A vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from H₁ to H₂ will be

The correct answer is Option A: (A/a){(√2/g)}[√H₁-√H₂].

Let us derive the expression step-by-step using fluid dynamics principles.

Let $y$ be the height of the liquid level in the vessel at any time $t$. The cross-sectional area of the vessel is $A$ and the cross-sectional area of the hole at the bottom is $a$.

According to Torricelli's law, the velocity $v$ of efflux of the liquid flowing out of the hole at depth $y$ is given by:
$v=\sqrt{2gy}$

Using the equation of continuity, the rate of decrease of volume of liquid in the vessel must equal the rate at which volume flows out of the hole:
$-A\frac{dy}{dt}=av$

Substituting the expression for $v$ into the equation:
$-A\frac{dy}{dt}=a\sqrt{2gy}$

Separating the variables $y$ and $t$:
$y^{-1/2}dy=-\frac{a}{A}\sqrt{2g}dt$

To find the total time $T$ taken to decrease the liquid level from $H_1$ to $H_2$, we integrate both sides with the limits of $y$ from $H_1$ to $H_2$ and $t$ from $0$ to $T$:
${\int }_{H_1}^{H_2}{y}^{-1/2}dy=-\frac{a}{A}\sqrt{2g}{\int }_0^{T}dt$

Integrating the left side:
${\left[2\sqrt{y}\right]}_{H_1}^{H_2}=-\frac{a}{A}\sqrt{2g}T$

Applying the limits:
$2\left(\sqrt{H_2}-\sqrt{H_1}\right)=-\frac{a}{A}\sqrt{2g}T$

Reversing the sign on the left side to eliminate the negative sign on the right side:
$2\left(\sqrt{H_1}-\sqrt{H_2}\right)=\frac{a}{A}\sqrt{2g}T$

Solving for $T$:
$T=\frac{A}{a}\frac{2}{\sqrt{2g}}\left[\sqrt{H_1}-\sqrt{H_2}\right]$

Simplifying the constant term $\frac{2}{\sqrt{2g}}=\frac{\sqrt{4}}{\sqrt{2g}}=\sqrt{\frac{2}{g}}$:
$T=\frac{A}{a}\sqrt{\frac{2}{g}}\left[\sqrt{H_1}-\sqrt{H_2}\right]$

Writing this in the format of the options:
$T=(A/a)\left\{\right(\surd 2/g\left)\right\}[\surd H_1-\surd H_2]$