Question

A large open tank has two holes in the wall. One is a square hole of side $2\text{cm}$ at a depth of $5\text{cm}$ from the top and the other is a circular hole of radius $R$ at a depth of $20\text{cm}$ from the top. When the tank is completely filled with water, the quantity of water flowing out per second from both the holes is same. Then, what is the value of $R$?

• A. $2\times {10}^{-3}\text{m}$
• B. $4\times {10}^{-3}\text{m}$
• C. $6\times {10}^{-3}\text{m}$
• D. $8\times {10}^{-3}\text{m}$

Answer 1

The correct option is D $8\times {10}^{-3}\text{m}$

Given that,
Side of square hole $\left(L\right)=2\text{cm}$
Side of circular hole $=R$
Depth of square hole from top $\left(h_1\right)=5\text{cm}$
Depth of circular hole $\left(h_2\right)=20\text{cm}$

The velocity of efflux is given by
$v=\sqrt{2gh}$
Since volume of water flowing out per second is same,
By using continuity equation,
$a_1{v}_1=a_2{v}_2$
$\Rightarrow L^2\left(\sqrt{2g{h}_1}\right)=\pi R^2\left(\sqrt{2g{h}_2}\right)$
$\Rightarrow \left(0.02{)}^2\right(\sqrt{2g\times 0.05})=\pi R^2\sqrt{2g\times 0.20}$
$\Rightarrow R^2=\frac{\left(0.02{)}^2\right(\sqrt{2g\times 0.05})}{\pi \sqrt{2g\times 0.20}}$
$\Rightarrow R=\frac{0.02}{\sqrt{2\pi }}=7.97\times {10}^{-3}\text{m}\cong 8\times {10}^{-3}\text{m}$.