A) \[16.95\times 10\]
B) \[1.695\times {{10}^{-2}}\]
C) \[\text{1}.0\text{17}\times \text{1}{{0}^{-\text{11}}}\text{ N}\]
D) \[\text{1}0\text{1}.\text{73}\times \text{1}{{0}^{-\text{9}}}\text{ N}\]
Correct Answer: D
Solution :
From Stokes formula \[F=6\pi \eta \,rv\] Given, \[\eta =18\times {{10}^{-5}}=18\times {{10}^{-6}}\] kg/mass, r = 0.3 mm \[=0.3\times {{10}^{-3}}m\], v = 1 m/s \[\therefore \] \[F=6\times 3.14\times 18\times {{10}^{6}}\times 0.3\times {{10}^{-3}}\times 1\] \[\Rightarrow \] \[F=101.73\times {{10}^{-9}}N\]
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