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JEE Main 2021 (26 Feb, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2021 (26 Feb, Shift 1)

Q47JEE Main 2021 · 26 FEB · SHIFT 1Medium

Applications of Dimensional Analysis

In a typical combustion engine the workdone by a gas molecule is given by W=α2βe−βx2kTW = \alpha^2\beta e^{\frac{-\beta x^2}{kT}}, where x is the displacement, k is the Boltzmann constant and T is the temperature. If α\alpha and β\beta are constants, dimensions of α\alpha will be:

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Answer: (a)

  1. Exponent is dimensionless: [β]=[kT][x2]=ML2T−2L2=MT−2[\beta] = \dfrac{[kT]}{[x^2]} = \dfrac{ML^2T^{-2}}{L^2} = MT^{-2}.
  2. [W]=[α2][β][W] = [\alpha^2][\beta], so [α2]=ML2T−2MT−2=L2[\alpha^2] = \dfrac{ML^2T^{-2}}{MT^{-2}} = L^2.
  3. [α]=L[\alpha] = L, so the answer is (a) [M0LT0][M^0LT^0].
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