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

Answer: (C)

Step 1: Given that:

The spacing between a droplet observed at $4^{th}$ second after its fall to the next droplet= $34.3m$

Acceleration due to gravity

$g=9.8m{s}^{-2}$

Step 2: Formula used:

The distance covered by a body while moving downward from a height is given by

$h=ut+\frac{1}{2}g{t}^2$

Where $u$ is the initial velocity of the body and $t$ is the time taken by the body.

For free fall, $u=0$

$h=\frac{1}{2}g{t}^2$

Step 3: Calculation of the rate at which the water droplets are coming from the tap:

The distance covered by the $1^{st}$ water droplet from the tap in 4 sec will be

$s_4=\frac{1}{2}\times 9.8m{s}^{-2}\times {\left(4s\right)}^2$

$s_4=4.9\times 16$

$s_4=78.4m$

The distance that would be covered by the $2^{nd}$ droplet

= The distance covered by the $1^{st}$ water droplet from the tap in 4 sec $-$ The spacing between the a droplet observed $4^{th}$ second after its fall to the next droplet

= $78.4m-34.3m$

= $44.1m$

Now, for the $2^{nd}$ droplet

$44.1m=\frac{1}{2}\times 9.8m{s}^{-2}\times {\left(t\right)}^2$

$4.9\times {\left(t\right)}^2=44.1$

$t^2=\frac{44.1}{4.9}$

$t^2=9$

$t=3sec$

Since the first droplet comes out in 4sec and the second droplet in 3 sec.

Therefore

The rate of the water droplet coming out of the tap is $1$ drop/sec.