Question

A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time $T$. If the tank is half-filled with water, it will be emptied in time:

• A. $\frac{T}{\sqrt{2}}$
• B. $\frac{T}{\sqrt{3}}$
• C. $\frac{T}{2}$
• D. $\frac{T}{2\sqrt{2}}$

Answer 1

The correct option is A $\frac{T}{\sqrt{2}}$

Let the actual Height of the tank be
$H$ and its uniform cross sectional area be $A$.

Now let us consider that at certain moment the height of water level be
$h$ and the velocity of water emerging through the orifice of cross sectional area $a$ at the bottom of the tank be $v$.As the surface of water and the orifice are in open atmosphere, then by Bernoulli's theorem we have

$v=\sqrt{2gh}$

Let $dh$ represents decrease in water level during infinitesimally small time interval $dt$ when the water level is at height $h$. So the rate of decrease in volume of water will be=
$-A\frac{dh}{dt}$

Again the rate of flow of water at this moment through the orifice is given by $v\times a=a\sqrt{2gh}$.

These two rate must be same by the principle of cotinuity.

Hence $a\sqrt{2gh}=-A\frac{dh}{dt}$

⇒$dt=-\frac{A}{a\sqrt{2g}}\times h^{\frac{-1}{2}}dh$

If the tank is filled to the brim then height of water level will be $H$ and time required to empty the tank $T$ can be obtained by integrating the above relation.

$T={\int }_0^{T_0}dt=-\frac{A}{a\sqrt{2g}}\cdot {\int }_H^0{h}^{\frac{-1}{2}}dh$

$T=\frac{A\sqrt{2H}}{a\sqrt{g}}$

If the tank be half filled with water and the time to empty it be $T^{{}^{\prime }}$ then

$T^{{}^{\prime }}=\frac{A\sqrt{2H}}{a\sqrt{2g}}$

So $\frac{T^{{}^{\prime }}}{T}=\frac{1}{\sqrt{2}}$

$T^{{}^{\prime }}=\frac{T}{\sqrt{2}}$