Question

A tank of height 5 m is full of water. There is a hole of cross-sectional area 1 $c{m^2}$ in its bottom. The initial volume of water that will come out from this hole per second is
A. ${10^{ - 3}}\dfrac{{{m^3}}}{s}$
B. ${10^{ - 4}}\dfrac{{{m^3}}}{s}$
C. $10\dfrac{{{m^3}}}{s}$
D. ${10^{ - 2}}\dfrac{{{m^3}}}{s}$

Answer 1

Hint: This problem can be solved by direct application of Bernoulli's principle. Bernoulli’s principle establishes a relationship between the various pressures acting on fluid during the incompressible flow of a fluid. It states that,
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.

Complete step by step answer:
Whenever a fluid is under motion, there are three kinds of pressure acting on it. They are:
i) Static pressure, which is related to the mechanical force.
ii) Pressure due to flow of the fluid
iii) Pressure due to the height at which the fluid is flowing from the ground.
As per Bernoulli's principle, the sum of the above three mentioned pressures is constant at every point in the fluid.
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.
Consider the tank of height 5m and a cross-sectional area of 1 $c{m^2}$. Let us apply Bernoulli’s principle for the surface of the tank and the hole.
${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$
The static pressure at both the surface and tank is the same since no external pressure is added.
So, ${P_1} = {P_2}$
Since, the water at the surface level of the tank is constant, ${v_1} = 0$
If we consider the level of the hole as datum, then ${h_2} = 0$ and the surface of water in the tank is at height of ${h_1} = h = 5m$
To calculate, Velocity at the hole of the tank – ${v_2}$
Applying the Bernoulli’s equation to this case by substituting, we get –
$\Rightarrow \dfrac{1}{2}\rho {v_2}^2 - \dfrac{1}{2}\rho {v_1}^2 = {P_2} - {P_1} + \rho g{h_2} - \rho g{h_1}$
$\Rightarrow \dfrac{1}{2}\rho {v_2}^2 = \rho gh$
$\Rightarrow {v_2}^2 = 2gh$
Taking g = $10m{s^{ - 2}}$,
$\Rightarrow {v_2}^2 = 2 \times 10 \times 5 = 100$
$\Rightarrow {v_2} = 10m{s^{ - 1}}$
The volume flow rate is the volume of water flowing through an area in one second.
Volume flow rate, ${V_2} = A{v_2}$
Area of flow in the hole, $A = 1c{m^2} = {10^{ - 4}}{m^2}$

Therefore, ${V_2} = 10 \times {10^{ - 4}} = {10^{ - 3}}{m^3}{s^{ - 1}}$. Hence, the correct option is Option A.

Note:
In the Bernoulli’s equation, if we multiply the whole equation by the volume of the fluid (V), we get
$PV + \dfrac{1}{2}\rho V{v^2} + \rho Vgh = C$
The product of volume and density is equal to mass.
$\Rightarrow PV + \dfrac{1}{2}m{v^2} + mgh = C$
As you can see, they get converted into energies – internal energy, kinetic energy, and potential energy. Their total sum at any instant is constant. This is nothing but the principle of conservation of energy. Hence, Bernoulli’s principle agrees to the fundamental law of conservation of energy.