Question

A cylindrical tank of Height H has a hole on its side. It is kept on a flat surface. Assuming that hole’s area is much smaller than the area of the tank, what should be the distance of the hole below the top surface so that water coming out of the hole travels the maximum horizontal distance at the instant when the height of water is H?

(a) H

(b) H/3

(c) H/2

(d) same for all positions

The question was asked during an internship interview.

Asked question is from Fluids Mechanical Properties topic in section Mechanical Properties of Fluids of Physics – Class 11

Answer 1

The correct choice is (c) H/2

For explanation: Let the hole be at a height x below the top surface.

The distance of hole from the ground will be H-x.

Speed of efflux ‘v’ = \(\sqrt{2gx}\).

Let the time taken for water to reach the ground be ’t’. H-x = 1/2gt^2.

∴ t = \(\sqrt{2/g(H – x)}\).

For maximum range, v*t should be maximum.

∴ d(\(\sqrt{2/g(H – x)}\)*\(\sqrt{2gx}\))/dx = 0

∴ d(\(\sqrt{Hx – x^2}\))/dx = 0

∴ (H-2x)/2 \(\sqrt{Hx – x^2}\) = 0

The numerator should be zero but denominator should be non-zero.

∴ H ≠ x (for denominator to be non-zero)

∴ H = 2x   OR x = H/2.