JEE PYQ: Units & Measurements - Question ID 534aabb4a599 (JEE Main 2021)

In physics, the principle of dimensional homogeneity states that every term in a physically meaningful equation must have the same dimensions. This principle is the foundation for determining the dimensions of unknown constants or variables in a given formula.

The work done ($W$) by a force is a form of energy. In the SI system, energy has the dimensions: $[W] = [M L^2 T^{-2}]$ where $M$ = mass, $L$ = length, and $T$ = time.

The Boltzmann constant ($k$) has dimensions of energy per unit temperature: $[k] = \frac{[W]}{[T]} = [M L^2 T^{-2} K^{-1}]$ where $K$ represents the dimension of temperature.

The exponential function $e^{y}$ is dimensionless, meaning its argument $y$ must also be dimensionless. This is a crucial point when analyzing terms inside exponentials.

Given the expression for work done: $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$ we analyze the dimensions of each component.

1. Dimensions of $W$: $[W] = [M L^2 T^{-2}]$
2. Dimensions of the exponential term: The exponential function $e^{-\frac{x^2}{\alpha k T}}$ is dimensionless, so its argument must be dimensionless: $\left[\frac{x^2}{\alpha k T}\right] = [M^0 L^0 T^0]$ Since $x$ is displacement, $[x] = [L]$, so: $\left[\frac{L^2}{\alpha k T}\right] = 1$ Rearranging: $[\alpha k T] = [L^2]$ We know $[k] = [M L^2 T^{-2} K^{-1}]$ and $[T] = [K]$, so: $[\alpha] \cdot [M L^2 T^{-2} K^{-1}] \cdot [K] = [L^2]$ Simplifying: $[\alpha] \cdot [M L^2 T^{-2}] = [L^2]$ Thus: $[\alpha] = \frac{[L^2]}{[M L^2 T^{-2}]} = [M^{-1} T^{2}]$
3. Analyzing the entire expression: The given expression is: $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$ Since the exponential term is dimensionless, the dimensions of $W$ must come from $\alpha \beta^2$: $[W] = [\alpha] [\beta]^2$ We already found $[\alpha] = [M^{-1} T^{2}]$, and $[W] = [M L^2 T^{-2}]$, so: $[M L^2 T^{-2}] = [M^{-1} T^{2}] [\beta]^2$ Solving for $[\beta]^2$: $[\beta]^2 = \frac{[M L^2 T^{-2}]}{[M^{-1} T^{2}]} = [M^{2} L^2 T^{-4}]$ Taking the square root: $[\beta] = [M L T^{-2}]$

The dimensions of $\beta$ are $[M L T^{-2}]$, which corresponds to option C.

Students often make the following mistakes in such questions:

• Ignoring the dimensionless nature of the exponential: Many forget that the argument of an exponential function must be dimensionless, leading to incorrect assumptions about $\alpha$.
• Incorrectly canceling dimensions: Some students mistakenly cancel dimensions without considering the full expression, especially when dealing with $\alpha$ and $\beta$ together.
• Confusing dimensions of $k$ and $T$: Misremembering the dimensions of the Boltzmann constant or temperature can lead to errors in the derivation.
• Overlooking the square on $\beta$: Forgetting that $\beta$ is squared in the expression can result in incorrect dimensional analysis.

Exam Tip: Always start by identifying the dimensions of the known quantities (like $W$, $k$, and $x$) and use the principle of dimensional homogeneity to systematically solve for the unknown dimensions. Double-check each step to ensure consistency.