JEE PYQ: Units & Measurements - Question ID 558d302ab8b4 (JEE Main 2021)

In a screw gauge (also called a micrometer screw gauge), the measurement consists of two parts:

1. Main scale reading (MSR): This is the reading on the linear scale (main scale) just before the thimble (circular scale) starts. It is usually in millimeters.
2. Circular scale reading (CSR): This is the reading on the circular scale where a division coincides with the reference line (datum line). The circular scale divides one complete rotation of the screw into a fixed number of divisions (here, 50). The pitch of the screw (distance moved per complete rotation) is given (here, 0.5 mm).

The least count (LC) of the screw gauge is the smallest measurement that can be made and is calculated as:

The true reading (TR) is then:

However, the screw gauge may have a zero error. In this question, when the ratchet is closed (jaws are in contact), the 5^th division of the circular scale coincides with the reference line. This indicates a positive zero error of:

The true reading must be corrected by subtracting this zero error from the observed reading.

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

• Pitch = 0.5 mm (distance moved per complete rotation)
• Number of divisions on circular scale = 50

Step 2: Determine the Zero Error
When the ratchet is closed, the 5^th division coincides with the reference line. This means the zero error is: $\text{Zero Error} = 5 \times \text{LC} = 5 \times 0.01 \text{ mm} = 0.05 \text{ mm}$ Since the 5^th division is ahead of the reference line, this is a positive zero error.

Step 3: Calculate the Observed Reading
For the given observation:

• Main scale reading (MSR) = 5 mm
• Circular scale reading (CSR) = 20^th division

Step 4: Apply Zero Error Correction
Since the zero error is positive, the true reading is obtained by subtracting the zero error from the observed reading: $\text{True Reading} = \text{Observed Reading} - \text{Zero Error} = 5.20 \text{ mm} - 0.05 \text{ mm} = 5.15 \text{ mm}$

1. Ignoring Zero Error: Many students forget to account for the zero error, leading them to choose 5.20 mm (Option D) instead of the correct 5.15 mm (Option C). Always check the zero error condition given in the question.
2. Sign of Zero Error: If the zero error is positive (as in this case), it must be subtracted from the observed reading. If it were negative, it would be added. Misinterpreting the sign leads to incorrect results.
3. Least Count Calculation: Some students mistakenly take the pitch as the least count or miscalculate the number of divisions. Always verify the least count using the formula: $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
4. Circular Scale Reading: Ensure that the circular scale reading is multiplied by the least count. For example, the 20^th division corresponds to $20 \times 0.01 \text{ mm} = 0.20 \text{ mm}$, not 20 mm.

Exam Tip: Always write down the zero error correction explicitly in your rough work. This helps avoid sign errors and ensures you don’t overlook it during time pressure.