Question

A tank is filled with a liquid upto a height $H$. A small hole is made at the bottom of this tank. Consider $t_1$ be the time taken to empty first half of the tank and $t_2$ be the time taken to empty rest half of the tank. Then, determine the ratio $\frac{t_1}{t_2}$?

• A. $1.33$
• B. $1.5$
• C. $2$
• D. $0.414$

Answer 1

Answer: (D)

$0.414$

The volume of liquid coming out from the hole in time $dt$

where, a is area c/s area of the small hole
$dV=av\times dt=a\sqrt{2gH}\cdot dt$ ($\because dH$ level falls in $dt$ time)

The volume of liquid coming out of hole must be equal $dV=A(-dH)$

$A(-dH)=a\sqrt{2gH}\cdot dt$
Since, ${\int }_0^{t}dt=\frac{A}{a\sqrt{2g}}{\int }_H^0{y}^{-1/2}dy$ ....(i)

Now, substituting the proper limits in equation (i), derived in the theory, we have
${\int }_0^{t_1}dt=\frac{A}{a\sqrt{2g}}{\int }_H^{H/2}{y}^{-1/2}dy$
$\Rightarrow t_1=\frac{2A}{a\sqrt{2g}}{\left[\sqrt{y}\right]}_{H/2}^H$
$\Rightarrow t_1=\frac{2A}{a\sqrt{2g}}[\sqrt{H}-\sqrt{\frac{H}{2}}]$
$\Rightarrow t_1=\frac{A}{a}\sqrt{\frac{H}{g}}(\sqrt{2}-1)$ .....(ii)
Similarly, ${\int }_0^{t_2}dt=\frac{A}{a\sqrt{2g}}{\int }_{H/2}^0{y}^{-1/2}dy$
$\Rightarrow t_2=\frac{A}{a}\sqrt{\frac{H}{g}}$ ....(iii)
From equations (ii) and (iii), we get
$\frac{t_1}{t_2}=\sqrt{2}-1\Rightarrow \frac{t_1}{t_2}=0.414$