A metallic sphere of radius 1.0 10^ -3 m and density 1.0 10^4 kg … - Vidyadip

Question

A metallic sphere of radius $1.0 \times 10^{-3} m$ and density $1.0 \times 10^4 kg m ^{-3}$ enters a tank of water, after a free fall through a distance of $h$ in the earth's gravitational field. If its velocity remains unchanged after entering water, determine the value of $h$. Given coefficient of viscosity of water $=1.0 \times 10^{-3} Nsm ^{-2}, g=10 ms^{-2}$ and density of water $=1.0 \times 10^3 kgm ^{-3}$.

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Answer

The velocity attained by the sphere after falling freely from height $h$ is $v =\sqrt{2 g h} \ldots$ (i)
After entering water, the velocity of the sphere does not change. So v is also the terminal velocity of the sphere. Hence $v =\frac{2}{9} \frac{r^2}{\eta}\left(\rho-\rho^{\prime}\right) g$
But $\rho=10^4 kgm ^{-3}, \rho^{\prime}=10^3 kgm ^{-3}, r =10^{-3} m, g =10 ms^{-2}, \eta=10^{-3} Nsm ^{-2}$
$
\therefore v=\frac{2}{9} \times \frac{\left(10^{-3}\right)^2 \times\left(10^4-10^3\right) \times 10}{10^{-3}}=20 ms^{-1}
$
From (i), h $=\frac{v^2}{2 g}=\frac{20 \times 20}{2 \times 10}=20 m$

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