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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2021 · 24 February · Morning Shift · Physics · Units and Dimensions

The work done by a gas molecule in an isolated system is given by,...

The work done by a gas molecule in an isolated system is given by, W=αβ2e−x2\alphakT,W = \alpha{\beta}^{2}{e}^{- \dfrac{{x}^{2}}{\alphakT}}, where xx is the displacement, kk is the Boltzmann constant and TT is the temperature. α\alpha and β\beta are constants. Then the dimensions of β\beta will be:

Options

  1. A[M2LT2]\left[{M}^{2}L{T}^{2}\right]
  2. B[ML2T−2]\left[{ML}^{2}{T}^{- 2}\right]
  3. C[MLT−2]\left[{MLT}^{- 2}\right]
  4. D[M0LT0]\left[{M}^{0}L{T}^{0}\right]

Correct answer

C. [MLT−2]\left[{MLT}^{- 2}\right]

Step-by-step solution

x2\alphakT→\dfrac{{x}^{2}}{\alphakT} \to dimensionless

⇒[α]=[x2][kT]=L2ML2T−2=M−1T2⇒ \left[\alpha\right] = \dfrac{\left[{x}^{2}\right]}{\left[kT\right]} = \dfrac{{L}^{2}}{{ML}^{2}{T}^{- 2}} = {M}^{- 1}{T}^{2}

Now \left[W\right] = \left[\alpha\right]{\left[\beta\right]}^{2}</p><p>[β]=ML2T−2M−1T2=M1L1T−2</p><p>\left[\beta\right] = \sqrt{\dfrac{{ML}^{2}{T}^{- 2}}{{M}^{- 1}{T}^{2}}} = {M}^{1}{L}^{1}{T}^{- 2}

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