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A rain drop of radius 0.3 mm has a terminal velocity in air $1 \mathrm{~ms}^{-1}$. The viscosity of air is $18 \times 10^{-5}$ poise. Find the viscous force on the rain drops.
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The correct answer is:
$1.018 \times 10^{-7} \mathrm{~N}$
Here, $\mathrm{r}=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}, \mathrm{v}=1 \mathrm{~ms}^{-1}$
$\eta=18 \times 10^{-5}$ poise $=18 \times 10^{-6}$ decapoise
Viscous force, $\mathrm{F}=6 \pi \eta \mathrm{rv}$
$\begin{aligned} & =6 \times \frac{22}{7} \times\left(18 \times 10^{-6}\right) \times\left(0.3 \times 10^{-3}\right) \times 1 \\ & =1.018 \times 10^{-7} \mathrm{~N}\end{aligned}$
$\eta=18 \times 10^{-5}$ poise $=18 \times 10^{-6}$ decapoise
Viscous force, $\mathrm{F}=6 \pi \eta \mathrm{rv}$
$\begin{aligned} & =6 \times \frac{22}{7} \times\left(18 \times 10^{-6}\right) \times\left(0.3 \times 10^{-3}\right) \times 1 \\ & =1.018 \times 10^{-7} \mathrm{~N}\end{aligned}$
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