Fluid Mechanics Viscosity Question 98
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} $
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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. $
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