At what speed, the velocity head of water is equal to pressure head of $40 \,cm$ of mercury is ........ $m / s$
Diffcult
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(b)
$\frac{1}{2} \rho_{\text {water }} V^2=\rho_{\text {mercury }} g h$
$V=\sqrt{2 \times \frac{\rho_{\text {mercury }}}{\rho_{\text {water }}} \times g \times h}$
$=\sqrt{2 \times 13.6 \times 9.8 \times \frac{40}{100}}$
$\Rightarrow V=10.32 \,m / s$
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