In a cylindrical water tank there are two small holes Q and P on the wall at a depth of $h_1$ from the upper level of water and at a height of $h_{2 }$from the lower end of the tank, respectively, as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of $h_1$ and $h_{2 }$is

The two stream strike at the same point on the ground.

${\mathrm{R}}_1={\mathrm{R}}_2=\mathrm{R}$

${\mathrm{V}}_1{\mathrm{t}}_1={\mathrm{V}}_2{\mathrm{t}}_2$

Where ${\mathrm{V}}_1=\text{ velocity of efflux at }\mathrm{Q}=\sqrt{2{\mathrm{gh}}_1}$

${\mathrm{V}}_2=\text{ velocity of efflux at }\mathrm{P}=\sqrt{2\mathrm{g}\left(\mathrm{H}-{\mathrm{h}}_2\right)}$.

${\mathrm{t}}_1=$ time of fall of water stream through

$\mathrm{Q}=\sqrt{\frac{2\left(\mathrm{H}-{\mathrm{h}}_1\right)}{\mathrm{g}}}$

${\mathrm{t}}_2=$ time of fall of water stream through

$P=\sqrt{\frac{2h_2}{\mathrm{g}}}$

Putting these values is equation (1) we get

$\left(\mathrm{H}-{\mathrm{h}}_1\right){\mathrm{h}}_1=\left(\mathrm{H}-{\mathrm{h}}_2\right){\mathrm{h}}_2$

$\left[\mathrm{H}-\left({\mathrm{h}}_1+{\mathrm{h}}_2\right)\right]\left[{\mathrm{h}}_1-{\mathrm{h}}_2\right]=0$

$\mathrm{H}={\mathrm{h}}_1+{\mathrm{h}}_2$ is irrelevant because the holes are at two different heights

${\mathrm{h}}_1={\mathrm{h}}_2\text{ or }{\mathrm{h}}_1/{\mathrm{h}}_2=1\text{. }$