JEE PYQ: Units & Measurements - Question ID d388307ac5ec (JEE Main 2024)

The resistivity $\rho$ of a metal wire is related to its resistance $R$, length $l$, and cross-sectional area $A$ by the formula: $\rho = R \cdot \frac{A}{l}$ Since the wire is cylindrical, its cross-sectional area is $A = \pi r^2$, where $r$ is the radius. Substituting, we get: $\rho = \frac{R \cdot \pi r^2}{l}$

The question asks for the percentage error in $\rho$. When a quantity is computed from measured values with uncertainties, the percentage error in the result is found using the rules of error propagation for multiplication and division:

• For a product or quotient of quantities, the relative error (fractional error) in the result is the sum of the relative errors of the individual quantities.
• If a quantity is raised to a power $n$, its relative error is multiplied by $|n|$.

Step 1: Express resistivity in terms of measured quantities
We have: $\rho = \frac{R \cdot \pi r^2}{l}$ Since $\pi$ is a constant with no error, we focus on $R$, $r$, and $l$. The relative error in $\rho$ is: $\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \cdot \frac{\Delta r}{r} + \frac{\Delta l}{l}$

Step 2: Compute relative errors of each measured quantity
Given:

• $r = 0.35 \pm 0.05$ cm → $\frac{\Delta r}{r} = \frac{0.05}{0.35} \approx 0.142857$
• $R = 100 \pm 10$ ohm → $\frac{\Delta R}{R} = \frac{10}{100} = 0.1$
• $l = 15 \pm 0.2$ cm → $\frac{\Delta l}{l} = \frac{0.2}{15} \approx 0.013333$

Step 3: Apply error propagation formula
Substitute the relative errors: $\frac{\Delta \rho}{\rho} = 0.1 + 2 \cdot 0.142857 + 0.013333$ $= 0.1 + 0.285714 + 0.013333$ $= 0.399047$

Step 4: Convert to percentage error
Multiply by 100: $\text{Percentage error} = 0.399047 \times 100 \approx 39.9\%$

Step 5: Match with given options
The closest option is D: 39.9%.

Students often make these mistakes:

• Forgetting the power rule: They add $\frac{\Delta r}{r}$ only once instead of twice, because $r$ is squared in the formula. This leads to an underestimation of error.
• Ignoring constant factors: They mistakenly include $\pi$ in error propagation, even though it has no uncertainty.
• Rounding errors prematurely: Rounding intermediate values (like $\frac{0.05}{0.35}$) too early can lead to inaccuracies. Keep more decimal places until the final step.
• Confusing absolute and relative error: Adding absolute errors directly instead of converting to relative errors first.