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 = (Assume R and L << y)
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 = (Assume R and L << y)
Solution
The problem is based on Torricelli's law, which states that the speed of efflux (v) is given by √(2gh), where g is the acceleration due to gravity and h is the height of the liquid column above the point of efflux.
The volume flow rate (Q) is given by the area of the hole (A) times the speed of efflux (v), or Q = Av.
For the square hole, the area A = L^2 and the height h = y. So, the volume flow rate Q1 = L^2 * √(2gy).
For the circular hole, the area A = πR^2 and the height h = 4y. So, the volume flow rate Q2 = πR^2 * √(2g*4y) = πR^2 * 2√(2gy).
The problem states that Q1 = Q2, so we can set the two equations equal to each other and solve for R:
L^2 * √(2gy) = πR^2 * 2√(2gy)
Divide both sides by √(2gy) to get:
L^2 = 2πR^2
Solve for R to get:
R = L/√(2π)
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