Question

A vessel of area of cross section A has liquid to a height H. There is a hole at the bottom of the vessel having area of cross-section a. The time taken to decrease the level from ${{H}_{1}}$to ${{H}_{2}}$ will be:
(A) $\dfrac{A}{a}\sqrt{\dfrac{2}{g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}]$
(B) $\sqrt{2gh}$
(C) $\sqrt{2gh({{H}_{1}}-{{H}_{2}})}$
(D) $\dfrac{A}{a}\sqrt{\dfrac{g}{2}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}]$

Answer 1

It is given that there is a vessel and it is filled with water up to a certain height. When a small hole is made at the bottom water will come out of it and we need to find the time for the level to decrease. So, we have to use Bernoulli’s principle.

Complete step by step answer:
Torricelli's law, gives us the velocity of the liquid coming out of the hole which is $\sqrt{2gh}$where h= ${{H}_{2}}-{{H}_{1}}$
So, the rate of discharge of liquid is calculated by multiplying the velocity with the area, $\sqrt{2gh}\times a$
Let in time dt level of water in the tank decreases by dh.
Volume is given by area$\times$height, so Equating volumes, we get
$Adh=dt\times \sqrt{2gh}\times a$
$dt=\dfrac{A}{a\sqrt{2g}}\dfrac{dh}{\sqrt{h}}$
This is an equation and if we integrate the LHS we will get the total time. So,

& t=\dfrac{A}{a\sqrt{2g}}\int\limits_{{{H}_{2}}}^{{{H}_{1}}}{\dfrac{dh}{\sqrt{h}}} \\\ & =\dfrac{A}{a\sqrt{2g}}\int\limits_{{{H}_{2}}}^{{{H}_{1}}}{{{h}^{-\dfrac{1}{2}}}dh} \\\ & =\dfrac{2A}{a\sqrt{2g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}] \\\ & =\dfrac{2A}{a\sqrt{2g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}] \\\ & =\dfrac{2A}{a\sqrt{2g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}] \\\ & =\dfrac{A\sqrt{2}}{a\sqrt{g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}] \\\ \end{aligned}$$ $$=\dfrac{A}{a}\sqrt{\dfrac{2}{g}}[\sqrt{{{H}_{1}}}-\sqrt{{{H}_{2}}}]$$ This is equal to the option given in part (D). So, the correct option is part(D) **Additional Information:** In Torricelli’s theorem the velocity of the liquid coming out of the hole of the container at the base is equal to the velocity of a body attained during the free fall. **Note:** During integration the limits are to be applied after doing the integration. Also if the area is exposed to the atmosphere then the pressure on that surface is equal to the atmospheric pressure.