Question

A cylindrical tank of height $H$ is open at the top end and it has a radius $r$. Water is filled in it up to a height of $h$. The time taken to empty the tank through a hole of radius $r'$ in its bottom is
A. $\sqrt {\dfrac{{2h}}{g}} \dfrac{{{r^2}}}{{{r^{'2}}}}$
B. $\sqrt {\dfrac{{2H}}{g}} \dfrac{{{r^2}}}{{{r^{'2}}}}$
C. $\sqrt h H$
D. none of these

Answer 1

Use the principle that the flow rate of the same liquid from two different holes is the same. Using this principle, derive the formula for the flow rate of lowering the water level in the tank and flow rate of the water from the hole at the bottom of the tank and equate them. Then integrate this equation to calculate the time required to empty the tank.

Formula used:
The expression for the flow rate of the liquid is
$Av = {\text{constant}}$ …… (1)
Here, $A$ is the area of the hole and $v$ is the velocity of the liquid.

Complete step by step answer:
We have given that the cylindrical tank of height $H$ is open at the top end and it has a radius $r$. Water is filled in it up to a height of $h$.The radius of the at the bottom of the tank is $r'$. We have asked to calculate the time required to empty the tank by flowing water from the hole.Let us first determine the velocity of the water from the hole at the bottom of the tank.According to the law of conservation of energy, the potential energy of the upper surface of the water is equal to the kinetic energy of the water coming out of the hole.
$mgh = \dfrac{1}{2}m{v^2}$
$\Rightarrow v = \sqrt {2gh}$

The flow rate of the water at which the water level in the tank is lowering from the tank is given by
${A_1}{v_1} = \left( {\pi {r^2}} \right)\left( { - \dfrac{{dh}}{{dt}}} \right)$
$\Rightarrow {A_1}{v_1} = - \dfrac{{dh}}{{dt}}\pi {r^2}$
The negative sign indicates that the water level in the tank is decreasing.The flow rate of the water from the hole at the bottom of the tank is given by
${A_2}{v_2} = \left( {\pi {r^{'2}}} \right)v$
$\Rightarrow {A_2}{v_2} = \left( {\pi {r^{'2}}} \right)\sqrt {2gh}$
According to equation (1), we can conclude that the rate of lowering the level of the water from the tank is equal to the rate of flow of water from the hole at the bottom of the tank.
${A_1}{v_1} = {A_2}{v_2}$
Substitute $- \dfrac{{dh}}{{dt}}\pi {r^2}$ for ${A_1}{v_1}$ and $\left( {\pi {r^{'2}}} \right)\sqrt {2gh}$ for ${A_2}{v_2}$ in the above equation.
$- \dfrac{{dh}}{{dt}}\pi {r^2} = \left( {\pi {r^{'2}}} \right)\sqrt {2gh}$
$\Rightarrow dt = - \dfrac{{{r^2}dh}}{{{r^{'2}}\sqrt {2gh} }}$
$\Rightarrow dt = - \dfrac{{{r^2}{h^{ - \dfrac{1}{2}}}dh}}{{{r^{'2}}\sqrt {2g} }}$

Integrate both sides of the above equation.
$\Rightarrow \int_0^t {dt} = - \int_0^h {\dfrac{{{r^2}{h^{ - \dfrac{1}{2}}}dh}}{{{r^{'2}}\sqrt {2g} }}}$
$\Rightarrow \int_0^t {dt} = - \dfrac{{{r^2}}}{{{r^{'2}}\sqrt {2g} }}\int_0^h {{h^{ - \dfrac{1}{2}}}dh}$
$\Rightarrow t = - \dfrac{{{r^2}}}{{{r^{'2}}\sqrt {2g} }}\left[ {\dfrac{{{h^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right]_h^0$
$\Rightarrow t = - \dfrac{{2{r^2}}}{{{r^{'2}}\sqrt {2g} }}\left[ {{h^{\dfrac{1}{2}}}} \right]_h^0$
$\Rightarrow t = - \dfrac{{\sqrt 2 {r^2}}}{{{r^{'2}}\sqrt g }}\left[ {0 - \sqrt h } \right]$
$\Rightarrow t = \dfrac{{\sqrt 2 {r^2}\sqrt h }}{{{r^{'2}}\sqrt g }}$
$\therefore t = \sqrt {\dfrac{{2h}}{g}} \dfrac{{{r^2}}}{{{r^{'2}}}}$
Therefore, the time required for the tank to empty is $\sqrt {\dfrac{{2h}}{g}} \dfrac{{{r^2}}}{{{r^{'2}}}}$.

Hence, the correct option is A.

Note: The students should not forget to use the negative sign for the flow rate of decreasing the water level from the cylindrical tank as the level of the water is decreasing. Also the students should keep in mind that the level of the water in the tank is decreasing. Hence, the students should integrate the equation from height h to zero.