Question

A laminar stream is flowing vertically down from a tap of cross-section area $1{cm}^2$. At a distance $10cm$ below the tap, the cross-section area of the stream has reduced to $1/2{cm}^2$. The volumetric flow rate of water from the tap in litre/min is

Answer 1

Given : $h=10cm=0.1m$ $A_1=1c{m}^2={10}^{-4}{m}^2$ $A_2=0.5c{m}^2=0.5\times {10}^{-4}{m}^2$

$\therefore$ Velocity of the stream at a distance h below the tap $v=\sqrt{\frac{2gh}{1-\frac{A_2^2}{A_1^2}}}$

$\therefore$ $v=\sqrt{\frac{2\left(9.8\right)\left(0.1\right)}{1-(0.5{)}^2}}=1.62m/s$

Volumetric flow of water at the tap is equal to that of at a distance h below the tap.

$\therefore$ $V_{tap}=v{A}_2=1.62\times 0.5\times {10}^{-4}=8.1\times {10}^{-5}{m}^3/s$

$⟹V_{tap}=8.1\times {10}^{-5}\times \frac{1000}{1/60}\approx 5$ Litre/min