JEE PYQ: Units And Dimensions Question 1
Question 1 - 2021 (24 Feb Shift 1)
The workdone by a gas molecule in an isolated system is given by, $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be -
(1) $[M^0 L T^0]$
(2) $[M^2 L T^2]$
(3) $[M L T^{-2}]$
(4) $[M L^2 T^{-2}]$
Show Answer
Answer: (3)
Solution
Given: $W = \alpha \beta^2 e^{-x^2/(\alpha k T)}$. Since the exponential term is dimensionless, $\frac{x^2}{\alpha k T} = [M^0 L^0 T^0]$. So $[\alpha] = \frac{L^2}{[M^1 L^2 T^{-2} K^{-1}] \cdot K} = [M^{-1} T^2]$. Then $W = \alpha \cdot \beta^2$ gives $[\beta^2] = \frac{[M^1 L^2 T^{-2}]}{[M^{-1} T^2]} = [M^2 L^2 T^{-4}]$, so $[\beta] = [M L T^{-2}]$.
BETA
