Question

A cylindrical tank stands on a frictionless surface. The seal over a circular hole of radius $0.5\,cm$ in the wall of the tank ruptures when the level of water above the hole is $1\,m$. The force that a person must apply on the cylinder to keep it from being set in motion is:
(A) $1.54 \times {10^{ - 3}}\,N$ forward
(B) $1.54 \times {10^{ - 3}}\,N$ backward
(C) $1.54\,N$ forward
(D) $1.54\,N$ backward

Answer 1

The force can be determined by using the force of the fluid formula. To find the force, the density of the liquid, the velocity of the liquid, and the area of the orifice is required, so the velocity of the liquid is determined by using Torricelli's theorem. Then, the force of the fluid is determined.

Formula used:
Torricelli's theorem,
$V = \sqrt {2 \times g \times H}$
Where, $V$ is the velocity of the fluid, $g$ is the acceleration due to gravity and $H$ is the height of the fluid
The force of the fluid,
$F = \rho \times A \times {V^2}$
Where, $F$ is the force, $\rho$ is the density of the fluid, $A$ is the area of the water jet and $V$ is the velocity of the fluid

Complete step by step answer:
Given that,
The circular hole of radius, $r = 0.5\,cm$,
The height of the fluid, $H = 1\,m$.
By using Torricelli's theorem,
$V = \sqrt {2 \times g \times H} \,...............\left( 1 \right)$
By substituting the acceleration due to gravity and the height of water in the equation (1), then,
$\Rightarrow V = \sqrt {2 \times 10 \times 1} \,m{s^{ - 1}}$
On multiplying the above equation is written as,
$\Rightarrow V = \sqrt {20} \,m{s^{ - 1}}$
The area of the cylinder is,
$A = \pi {r^2}\,..................\left( 2 \right)$
By substituting the radius in the above equation, then
$A = \pi \times {\left( {0.5 \times {{10}^{ - 2}}} \right)^2}$
On multiplying, then the above equation is written as,
$\Rightarrow A = 0.25 \times \pi \times {10^{ - 4}}\,{m^2}$
The force of the fluid,
$F = \rho \times A \times {V^2}\,..................\left( 3 \right)$
On substituting the density of the water, area, and the velocity of the water in the above equation, then
$\Rightarrow F = 1000 \times 0.25 \times \pi \times {10^{ - 4}} \times {\left( {\sqrt {20} } \right)^2}$
On multiplying the above equation, then the above equation is written as,
$\Rightarrow F = 1.539\,N$
$\Rightarrow F \simeq 1.54\,N$
Thus, the above equation shows the force that is applied by a person.

$\therefore$ The force is 1.54N in forward direction. Hence, option (C) is the correct answer.

Note:
Here the direction of force is forward because the force value is positive, if the force value is negative then the direction is backward. While substituting the radius value in the area formula, it is converted to meter because the height of the water is given in meters.