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JEE Main 2019 (11 Jan, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2019 (11 Jan, Shift 1)

Q60JEE Main 2019 · 11 JAN · SHIFT 1Medium

Applications of Dimensional Analysis

The force of interaction between two atoms is given by F = $\alpha\beta$ exp $\left(-\frac{x^2}{\alpha kT}\right)$ ; where x is the distance, k is the Boltzmann constant and T is temperature and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is:

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Answer: (b)

Exponent is dimensionless: $[\alpha] = \dfrac{L^2}{ML^2T^{-2}} = M^{-1}T^2$ .
$[F] = [\alpha][\beta]$ , so $[\beta] = \dfrac{MLT^{-2}}{M^{-1}T^2}$ .
$[\beta] = M^2LT^{-4}$ , so the answer is (b).

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