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A small copper ball of mass $m$ falls in a viscous liquid with terminal velocity $v$ . Another spherical copper ball of mass $M$ falls in the same liquid with terminal velocity $n v$ where $n$ is a number. The ratio $\frac{M}{m}$ is

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$\sqrt{n}$

$n$

$n^{3 / 2}$

$n^{2}$

answer is C.

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Detailed Solution

Let r be the radius of the ball of mass m and R be the radius of the ball with mass M.

Terminal velocity $\propto r^{2}$ . Therefore

$\frac{n v}{v} = \frac{R^{2}}{r^{2}} \Rightarrow \frac{R}{r} = n^{1 / 2}$

Now $\frac{M}{m} = \frac{\frac{4 π}{3} ρ R^{3}}{\frac{4 π}{3} ρ r^{3}} = \frac{R^{3}}{r^{3}} = {(\frac{R}{r})}^{3} = n^{3 / 2}$

So the correct choice is (3 ).

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