Question

Water contained in a tank flows through an orifice of diameter $2\text{cm}$. The rate of flow of water through the orifice is

• A. $\pi \sqrt{10}\times {10}^{-4}{\text{m}}^3/\text{s}$
• B. $2\pi \sqrt{5}\times {10}^{-4}{\text{m}}^3/\text{s}$
• C. $\sqrt{10}\times {10}^{-3}{\text{m}}^3/\text{s}$
• D. $3\sqrt{10}\times {10}^{-3}{\text{m}}^3/\text{s}$

Answer 1

The correct option is A $\pi \sqrt{10}\times {10}^{-4}{\text{m}}^3/\text{s}$

We know,
Velocity of efflux,
$V=\sqrt{2gh}=\sqrt{2\times 10\times 0.5}$
$V=\sqrt{10}\text{m/s}$

Also,
Rate of flow of water $=A\times V$
where $A$ is the area of orifice
$\because A=\pi r^2=\pi {\left(\frac{D}{2}\right)}^2=\frac{\pi }{4}\times \left(D^2\right)$
$\Rightarrow A=\frac{\pi }{4}\times (0.02{)}^2=\pi \times {10}^{-4}{\text{m}}^2$
Now, rate of flow of water
$=\pi \times {10}^{-4}\times \sqrt{10}{\text{m}}^3/\text{s}$
$=\pi \sqrt{10}\times {10}^{-4}{\text{m}}^3/\text{s}$