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Q.
Two solid spheres of same metal but of mass $M$ and $8\,M$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $nv$, then value of $n$ is
The terminal velocity is given by $v$, where $v \propto r^{2}(r=$ radius of the sphere).
The mass of the sphere can be given by $m=\frac{4}{3} \pi r^{3} \rho$, thus $m \propto r^{3}$.
So, according to the question, $\frac{m_{1}}{m_{2}}$
$=\frac{1}{8}=\left(\frac{r_{1}}{r_{2}}\right)^{3}$
$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2}$
and since, $\frac{v_{1}}{v_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$
$\Rightarrow \frac{v}{n v}=\frac{1}{4}$
$ \Rightarrow n=4$