A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water

The correct option is 7 minute.

To find the time taken to empty the vessel when it is half filled with water, we can use Torricelli's Law of Efflux.

According to Torricelli's Law, the velocity of efflux of a liquid flowing out of an orifice at a depth $h$ below the free surface is given by:
$v=\sqrt{2gh}$
where $g$ is the acceleration due to gravity.

Let $A$ be the cross-sectional area of the rectangular vessel, and $a$ be the area of the orifice at the bottom. The rate of decrease of water volume in the vessel is equal to the volume of water flowing out through the orifice per unit time:
$-A\frac{dh}{dt}$
Substituting the velocity of efflux $v$ into this expression:
$-A\frac{dh}{dt}$

Rearranging the terms to integrate with respect to height and time:
$h^{-1/2}dh=-\frac{a\sqrt{2g}}{A}dt$

Let $t$ be the time taken to empty the vessel from an initial height $H$ to a final height of $0$. Integrating both sides with these limits:
${\int }_H^0{h}^{-1/2}dh=-\frac{a\sqrt{2g}}{A}{\int }_0^{t}dt$
Evaluating the integrals:
${\left[2\sqrt{h}\right]}_H^0=-\frac{a\sqrt{2g}}{A}t$
$-2\sqrt{H}=-\frac{a\sqrt{2g}}{A}t$
Solving for $t$:
$t=\frac{2A\sqrt{H}}{a\sqrt{2g}}$

From this equation, we can establish that the time taken to empty the vessel is directly proportional to the square root of the initial height of the liquid:
$t\propto \sqrt{H}$

Let $t_1$ be the time taken to empty the vessel when it is full (initial height $H$):
$t_1=10\text{minutes}$

Let $t_2$ be the time taken to empty the vessel when it is half full (initial height $H/2$):
$\frac{t_2}{t_1}=\sqrt{\frac{H/2}{H}}=\frac{1}{\sqrt{2}}$
$t_2=\frac{t_1}{\sqrt{2}}$

Substituting the value of $t_1=10\text{minutes}$:
$t_2=\frac{10}{1.414}\approx 7.07\text{minutes}$

Therefore, it will take approximately 7 minute to empty the vessel when it is half filled with water.