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