JEE PYQ: Units & Measurements - Question ID f28e0baca1d3 (JEE Main 2022)

Core Formula & Concept:

In problems involving conversion between two systems of units, the fundamental principle is dimensional consistency. Every physical quantity can be expressed as a product of powers of the base dimensions: mass ($M$), length ($L$), and time ($T$). The key formulas we use are:

• Velocity: $v = \frac{L}{T} \implies [v] = L^1 T^{-1}$
• Acceleration: $a = \frac{v}{T} = \frac{L}{T^2} \implies [a] = L^1 T^{-2}$

Given the relations between $v_1, v_2$ and $a_1, a_2$, we can express the conversion factors for length ($L$) and time ($T$) in terms of the constants $m$ and $n$.

Step-by-Step Derivation:

Step 1: Express velocity conversion in terms of $L$ and $T$

Given: $v_2 = \frac{n}{m^2} v_1$ Since $v = \frac{L}{T}$, we can write: $\frac{L_2}{T_2} = \frac{n}{m^2} \frac{L_1}{T_1}$ Let the conversion factors for length and time be: $L_2 = k_L L_1, \quad T_2 = k_T T_1$ Substituting: $\frac{k_L L_1}{k_T T_1} = \frac{n}{m^2} \frac{L_1}{T_1}$ Cancel $L_1$ and $T_1$: $\frac{k_L}{k_T} = \frac{n}{m^2} \quad \text{(Equation 1)}$

Step 2: Express acceleration conversion in terms of $L$ and $T$

Given: $a_2 = \frac{a_1}{mn}$ Since $a = \frac{L}{T^2}$, we can write: $\frac{L_2}{T_2^2} = \frac{1}{mn} \frac{L_1}{T_1^2}$ Substitute $L_2 = k_L L_1$ and $T_2 = k_T T_1$: $\frac{k_L L_1}{k_T^2 T_1^2} = \frac{1}{mn} \frac{L_1}{T_1^2}$ Cancel $L_1$ and $T_1^2$: $\frac{k_L}{k_T^2} = \frac{1}{mn} \quad \text{(Equation 2)}$

Step 3: Solve for $k_L$ and $k_T$

From Equation 1: $k_L = \frac{n}{m^2} k_T$ Substitute into Equation 2: $\frac{\frac{n}{m^2} k_T}{k_T^2} = \frac{1}{mn}$ Simplify: $\frac{n}{m^2 k_T} = \frac{1}{mn}$ Cross-multiply: $n^2 m = m^2 k_T$ Solve for $k_T$: $k_T = \frac{n^2}{m}$ Now substitute $k_T$ back into Equation 1 to find $k_L$: $k_L = \frac{n}{m^2} \cdot \frac{n^2}{m} = \frac{n^3}{m^3}$

Step 4: Write the final conversion relations

Thus, the relations for distance and time are: $L_2 = \frac{n^3}{m^3} L_1 \quad \text{or} \quad \frac{n^3}{m^3} L_1 = L_2$ $T_2 = \frac{n^2}{m} T_1 \quad \text{or} \quad \frac{n^2}{m} T_1 = T_2$

Step 5: Match with the given options

Comparing with the options:

• Option A: $\frac{n^3}{m^3} L_1 = L_2$ and $\frac{n^2}{m} T_1 = T_2$ → Correct
• Options B, C, D do not match the derived relations.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Incorrectly relating dimensions: Some confuse the exponents of $L$ and $T$ when converting velocity and acceleration. Remember: velocity is $L^1 T^{-1}$ and acceleration is $L^1 T^{-2}$.
• Algebraic errors: Misplacing $m$ and $n$ or their powers while solving the two equations. Always cross-verify by substituting back.
• Sign errors in exponents: Forgetting that acceleration involves $T^2$ in the denominator, leading to $k_T^2$ in the conversion.

Exam Tip: Always write down the dimensional formula of each quantity involved. Then express the given relations in terms of $L$ and $T$ conversion factors. Solve systematically using substitution. This approach minimizes errors and ensures full marks.