A) \[[{{M}^{0}}{{L}^{3}}{{T}^{-2}}]\]
B) \[[{{M}^{0}}{{L}^{3}}{{T}^{0}}]\]
C) \[[M{{L}^{-2}}{{T}^{-2}}]\]
D) \[[M{{L}^{5}}{{T}^{-2}}]\]
Correct Answer: D
Solution :
We know that only quantities with similar dimensions can be added or subtracted. Given equation is \[\left( P+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\] Therefore, dimensions of P = dimensions of\[\frac{a}{{{V}^{2}}}\] \[\Rightarrow \]dimensions of a = dimensions of\[P{{V}^{2}}\] \[=[M{{L}^{5}}{{T}^{-2}}]\]
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