A lead sphere of mass' \(m\) ' falls in a viscous liquid with terminal velocity ' \(V\) ' Another lead sphere of mass ' \(8 \mathrm{~m}\) ' will fall through the same liquid with terminal velocity

The terminal velocity is given by,
\(\begin{aligned} & \mathrm{V}_{\mathrm{T}}=\frac{2}{9} \mathrm{r}^2 \frac{(\rho-\sigma)}{\eta} \\ & \Rightarrow \mathrm{V}_{\mathrm{T}} \propto \mathrm{r}^2\end{aligned}\)
where \(r\) radius sphere
\(\begin{aligned} & \text { Mass } \propto \text { volume } \Rightarrow \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\left(\frac{4}{3} \pi \mathrm{r}_1^3\right)}{\left(\frac{4}{3} \pi \mathrm{r}_2^3\right)} \Rightarrow \frac{\mathrm{m}}{8 \mathrm{~m}}=\frac{\mathrm{r}_1^3}{\mathrm{r}_2^3} \\ & \Rightarrow \frac{1}{8}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3 \Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{2} \Rightarrow \mathrm{r}_2=2 \mathrm{r}_1 \\ & \text { So } \frac{\mathrm{V}_{\mathrm{T} 2}}{\mathrm{~V}_{\mathrm{T} 1}}=\frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=4 \Rightarrow \mathrm{V}_{\mathrm{T} 2}=4 \mathrm{~V}_{\mathrm{T} 1}=4 \mathrm{~V}_{\mathrm{T}}\end{aligned}\)