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**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 a 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 solution:-**

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 filled with water and a hole is made in the bottom. Let us apply Bernoulli’s principle for the surface of the tank and the hole.

${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

Since the fluid is incompressible, there will be no change in the static pressures of the fluid. This means that:

${P_1} = {P_2}$

Since, the water at the surface level of the tank is constant, ${v_1} = 0$

To calculate, Velocity at the hole of the tank – ${v_2}$

Applying the Bernoulli’s equation to this case by substituting, we get –

$\dfrac{1}{2}\rho {v_2}^2 - \dfrac{1}{2}\rho {v_1}^2 = {P_2} - {P_1} + \rho g{h_2} - \rho g{h_1}$

$\dfrac{1}{2}\rho {v_2}^2 = \rho gh$

${v_2}^2 = 2gh$

To find h –

The height difference can be calculated with the help of pressure difference between the surface and at the hole by using the formula below –

$p = \rho gh$

Pressure at surface will be equal to the atmospheric pressure

${p_1} = \rho g{h_1} = 1atm = {10^5}N{m^{ - 2}}$

Pressure at hole as given

${p_2} = \rho g{h_2} = 3atm = 3 \times {10^5}N{m^{ - 2}}$

Density of water, $\rho = 1000kg{m^{ - 3}}$

Assuming the acceleration due to gravity, $g = 10m{s^{ - 2}}$

${p_2} - {p_1} = \rho g\left( {{h_2} - {h_1}} \right) = \rho gh$

Hence,

$h = \dfrac{{{p_2} - {p_1}}}{{\rho g}} = \dfrac{{3 \times {{10}^5} - {{10}^5}}}{{1000 \times 10}} = 20m$

Substituting in the equation for velocity,

${v_2}^2 = 2gh$

${v_2}^2 = 2 \times 10 \times 20 = 400$

$ \Rightarrow {v_2} = \sqrt {400} m{s^{ - 1}}$

Hence, the correct option is Option A.

**Note:-**This is the most fundamental method to solve the problem. However, there is a shortcut to solve this problem. For that, the concept of Torricelli’s law should be applied –

The Torricelli’s law gives the velocity of efflux of water jet from a hole which is at a height h from the surface.

Velocity, $v = \sqrt {2gh} $

So, you can see that instead of deriving the above equation from Bernoulli’s principle, you can directly use Torricelli's equation to find the velocity of efflux.

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