JEE Main 2021 · 24 Feb · Shift 1 · Physics Q1

← JEE Main 2021 · 24 Feb · Shift 1

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The workdone by a gas molecule in an isolated system is given by, $W=\alpha\beta^2e^{-\frac{x^2}{\alpha kT}}$ , where $x$ is the displacement, k is the Boltzmann constant and T is the temperature. α and β are constants. Then the dimensions of β will be :

[M^2LT^2]

[M^0LT^0]

[MLT^{-2}]

[ML^2T^{-2}]

Answer and worked solution

Official answer: C

Exponent is dimensionless. $[\alpha]=\dfrac{L^2}{[kT]}=\dfrac{L^2}{ML^2T^{-2}}$ . Then $\beta^2=\dfrac W\alpha$ has $(ML^2T^{-2})^2/L^2$ , so $[\beta]=MLT^{-2}$ .

(C)

Topics:Dimensional analysis

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JEE Main 2021 · 24 Feb · Shift 1, Physics question 1