Question

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

(a) L/√2π
(b) 2πL
(c) L
(d) L/2π

Answer 1

Correct option: (a) R = L / √(2π)

The relation between the radius of the circular hole and the side of the square hole in a container, when the flow rates are equal, is found using Torricelli's law. By equating the flow rates and simplifying, the correct relationship is that the radius of the circular hole is equal to the side of the square hole divided by the square root of 2π.

Step-by-step explanation:

The question relates to the topic of fluid dynamics in Physics, specifically to Torricelli's law which describes the flow of water through holes in a container.

To find the relation between the radius R of the circular hole and the side L of the square hole, we can equate the flow rates by using the equation:

Flow rate (Q) = Area of the hole (A) × Velocity of water (v)
For the square hole, Q = L2 × √(2gy)
For the circular hole, Q = πR2 × √(2g×4y)

Since the flow rates are equal, L2√(2gy) = πR2√(2g×4y).

We can simplify this to obtain the relationship between R and L.

By carrying out the simplification, we find that:
R = √t(L2 / (4 πR2))
R = L / √(2π)

Therefore, the correct answer is: (a) R = L / √(2π)