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Neglecting the density of air, the terminal velocity obtained by a raindrop of radius 0.3mm falling through air of viscosity $1.8 \times {10^{ - 5}}Ns{m^{ - 3}}$ will be:A. $10.9m{s^{ - 1}}$B. $7.48m{s^{ - 1}}$C. $3.7m{s^{ - 1}}$D. $12.8m{s^{ - 1}}$

Last updated date: 23rd Jun 2024
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Hint: We have a raindrop that is freely falling in air. We can use Stokes’ Law which determines the terminal velocity of small spherical objects which are freely falling in a fluid medium. Substituting the given values in the Stokes’ Law equation, we will have the velocity of the raindrop.
Formula used:
${v_t} = \dfrac{2}{9}\dfrac{{{r^2}\left( {\rho - \rho '} \right)g}}{\eta }$

In the question, they’ve given us a raindrop whose radius is $0.3mm$ and viscosity of the $1.8 \times {10^{ - 5}}Ns{m^{ - 3}}$.
From Stokes’ Law we have
${V_t} = \dfrac{2}{9}\dfrac{{{r^2}\left( {\rho - \rho '} \right)g}}{\eta }$
Where,
$V_t$ is the terminal velocity of the raindrop
r is the radius of the raindrop
ρ is the density of the raindrop
ρ’ is the density of the medium in which the raindrop is traveling
g is the acceleration due to gravity
η is the viscosity of the medium
We’ll be ignoring the ρ’ as the raindrop is traveling in the air and they’ve given that the density of air is neglected. Substituting the values given in the question, we get
\eqalign{ & {v_t} = \dfrac{2}{9}\dfrac{{{r^2}\rho g}}{\eta } \cr & \Rightarrow {v_t} = \dfrac{2}{9} \times \dfrac{{{{\left( {0.3 \times {{10}^{ - 3}}} \right)}^2} \times 1000 \times 9.8}}{{1.8 \times {{10}^{ - 5}}}} \cr & \Rightarrow {v_t} = \dfrac{2}{9} \times 49 = 10.89m{s^{ - 1}} \cr & \therefore {v_t} = 10.89m{s^{ - 1}} \sim 10.9m{s^{ - 1}} \cr}
Therefore, the terminal velocity of the raindrop is 10.9m/s.

Thus, the correct option is A.

Note:
The terminal velocity is the velocity attained by the particle that is freely falling in a fluid medium. Here, the fluid medium is the air, which is a gas. The density of the air is very small compared to that of water. That is why it is mostly ignored. One might notice that Stokes’ Law is mostly applicable only to the small spherical objects.