A) \[16.95\times {{10}^{-9}}N\]
B) \[1.695\times {{10}^{-9}}N\]
C) \[10.173\times {{10}^{-9}}N\]
D) \[101.73\times {{10}^{-9}}N\]
Correct Answer: D
Solution :
For a drop of radius r and terminal velocity v, the viscous force is given by \[F=6\pi \eta rv\] where, \[\eta \] ( is coefficient of viscosity, Putting the numerical values from the question, we have \[\eta =18\times {{10}^{-5}}\,\] \[poise=18\times {{10}^{-6}}\,kg/m-s\,\] \[r=0.3\,mm\,=0.3\times {{10}^{-3}}\,m,\,v=1\,m/s\] \[\therefore \] \[F=6\times 3.14\times 18\times {{10}^{-6}}\times 0.3\times {{10}^{-3}}\times 1\] \[F=101.73\times {{10}^{-9}}\,N\]
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