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A spherical ball of radius $1 x 10^{- 4} m$ and density $10^{4} k g / m^{3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If the velocity of the ball does not change, after entering the water find h. Viscosity of water is $9.8 x 10^{- 6} N - s / m^{2}$ .

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$10.41 m$

$15.41 m$

$18.41 m$

$20.41 m$

answer is D.

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Detailed Solution

$v = \sqrt{2 g h} . \sqrt{2 g h} = \frac{2}{9} r^{2} [\frac{ρ - σ}{η}] g 2 g h = {[\frac{2}{9} x {(10^{- 4})}^{2} x \frac{(10^{4} - 10^{3}) x 9.8}{9.8 x 10^{- 6}}]}^{2} \Rightarrow h = \frac{20 x 20}{2 x 9.8} = 20.41 m$

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