In the figure shown,

find the range R.

Applying Bernoulli's equation at points 1 and 2 .

$$\begin{gathered} {\mathrm{p}}_1+\frac{1}{2}\left(2\mathrm{\rho }\right){\mathrm{v}}_1^2+\left(2\mathrm{\rho }\right){\mathrm{gh}}_1={\mathrm{p}}_2+\frac{1}{2}\left(2\mathrm{\rho }\right){\mathrm{v}}_2^2+\left(2\mathrm{\rho }\right){\mathrm{gh}}_2 \\ \mathrm{Here}, \\ {\mathrm{v}}_1=0,{\mathrm{v}}_2=\mathrm{v},{\mathrm{h}}_1={\mathrm{h}}_2,{\mathrm{p}}_2={\mathrm{p}}_0 \\ \mathrm{and} \\ {\mathrm{p}}_1={\mathrm{p}}_0+\mathrm{\rho gH}+2\mathrm{\rho gH} \\ ={\mathrm{p}}_0+3\mathrm{\rho gH} \end{gathered}$$

Substituting in the above equation we have,

$$\begin{gathered} {\mathrm{p}}_0+3\mathrm{\rho gH}={\mathrm{p}}_0+{\mathrm{\rho v}}^2 \\ \mathrm{v}=\sqrt{3\mathrm{gH}} \end{gathered}$$

Now, this velocity is horizontal. So, time taken by the liquid to fall to the ground is free fall time or

$\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$

(where, h=H )

or

$$\begin{gathered} \mathrm{t}=\sqrt{\frac{2\mathrm{H}}{\mathrm{g}}} \\ \mathrm{R}=\mathrm{vt}=\left(\sqrt{3\mathrm{gH}}\right)\left(\sqrt{\frac{2\mathrm{H}}{\mathrm{g}}}\right) \\ =\sqrt{6}\mathrm{H} \end{gathered}$$