Question

A small hole in an area of cross-section $2m{m}^2$ is present near the bottom of a fully filled open tank of height $2m$. Taking $g=10m/s^2$, the rate of flow of water through the open hole would be nearly

Answer 1

Step 1: Given data

1. The area of the cross-section of the hole is $a=2\times {10}^{-6}{m}^2.$
2. The height of the tank is $h=2m.$
3. Acceleration due to gravity $g=10m/s^2$.

Diagram

Step 2: Formulae and concept

1. From the concept of fluid dynamics, the velocity of liquid flow from a container is given by, $v=\sqrt{2gh}$, where, h is the height of the liquid level, g is the acceleration due to gravity, and v is the velocity of liquid flow.
2. We know that the rate of liquid flow is $r=Areaofcrosssection\left(a\right)\times velocityofliquidflow\left(v\right)$, i.e. $r=av.$

Step 3: Finding the rate of liquid flow

As we know, the rate of liquid flow is $r=av.$ So,

$$\begin{gathered} r=a\sqrt{2gh}=\left(2\times {10}^{-6}\right)m^2\times \sqrt{\left(2\times 10\times 2\right)}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ orr=\left(2\times {10}^{-6}\right)\times \sqrt{40} \\ orr=12.6\times {10}^{-6}{m}^3/sec. \end{gathered}$$

Therefore, the rate of flow of water through the open hole is $12.6\times {10}^{-6}{m}^3/sec.$