In a cylindrical water tank there are two small holes Q and P on the wall at a depth of h₁ from the upper level of water and at a height of h₂ 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 h₁ and h₂ is

 

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

$R_1=R_2=R$

$u_1{t}_1=u_2{t}_2........\left(i\right)$

$\text{ where }{u}_1=\text{ velocity of efflux at }Q=\sqrt{\left(2g{h}_1\right)}\text{ and }{u}_2=\text{ velocity of efflux at }P=\sqrt{\left[2g\left(H-h_2\right)\right]}$

$t_1=\text{ time of fall of water stream through }Q\text{ is }=\sqrt{\frac{2\left(H-h_1\right)}{g}}$

$t_2=\text{ time of fall of the water stream through }P=\sqrt{\frac{2h_2}{g}}$

Putting these values is Eq. (i), we get

$\left(H-h_1\right)h_1=\left(H-h_2\right)h_2$

$\text{ or }\left[H-\left(h_1+h_2\right)\right]\left[h_1-h_2\right]=0$

$H=h_1+h_2\text{ is irrelevant because the holes are at two different heights. Therefore, }{h}_1=h_2\text{ or }{h}_1/h_2=1$