First slide

Fluid dynamics

Question

Two holes are made in the side of the tank such that the jets of water flowing out of them meet at the same point on the ground. If one hole is at a height of 3 cm above the bottom, them the distance of the other hole from the top surface of water is

Moderate

A

$\frac{3}{2} cm$
B

$\sqrt{6} cm$
C

$\sqrt{3} c m$
D

3 cm

Solution

x = $\sqrt{2 {gh}_{1}} \times \sqrt{\frac{2 h_{2}}{g}} or x = 2 \sqrt{h_{1} h_{2}}$
Now, imagine a hole at a depth $h_{2} below the free surface of the liquid . The height of this hole will be h_{1}$
Clearly, x remains the same.

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