Water is filled in a container upto height $3\text{ m}$. A small hole of area $a$ is punched in the wall of the container at a height $52.5\text{ cm}$ from the bottom. The cross sectional area of the container is $A$. If $a/A = 0.1$ then $v^2$ is (where $v$ is the velocity of water coming out of the hole).

Step 1: Identify the given parameters.

Convert all quantities to SI units.

$$H = 3\,\text{m}$$ $$h = 52.5\,\text{cm} = 0.525\,\text{m}$$ $$\frac{a}{A} = 0.1$$

Step 2: Apply the equation of continuity.

Relate the velocity of the falling free surface $V$ to the velocity of efflux $v$.

$$A V = a v$$ $$V = \frac{a}{A}v = 0.1v$$

Step 3: Apply Bernoulli's equation.

Apply Bernoulli's equation at the free surface of the water and at the hole.

$$P_0 + \rho g H + \frac{1}{2}\rho V^{2} = P_0 + \rho g h + \frac{1}{2}\rho v^{2}$$

Step 4: Simplify the equation.

Subtract the atmospheric pressure $P_0$ from both sides and divide by the density $\rho$.

$$g(H - h) = \frac{1}{2}\left(v^{2} - V^{2}\right)$$

Step 5: Substitute the relation for $V$ and solve for $v^2$.

Substitute $V = \frac{a}{A}v$ into the simplified equation.

$$g(H - h) = \frac{1}{2}\left[v^{2} - \left(\frac{a}{A}v\right)^{2}\right]$$ $$v^{2}\left[1 - \left(\frac{a}{A}\right)^{2}\right] = 2g(H - h)$$ $$v^{2} = \frac{2g(H - h)}{1 - \left(\frac{a}{A}\right)^{2}}$$

Step 6: Calculate the numerical value of $v^2$.

Substitute the given values using $g = 10\,\text{m/s}^{2}$.

$$H - h = 3 - 0.525 = 2.475\,\text{m}$$ $$1 - \left(\frac{a}{A}\right)^{2} = 1 - (0.1)^{2} = 0.99$$ $$v^{2} = \frac{2(10)(2.475)}{0.99}$$ $$v^{2} = \frac{49.5}{0.99} = 50\,\text{m}^{2}/\text{s}^{2}$$

Answer: $v^{2} = 50\,\text{m}^{2}/\text{s}^{2}$