A) \[x=1,y=1,z=-1\]
B) \[x=1,y=-1,z=1\]
C) \[x=-1,y=1,z=1\]
D) \[x=1,y=1,z=1\]
Correct Answer: B
Solution :
\[[P]=\frac{[F]}{[A]}=[M{{L}^{-1}}{{T}^{-2}}],[c]=[L{{T}^{-1}}]\] \[[Q]=\frac{[E]}{[A][T]}=[M{{T}^{-3}}]\] As given, \[{{P}^{x}}{{Q}^{y}}{{c}^{z}}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[{{[M{{L}^{-1}}{{T}^{2}}]}^{x}}{{[L{{T}^{-1}}]}^{2}}{{[M{{T}^{-3}}]}^{y}}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]\[{{M}^{x+y}}{{L}^{-x+z}}{{T}^{-2x-z-3y}}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[\therefore \] \[x+y=0\] \[-x+z=0\] \[-2x-z-3y=0\] Solving, we get \[x=1,y=-1,z=1\]You need to login to perform this action.
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