Question

The area of cross-section of a large tank is $0.5m^2$ It has an opeoing near the bottom having area ofcross-section $1c{m}^2$, A load of 20 kg is applied on the water at the top. Find the velocity of the water coming out of the opening at the time when the height of water level is 50 cm above the bottom. (Takeg$=10m/s^2)$

Answer 1

$\left(\frac{Mg}{A}\right)+\rho gh=\frac{1}{2}\rho v^2$
$\therefore$ $v=\sqrt{\frac{2Mg}{\rho A}+2gh}$
$=\sqrt{\frac{2\times 20\times 10}{0.5\times {10}^3}+2\times 10\times 0.5}$
$=3.28$ m/s