Q. A cylindrical tank is filled with water to level of $3 \,m$. A hole is opened at 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 $0.1$. The square of the speed with which water is coming out from the orifice is $(Take \,g \,= \,10\, ms^{-2})$
Solution:
Suppose A be the area of cross section of tank, a be the area of hole, v$_{e}$ be the velocity of efflux, h be the height of liquid above the hole,
Let v be the speed with which the level decreases in the container. Using equation ofcontinuity, we get
av$_{e}$ Av or $v=\frac{av_{e}}{A}$
Using Bernoulli’s theorem, we have
$P_{0}+h\rho g+\frac{1}{2}\rho v^{2}=P_{0}+\frac{1}{2}\rho v^{2}_{e}$
$h\rho g+\frac{1}{2}\rho\left(\frac{av_{e}}{A}\right)^{^2}=\frac{1}{2}\rho v^{2}_{e}$
or $\quad v^{2}_{e}=\frac{2hg}{1-\left(a^{2}/A^{2}\right)}=\frac{2\times\left(3-0.525\right)\times10}{1-\left(0.1\right)^{2}}$
= 50 m$^{2}$ s$^{-2}$
