JEE PYQ: Units & Measurements - Question ID 5561befb0f22 (JEE Main 2026)

Core Formula & Concept:

A vernier calliper measures lengths with higher precision than a simple main scale. Its key features are:

• Main scale (MS): Graduated in whole millimetres (1 mm per division).
• Vernier scale (VS): A sliding scale with $n$ divisions that span the same length as $(n-1)$ divisions on the main scale. The difference between one main-scale division and one vernier-scale division is the least count (LC).
• Least count (LC): $\text{LC} = \frac{\text{Value of 1 main-scale division}}{\text{Number of vernier divisions}} = \frac{1\ \text{mm}}{10} = 0.1\ \text{mm}.$
• Zero error: When the jaws touch, the vernier zero does not align with the main-scale zero. If the vernier zero lies to the right of the main-scale zero, the error is positive and must be subtracted from the observed reading.
• Reading formula: $\text{Actual length} = (\text{Main-scale reading}) + (\text{Vernier coincidence}\times\text{LC}) - \text{Zero error}.$

Step-by-Step Derivation:

Step 1: Determine the zero error.

When the jaws touch, the vernier zero is to the right of the main-scale zero, and the $4^{\text{th}}$ vernier division coincides with a main-scale mark. Since the least count is $0.1\ \text{mm}$, the zero error is $\text{Zero error} = +\,(4 \times 0.1\ \text{mm}) = +\,0.4\ \text{mm}.$ This positive error must be subtracted from any observed reading.

Step 2: Record the observed reading.

While measuring the cylinder:

• Main-scale reading = $15$ divisions = $15\ \text{mm}$.
• $5^{\text{th}}$ vernier division coincides with a main-scale mark.

The vernier contribution is $5 \times 0.1\ \text{mm} = 0.5\ \text{mm}.$ Hence the observed length is $15\ \text{mm} + 0.5\ \text{mm} = 15.5\ \text{mm}.$

Step 3: Apply the zero-error correction.

Actual length = Observed length $-\, \text{Zero error}$ $= 15.5\ \text{mm} - 0.4\ \text{mm} = 15.1\ \text{mm}.$

Common Traps & Exam Tip:

1. Sign of zero error: Many students forget that a vernier zero to the right of the main-scale zero gives a positive error that must be subtracted. Reversing the sign leads to $15.9\ \text{mm}$ (option C).
2. Least-count confusion: Some miscalculate the least count as $0.01\ \text{mm}$ or $1\ \text{mm}$, producing wrong vernier contributions.
3. Coincidence miscount: Counting the $4^{\text{th}}$ or $5^{\text{th}}$ division incorrectly shifts the final answer by $\pm0.1\ \text{mm}$.

Always double-check the zero-error sign and the coincidence number before applying the correction.