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