A small spherical ball of radius 0.1 ,mm and | vedclass

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(D) The velocity of the ball after falling through height $h$ is given by $v = \sqrt{2gh}$.Since the velocity remains constant inside the water,this v

Vedclass · Accepted Answer

$20$

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(D) The velocity of the ball after falling through height $h$ is given by $v = \sqrt{2gh}$.Since the velocity remains constant inside the water,this velocity must be equal to the terminal velocity $v_t$ of the ball in water.The formula for terminal velocity is $v_t = \frac{2}{9} \frac{r^2 (ho - ho_w) g}{\eta}$.Given: $r = 0.1 \,mm = 10^{-4} \,m$,$ho = 10^4 \,kg \,m^{-3}$,$ho_w = 10^3 \,kg \,m^{-3}$,$\eta = 1.0 	imes 10^{-5} \,N \,s \,m^{-2}$,$g = 10 \,m \,s^{-2}$.Equating $v = v_t$:$\sqrt{2gh} = \frac{2}{9} \frac{(10^{-4})^2 (10^4 - 10^3) 	imes 10}{10^{-5}}$$\sqrt{2gh} = \frac{2}{9} \frac{10^{-8} 	imes 9 	imes 10^3 	imes 10}{10^{-5}}$$\sqrt{2gh} = \frac{2}{9} 	imes 9 	imes 10^{-8+4+5} = 2 	imes 10^1 = 20 \,m/s$.Squaring both sides: $2gh = 400$.$2 	imes 10 	imes h = 400$.$20h = 400 \implies h = 20 \,m$.

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