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The viscous drag acting on a metal sphere of diameter 1 mm, falling through a fluid of viscosity 0.8 Pa s with a velocity of 2 m s$^{-1}$ is equal to:
A
$15\times10^{-3}$ N
B
$30\times10^{-3}$ N
C
$1.5\times10^{-3}$ N
D
$20\times10^{-3}$ N
Detailed Solution
$F=6\pi\eta rv=(6)(3.14)\left(\dfrac{1\times10^{-3}}{2}\right)(0.8\times10^{-1})... $ Using $r=0.5\times10^{-3}$ m: $F=6\pi\times0.8\times0.5\times10^{-3}\times2=15\times10^{-3}$ N.