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 \[\eta \] the value of h is given by 
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) \[\frac{2}{9}\,{{r}^{4}}\,{{\left( \frac{\rho -1}{\eta } \right)}^{2}}\,g\]
Correct Answer: C
Solution :
Velocity of ball when it strikes the water surface \[\operatorname{v}=\,\sqrt{2gh}\] ... (i) Terminal velocity of ball inside the water \[v=\frac{2}{9}{{r}^{2}}g\frac{(\rho -1)}{\eta }\] ? (ii) Equation (i) and (ii) we get \[\sqrt{2\,gh}=\frac{2}{9}\frac{{{r}^{2}}g}{\eta }(\rho -1)\] \[\Rightarrow \,\,\,h=\frac{2}{81}{{r}^{4}}{{\left( \frac{\rho -1}{\eta } \right)}^{2}}g\]
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