A ball of radius r and density rho falls free | vedclass

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(C) The velocity of the ball when it strikes the water surface is given by $v = \sqrt{2gh}$ ... $(i)$The terminal velocity of a ball inside a viscous

Vedclass · Accepted Answer

$\frac{2}{{81}}{r^4}{\left( {\frac{{\rho - 1}}{\eta }} \right)^2}g$

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(C) The velocity of the ball when it strikes the water surface is given by $v = \sqrt{2gh}$ ... $(i)$The terminal velocity of a ball inside a viscous liquid is given by $v = \frac{2}{9}r^2 g \frac{(\rho - \sigma)}{\eta}$. Assuming the density of water $\sigma = 1$,we have $v = \frac{2}{9}r^2 g \frac{(\rho - 1)}{\eta}$ ... (ii)Since the velocity does not change upon entering the water,we equate $(i)$ and (ii):$\sqrt{2gh} = \frac{2}{9} \frac{r^2 g}{\eta} (\rho - 1)$Squaring both sides:$2gh = \left( \frac{2}{9} \right)^2 \frac{r^4 g^2}{\eta^2} (\rho - 1)^2$Solving for $h$:$h = \frac{1}{2g} \cdot \frac{4}{81} \cdot \frac{r^4 g^2}{\eta^2} (\rho - 1)^2$$h = \frac{2}{81} r^4 \left( \frac{\rho - 1}{\eta} \right)^2 g$

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