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Q.
A tank is filled with water upto height $H$. When a hole is made at a distance $h$ below the level of water, what will be the horizontal range of water jet?
BHUBHU 2003
Solution:
The sum of the pressure and the total energy per unit volume of the liquid must be the same at the surface of liquid and at every point of orifice.
From Bernoulli's theorem,
$P+0+\rho g H=P+\frac{1}{2} \rho v^{2}+\rho g(H-h)$
where $P$ is atmospheric pressure $\rho$ is density and $v$ is velocity of efflux.
$\therefore \frac{1}{2} \rho v^{2}=\rho g h$
$\Rightarrow v=\sqrt{2 g h}$
After emerging from orifice, liquid falls along a parabolic path.
$ \therefore s =\frac{1}{2} g t^{2} $
$H-h =\frac{1}{2} g t^{2} $
$\Rightarrow t =\sqrt{\frac{2(H-h)}{g}}$
The horizontal distance is
$x =$ horizontal velocity $ \times$ time
$=v \times t$
$=\sqrt{2 g h} \times \sqrt{\frac{2(H-h)}{g}}$
$=2 \sqrt{h(H-h)}$
Note : When the orifice is exactly in the middle of the wall of the vessel, the stream of liquid will fall at a maximum distance equal to the height of liquid in the vessel.
