First slide

Fluid dynamics

Question

A water tank which is on ground has an arrangement to maintain a constant water level of depth 60 cm. Through a hole on its vertical wall at a depth of 20cm from the free surface water comes out and reaches the ground at a certain distance. To have the same horizontal range another hole can be made at a depth of

Easy

A

30 cm
B

10 cm
C

40 cm
D

50 cm

Solution

If h be the depth of hole below the water surface, the height of hole above the ground y = (H - h) where H is the total depth of water.
$\large \therefore$ Velocity of efflux, $\large V=\sqrt {2gh}$ time taken by a water particle to strike the ground, $\large t=\sqrt {2y/g}=\sqrt {\frac {2(H-h)}{g}}$
$\large \therefore$ Range R = V.t = $\large 2\sqrt {h(H-h)}$
$\large \therefore 2\sqrt {h_1(H-h_1)}=2\sqrt {h_2(H-h_2)}$
$\large h_1+h_2=H\Rightarrow h_2=H-h_1=(60-20)=40cm$

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