Fluid Mechanics Viscosity Question 296

In a cylindrical water tank, there are two small holes A and B on the wall at a depth of$ h_{1} $ from the surface of water and at a height of$ h_{2} $ from the bottom of water tank. Surface of water is at height H from the bottom of water tank. Water coming out from both holes strikes the ground at the same point S. Find the ratio of$ h_{1} $ and $ h_{2} $

Options:

A) Depends on H

B) $ 1:1 $

C) $ 2:2 $

D) $ 1:2 $

Show Answer

Answer:

Correct Answer: C

Solution:

Range is same for both holes $ \therefore ,,,,,,2\sqrt{(H-h_{1})h_{1}}=1\sqrt{(H-h_{2})h_{2}} $ Squaring both sides, $ 4(H-h_{1})h_{1}=4(H-h_{2})h_{2} $ $ Hh_{1}-h_{1}^{2}=Hh_{2}-h_{2}^{2} $ On solving we get, $ H=h_{1}+h_{2} $ (not possible) and $ h_{1}-h_{2}=0 $ Hence, the ratio of $ \frac{h_{1}}{h_{2}}=1:1 $



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