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An iron sphere of mass $20 \times 10^{-3} \mathrm{~kg}$ falls through a viscous liquid with terminal velocity $0.5 \mathrm{~ms}^{-1}$. The terminal velocity (in $\mathrm{ms}^{-1}$ ) of another iron sphere of mass $54 \times 10^{-2} \mathrm{~kg}$ is
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The correct answer is:
4.5
Terminal velocity, $v \propto r^2$
$\begin{aligned}
\text { or } \quad \frac{v_1}{v_2} & =\left(\frac{r_1}{r_2}\right)^2=\left[\left(\frac{r_1}{r_2}\right)^3\right]^{\frac{2}{3}} \\
\frac{v_1}{v_2} & =\left(\frac{M_1}{M_2}\right)^{2 / 3} \\
\therefore \quad \frac{0.5}{v_2} & =\left(\frac{20 \times 10^{-3}}{54 \times 10^{-2}}\right)^{2 / 3} \\
\frac{0.5}{v_2} & =\frac{1}{9} \Rightarrow v_2=4.5 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\begin{aligned}
\text { or } \quad \frac{v_1}{v_2} & =\left(\frac{r_1}{r_2}\right)^2=\left[\left(\frac{r_1}{r_2}\right)^3\right]^{\frac{2}{3}} \\
\frac{v_1}{v_2} & =\left(\frac{M_1}{M_2}\right)^{2 / 3} \\
\therefore \quad \frac{0.5}{v_2} & =\left(\frac{20 \times 10^{-3}}{54 \times 10^{-2}}\right)^{2 / 3} \\
\frac{0.5}{v_2} & =\frac{1}{9} \Rightarrow v_2=4.5 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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