Fluid Mechanics Viscosity Question 296
Question: 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 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:
[c] 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 $
BETA
