$\Rightarrow 78.4-34.3=44.1\, m$.
Time for $2^{\text {nd }}$ drop
$\frac{1}{2}(9.8) t^{2}=44.1$
$t=3\, \sec$
$\therefore$ each drop have time gap of $1\, \sec$
$\therefore 1$ drop per $\sec$