A) 1/2, 1/2
B) -1/2, -1/2
C) 1/2, -1/2
D) -1/2, ½
Correct Answer: D
Solution :
\[f=c{{m}^{x}}{{k}^{y}}\] \[\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]={{\left[ {{M}^{1}} \right]}^{x}}{{\left[ {{M}^{0}}{{L}^{0}}{{T}^{-2}} \right]}^{y}}\] \[{{M}^{0}}{{L}^{0}}{{T}^{-1}}={{M}^{x}}^{+y}{{L}^{0}}{{T}^{-2y}}\] x + y =0 ... (1) \[-2y=-1\] ... (2)You need to login to perform this action.
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