Question

A tank full of water has a small hole at its bottom. Let $t_1$ be the time taken to empty first half of the tank and $t_2$ be the time needed to empty rest half of the tank, then

• A. $t_1=t_2$
• B. $t_1>t_2$
• C. $t_1=0.414t_2$
• D. $t_1=\frac{t_2}{\sqrt{2}}$

Answer 1

The correct option is C $t_1=0.414t_2$

We know that
time take to empty tank of height $H$ is $T_{H\to 0}=\frac{A}{a}\sqrt{\frac{2H}{g}}$
time take to empty half filled tank of height $H$ is $T_{H/2\to 0}=t_2=\frac{A}{a}\sqrt{\frac{H}{g}}$


Now,
$T_{H\to 0}=T_{H\to H/2}+T_{H/2\to 0}$
$\frac{A}{a}\sqrt{\frac{2H}{g}}=t_1+\frac{A}{a}\sqrt{\frac{H}{g}}$
$t_1=(\sqrt{2}-1)t_2=0.414t_2$