Question

A water tap leaks such that water drops fall at regular intervals. Tap is fixed $5m$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

• A. $3m$
• B. $4.5m$
• C. $3.75m$
• D. $2.5m$

Answer 1

Answer: (C)

Velocity of drop 5m below the tap can be obtained by energy conservation.
Assume that potential energy is zero 5m below the tap.
Now, $v=\sqrt{2gh}=10m/s$
Now, time taken to achieve this velocity$=\frac{10}{10}=1s$
it is given that drops leave the tap at a regular time interval. Thus, second drop will leave at 0.5s as the third drop is leaving at 1s.
Now, total time of motion of second drop is 0.5s
Thus, distance moved by second drop in 0.5s is
$s=\frac{1}{2}\times 10\times {\left(0.5\right)}^2=1.25$
Thus height from ground level is $5-1.25=3.75m$