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