A metal ball of radius 9 times 10^{-4} m and | vedclass

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(A) The velocity of the ball when it hits the surface of the water is given by $v = \sqrt{2gh}$.The terminal velocity of the ball in water is given by

Vedclass · Accepted Answer

$20$

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(A) The velocity of the ball when it hits the surface of the water is given by $v = \sqrt{2gh}$.The terminal velocity of the ball in water is given by Stokes' Law: $v = \frac{2}{9} r^2 g \frac{(ho - \sigma)}{\eta}$,where $ho$ is the density of the ball,$\sigma$ is the density of water,and $\eta$ is the coefficient of viscosity.Equating the two expressions for velocity:$\sqrt{2gh} = \frac{2}{9} \frac{r^2 g}{\eta} (ho - \sigma)$Squaring both sides:$2gh = \left( \frac{2}{9} \frac{r^2 g}{\eta} (ho - \sigma) ight)^2$$h = \frac{1}{2g} \left( \frac{2}{9} \frac{r^2 g}{\eta} (ho - \sigma) ight)^2 = \frac{2}{81} \frac{r^4 g}{\eta^2} (ho - \sigma)^2$Substituting the given values:$r = 9 	imes 10^{-4} \ m$,$ho = 10^4 \ kg/m^3$,$\sigma = 10^3 \ kg/m^3$,$\eta = 8.1 	imes 10^{-4} \ Pa \cdot s$,$g = 10 \ m/s^2$.$h = \frac{2}{81} 	imes \frac{(9 	imes 10^{-4})^4 	imes 10}{(8.1 	imes 10^{-4})^2} 	imes (10^4 - 10^3)^2$$h = \frac{2}{81} 	imes \frac{6561 	imes 10^{-16} 	imes 10}{65.61 	imes 10^{-8}} 	imes (9000)^2$$h = \frac{2}{81} 	imes 10^{-7} 	imes 100 	imes 81 	imes 10^6 = 20 \ m$.

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