JEE PYQ: Units & Measurements - Question ID a18339500b1a (JEE Main 2026)

Core Formula & Concept:

In experiments involving Ohm’s law, the resistance $R$ of a wire is determined by measuring the potential difference $V$ across it and the current $I$ flowing through it. The fundamental relation is: $R = \frac{V}{I}.$

When instruments (voltmeter and ammeter) are used, each reading carries an inherent uncertainty due to the least count of the instrument. The least count is the smallest division that can be read on the instrument’s scale. The absolute error in a measurement is typically taken as half the least count (assuming uniform distribution of error), though in many practical contexts, especially in JEE, the least count itself is treated as the maximum absolute error.

The relative error in a quantity $x$ is given by: $\frac{\Delta x}{x},$ where $\Delta x$ is the absolute error in $x$. For a function $R = \frac{V}{I}$, the relative error in $R$ is obtained using the formula for error propagation in division: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}.$ The absolute error in $R$ is then: $\Delta R = R \left( \frac{\Delta V}{V} + \frac{\Delta I}{I} \right).$

Step-by-Step Derivation:

Step 1: Record the given data
- Voltmeter reading: $V = 10 \, \text{V}$ - Ammeter reading: $I = 5 \, \text{A}$ - Least count of voltmeter: $500 \, \text{mV} = 0.5 \, \text{V}$ - Least count of ammeter: $200 \, \text{mA} = 0.2 \, \text{A}$

Step 2: Compute the nominal resistance
Using Ohm’s law: $R = \frac{V}{I} = \frac{10}{5} = 2 \, \Omega$

Step 3: Determine absolute errors in $V$ and $I$
The least count is treated as the maximum absolute error: - $\Delta V = 0.5 \, \text{V}$ - $\Delta I = 0.2 \, \text{A}$

Step 4: Compute relative errors
- Relative error in $V$: $\frac{\Delta V}{V} = \frac{0.5}{10} = 0.05$ - Relative error in $I$: $\frac{\Delta I}{I} = \frac{0.2}{5} = 0.04$

Step 5: Propagate errors to find $\Delta R$
Using the error propagation formula for division: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = 0.05 + 0.04 = 0.09$ Thus, $\Delta R = R \times 0.09 = 2 \times 0.09 = 0.18 \, \Omega$

Common Traps & Exam Tip:

Trap 1: Students often confuse the least count with half the least count. While in some contexts, the error is taken as half the least count, in JEE and most standard experiments, the least count itself is used as the absolute error unless specified otherwise.

Trap 2: Misapplying error propagation rules. Some students multiply the relative errors instead of adding them when dealing with division. Remember: for $R = \frac{V}{I}$, the relative errors add.

Trap 3: Forgetting to convert units. The least counts are given in millivolts and milliamperes, so they must be converted to volts and amperes before use.

Exam Tip: Always write down the error propagation formula explicitly. Even if you remember the rule, showing the formula helps avoid sign errors and ensures clarity in reasoning.

Thus, the estimated error in the resistance measurement is $0.18 \, \Omega$, which corresponds to option D.