A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, $R$ is equal to

Step 1: Determine the velocity of efflux for both holes.

According to Torricelli's Law, the velocity of efflux $v$ from a hole at a depth $h$ from the top of a completely filled tank is given by:

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

For the square hole at depth $y$, the velocity $v_1$ is:

$$v_1 = \sqrt{2gy}$$

For the circular hole at depth $4y$, the velocity $v_2$ is:

$$v_2 = \sqrt{2g(4y)}$$ $$v_2 = 2\sqrt{2gy}$$ $$v_2 = 2v_1$$

Step 2: Express the volume flow rates.

The volume flow rate $Q$ is the product of the cross-sectional area of the hole $A$ and the velocity of efflux $v$.

For the square hole of side $L$:

$$Q_1 = L^2 v_1$$

For the circular hole of radius $R$:

$$Q_2 = \pi R^2 v_2$$

Step 3: Equate the flow rates to find $R$.

Since the quantities of water flowing out per second from both holes are the same, we set $Q_1 = Q_2$.

$$L^2 v_1 = \pi R^2 v_2$$

Substitute $v_2 = 2v_1$ into the equation.

$$L^2 v_1 = \pi R^2 (2v_1)$$ $$L^2 = 2\pi R^2$$

Solve for $R$.

$$R^2 = \frac{L^2}{2\pi}$$ $$R = \frac{L}{\sqrt{2\pi}}$$

Answer: The radius $R$ is equal to $\dfrac{L}{\sqrt{2\pi}}$, which corresponds to option (1).