A) \[x\] and \[B\]
B) \[C\] and \[{{z}^{-1}}\]
C) \[y\] and \[B/A\]
D) \[x\] and \[A\]
Correct Answer: D
Solution :
\[x=Ay+B\,\tan Cz\] From the dimensional homogenity \[[x]=[Ay]=[B]\Rightarrow \left[ \frac{x}{A} \right]=[y]=\left[ \frac{B}{A} \right]\] \[[Cz]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]=\]Dimension less \[x\] and \[B\]; \[C\] and \[{{Z}^{-1}};y\] and \[\frac{B}{A}\] have the same dimension but \[x\] and \[A\] have the different dimensions.
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