Question

A cylinder of height $h=1m$ has a narrow vertical slit along its wall running up to a length $\ell =40cm$ from the bottom. The width of the slit $b=1mm.$ The cylinder is filled with water with the slit closed. If the force experienced by the vessel immediately after the slit is opened is $\frac{x}{10}N$ , what is the value of $x$ ?

Answer 1

Consider an element of the slit of length $dx$ at a depth $x$ from the top.
Velocity of efflux at $x=\sqrt{2 g x}.$
Force on elementary area $=$ rate of change of momentum $=\left(\right.bdx\rho v\left.\right)v=b\rho dx\left(2 g x\right)$
$\therefore F$ (total force experienced)
$=2b\rho g\displaystyle \int _{h - l}^{h}xdx=b\rho g\ell \left(2 h - \ell \right)$
or or $F=10^{-3} \times 1000 \times 10 \times 0.4(2-0.4)=6.40 N$