JEE PYQ: Units And Dimensions Question 4
Question 4 - 2021 (26 Feb Shift 1)
In a typical combustion engine the workdone by a gas molecule is given by $W = \alpha^2 \beta e^{-\frac{\beta x^2}{kT}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be:
(1) $[M^0 L T^0]$
(2) $[M^2 L T^{-2}]$
(3) $[M L T^{-2}]$
(4) $[M L T^{-1}]$
Show Answer
Answer: (1)
Solution
$\frac{\beta x^2}{kT}$ is dimensionless. So $KT = \beta x^2 \Rightarrow M^1 L^2 T^{-2} = \beta \cdot L^2$, giving $\beta = M^1 T^{-2}$. Then $M^1 L^2 T^{-2} = \alpha^2 M^1 T^{-2}$, so $\alpha^2 = L^2$, $\alpha = L$, i.e., $\alpha = [M^0 L^1 T^0]$.
BETA
