A) \[16.95\times {{10}^{-9}}\,N\]
B) \[1.695\times {{10}^{-9}}\,N\]
C) \[10.17\times {{10}^{-9}}\,N\]
D) \[101.17\times {{10}^{-9}}\,N\]
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 \[\eta =18\times {{10}^{-5}}poise=18\times {{10}^{-6}}kg/m-s\] \[r=0.3\,mm=0.3\times {{10}^{-3}}m,\,v=1\,m/s\] \[\therefore \]\[F=6\times 3.14\times 18\times {{10}^{-6}}\times 0.3\times {{10}^{-3}}\times 1\] \[=101.74\times {{10}^{-9}}N\]
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