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Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 5 April, Shift 2 — Question 41

What is the dimensional formula of ab−1\mathrm{ab}^{-1} in the equation (P+aV2)(V−b)=RT\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^{2}}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}, where letters have their usual

meaning.

  1. Option A:

    [M0 L3 T−2]\left[\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]

  2. Option B:

    [ML2 T−2]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]

    Correct
  3. Option C:

    [M−1 L5 T3]\left[M^{-1} \mathrm{~L}^{5} \mathrm{~T}^{3}\right]

  4. Option D:

    [M6L7T4]\left[M^{6} L^{7} T^{4}\right]

Answer: B

Step-by-step solution

∵[V]=[b]\because[\mathrm{V}]=[\mathrm{b}]

∴\therefore Dimension of b=[L3]\mathrm{b}=\left[\mathrm{L}^{3}\right]

&[P]=[aV2]\&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^{2}}\right] [a]=[PV2]=[ML−1 T−2][L6][\mathrm{a}]=\left[\mathrm{PV}^{2}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{6}\right]

Dimension of a=[ML5 T−2]\mathrm{a}=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]

∴ab−1=[ML5 T−2][L3]=[ML2 T−2]\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]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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