Physics - Fluid Question with Solution | TestHub

The rate of flow of water is $Q=\pi \times 10^{-1}{\mathrm{m}}^3/\mathrm{s}$.

The radius of the pipe is $r=10\mathrm{cm}=0.1\mathrm{m}$.

The cross-sectional area is $A=\pi r^2=\pi (0.1{)}^2=\pi \times 10^{-2}{\mathrm{m}}^2$.

Using the relation for the rate of flow, $Q=Av$.

$v=\frac{Q}{A}$

$v=\frac{\pi \times 10^{-1}}{\pi \times 10^{-2}}$

$v=10\mathrm{m}/\mathrm{s}$

Answer: $10\mathrm{m}/\mathrm{s}$