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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2024
  • Hard
  • MCQ

What is the dimensional formula of \(a b^{-1}\) in the equation? \(\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}\), where letters have their usual meaning.

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Answer & Solution

Correct Answer

(B) \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)

Step-by-step Solution

Detailed explanation

\(\because[\mathrm{V}]=[\mathrm{b}]\)

\(\therefore \text { Dimension of } \mathrm{b}=\left[\mathrm{L}^3\right]\)

\(\&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right]\)

\({[\mathrm{a}]=\left[\mathrm{PV}^2\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^6\right]}\)

\(\text { Dimension of } \mathrm{a}=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]\)

\(\therefore \mathrm{ab}^{-1}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)

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