Properties Of Solids And Liquids Question 214
Question: A large tank filled with water to a height ?h? is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to $ \frac{h}{2} $ and from $ \frac{h}{2} $ to zero is [EAMCET (Engg.) 2003]
Options:
A)$ \sqrt{2} $
B)$ \frac{1}{\sqrt{2}} $
C)$ \sqrt{2}-1 $
D)$ \frac{1}{\sqrt{2}-1} $
Show Answer
Answer:
Correct Answer: C
Solution:
Time taken for the level to fall from H to $ H’ $
$ t=\frac{A}{A _{0}}\sqrt{\frac{2}{g}}\left[ \sqrt{H}-\sqrt{H’} \right] $
According to problem- the time taken for the level to fall from h to $ \frac{h}{2} $
$ t _{1}=\frac{A}{A _{0}}\sqrt{\frac{2}{g}}\ \ \left[ \sqrt{h}-\sqrt{\frac{h}{2}} \right] $
and similarly time taken for the level to fall from $ \frac{h}{2} $ to zero $ t _{2}=\frac{A}{A _{0}}\sqrt{\frac{2}{g}}\ \left[ \sqrt{\frac{h}{2}}-0 \right] $
$ \therefore \ \frac{t _{1}}{t _{2}}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1. $
