A) \[[M{{L}^{5}}{{T}^{-2}}]\]
B) \[[{{M}^{-1}}{{L}^{5}}{{T}^{-2}}]\]
C) \[[M{{L}^{-1}}{{T}^{-2}}]\]
D) \[[M{{L}^{-5}}{{T}^{-2}}]\]
Correct Answer: A
Solution :
In the equation p, V and T are pressure, volume and temperature respectively. \[\left( p+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\] Dimensions of \[\frac{a}{{{V}^{2}}}\]will be same as that of pressure. \[\therefore \]Dimensions of \[\frac{a}{{{V}^{2}}}=\] dimensions of p \[\Rightarrow \]Dimensions of \[a=\]dimensions of \[p\,\,\times \]dimensions of \[{{V}^{2}}\] \[=[M{{L}^{-1}}{{T}^{-2}}][{{L}^{6}}]\] \[=[M{{L}^{5}}{{T}^{-2}}]\]
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