A) 0.4 m/s
B) 0.133 m/s
C) 0.1 m/s
D) 0.2 m/s
Correct Answer: C
Solution :
Terminal speed of spherical body in a viscous liquid is given by \[{{v}_{T}}=\frac{2{{r}^{2}}(\rho -\sigma )g}{9\eta }\] \[\Rightarrow \] \[{{v}_{T}}\propto (\rho -\sigma )\] where,\[\rho =\]density of substance of body \[\sigma =\]density of liquid. Thus, from given data \[\frac{{{v}_{T}}(Ag)}{{{v}_{T}}(Gold)}=\frac{{{\rho }_{Ag}}-{{\sigma }_{l}}}{{{\rho }_{Gold}}-{{\sigma }_{l}}}\] \[\Rightarrow \]\[{{v}_{T}}(Ag)=\frac{10.5-1.5}{19.5-1.5}\times 0.2\] \[=\frac{9}{18}\times 0.2\] \[=0.1\text{ }m/s\]
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