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A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5$ $\mathrm{kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:
(The coefficient of viscosity of water is $9.8 \times 10^{-6}$ $\left.\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$
$2296 \mathrm{~m}$
$2249 \mathrm{~m}$
$2518 \mathrm{~m}$
$2396 \mathrm{~m}$
Solution
$V_T=\frac{2 g}{9} \frac{R^2\left[\rho_B-\rho_L\right]}{\eta}$
$\Rightarrow V_T=\frac{2}{9} \times \frac{10 \times\left(10^{-4}\right)^2}{9.8 \times 10^{-6}}\left[10^5-10^3\right]$
$\Rightarrow V_T=224.5$
when ball fall from height ($h$)
$\mathrm{V}=\sqrt{2 \mathrm{gh}}$
$\mathrm{h}=\left(\frac{\mathrm{V}^2}{2 \mathrm{~g}}\right)=2518 \mathrm{~m}$
