JEE PYQ: Units & Measurements - Question ID b018c9692a45 (JEE Main 2024)

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.

Given a force $F$ expressed as: $F = a x^2 + b t^{\frac{1}{2}}$ where $x$ is distance (dimension $[L]$) and $t$ is time (dimension $[T]$), we are to find the dimensions of the ratio $\frac{b^2}{a}$.

We know that:

• Force $F$ has dimensions $[M L T^{-2}]$.
• The term $a x^2$ must have the same dimensions as $F$.
• The term $b t^{\frac{1}{2}}$ must also have the same dimensions as $F$, since they are added.

From these, we can deduce the dimensions of $a$ and $b$, and then compute $\frac{b^2}{a}$.

Step-by-Step Derivation:

Step 1: Determine the dimensions of $a$

The term $a x^2$ must have the dimensions of force: $[a x^2] = [F] = [M L T^{-2}]$ Since $x$ has dimension $[L]$, $x^2$ has dimension $[L^2]$. Thus: $[a] \cdot [L^2] = [M L T^{-2}]$ Solving for $[a]$: $[a] = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$

Step 2: Determine the dimensions of $b$

The term $b t^{\frac{1}{2}}$ must also have the dimensions of force: $[b t^{\frac{1}{2}}] = [F] = [M L T^{-2}]$ Since $t$ has dimension $[T]$, $t^{\frac{1}{2}}$ has dimension $[T^{\frac{1}{2}}]$. Thus: $[b] \cdot [T^{\frac{1}{2}}] = [M L T^{-2}]$ Solving for $[b]$: $[b] = \frac{[M L T^{-2}]}{[T^{\frac{1}{2}}]} = [M L T^{-\frac{5}{2}}]$

Step 3: Compute the dimensions of $b^2$

Using the dimensions of $b$: $[b^2] = [b]^2 = \left([M L T^{-\frac{5}{2}}]\right)^2 = [M^2 L^2 T^{-5}]$

Step 4: Compute the dimensions of $\frac{b^2}{a}$

We have: $\left[\frac{b^2}{a}\right] = \frac{[b^2]}{[a]} = \frac{[M^2 L^2 T^{-5}]}{[M L^{-1} T^{-2}]}$ Simplify the expression by subtracting exponents:

• For $M$: $2 - 1 = 1$
• For $L$: $2 - (-1) = 3$
• For $T$: $-5 - (-2) = -3$

Thus: $\left[\frac{b^2}{a}\right] = [M L^3 T^{-3}]$

Step 5: Match with the given options

The derived dimensions $[M L^3 T^{-3}]$ correspond to option B.

Common Traps & Exam Tip:

Students often make the following mistakes:

1. Incorrectly handling fractional exponents: Some confuse $t^{\frac{1}{2}}$ with $t^{-2}$ or misapply exponent rules, leading to wrong dimensions for $b$.
2. Forgetting to square $b$: The question asks for $\frac{b^2}{a}$, not $\frac{b}{a}$. Squaring $b$ affects the exponents of $M$, $L$, and $T$.
3. Sign errors in division: When dividing dimensions, students sometimes add exponents instead of subtracting them, especially for $L$ in $\frac{b^2}{a}$.
4. Mixing up dimensions of $a$ and $b$: Since $a$ and $b$ appear in different terms, students may incorrectly assume they have the same dimensions.

Exam Tip: Always verify the dimensions of each term separately before combining them. Double-check the arithmetic of exponents, especially when dealing with fractional powers or division of dimensions.