JEE PYQ: Units & Measurements - Question ID e3e7e4c9493a (JEE Main 2025)

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), and time ($T$). The principle of homogeneity states that the dimensions on both sides of a physically meaningful equation must be identical. This means:

• Each term in an equation must have the same dimensions.
• Arguments of transcendental functions (like $\sin$, $\log$, etc.) must be dimensionless.
• Addition or subtraction of quantities is only possible if they have the same dimensions.

Given the velocity expression: $v = At^2 + \frac{Bt}{C + t}$ we need to determine the dimensions of the product $ABC$. Since $v$ is velocity, its dimension is $[L T^{-1}]$. We will use the homogeneity principle to find the dimensions of $A$, $B$, and $C$ individually, and then compute $[ABC]$.

Step-by-Step Derivation:

Step 1: Analyze the first term $At^2$

The first term in the expression is $At^2$. Since $v$ has dimensions $[L T^{-1}]$, the term $At^2$ must also have the same dimensions: $[At^2] = [L T^{-1}]$ We know that $[t] = [T]$, so: $[A] \cdot [T]^2 = [L T^{-1}]$ Solving for $[A]$: $[A] = \frac{[L T^{-1}]}{[T]^2} = [L T^{-3}]$

Step 2: Analyze the second term $\frac{Bt}{C + t}$

The second term is $\frac{Bt}{C + t}$. Again, this term must have the same dimensions as $v$, which is $[L T^{-1}]$: $\left[\frac{Bt}{C + t}\right] = [L T^{-1}]$ Since the denominator $(C + t)$ involves addition, $C$ and $t$ must have the same dimensions (homogeneity principle). Thus: $[C] = [t] = [T]$ Now, rewrite the term with known dimensions: $\left[\frac{Bt}{C + t}\right] = \frac{[B] \cdot [T]}{[T]} = [B]$ But we know this must equal $[L T^{-1}]$, so: $[B] = [L T^{-1}]$

Step 3: Compute the dimensions of $ABC$

Now that we have the dimensions of $A$, $B$, and $C$:

• $[A] = [L T^{-3}]$
• $[B] = [L T^{-1}]$
• $[C] = [T]$

The dimension of the product $ABC$ is: $[ABC] = [A] \cdot [B] \cdot [C] = [L T^{-3}] \cdot [L T^{-1}] \cdot [T]$ Simplify the expression: $[ABC] = [L^2 T^{-3 -1 + 1}] = [L^2 T^{-3}]$ This matches option B: $[M^0 L^2 T^{-3}]$.

Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Ignoring homogeneity in the denominator: Some students overlook that $C$ and $t$ must have the same dimensions because they are added. This leads to incorrect dimensions for $C$ and, consequently, $ABC$.
2. Incorrectly simplifying the second term: A common error is to assume $\frac{Bt}{C + t}$ simplifies to $\frac{Bt}{C}$ without considering the dimensional consistency of the denominator. This results in wrong dimensions for $B$.
3. Miscounting exponents: When multiplying dimensions, students sometimes miscount the exponents of $T$, leading to incorrect final dimensions (e.g., $[L^2 T^{-2}]$ instead of $[L^2 T^{-3}]$).
4. Forgetting the dimensionless nature of constants: Some students mistakenly assign dimensions to constants like $A$, $B$, or $C$ based on their position in the equation, rather than deriving them from homogeneity.

Exam Tip: Always ensure that every term in an equation has the same dimensions. For terms involving addition or subtraction, the quantities being added or subtracted must have identical dimensions. This is a powerful tool to quickly verify or derive dimensions in complex expressions.