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

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Aicharya · Accepted Answer

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

A lead sphere of mass' $m$ ' falls in a viscous liquid with terminal velocity ' $V$ ' Another lead sphere of…

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