A metal ball of radius $9 imes 10^{-4} \mathrm{~m}$ and density $10^4 \mathrm{~kg} / \mathrm{m}^3$ falls freely under gravity through a distance 'h' and erfters a tank of water. Considering that the metal ball has constant velocity, the value of h is [coefficient of viscosity of water $=8.1…

Question

A metal ball of radius $9 	imes 10^{-4} \mathrm{~m}$ and density $10^4 \mathrm{~kg} / \mathrm{m}^3$ falls freely under gravity through a distance 'h' and erfters a tank of water. Considering that the metal ball has constant velocity, the value of h is [coefficient of viscosity of water $=8.1 	imes 10^{-4} \mathrm{pa}-\mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ density of water $\left.=10^3 \mathrm{~kg} / \mathrm{m}^3ight]$

Aicharya · Accepted Answer

Velocity of the ball when the ball hits the surface of water $v=\sqrt{2 g h}$
Terminal velocity of the ball, $v=\frac{2}{9} r^2 g \frac{(P-\sigma)}{\eta}$
Equating (i) and (ii), $\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(P-\sigma)$
Solving for h , $h=\frac{2}{81} r^4\left(\frac{P-\sigma}{\eta}\right)^2 g$
Substituting the given values in the above equation, we have $\mathrm{h}=\frac{2}{81} \times\left(9 \times 10^{-4}\right)^4 \times\left(\frac{10^4-10^3}{8.1 \times 10^{-4}}\right)^2 \mathrm{~g}=20 \mathrm{~m}$

A metal ball of radius $9 \times 10^{-4} \mathrm{~m}$ and density $10^4 \mathrm{~kg} / \mathrm{m}^3$ falls…

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