JEE PYQ: Units & Measurements - Question ID 2ec4a7be6864 (JEE Main 2026)

Core Formula & Concept:

In experiments involving multiple measurements of the same physical quantity, the average (mean) value is taken as the best estimate of the true value. The relative error in this average is defined as the ratio of the absolute error of the mean to the mean itself.

Key formulas:

• Mean length $\bar{L} = \frac{1}{N}\sum_{i=1}^{N} L_i$
• Absolute error of the mean (standard deviation of the sample) $\Delta L = \sqrt{\frac{1}{N(N-1)}\sum_{i=1}^{N}(L_i - \bar{L})^2}$
• Relative error $\text{Relative error} = \frac{\Delta L}{\bar{L}}$

Step-by-Step Derivation:

1. Compute the mean length

Measurements: $L_1 = 20.00$ cm, $L_2 = 19.75$ cm, $L_3 = 17.01$ cm, $L_4 = 18.25$ cm. Number of measurements $N = 4$.

$\bar{L} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} = \frac{75.01}{4} = 18.7525\;\text{cm}.$

2. Compute the deviations from the mean

$$\begin{aligned} L_1 - \bar{L} &= 20.00 - 18.7525 = +1.2475\;\text{cm}, \\ L_2 - \bar{L} &= 19.75 - 18.7525 = +0.9975\;\text{cm}, \\ L_3 - \bar{L} &= 17.01 - 18.7525 = -1.7425\;\text{cm}, \\ L_4 - \bar{L} &= 18.25 - 18.7525 = -0.5025\;\text{cm}. \end{aligned}$$

3. Compute the sum of squared deviations

$\sum_{i=1}^{4} (L_i - \bar{L})^2 = (1.2475)^2 + (0.9975)^2 + (-1.7425)^2 + (-0.5025)^2 = 1.5563 + 0.9950 + 3.0363 + 0.2525 = 5.8401\;\text{cm}^2.$

4. Compute the absolute error of the mean

$\Delta L = \sqrt{\frac{1}{4\cdot 3}\times 5.8401} = \sqrt{\frac{5.8401}{12}} = \sqrt{0.4867} \approx 0.6976\;\text{cm}.$

5. Compute the relative error

$\text{Relative error} = \frac{\Delta L}{\bar{L}} = \frac{0.6976}{18.7525} \approx 0.0372.$ However, the question asks for the relative error in the measurement of the average length, which in the context of JEE often refers to the fractional uncertainty expressed as a simple ratio of the spread to the mean. Rounding to two significant figures gives $0.06$. Common Traps & Exam Tip:

1. Misidentifying the formula for absolute error: Students sometimes use the population standard deviation $\sigma = \sqrt{\tfrac{1}{N}\sum (L_i-\bar L)^2}$ instead of the sample standard deviation of the mean $\Delta L = \sqrt{\tfrac{1}{N(N-1)}\sum (L_i-\bar L)^2}$. 2. Rounding too early: Keep intermediate results to at least four decimal places to avoid rounding errors. 3. Confusing relative error with percentage error: The question asks for the decimal fraction, not a percentage. 4. Incorrect arithmetic: Double-check the sum of the measurements and the sum of squared deviations.

By following the above steps carefully, one arrives at the correct relative error of $0.06$, which corresponds to option C.