Alaminar stream is flowing vertically down from a tap of cross-section area $1$ $cm^2$. At a distance $10 $ $cm$ 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 must be about ........ $litre/\min$
Diffcult
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By Bernoulli's equation
$\mathrm{P}_{0}+\rho \mathrm{gh}+1 / 2 \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{0}+1 / 2 \rho \mathrm{v}_{2}^{2}$
By continuity equation
$A_{1} v_{1}=A_{2} v_{2}$
$v_{1}=\frac{v_{2}}{2}$
$g h+1 / 2\left(\frac{v_{2}}{2}\right)^{2}=\frac{1}{2} v_{2}^{2}$
$g h=\frac{v_{2}^{2}}{2}[1-1 / 4]=3 / 8 v_{2}^{2}$
$v_{2}=\sqrt{\frac{8 g h}{3}} \Rightarrow v_{1}=\sqrt{\frac{2 g h}{3}}=\sqrt{\frac{2}{3}}$
volume flow rate $=\sqrt{\frac{2}{3}} \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}$
$=60 \sqrt{\frac{2}{3}} \times 10^{-4} \mathrm{m}^{3} / \mathrm{min}$
$=60 \sqrt{\frac{2}{3}} \times \frac{1}{10} \mathrm{litre} / \mathrm{min}$
$=4.9$ $litre/min$
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