A) \[\operatorname{x}=\frac{1}{2}, y=\frac{1}{2}\]
B) \[\operatorname{x}=\frac{1}{2}, z=\frac{1}{2}\]
C) \[y=\frac{3}{2}, z=\frac{1}{2}\]
D) \[y=\frac{1}{2}, z=\frac{3}{2}\]
Correct Answer: C
Solution :
\[\operatorname{Length} \propto {{\left[ G \right]}^{x}}{{\left[ C \right]}^{y}}{{\left[ h \right]}^{z}}\] \[{{\operatorname{M}}^{o}}{{L}^{1}}{{T}^{o}}\propto {{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]}^{x}}{{\left[ L{{T}^{-1}} \right]}^{y}} {{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{z}}\] \[{{\operatorname{M}}^{o}}{{L}^{1}}{{T}^{o}}\propto {{\left[ M \right]}^{z-x}}{{\left[ L \right]}^{3x+y+2z}} {{\left[ T \right]}^{-3x-y-Z}}\] On comparing we get \[\operatorname{x} =z=\frac{1}{2} \,\,\,,y=\frac{-3}{2}\]
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