First slide

Fluid dynamics

Question

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

Moderate

A

2
B

1
C

>1
D

<1

Solution

The two streams strike at the same point on the ground.
$R_{1} = R_{2} = R$
$u_{1} t_{1} = u_{2} t_{2} . . . . . . . . (i)$
$where u_{1} = velocity of efflux at Q = \sqrt{(2 g h_{1})} and u_{2} = velocity of efflux at P = \sqrt{[2 g (H - h_{2})]}$
$t_{1} = time of fall of water stream through Q is = \sqrt{\frac{2 (H - h_{1})}{g}}$
$t_{2} = time of fall of the water stream through P = \sqrt{\frac{2 h_{2}}{g}}$
Putting these values is Eq. (i), we get
$(H - h_{1}) h_{1} = (H - h_{2}) h_{2}$
$or [H - (h_{1} + h_{2})] [h_{1} - h_{2}] = 0$
$H = h_{1} + h_{2} is irrelevant because the holes are at two different heights. Therefore, h_{1} = h_{2} or h_{1} / h_{2} = 1$

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