Q95 / 120 · VITEEE 2019 · Physics
A rain drop of radius $0.3 \mathrm{~mm}$ has a terminal velocity in air is $1 \mathrm{~m} / \mathrm{s}$. The viscosity of air is $8 \times$ $10^{-5}$ poise. The viscous force on it is
- Correct answerA$45.2 \times 10^{-4}$ dyne
- B$101.73 \times 10^{-5}$ dyne
- C$16.95 \times 10^{-4}$ dyne
- D$16.95 \times 10^{-5}$ dyne
Solution
Radius of drop, $\mathrm{r}=0.3 \mathrm{~mm}=0.03 \mathrm{~cm}$
Terminal velocity, $\mathrm{v}=1 \mathrm{~m} / \mathrm{s}=100 \mathrm{~cm} / \mathrm{sec}$
Viscosity of air, $\eta=8 \times 10^{-5}$ poise
Viscous force, $\mathrm{f}=6 \pi \eta \mathrm{rv}$
$$
\begin{array}{l}
\mathrm{F}=6 \times 3.14 \times\left(8 \times 10^{-5}\right) \times 0.03 \times 100=4.52 \\
\times 10^{-3} \text { dyne }
\end{array}
$$
Hence, the correct answer is option: 'A'.