First slide

Fluid dynamics

Question

A water tank placed on the floor has two small holes, punched in the vertical wall, one above the other. The holes are 3.3 cm and 4.7 cm above the floor. If the jets of water issuing out from the holes hit the floor at the same point on the floor, then the height of water in the tank is

Moderate

A

3 cm
B

6 cm
C

8 cm
D

9 cm

Solution

R = 2 $\sqrt{h (H - h)}$
Now, 3.3[H-3.3] = 4.7[H-4.7]
or $(4.7 - 3.3) H = 4.7 \times 4.7 - 3.3 \times 3.3$
or H = $\frac{22.09 - 10.89}{1.4} cm$
= $\frac{11.2}{1.4} cm = 8 cm$

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