Properties Of Solids And Liquids Question 234
Question: A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes $ T _{1} $ time to decrease the height of water to $ \frac{H}{\eta }(\eta >1) $ ; and it takes $ T _{2} $ time to take out the rest of water. If $ T _{1}=T _{2} $ , then the value of $ \eta $ is
Options:
A) $ \sqrt{\frac{2h}{g}} $
B) $ \sqrt{\frac{2h}{g}.\frac{D}{d}} $
C) $ \sqrt{\frac{2h}{g}.\frac{d}{D}} $
D) $ \sqrt{\frac{2h}{g}}\left( \frac{d}{D-d} \right) $
Show Answer
Answer:
Correct Answer: D
Solution:
Upthrust ? weight of body = apparent weight $ VDg-Vdg=Vda, $
Where a = retardation of body \ $ a=\left( \frac{D-d}{d} \right)\ g $
The velocity gained after fall from h height in air, $ v=\sqrt{2gh} $
Hence, time to come in rest, $ t=\frac{v}{a}=\frac{\sqrt{2gh}\times d}{(D-d)g}=\sqrt{\frac{2h}{g}}\times \frac{d}{(D-d)} $
