Question

A rectangular vessel when full of water, takes $10\min $ to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water?
A. $9\min $
B. $7\min $
C. $5\min $
D. $3\min $

Answer 1

Hint: First of all, we will draw the diagram as per the question for better understanding of the question and then we will find the height of water as well we will find the height at which orifice is placed and then on basis of it we will solve the problem and find the answer by using the equation of continuity which can be given as, ${{A}_{1}}{{V}_{1}}={{A}_{2}}{{V}_{2}}$.

Formula used:
${{A}_{1}}{{V}_{1}}={{A}_{2}}{{V}_{2}}$

Complete answer:
In question it is given that a rectangular vessel when full of water, takes $10\min $ to be emptied through an orifice in its bottom and we are asked to find the time in which half filled tank can be emptied. So, first of all, we will draw the diagram as per the question,
Figure
Now, as shown in the figure,
${{A}_{1}}$ is the area of fluid or liquid in the vessel
${{V}_{1}}$ is the velocity of liquid at point 1 or height h.
${{A}_{2}}$ is the area of the vessel
${{V}_{2}}$ is the velocity of liquid at point 2 or height of the orifice.
Now, the relation between area and velocity can be given by the formula of equation of continuity which can be given as,
${{A}_{1}}{{V}_{1}}={{A}_{2}}{{V}_{2}}$ ……………………..(i)
Now, as the orifice is at the height h then the velocity of water coming out from the orifice i.e. ${{V}_{2}}$ can be given as,
${{V}_{2}}=\sqrt{2gh}$
Now, by substituting the value of ${{V}_{2}}$ in equation (i) we will get,
\[{{A}_{1}}=\dfrac{{{A}_{2}}{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{A}_{2}}}{{{V}_{1}}}\times \sqrt{2gh}\]…………………(i)

Now, the height of liquid changes per second the water is filled and due to that velocity also changes so taking a small component of height we will find the velocity of liquid in vessel which can be given mathematically as,
${{V}_{1}}=\dfrac{dh}{dt}$
Substituting this value in equation (i) we will get,

\[{{A}_{1}}=\dfrac{{{A}_{2}}}{\dfrac{dh}{dt}}\times \sqrt{2gh}\Rightarrow \dfrac{dh}{dt}=\dfrac{{{A}_{2}}}{{{A}_{1}}}\times \sqrt{2gh}\] ……………….(iii)

Now, we will separate the variables which can be given as,

\[\dfrac{dh}{\sqrt{h}}=\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}dt\]
Now, on integrating on both the sides we will get,

\[\int{\dfrac{dh}{\sqrt{h}}}=\int{\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}dt}\]
Now, the limit of h is from 0 to h and the time will be 0 to t, substituting these limits we will get,
\[\int\limits_{0}^{h}{\dfrac{dh}{\sqrt{h}}}=\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}\int\limits_{0}^{t}{dt}\]
Solving it further we will get,

\[\left[ \dfrac{{{h}^{-\dfrac{1}{2}+1}}}{-\dfrac{1}{2}+1} \right]_{0}^{h}=\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}\left[ t \right]_{0}^{t}\]
\[\Rightarrow \left[ \dfrac{{{h}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right]=\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}\left[ t-0 \right]\]
\[\Rightarrow 2\sqrt{h}=\dfrac{{{A}_{2}}}{{{A}_{1}}}\sqrt{2g}t\]

Now, making t as main in the equation and transferring rest of the quantities we will get,
\[t=\dfrac{A}{a}\sqrt{\dfrac{2h}{g}}\] ……………….(iv)

Now, the time for half height can be given as,

\[t'=\dfrac{A}{a}\sqrt{\dfrac{\dfrac{2h}{2}}{g}}\Rightarrow t'=\dfrac{1}{\sqrt{2}}\dfrac{A}{a}\sqrt{\dfrac{gh}{g}}\]
Now, substituting t from expression (iv) we will get,
\[t'=\dfrac{1}{\sqrt{2}}t\]
Now, t is 10 min as per the question so replacing the value of t we will get,
\[t'=\dfrac{1}{\sqrt{2}}10=7.092\cong 7\min \]
Hence, it can be said that a vessel will take $7\min $ to be emptied if the vessel is half filled.

So, the correct answer is “Option B”.

Note:
Students might make mistakes in taking the integration and also in considering the limits of the integration and due to that the sum might go wrong, so students must be careful before taking the limits and solve the problem carefully.