A raindrop with radius \(R=0.2\, {mm}\) fells 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 attainde by the raindrop is : (In \({ms}^{-1}\)) [Density of water \(f_{{w}}=1000\;{kg} {m}^{-3}\) and density of air \(f_{{a}}=1.2\; {kg} {m}^{-3}, {g}=10 \;{m} / {s}^{2}\) Coefficient of viscosity of air \(=18 \times 10^{-5} \;{Nsm}^{-2}\) ]

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(D) \(4.94\)

Detailed explanation

At terminal speed

\({F}_{\text {net }} =0\)

\({Mg} ={F}_{{v}}=6 \pi {\eta} {Rv}\)

\({V} =\frac{{mg}}{6 \pi \eta {Rv}}\)

\({V} =\frac{\rho_{{w}} \frac{4 \pi}{3} {R}^{3} {g}}{6 \pi \eta {R}}\)

\(=\frac{2 p_{{w}} {R}^{2} {g}}{9 {\eta}}\)

\(=\frac{400}{81}\, {m} / {s}\)

\(=4.94\, {m} / {s}\)

Same subject

From JEE Mains