JEE PYQ: Units & Measurements - Question ID 7df2fe0ee3f6 (JEE Main 2026)

Core Formula & Concept:

In experiments involving measurements, the least count (LC) of an instrument is the smallest value that can be measured or read using that instrument. It determines the precision of the measurement.

When a set of readings is given, the least count can be inferred from the difference between consecutive readings. Specifically:

• The least count must be smaller than or equal to the smallest difference between any two recorded values.
• It must also be consistent with the precision of the readings (i.e., the number of decimal places).

Mathematically, if readings are given as $x_1, x_2, x_3, \dots$, then: $\text{LC} \leq \min |x_i - x_j| \quad \text{for} \quad i \neq j$ Additionally, the least count must match the smallest unit in which the readings are expressed.

--- Step-by-Step Derivation:

Step 1: List the given readings

The readings obtained are: $1.24\ \text{mm}, \quad 1.25\ \text{mm}, \quad 1.23\ \text{mm}, \quad 1.21\ \text{mm}$

Step 2: Compute differences between consecutive readings

We compute the absolute differences between each pair of readings:

• $|1.25 - 1.24| = 0.01\ \text{mm}$
• $|1.24 - 1.23| = 0.01\ \text{mm}$
• $|1.23 - 1.21| = 0.02\ \text{mm}$
• $|1.25 - 1.23| = 0.02\ \text{mm}$
• $|1.25 - 1.21| = 0.04\ \text{mm}$
• $|1.24 - 1.21| = 0.03\ \text{mm}$

The smallest difference observed is $0.01\ \text{mm}$.

Step 3: Analyze the precision of the readings

All readings are given to two decimal places in millimeters. This implies that the instrument can measure up to $0.01\ \text{mm}$.

If the least count were smaller (e.g., $0.001\ \text{mm}$), the readings would have been recorded with three decimal places. Since they are not, the least count cannot be $0.001\ \text{mm}$.

Step 4: Apply the least count condition

The least count must satisfy: $\text{LC} \leq \text{smallest difference} = 0.01\ \text{mm}$ and must be consistent with the precision of the readings.

Among the options: - $0.01\ \text{mm}$ (Option A) satisfies both conditions. - $0.05\ \text{mm}$ and $0.1\ \text{mm}$ are too large (they exceed the smallest difference). - $0.001\ \text{mm}$ is too small (readings would have more decimal places).

Conclusion:

The expected least count of the instrument is $\boxed{0.01\ \text{mm}}$, which corresponds to Option A.

--- Common Traps & Exam Tip:

Trap 1: Confusing least count with smallest difference
Some students assume the least count is equal to the smallest difference between readings. While the least count must be ≤ the smallest difference, it must also match the precision of the readings. For example, if readings are given to two decimal places, the least count cannot be $0.001$ (three decimal places).

Trap 2: Ignoring the number of decimal places
Students often overlook the fact that the number of decimal places in the readings directly hints at the least count. If readings are given as $1.24, 1.25$, etc., the instrument must have a least count of $0.01$, not $0.1$ or $0.001$.

Exam Tip:
Always check: 1. The smallest difference between readings. 2. The number of decimal places in the readings. The least count must satisfy both conditions. In this case, $0.01\ \text{mm}$ is the only option that does.