JEE PYQ: Units & Measurements - Question ID b2a3b3faa830 (JEE Main 2019)

To solve this problem, we need to relate the electrical resistance of a wire to its physical dimensions. The key formula is:

where:

• $R$ = electrical resistance of the wire
• $\rho$ = resistivity of the material (copper, assumed constant)
• $L$ = length of the wire
• $A$ = cross-sectional area of the wire

The problem states that the volume of the wire remains unchanged. The volume $V$ of a cylindrical wire is: $V = A \cdot L$ Since $V$ is constant, any change in length $L$ must be compensated by an inverse change in area $A$ to keep $V$ the same.

Step 1: Express the new length and area after stretching
The wire is stretched to become 0.5% longer. Thus, the new length $L'$ is: $L' = L + 0.5\% \text{ of } L = L \left(1 + \frac{0.5}{100}\right) = 1.005 L$

Step 2: Use volume conservation to find the new area
Since volume $V = A \cdot L$ remains constant: $A' \cdot L' = A \cdot L$ Substitute $L' = 1.005 L$: $A' \cdot 1.005 L = A \cdot L$ $A' = \frac{A}{1.005}$

Step 3: Compute the new resistance $R'$
Using the resistance formula: $R' = \rho \frac{L'}{A'}$ Substitute $L' = 1.005 L$ and $A' = \frac{A}{1.005}$: $R' = \rho \frac{1.005 L}{\frac{A}{1.005}} = \rho \frac{(1.005)^2 L}{A}$ $R' = (1.005)^2 \cdot \rho \frac{L}{A} = (1.005)^2 R$

Step 4: Calculate the percentage change in resistance
The percentage change in resistance is: $\text{Percentage Change} = \left(\frac{R' - R}{R}\right) \times 100\%$ Substitute $R' = (1.005)^2 R$: $\text{Percentage Change} = \left((1.005)^2 - 1\right) \times 100\%$ Expand $(1.005)^2$ using the binomial approximation $(1 + x)^2 \approx 1 + 2x$ for small $x$: $(1.005)^2 \approx 1 + 2 \times 0.005 = 1.01$ Thus: $\text{Percentage Change} = (1.01 - 1) \times 100\% = 1\%$

Conclusion: The resistance increases by 1.0%, so the correct answer is Option C.

1. Ignoring volume conservation: Students often forget that stretching the wire changes both its length and cross-sectional area. If they assume only $L$ changes, they might incorrectly conclude that resistance increases by 0.5% (Option D).
2. Incorrect binomial approximation: Some students use $(1 + x)^2 \approx 1 + x$ instead of $(1 + x)^2 \approx 1 + 2x$, leading to a 0.5% change (Option D). The correct approximation is crucial for small percentages.
3. Sign errors: The resistance increases, not decreases. A negative percentage change (e.g., -1%) is incorrect.
4. Assuming resistivity changes: Resistivity $\rho$ is a material property and remains constant unless temperature changes are specified. Do not assume $\rho$ varies.

Exam Tip: Always check if volume, mass, or other conserved quantities are mentioned. In such cases, relate the changing dimensions using conservation laws before applying the resistance formula.