Question

A large open tank is filled with water upto a height H. A small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $\frac{H}{n}(n>1)$ and it takes $T_2$ time to take out the remaining water. If $T_1=T_2,$ then the value of $n$ is :

• A. $2$
• B. $3$
• C. $4$
• D. $2\sqrt{2}$

Answer 1

The correct option is C $4$

If $A_0$ is the area of the orifice at a depth y below the free surface and A that of the container, the volume of the liquid coming out of the orifice(hole) per second is
$\frac{dV}{dt}=v{A}_0=A_0\sqrt{2gy}$ ....(1)
Since $dV=-Ady$, substituting in (1) we get,

$-A\frac{dy}{dt}=A_0\sqrt{2gy}$

$\therefore {\int }_0^{T_1}dt=-\frac{A}{A_0}(2g{)}^{-\frac{1}{2}}{\int }_H^{\frac{H}{n}}{y}^{-}\frac{1}{2}dy$

$\Rightarrow T_1=\frac{A}{A_0}\sqrt{\frac{2}{g}}(\sqrt{H}-\sqrt{\frac{H}{n}})$ ...(4)

For the liquid to go from height $\frac{H}{n}$ to 0 time $T_2$ ia taken as given.

$\therefore T_2=\frac{A}{A_0}\sqrt{\frac{2}{g}}(\sqrt{\frac{H}{n}}-0)$ .....(5)

Given $T_1=T_2$

$\Rightarrow \frac{A}{A_0}\sqrt{\frac{2}{g}}(\sqrt{H}-\sqrt{\frac{H}{n}})=\frac{A}{A_0}\sqrt{\frac{2}{g}}(\sqrt{\frac{H}{n}}-0)$

$\therefore 2\sqrt{\frac{H}{n}}=\sqrt{H}$

$\Rightarrow \frac{4}{n}=1$

$\Rightarrow n=4$