Fluid Mechanics Viscosity Question 120
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) $
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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 to rest, $ t=\frac{v}{a}=\frac{\sqrt{2gh}\times d}{(D-d)g}=\sqrt{\frac{2h}{g}}\times \frac{d}{(-(D-d))} $
BETA
