A) 0.67
B) 0.77
C) 0.87
D) 0.97
Correct Answer: C
Solution :
Terminal velocity \[({{v}_{T}})\] is given by \[{{v}_{T}}=\frac{2}{9}.\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] where, \[\eta \]is coefficient of viscosity,\[\rho \] is density of silt,\[\sigma \]is density of lake .water and g is acceleration due to gravity. Given, \[r=20\mu m=20\times {{10}^{-6}}m,\] \[\rho =2000\,kg/{{m}^{3}},\] \[\sigma =1000\,kg/{{m}^{3}},g=9.8\,m/{{s}^{2}},\,\eta =1\times {{10}^{-3}}Pa\] \[\therefore \] \[{{v}_{T}}=\frac{2}{9}\frac{{{(20\times {{10}^{-6}})}^{2}}(2000-1000)\times 9.8}{1\times {{10}^{-3}}}\] \[\Rightarrow \] \[{{v}_{T}}=8.7\times {{10}^{-4}}m/s\] \[\Rightarrow \] \[{{v}_{T}}=0.87\,mm/s\]You need to login to perform this action.
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