JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(=18 \times 10^{-5} \;{Nsm}^{-2}\) ]
- A \(14.4\)
- B \(2.47\)
- C \(43.56\)
- D \(4.94\)
Answer & Solution
Correct Answer
(D) \(4.94\)
Step-by-step Solution
Detailed explanation
At terminal speed \({F}_{\text {net }} =0\) \({Mg} ={F}_{{v}}=6 \pi {\eta} {Rv}\) \({V} =\frac{{mg}}{6 \pi \eta {Rv}}\) \({V} =\frac{\rho_{{w}} \frac{4 \pi}{3} {R}^{3} {g}}{6 \pi \eta {R}}\) \(=\frac{2 p_{{w}} {R}^{2} {g}}{9 {\eta}}\) \(=\frac{400}{81}\, {m} / {s}\)…
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