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If the terminal speed of a sphere of gold (density $=19.5 \mathrm{~kg} / \mathrm{m}^{3}$ ) is $0.2 \mathrm{~m} / \mathrm{s}$ in a viscous liquid (density $=1.5 \mathrm{~kg} / \mathrm{m}^{3}$ ), find the terminal speed of a sphere of silver (density $=10.5 \mathrm{~kg} / \mathrm{m}^{3}$ ) of the same size in the same liquid
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The correct answer is:
$0.1 \mathrm{~m} / \mathrm{s}$
Terminal velocity,
$$
\begin{array}{l}
v_{\mathrm{T}}=\frac{2 r^{2}\left(\mathrm{~d}_{1}-\mathrm{d}_{2}\right) \mathrm{g}}{9 \eta} \\
\frac{\mathrm{v}_{\mathrm{T}_{2}}}{0.2}=\frac{(10.5-1.5)}{(19.5-1.5)} \Rightarrow \mathrm{v}_{\mathrm{T}_{2}}=0.2 \times \frac{9}{18} \\
\therefore \mathrm{v}_{\mathrm{T}_{2}}=0.1 \mathrm{~m} / \mathrm{s}
\end{array}
$$
$$
\begin{array}{l}
v_{\mathrm{T}}=\frac{2 r^{2}\left(\mathrm{~d}_{1}-\mathrm{d}_{2}\right) \mathrm{g}}{9 \eta} \\
\frac{\mathrm{v}_{\mathrm{T}_{2}}}{0.2}=\frac{(10.5-1.5)}{(19.5-1.5)} \Rightarrow \mathrm{v}_{\mathrm{T}_{2}}=0.2 \times \frac{9}{18} \\
\therefore \mathrm{v}_{\mathrm{T}_{2}}=0.1 \mathrm{~m} / \mathrm{s}
\end{array}
$$
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