$\begin{array}{1 1}(A)\;4cm/s,4.2\times 10^{-10}N\\(B)\;4.8cm/s,8\times 10^{-9}N\\(C)\;5.8cm/s,3.9\times 10^{-10}N\\(D)\;5cm/s,3.8\times 10^{-9}N\end{array} $

Terminal speed =$5.8cm/s$

Viscous force =$3.9\times 10^{-10}N$

Radius of drop $r=2\times 10^{-5}m$

$\rho=1.2\times 10^3Kgm^{-3}$

Viscosity of air $\eta=1.8\times 10^{-5}pas$

Terminal velocity $V=\large\frac{2r^2-(\rho-\rho_0)g}{g\eta}$

$\Rightarrow \large\frac{2\times (2\times 10^{-5})^2(1.2\times 10^3-0)\times 9.8}{9\times 1.8\times 10^{-5}}$

$\Rightarrow 5.8cms^{-1}$

Here $\rho_0$(density of air)=0 to neglect byoyancy

$\therefore$ Terminal velocity =$5.8cms^{-1}$

$F=6\pi \eta r V$

$\;\;\;=3.9\times 10^{-10}N$

Hence (C) is the correct answer.

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