JEE PYQ: Units & Measurements - Question ID 426bb11f17fd (JEE Main 2025)

Core Formula & Concept:

In dimensional analysis, every term in a physically meaningful equation must have the same dimensions. The given position function is: $x(t) = A \sin t + B \cos^2 t + C t^2 + D$ Here, $x(t)$ represents position, so its dimension is length, denoted by $\mathrm{L}$. The arguments of trigonometric functions ($\sin t$ and $\cos t$) must be dimensionless, meaning $t$ must have the dimension of time $\mathrm{T}$ (since $\sin t$ and $\cos t$ are dimensionless only if $t$ is dimensionless or has the dimension of an angle, but in physics, $t$ is time). However, since $t$ appears inside $\sin t$ and $\cos t$, it must be dimensionless. But in the term $C t^2$, $t$ has the dimension of time $\mathrm{T}$. This apparent contradiction is resolved by recognizing that the argument of trigonometric functions must be dimensionless, implying that $t$ is actually a dimensionless quantity (e.g., $t$ is time measured in seconds, but the argument is $\omega t$, where $\omega$ has dimension $\mathrm{T}^{-1}$). For simplicity, we assume the given $t$ is dimensionless (as it is the argument of $\sin$ and $\cos$), but the term $C t^2$ suggests $t$ has dimension $\mathrm{T}$. To resolve this, we infer that the actual equation is likely: $x(t) = A \sin(\omega t) + B \cos^2(\omega t) + C t^2 + D$ where $\omega$ has dimension $\mathrm{T}^{-1}$. However, since $\omega$ is not given, we proceed by assuming the given $t$ is dimensionless in the trigonometric terms but has dimension $\mathrm{T}$ in the $C t^2$ term. This is a common simplification in such problems, where the dimension of $t$ is treated as $\mathrm{T}$ for all terms except the trigonometric arguments (which are dimensionless).

The key concept is that the dimensions of $A$, $B$, $C$, and $D$ must be such that every term in $x(t)$ has the dimension of length $\mathrm{L}$. We will use this to find the dimensions of $A$, $B$, $C$, and $D$ individually and then compute the dimension of $\frac{A B C}{D}$.

Step-by-Step Derivation:

Step 1: Analyze each term in $x(t)$ for dimensional consistency. The position $x(t)$ has dimension $\mathrm{L}$. Therefore, each term on the right-hand side must also have dimension $\mathrm{L}$: 1. $A \sin t$: Since $\sin t$ is dimensionless, $A$ must have dimension $\mathrm{L}$. 2. $B \cos^2 t$: Since $\cos^2 t$ is dimensionless, $B$ must have dimension $\mathrm{L}$. 3. $C t^2$: Here, $t$ has dimension $\mathrm{T}$, so $t^2$ has dimension $\mathrm{T}^2$. For $C t^2$ to have dimension $\mathrm{L}$, $C$ must have dimension $\mathrm{L} \mathrm{T}^{-2}$. 4. $D$: This is a constant term, so $D$ must have dimension $\mathrm{L}$.

Step 2: Write the dimensions of $A$, $B$, $C$, and $D$. From the above analysis: - $[A] = \mathrm{L}$ - $[B] = \mathrm{L}$ - $[C] = \mathrm{L} \mathrm{T}^{-2}$ - $[D] = \mathrm{L}$

Step 3: Compute the dimension of $\frac{A B C}{D}$. Multiply the dimensions of $A$, $B$, and $C$, and divide by the dimension of $D$: $$\left[ \frac{A B C}{D} \right] = \frac{[A] \cdot [B] \cdot [C]}{[D]} = \frac{\mathrm{L} \cdot \mathrm{L} \cdot \mathrm{L} \mathrm{T}^{-2}}{\mathrm{L}} = \frac{\mathrm{L}^3 \mathrm{T}^{-2}}{\mathrm{L}} = \mathrm{L}^2 \mathrm{T}^{-2}$$

Step 4: Match the result with the given options. The dimension $\mathrm{L}^2 \mathrm{T}^{-2}$ corresponds to option C.

Common Traps & Exam Tip:

1. Ignoring the dimension of $t$ in trigonometric functions: Students often forget that the argument of $\sin$ or $\cos$ must be dimensionless. If they treat $t$ as having dimension $\mathrm{T}$ in $\sin t$ or $\cos t$, they will incorrectly assign dimensions to $A$ and $B$. Always ensure that trigonometric arguments are dimensionless.

2. Miscounting dimensions in the product: When computing $\frac{A B C}{D}$, students may incorrectly cancel dimensions or misapply exponents. For example, they might write $\mathrm{L}^3 \mathrm{T}^{-2} / \mathrm{L} = \mathrm{L}^3 \mathrm{T}^{-2}$ (forgetting to subtract exponents) or $\mathrm{L}^2 \mathrm{T}^{-1}$ (incorrectly handling the exponents). Always perform dimensional arithmetic carefully.

3. Assuming $D$ is dimensionless: Some students assume that $D$ is a dimensionless constant, leading to incorrect dimensions for $\frac{A B C}{D}$. Remember that $D$ is a position offset, so it must have the dimension of length $\mathrm{L}$.

Exam Tip: When dealing with such problems, always write down the dimension of each term explicitly before combining them. This minimizes errors and makes the solution clearer.