JEE PYQ: Units & Measurements - Question ID c259526bd2d7 (JEE Main 2022)

Core Formula & Concept:

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 a physically valid equation must match. Additionally, the argument of a trigonometric function (like $\sin$) must be dimensionless.

Given the expression for energy density: $u = \frac{\alpha}{\beta} \sin\left(\frac{\alpha x}{k t}\right)$ we know:

• Energy density ($u$) has dimensions of energy per unit volume, i.e., $\left[M L^2 T^{-2}\right] / \left[L^3\right] = \left[M L^{-1} T^{-2}\right]$.
• The Boltzmann constant ($k$) has dimensions $\left[M L^2 T^{-2} \theta^{-1}\right]$.
• $x$ (displacement) has dimensions $\left[L\right]$.
• $t$ (temperature) has dimensions $\left[\theta\right]$.

Step-by-Step Derivation:

Step 1: Analyze the argument of the sine function.
The argument of $\sin$ must be dimensionless. Thus: $\frac{\alpha x}{k t} \quad \text{must be dimensionless.}$ Let the dimensions of $\alpha$ be $\left[\alpha\right]$. Then: $\left[\alpha\right] \cdot \left[L\right] = \left[k\right] \cdot \left[\theta\right]$ Substitute the dimensions of $k$: $\left[\alpha\right] \cdot \left[L\right] = \left[M L^2 T^{-2} \theta^{-1}\right] \cdot \left[\theta\right] = \left[M L^2 T^{-2}\right]$ Solve for $\left[\alpha\right]$: $\left[\alpha\right] = \frac{\left[M L^2 T^{-2}\right]}{\left[L\right]} = \left[M L T^{-2}\right]$ Step 2: Analyze the dimensions of the entire expression for $u$.
The given expression is: $u = \frac{\alpha}{\beta} \sin\left(\frac{\alpha x}{k t}\right)$ Since $\sin$ is dimensionless, the dimensions of $u$ are determined by $\frac{\alpha}{\beta}$: $\left[u\right] = \frac{\left[\alpha\right]}{\left[\beta\right]}$ We know $\left[u\right] = \left[M L^{-1} T^{-2}\right]$ and $\left[\alpha\right] = \left[M L T^{-2}\right]$. Substitute these: $\left[M L^{-1} T^{-2}\right] = \frac{\left[M L T^{-2}\right]}{\left[\beta\right]}$ Solve for $\left[\beta\right]$: $\left[\beta\right] = \frac{\left[M L T^{-2}\right]}{\left[M L^{-1} T^{-2}\right]} = \left[L^2\right]$ Thus, the dimensions of $\beta$ are $\left[M^0 L^2 T^0\right]$. Step 3: Match with the given options.
The derived dimensions of $\beta$ are $\left[M^0 L^2 T^0\right]$, which corresponds to option D.

Common Traps & Exam Tip:

1. Ignoring the dimensionless nature of trigonometric arguments: Students often forget that the argument of $\sin$ must be dimensionless, leading to incorrect dimensions for $\alpha$ or $\beta$.
2. Miscounting dimensions of energy density: Energy density is energy per unit volume, not just energy. Confusing this with $\left[M L^2 T^{-2}\right]$ instead of $\left[M L^{-1} T^{-2}\right]$ will lead to errors.
3. Overcomplicating the problem: Some students try to assign dimensions to $\sin$ or its output, which is unnecessary since $\sin$ is always dimensionless.
4. Misapplying the Boltzmann constant: The Boltzmann constant has dimensions $\left[M L^2 T^{-2} \theta^{-1}\right]$, not $\left[M L^2 T^{-2}\right]$. Forgetting the temperature dimension will lead to incorrect results.

Exam Tip: Always start by ensuring the argument of trigonometric/exponential functions is dimensionless. This is a quick sanity check and often simplifies the problem significantly.