Properties Of Solids And Liquids Question 260
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}}) $
