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The terminal velocity of a copper ball of radius $5 \mathrm{~mm}$ falling through a tank of oil at room temperature is $10 \mathrm{~cm} \mathrm{~s}^{-1}$. If the viscosity of oil at room temperature is $0.9 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$, the viscous drag force is:
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Verified Answer
The correct answer is:
$8.48 \times 10^{-3} \mathrm{~N}$
By Stobal's law, $\mathrm{F}=6 \pi \eta r v$
(where, the symbols have their usual meanings)
$\begin{aligned}
\mathrm{F} & =6 \times 3.14 \times 0.9 \times 5 \times 10^{-3} \times 10 \times 10^{-2} \\
\mathrm{~F} & =84.78 \times 10^{-4} \\
& =8.478 \times 10^{-3} \mathrm{~N}=8.48 \times 10^{-3} \mathrm{~N}
\end{aligned}$
(where, the symbols have their usual meanings)
$\begin{aligned}
\mathrm{F} & =6 \times 3.14 \times 0.9 \times 5 \times 10^{-3} \times 10 \times 10^{-2} \\
\mathrm{~F} & =84.78 \times 10^{-4} \\
& =8.478 \times 10^{-3} \mathrm{~N}=8.48 \times 10^{-3} \mathrm{~N}
\end{aligned}$
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