Question

A cylindrical tank is filled with water to a level of 3 m. A hole is opened at a height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross sectional area of the cylinder is 0.1. Find the square of the velocity (in $m^2/{\sec }^2$) with which water is coming out.
$(g=10m/{\sec }^2)$

Answer 1

Applying Bernoulli's theorem and equation of continuity for the stated case –
$a{v}_e=AV$ or $V=\frac{a{v}_e}{A}$...(1)
and $P_0+H\rho g+\frac{1}{2}\rho {\left(\frac{a{v}_e}{A}\right)}^2=P_0+\frac{1}{2}\rho v_e^2$
$v_e^2=\frac{2Hg}{1-{\left(\frac{a}{A}\right)}^2}$
$=\frac{2\times (3-0.525)\times 10}{1-(0.1{)}^2}$
$=50m^2/{\sec }^2$

Why this question? Caution: The formula $v=\sqrt{2gh}$ can only be used when area of hole is negligible as compared to area of cross-section of container.