Fluid Mechanics Viscosity Question 130
Question: A streamlined body falls through air from a height $ h $ on the surface of a liquid. If $ d $ and $ D(D>d) $ represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, 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:
[d] Upthrust-weight of body=apparent weight $ ~VDg-Vdg=Vda, $ Where a=retardation of body $ \therefore 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)} $
BETA
