A) \[x=\frac{1}{2},\,y=\frac{1}{2}\]
B) \[x=\frac{1}{2},\,z=\frac{1}{2}\]
C) \[y=\frac{1}{2},\,z=\frac{3}{2}\]
D) \[y=\frac{3}{2},\,z=\frac{1}{2}\]
Correct Answer: B
Solution :
[b] Length \[L\propto {{G}^{x}}{{c}^{y}}{{h}^{z}}\] \[\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]={{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]}^{x}}{{[L{{T}^{-1}}]}^{y}}{{[M{{L}^{2}}{{T}^{-1}}]}^{z}}\] \[\Rightarrow \,\,\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]=[{{M}^{-x+z}}{{L}^{3x+y+2z}}{{T}^{-2x-y-z}}]\] By comparing the power of M, L and T in both sides we get \[-x+z=0,\] \[3x+y+2z=1\] and \[-2x-y-z=0\] By solving above, three equation we get \[x=\frac{1}{2},\,\,y=-\frac{3}{2},\,\,z=\frac{1}{2}\]
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