JEE PYQ: Units & Measurements - Question ID 63f0d0d0aa7a (JEE Main 2024)

Core Formula & Concept:

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

• A linear (main) scale marked in millimeters (or other units) along the spindle.
• A circular (thimble) scale with fine divisions, rotating around the linear scale.

Key definitions:

• Pitch ($p$): The distance moved by the spindle per complete rotation of the circular scale. Here, $p = 1 \text{ mm}$.
• Least Count (LC): The smallest measurement possible with the screw gauge, given by: $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}.$
• Zero Error: A systematic error when the zero of the circular scale does not align with the reference line (linear scale) when the jaws are closed. Here, the zero is 5 divisions below the reference line, indicating a positive zero error of: $\text{Zero Error} = +5 \times \text{LC} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}.$

The observed reading is the sum of the linear scale reading and the circular scale reading, corrected for zero error: $\text{Actual Diameter} = (\text{Linear Scale Reading} + \text{Circular Scale Reading}) - \text{Zero Error}.$

--- Step-by-Step Derivation:

1. Calculate the Linear Scale Reading:

   The linear scale shows 4 divisions clearly visible. Since the pitch is $1 \text{ mm}$, each division corresponds to $1 \text{ mm}$. Thus: $\text{Linear Scale Reading} = 4 \times 1 \text{ mm} = 4 \text{ mm}.$

2. Calculate the Circular Scale Reading:

   The circular scale has 60 divisions coinciding with the reference line. The least count is $0.01 \text{ mm}$, so: $\text{Circular Scale Reading} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}.$

3. Compute the Observed Diameter:

   The observed diameter (before zero error correction) is: $\text{Observed Diameter} = \text{Linear Scale Reading} + \text{Circular Scale Reading} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}.$

4. Apply Zero Error Correction:

   The zero error is $+0.05 \text{ mm}$ (since the zero is below the reference line, the instrument over-reads). To get the actual diameter, subtract the zero error: $\text{Actual Diameter} = \text{Observed Diameter} - \text{Zero Error} = 4.60 \text{ mm} - 0.05 \text{ mm} = 4.55 \text{ mm}.$

5. Match with Given Options:

   The calculated diameter is $4.55 \text{ mm}$, which corresponds to Option C.

--- Common Traps & Exam Tip:

1. Misinterpreting Zero Error Direction:

   Students often confuse whether to add or subtract the zero error. Remember:

   • If the zero is above the reference line, the error is negative (instrument under-reads), so add the error.
   • If the zero is below the reference line, the error is positive (instrument over-reads), so subtract the error.

2. Ignoring Least Count Calculation:

   Some students assume the circular scale divisions are in $0.1 \text{ mm}$ instead of calculating the least count ($0.01 \text{ mm}$ here). Always compute LC as $\frac{\text{Pitch}}{\text{Number of divisions}}$.

3. Linear Scale Misreading:

   The linear scale reading is the number of fully visible divisions (4 here), not the position of the reference line. Do not confuse it with the circular scale reading.

4. Sign Errors in Final Calculation:

   Ensure the zero error correction is applied with the correct sign. A quick sanity check: If the zero is below the line, the actual diameter should be less than the observed reading.

Exam Tip: Always sketch the screw gauge readings (linear and circular scales) to visualize the zero error and avoid sign mistakes.