JEE PYQ: Units & Measurements - Question ID 0146a500c919 (JEE Main 2024)

Core Formula & Concept:

The question involves the propagation of percentage errors in derived physical quantities. The key concept here is that when a quantity is calculated using multiplication or division of measured variables, the relative (percentage) errors in those variables add up to give the relative error in the final result.

The resistance \( R \) of a cylindrical wire is given by the formula: $R = \rho \frac{L}{A}$ where:

• \( \rho \) = resistivity of the material (assumed constant, so no error in \( \rho \)),
• \( L \) = length of the wire,
• \( A \) = cross-sectional area of the wire.

For a wire of diameter \( D \), the cross-sectional area is: $A = \pi \frac{D^2}{4}$ Thus, the resistance can be rewritten as: $R = \rho \frac{L}{\pi D^2 / 4} = \frac{4 \rho L}{\pi D^2}$

Since \( \rho \) and \( \pi \) are constants, the percentage error in \( R \) depends only on the percentage errors in \( L \) and \( D \). The rule for error propagation in products/quotients is: $\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2 \frac{\Delta D}{D}$ where \( \frac{\Delta L}{L} \) and \( \frac{\Delta D}{D} \) are the relative errors in \( L \) and \( D \), respectively. The factor of 2 arises because \( D \) is squared in the formula for \( R \).

Step-by-Step Derivation:

Given:

• Percentage error in length \( L \): \( \frac{\Delta L}{L} \times 100 = 0.1\% \), so \( \frac{\Delta L}{L} = 0.001 \).
• Percentage error in diameter \( D \): \( \frac{\Delta D}{D} \times 100 = 0.1\% \), so \( \frac{\Delta D}{D} = 0.001 \).

The relative error in \( R \) is: $\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2 \frac{\Delta D}{D}$ Substitute the given values: $\frac{\Delta R}{R} = 0.001 + 2 \times 0.001 = 0.001 + 0.002 = 0.003$

Convert the relative error to a percentage: $\frac{\Delta R}{R} \times 100 = 0.003 \times 100 = 0.3\%$

Thus, the percentage error in measuring the resistance is 0.3%, which corresponds to option D.

Common Traps & Exam Tip:

1. Ignoring the exponent in \( D^2 \): Many students forget that the error in \( D \) is doubled because \( D \) appears squared in the formula for \( R \). This leads them to incorrectly add the errors as \( 0.1\% + 0.1\% = 0.2\% \), choosing option B.
2. Assuming errors subtract: Some students mistakenly think errors can cancel out, leading to answers like \( 0.1\% \) (option C). Errors always add in multiplication/division, never subtract.
3. Confusing absolute and percentage errors: The question gives percentage errors, but students sometimes treat them as absolute errors, leading to incorrect calculations.
4. Forgetting constants: While \( \rho \) and \( \pi \) are constants with no error, students might unnecessarily include them in error propagation, complicating the problem.

Exam Tip: Always write down the formula for the derived quantity first, then apply the error propagation rules systematically. For products/quotients, add the relative errors of each variable, multiplying by the exponent if the variable is raised to a power.