Question

There is a hole at the side-bottom of a big water tank. The area of the hole is $2m{m}^2$. Through it, a pipe is connected. The upper surface of the water is $5m$ above the hole. The rate of flow of water through the pipe is ( in $m^3{s}^{-1}$) ( $g=10m{s}^{-2}$)

• A. $4\times {10}^{-5}$
• B. $4\times {10}^5$
• C. $4\times {10}^{-6}$
• D. $28\times {10}^{-6}$

Answer 1

Answer: (A)

$4\times {10}^{-5}$

Velocity of water at the side bottom hole $=\sqrt{2gh}$$=\sqrt{2\times 10\times 5}=10m/s$
Area of hole = $4m{m}^2=4\times {10}^{-6}{m}^2$
So, rate of flow of water through pipe = $4\times {10}^{-6}\times 10m^3/s$$=4\times {10}^{-5}{m}^{-3}/s$