Properties Of Solids And Liquids Question 223
Question: A ball of radius r and density r falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is h, the value of h is given by
Options:
A)$ \frac{2}{9}r^{2}\left( \frac{1-\rho }{\eta } \right)g $
B)$ \frac{2}{81}r^{2}\left( \frac{\rho -1}{\eta } \right)g $
C)$ \frac{2}{81}r^{4}{{\left( \frac{\rho -1}{\eta } \right)}^{2}}g $
D)$ \frac{2}{9}r^{4}{{\left( \frac{\rho -1}{\eta } \right)}^{2}}g $
Show Answer
Answer:
Correct Answer: C
Solution:
Velocity of ball when it strikes the water surface $ v=\sqrt{2gh} $ -(i)
Terminal velocity of ball inside the water $ v=\frac{2}{9}r^{2}g\frac{\left( \rho -1 \right)}{\eta } $ -(ii)
Equating(i) and (ii) we get $ \sqrt{2gh}=\frac{2}{9}\frac{r^{2}g}{\eta }(\rho -1) $
therefore $ h=\frac{2}{81}r^{4}{{\left( \frac{\rho -1}{\eta } \right)}^{2}}g $
