A) \[{{C}_{w}}={{C}_{g}}=0\]
B) \[{{C}_{w}}={{C}_{g}}\]
C) \[{{C}_{w}}>{{C}_{g}}\]
D) \[{{C}_{w}}<{{C}_{g}}\]
Correct Answer: C
Solution :
We know that the critical angle for water and glass are given by, \[\sin {{C}_{w}}=\frac{1}{{{\mu }_{w}}}\] \[\Rightarrow \] \[{{C}_{w}}{{\sin }^{-1}}\left( \frac{1}{{{\mu }_{w}}} \right)\] \[\sin {{C}_{g}}=\frac{1}{{{\mu }_{g}}}\] \[\Rightarrow \] \[{{C}_{g}}{{\sin }^{-1}}\left( 1 \right){{\mu }_{g}}\] since \[{{\mu }_{w}}<{{\mu }_{g}}\] \[\Rightarrow \] \[\frac{1}{{{\mu }_{w}}}>\frac{1}{{{\mu }_{g}}}\] \[\Rightarrow \] \[{{\sin }^{-1}}\left( \frac{1}{{{\mu }_{w}}} \right)>{{\sin }^{-1}}\left( \frac{1}{{{\mu }_{g}}} \right)\] \[\therefore \] \[{{C}_{w}}>{{C}_{g}}\]
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