A ball of radius r and relative density $ρ$ 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 $η$ the value of h is given by

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$\frac{2}{9} r^{2} (\frac{1 - ρ}{η}) g$

$\frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

$\frac{2}{9} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

$\frac{2}{81} r^{2} (\frac{ρ - 1}{η}) g$

answer is C.

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Detailed Solution

$ϑ = \sqrt{2 gh} : ϑ_{T} = ϑ = \frac{2}{9} r^{2} (\frac{ρ_{b} - ρ_{ℓ}}{η}) g$

$h = \frac{2}{81} r^{4} {(\frac{ρ_{b} - ρ_{l}}{η})}^{2} g$

$ρ_{b} = ρ, ρ_{1} = ρ_{water} = 1 gm / cc$

$\Rightarrow h = \frac{2}{81} r^{4} {(\frac{ρ - 1}{η})}^{2} g$

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