Fluid Mechanics Viscosity Question 148
Question: A tank has a hole at its bottom. The time needed to empty the tank from level $ h_{1} $ to $ h_{2} $ will be proportional to
Options:
A) $ h_{1}-h_{2} $
B) $ h_{1}+h_{2} $
C) $ \sqrt{h_{1}}-\sqrt{h_{2}} $
D) $ \sqrt{h_{1}}+\sqrt{h_{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ Let-\frac{dh}{dt} $ represent the rate of descent of water level, let A and a represent the cross-sectional areas of the container and hole respectively Then, $ -A\frac{dh}{dt}=a\sqrt{2gh} $ Or $ dt=-k\frac{dh}{\sqrt{n}}dt $ Or $ \int_{0}^{t}{dt=-k\int_{h_{1}}^{h_{2}}{\frac{1}{\sqrt{h}}dh}} $ Or $ t=-k{{\left| \frac{{{h}^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1} \right|}{h{1}}}^{h_{2}} $ Or $ t=-2k[\sqrt{h_{2}}-\sqrt{h_{1}}] $ Or $ t\propto (\sqrt{h_{1}}-\sqrt{h_{2}}) $
BETA
