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. $\dfrac{{\sqrt 2 A}}{{\pi {a^2}}}\sqrt {\dfrac{h}{g}}$
D. $\dfrac{A}{{\sqrt 2 \pi {a^2}}}\sqrt {\dfrac{h}{g}}$

Answer 1

Total energy is constant in any process. Therefore we can apply conservation of energy. The potential energy of the water in the tank gets converted into the kinetic energy of the water flowing out of the hole. We can apply the equation of continuity to get the velocity of the water running out of the hole. We know that velocity is distance travelled per unit time, therefore we get time as $t = \dfrac{h}{v}$, where h is the distance to be travelled and v the velocity of water coming out of the hole.

Complete step by step answer:
Let v be the velocity of water coming out of the hole. M is the mass of water in the tank and m is the mass of the water coming out of the hole. Applying conservation of energy,
PE$= KE$
Mgh=$\dfrac{1}{2}m{v^2}$
At the time when the tank is empty, m$= M$
$\Rightarrow v = \sqrt {2gh}$
Let v’ velocity of water in the tank.
Applying principle of continuity we get,
Av’$= \pi {a^2}v$
Substituting the velocity of water coming out of the hole v.
Av’$= \pi {a^2}\sqrt {2gh}$
v’$= \dfrac{{\pi {a^2}\sqrt {2gh} }}{A}$
We know that time is the ratio of distance to speed.
Therefore t$= \dfrac{h}{{v'}}$
$t = \dfrac{h}{{\dfrac{{\pi {a^2}\sqrt {2gh} }}{A}}} = \dfrac{A}{{\sqrt 2 \pi {a^2}}}\sqrt {\dfrac{h}{g}}$

Note: The principle of continuity is most commonly used in questions from fluid. It is simply a mass balance of a fluid flowing through a stationary volume element. It states that the rate of mass accumulation in this volume element equals the rate of mass in minus the rate of mass out.