JEE PYQ: Units & Measurements - Question ID 57df0c492e35 (JEE Main 2019)

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), temperature ($\Theta$), etc. The key principle is that the dimensions on both sides of any physically meaningful equation must be identical. This is known as the principle of homogeneity of dimensions.

Given a formula involving multiple variables, we can determine the dimensions of an unknown constant by ensuring that the argument of any transcendental function (like exponential, logarithmic, or trigonometric functions) is dimensionless. Additionally, the dimensions of the entire expression must match the dimensions of the physical quantity it represents (in this case, force).

The force $F$ has dimensions of $MLT^{-2}$. The Boltzmann constant $k$ has dimensions of $ML^2T^{-2}\Theta^{-1}$.

The given force equation is: $F = \alpha \beta \exp \left( -\frac{x^2}{\alpha k T} \right)$

1. Analyze the exponential term: The argument of the exponential function must be dimensionless. Thus: $\left[ \frac{x^2}{\alpha k T} \right] = M^0 L^0 T^0$ Since $x$ is distance, $[x] = L$. The Boltzmann constant $k$ has dimensions $[k] = ML^2T^{-2}\Theta^{-1}$, and temperature $T$ has $[\Theta]$. Thus: $\left[ \frac{x^2}{\alpha k T} \right] = \frac{L^2}{[\alpha] \cdot ML^2T^{-2}\Theta^{-1} \cdot \Theta} = \frac{L^2}{[\alpha] \cdot ML^2T^{-2}} = M^0 L^0 T^0$ Simplifying: $\frac{1}{[\alpha] \cdot M T^{-2}} = 1 \implies [\alpha] = M^{-1} T^2$ So, the dimension of $\alpha$ is $M^{-1}T^2$.
2. Analyze the entire force equation: The exponential term is dimensionless, so the dimensions of $F$ come entirely from the product $\alpha \beta$: $[F] = [\alpha] \cdot [\beta]$ We know $[F] = MLT^{-2}$ and $[\alpha] = M^{-1}T^2$, so: $MLT^{-2} = M^{-1}T^2 \cdot [\beta]$ Solving for $[\beta]$: $[\beta] = \frac{MLT^{-2}}{M^{-1}T^2} = MLT^{-2} \cdot M T^{-2} = M^2 L T^{-4}$ Thus, the dimension of $\beta$ is $M^2 L T^{-4}$.

Students often make the following mistakes in such questions:

1. Ignoring the dimensionless nature of the exponential argument: Many students forget that the argument of the exponential function must be dimensionless, leading to incorrect dimensions for $\alpha$ and consequently $\beta$.
2. Miscounting dimensions of the Boltzmann constant: The Boltzmann constant $k$ has dimensions $ML^2T^{-2}\Theta^{-1}$, not just $ML^2T^{-2}$. Forgetting the temperature dimension ($\Theta$) leads to errors.
3. Incorrect simplification of dimensions: When solving for $[\beta]$, students may incorrectly cancel or combine dimensions, especially with negative exponents. Always double-check each step.
4. Confusing the dimensions of $\alpha$ and $\beta$: Some students assume $\alpha$ is dimensionless or has the same dimensions as $\beta$. Always derive dimensions systematically.

Exam Tip: Always start by ensuring the argument of any transcendental function is dimensionless. This is a powerful tool to simplify dimensional analysis problems.