JEE PYQ: Units & Measurements - Question ID 689e443eceea (JEE Main 2026)

Core Formula & Concept:

A screw gauge is a precision instrument used to measure small lengths (e.g., diameters of wires or spheres). It consists of:

• A main scale (linear scale) marked along the sleeve.
• A circular scale (rotating thimble) with divisions that move past a reference line on the main scale.

Key definitions:

• Pitch ($p$): The distance moved by the screw along the main scale for one complete rotation of the circular scale. Here, $p = 0.1$ mm.
• Least Count ($LC$): The smallest length measurable by the screw gauge, given by: $LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{p}{N}$ where $N = 100$ divisions.
• Zero Error: If the zero of the circular scale does not coincide with the reference line when the studs are in contact, the instrument has a zero error. Here, the zero of the main scale coincides with the 5^th division of the circular scale, indicating a positive zero error of $+5 \times LC$.

The measured diameter ($D$) is calculated as: $D = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC) - \text{Zero Error}$

--- Step-by-Step Derivation:

Step 1: Calculate the Least Count ($LC$)

Given:

• Pitch ($p$) = 0.1 mm
• Number of divisions on circular scale ($N$) = 100

The least count is: $LC = \frac{p}{N} = \frac{0.1 \text{ mm}}{100} = 0.001 \text{ mm}$

Step 2: Determine the Zero Error

When the studs are in contact, the zero of the main scale coincides with the 5^th division of the circular scale. This means the zero error is: $\text{Zero Error} = +5 \times LC = +5 \times 0.001 \text{ mm} = +0.005 \text{ mm}$ (Positive because the circular scale reading is ahead of the main scale zero.)

Step 3: Record the Observed Reading

For the sphere:

• Main scale reading = 5 mm
• Circular scale reading = 50^th division

The observed diameter (before correcting for zero error) is: $$D_{\text{observed}} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC) \\ = 5 \text{ mm} + (50 \times 0.001 \text{ mm}) \\ = 5 \text{ mm} + 0.05 \text{ mm} \\ = 5.05 \text{ mm}$$

Step 4: Correct for Zero Error

The actual diameter ($D$) is the observed reading minus the zero error: $$D = D_{\text{observed}} - \text{Zero Error} \\ = 5.05 \text{ mm} - 0.005 \text{ mm} \\ = 5.045 \text{ mm}$$

Step 5: Match with Given Options

The calculated diameter is $5.045$ mm, which corresponds to Option A.

--- Common Traps & Exam Tip:

1. Ignoring Zero Error: Many students forget to account for the zero error, leading to an incorrect answer of $5.05$ mm (which is not among the options). Always check for zero error in screw gauge questions.
2. Sign of Zero Error: Students often confuse whether to add or subtract the zero error. Here, since the circular scale reading is ahead of the main scale zero, the zero error is positive and must be subtracted from the observed reading.
3. Least Count Calculation: Misinterpreting the pitch or number of divisions can lead to an incorrect least count. Double-check that $LC = \frac{p}{N}$.
4. Circular Scale Reading: Ensure the circular scale reading is multiplied by the least count (not the pitch). Here, $50 \times 0.001$ mm, not $50 \times 0.1$ mm.

Exam Tip: Always write down the zero error correction explicitly in your solution to avoid sign errors. For screw gauge problems, the mantra is: "Main Scale + (Circular Scale × LC) – Zero Error".