JEE PYQ: Units And Dimensions Question 21
Question 21 - 2019 (11 Jan Shift 1)
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:
(1) $M^0L^2T^{-4}$
(2) $M^2LT^{-4}$
(3) $MLT^{-2}$
(4) $M^2L^2T^{-2}$
Show Answer
Answer: (2)
Solution
Since the exponential is dimensionless: $[\alpha] = M^{-1}T^2$. Then $[F] = [\alpha][\beta]$ gives $MLT^{-2} = M^{-1}T^2 \cdot [\beta]$, so $[\beta] = M^2LT^{-4}$.
BETA
