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Consider a water jar of radius $R$ that has water filled up to height $H$ and is kept on a stand of height $h$ (see figure). Through a hole of radius $r(r < < R)$ at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is $x$. Then
$x = r{\left( {\frac{H}{{H + h}}} \right)^2}$
$x = r{\left( {\frac{H}{{H + h}}} \right)^{\frac{1}{2}}}$
$x = r\left( {\frac{H}{{H + h}}} \right)$
$x = r{\left( {\frac{H}{{H + h}}} \right)^{\frac{1}{4}}}$
Solution
$\frac{1}{2} \rho v_{1}^{2}+\rho g h=\frac{1}{2} \rho v_{2}^{2}$
$v_{1}^{2}+2 g h=v_{2}^{2}$
$2 g H+2 g h=v_{2}^{2}$
$a_{1} v_{1}=a_{2} v_{2}$
$\pi r^{2} \sqrt{2 g h}=\pi \lambda^{2} v_{2}$
$\frac{r^{2}}{x^{2}} \sqrt{2 g h}=v_{2}$
$2 g H+2 g h=\frac{r^{4}}{x^{4}} 2 g h$
$x=r\left[\frac{H}{H+h}\right]^{\frac{1}{4}}$
