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JEE Main 2021 (24 Feb, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2021 (24 Feb, Shift 1)

Q48JEE Main 2021 · 24 FEB · SHIFT 1Medium

Applications of Dimensional Analysis

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

given by, W=αβ2e−x2αkTW = \alpha\beta^2 e^{-\frac{x^2}{\alpha kT}}, where x is the displacement, k is the Boltzmann constant and T is the temperature. α\alpha and β\beta are constants. Then the dimensions of β\beta will be:

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Answer: (c)

  1. Exponent is dimensionless: [α]=[x2][kT]=L2ML2T−2=M−1T2[\alpha] = \dfrac{[x^2]}{[kT]} = \dfrac{L^2}{ML^2T^{-2}} = M^{-1}T^2.
  2. [W]=[α][β2][W] = [\alpha][\beta^2], so [β2]=ML2T−2M−1T2=M2L2T−4[\beta^2] = \dfrac{ML^2T^{-2}}{M^{-1}T^2} = M^2L^2T^{-4}.
  3. [β]=MLT−2[\beta] = MLT^{-2}, so the answer is (c).
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