Question

A cylindrical vessel of cross-section $A$ contains water to a height $h$. There is a hole in the bottom of radius ‘$a$’. The time in which it will be emptied is:
(A) $\dfrac{2A}{\pi {{a}^{2}}}\sqrt{\dfrac{h}{g}}$
(B) $\dfrac{\sqrt{2}A}{\pi {{a}^{2}}}\sqrt{\dfrac{h}{g}}$
(C) $\dfrac{2\sqrt{2}A}{\pi {{a}^{2}}}\sqrt{\dfrac{h}{g}}$
(D) $\dfrac{A}{\sqrt{2} \pi a^{2}} \sqrt{\dfrac{h}{g}}$

Answer 1

Hint
We should know that the law of conservation of energy states that the energy cannot be created nor can it be destroyed but it can only be changed from one form to another. Based on this concept we have to solve this question.

Complete step by step answer
Let the velocities of water in the tank and that coming out of the hole be $\mathrm{V}$ and $\mathrm{V}_{\mathrm{o}}$ respectively. From the conservation of energy, we can determine that-
$\Rightarrow \mathrm{PE}=\mathrm{KE}$
Where, $PE$ is the potential energy while $KE$ is the kinetic energy
$\Rightarrow \text{Mgh}=\dfrac{1}{2\text{mV}_{\text{o}}^{2}}$
where, Mass of water in the tank is denoted by the variable $M$ and the total mass coming out from the hole is denoted as $m$. This means that for time ($t$) $\text{M}=\text{m}$.
Moreover, from the continuity equation we get that,
$\Rightarrow \mathrm{AV}=\pi \mathrm{a}^{2} \mathrm{V}_{\mathrm{o}}$
From the above equation, we get,
$\Rightarrow 2 \mathrm{gh}=\dfrac{\mathrm{A}^{2} \mathrm{V}^{2}}{\left(\pi \mathrm{a}^{2}\right)^{2}}$
$\Rightarrow \text{V}=\dfrac{\pi {{\text{a}}^{2}}\sqrt{2\text{gh}}}{\text{A}}$
Time, $t=\dfrac{h}{V}$
$\therefore \text{t}=\dfrac{\text{A}}{\sqrt{2}\pi {{\text{a}}^{2}}}\sqrt{\dfrac{\text{h}}{\text{g}}}$
Therefore, the correct answer is Option (D).

Note
We know that potential energy is defined as the amount of energy that is stored or conserved in an object or a substance. This energy is stored on the basis of the position, arrangement or the state of the object or the substance. Kinetic energy is defined as the energy that is possessed in an object due to the motion.