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(C) When the sphere is dropped from height $h$,its velocity $v$ just before entering the water is given by the equation of motion: $v = \sqrt{2gh}$ $(

Vedclass · Accepted Answer

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

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(C) When the sphere is dropped from height $h$,its velocity $v$ just before entering the water is given by the equation of motion: $v = \sqrt{2gh}$ $(i)$The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law: $v_t = \frac{2}{9}r^2 g \frac{(\rho - \sigma)}{\eta}$. Assuming the density of water $\sigma = 1$,we have: $v = \frac{2}{9}r^2 g \frac{(\rho - 1)}{\eta}$ (ii)Equating the velocity from $(i)$ and (ii): $\sqrt{2gh} = \frac{2}{9}r^2 g \frac{(\rho - 1)}{\eta}$Squaring both sides: $2gh = \left( \frac{2}{9} \right)^2 r^4 g^2 \frac{(\rho - 1)^2}{\eta^2}$$2gh = \frac{4}{81} r^4 g^2 \frac{(\rho - 1)^2}{\eta^2}$$h = \frac{2}{81} r^4 g \left( \frac{\rho - 1}{\eta} \right)^2$

$A$ sphere of radius $r$ and density $\rho$ is dropped from a height $h$. When it falls into water,it attains a terminal velocity. If the coefficient of viscosity of water is $\eta$,then $h =$

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