Question

A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in ${t_1}$ seconds and the remaining three-fourths of the tank is emptied in ${t_2}$ seconds . Then the ratio $\dfrac{{{t_1}}}{{{t_2}}}$ is:
A. $\sqrt 3$
B. $\sqrt 2$
C. $\dfrac{1}{{\sqrt 2 }}$
D. None of these

Answer 1

In this question we will simply put the given data in the general formula for time taken to empty the tank from water level $H$ to $h$ and then we will compare both the time to get the required ratio. Or in other way first we will have to calculate the initial velocity and then put the entire value in the equation of motion to get the time and then compare.

Formula used:
$T = \dfrac{{2A}}{{a\sqrt {2g} }}(\sqrt H - \sqrt h )$
Where, $A$ and $a$ are the area of the tank and opening in the tank, $H$ is the initial liquid level in the tank and $h$ is the final water level in the tank.

Complete step by step answer:
Let, $H$ the total height of liquid, time taken to empty the $\dfrac{H}{4}$ is ${t_1}$. We know that the general equation for time required to empty the tank from water level $H$ to $h$ is given by,
$T = \dfrac{{2A}}{{a\sqrt {2g} }}(\sqrt H - \sqrt h )$
Therefore, one-fourth of the tank is emptied in ${t_1}$ seconds is,
${t_1} = \dfrac{{2A}}{{a\sqrt {2g} }}\left( {\sqrt H - \sqrt {\dfrac{{3H}}{4}} } \right) \\\ \Rightarrow {t_1} = \dfrac{{2A}}{{a\sqrt {2g} }}\sqrt H \left( {\dfrac{{2 - \sqrt 3 }}{2}} \right) \\\$
Similarly, the remaining three-fourths of the tank is emptied in ${t_2}$ seconds is,
${t_2} = \dfrac{{2A}}{{a\sqrt {2g} }}\left( {\sqrt {\dfrac{{3H}}{4}} - 0} \right) \\\ \Rightarrow {t_2} = \dfrac{{2A}}{{a\sqrt {2g} }}\left( {\dfrac{{\sqrt {3H} }}{2}} \right) \\\$
So, now the ratio of $\dfrac{{{t_1}}}{{{t_2}}}$ is,
$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{\dfrac{{2A}}{{a\sqrt {2g} }}\sqrt H \left( {\dfrac{{2 - \sqrt 3 }}{2}} \right)}}{{\dfrac{{2A}}{{a\sqrt {2g} }}\left( {\dfrac{{\sqrt {3H} }}{2}} \right)}} \\\ \therefore \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}$
So, now the ratio of $\dfrac{{{t_1}}}{{{t_2}}}$ is $\dfrac{{2 - \sqrt 3 }}{{\sqrt 3 }}$.

Hence, the correct option is D.

Note: Note that initial velocity of liquid when it will start flow is given by $\sqrt {2gx}$ where $x$ is the complete height of liquid. And if there will be two fluids then they will have two different densities and then in that case we can use Bernoulli’s equation to answer that and in that case the liquid with denser volume will come first.