Question

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

Answer 1

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

R₁ = R₂ = R

u₁t₁ = u₂t_{2 $-\left(i\right)$}

where,

u₁ = velocity of efflux at Q = $\sqrt{2g{h}_1}$ and

u₂ = velocity of efflux at P = $\sqrt{2g(H–h_2)}$

t₁ = time of fall of water stream through Q = $\sqrt{\frac{2(H-h_1)}{g}}$

t₂ = time of fall of the water stream through P = $\sqrt{\frac{2h_2}{g}}$

Putting these values in the equation (i), we get

(H – h₁)h₁ = (H – h₂)h₂

[H – (h₁ + h₂)][h₁ –h₂] = 0

H = h₁ + h₂ is irrelevant because the holes are at two different heights. Therefore, h₁ = h₂ and h₁/h₂ =1.