Question

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{\text{th}}\text{second}$ after its fall, to the next droplet, is $34.3\text{m}.$ At what rate the droplets are coming from the tap ? (Take $g=9.8{\text{m/s}}^2$)

• A. $3\text{drops/s}$
• B. $2\text{drops/s}$
• C. $1\text{drop/s}$
• D. $\frac{1}{7}\text{drops/s}$

Answer 1

The correct option is C $1\text{drop/s}$

The distance travelled by a freely falling drop is,
$h=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2(\because u=0)$

In $4\text{sec,}{1}^{\text{st}}$ drop will travel,

$h_1=\frac{1}{2}\times \left(9.8\right)\times (4{)}^2=78.4\text{m}$

$\therefore 2^{\text{nd}}$ drop would have travel,

$h_2=78.4-34.3=44.1\text{m}$

So, time taken by $2^{\text{nd}}$ drop is,

$h_2=\frac{1}{2}\left(9.8\right)t^2=44.1$

$\therefore t=3\text{sec}$

It means each drop have time gap of $1\text{sec}$.

So, drops are falling at a rate of $1\text{drop/s}$

Hence, $\left(\text{C}\right)$ is the correct answer.