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If the terminal speed of a sphere A [density $\rho_A=7.5 \mathrm{~kg} \mathrm{~m}^{-3}$ ] is $0.4 \mathrm{~ms}^{-1}$, in a viscous liquid [density $\rho_{\mathrm{L}}=1.5 \mathrm{~kg} \mathrm{~m}^{-3}$ ], the terminal speed of sphere $B$ [density $\rho_B=3 \mathrm{~kg} \mathrm{~m}^{-3}$ ] of the same size in the same liquid is
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Verified Answer
The correct answer is:
$0.1 \mathrm{~ms}^{-1}$
Terminal speed is given by
$$
\begin{aligned}
& \mathrm{V}=\frac{2}{9} \frac{\mathrm{r}^2\left(\rho-\rho_{\mathrm{L}}\right)}{\eta} \\
& \therefore \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{\rho_{\mathrm{A}}-\rho_{\mathrm{L}}}{\rho_{\mathrm{B}}-\rho_{\mathrm{L}}}=\frac{7.5-1.5}{3-1.5}=\frac{6}{1.5}=4 \\
& \therefore \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{V}_{\mathrm{A}}}{4}=\frac{0.4}{4}=0.1 \mathrm{~ms}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{V}=\frac{2}{9} \frac{\mathrm{r}^2\left(\rho-\rho_{\mathrm{L}}\right)}{\eta} \\
& \therefore \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{\rho_{\mathrm{A}}-\rho_{\mathrm{L}}}{\rho_{\mathrm{B}}-\rho_{\mathrm{L}}}=\frac{7.5-1.5}{3-1.5}=\frac{6}{1.5}=4 \\
& \therefore \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{V}_{\mathrm{A}}}{4}=\frac{0.4}{4}=0.1 \mathrm{~ms}^{-1}
\end{aligned}
$$
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