Question

The level of water in a tank is 5m high. A hole of area of cross section 1$c{m^2}$is made at the bottom of the tank. The rate of leakage of water from the hole in ${m^3}{s^{ - 1}}$ is ($g = 10m{s^{ - 2}}$):
(A) ${10^{ - 3}}$
(B) ${10^{ - 4}}$
(C) $10$
(D) ${10^{ - 2}}$

Answer 1

Hint: The volume rate of leakage of water coming out from the hole will be the product of velocity of water coming out of the hole and the area of cross section of the hole. Use Bernoulli’s equation.

Complete step by step solution:
We will begin by first determining what the velocity of water (let’s say, v) coming out from the hole would be. For this we are given the height of the water level in the tank, i.e. $h = 5m$and the acceleration due to gravity, i.e.$g = 10m{s^{ - 2}}$.
Using Bernoulli’s equation (here 1 represents the top and 2 represents the bottom where the hole is present)
$ \Rightarrow {P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$
Since air pressure doesn’t change significantly in such small height differences.
Bottom height \[{h_2} = 0\] and velocity of water particles at the top is almost negligible \[{v_1} = 0.\]
So that way we get
$ \Rightarrow $ Velocity of water coming out of the hole: ${v_2} = {\sqrt {2gh} _1} = \sqrt {2 \times 10 \times 5} = \sqrt {100} = 10m{s^{ - 1}}$.
Area of the cross section of the hole is given to be, $A = 1c{m^2} = {10^{ - 4}}{m^2}$.
Rate of leakage of water from the hole will be, $Q = Av$.
$
   \Rightarrow Q = {10^{ - 4}}{m^2} \times 10m{s^{ - 1}} \\
   \Rightarrow Q = {10^{ - 3}}{m^3}{s^{ - 1}} \\
 $
Therefore, option (A) is correct.

Note: The expression from velocity used in the above question comes from the assertion that the kinetic energy coming out of water coming out the hole will be equal to the potential energy of water stored at some height. This principle in fluid mechanics is called Bernoulli’s Equation.