A raindrop with radius R=0.2|FLUID MECHANICS

JEE MAIN 2021

27-07-2021 S2

Question

A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water $f_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and Density of air $f_a=1.2 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} / \mathrm{s}^2$ Coefficient of viscosity of air = $\left.1.8 \times 10^{-5} \mathrm{Nsm}^{-2}\right]$

Select the correct option:

$250.6 \mathrm{~ms}^{-1}$

$43.56 \mathrm{~ms}^{-1}$

$4.94 \mathrm{~ms}^{-1}$

$14.4 \mathrm{~ms}^{-1}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ \begin{aligned} &\text { At terminal speed }\\ &\begin{aligned} & a=0 \\ & F_{\text {net }}=0 \\ & m g=F_v=6 \pi \eta R v \end{aligned}\\ &\begin{aligned} & v=\frac{m g}{6 \pi \eta R v} \\ & v=\frac{\rho_w \frac{4 \pi}{3} R^3 g}{6 \pi \eta R}=\frac{2 \rho_w R^2 g}{9 \eta}=\frac{400}{81} \mathrm{~m} / \mathrm{s} \\ & =4.94 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned} $$

Question Tags

JEE Main

Physics

Easy

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