JEE PYQ: Units & Measurements - Question ID b7eee75c9519 (JEE Main 2021)

A screw gauge is a precision instrument used to measure small lengths (e.g., diameter of wires, thickness of sheets). It consists of two scales:

• Linear (Main) Scale: A scale along the axis of the screw, with markings typically in millimeters (mm). The pitch of the screw is the distance moved by the screw per complete rotation of the circular scale. Here, the pitch is given as $1 \text{ mm}$.
• Circular Scale: A rotating scale with divisions (here, 100 divisions). Each division on the circular scale represents a fraction of the pitch. The least count (LC) of the screw gauge is the smallest length measurable and is calculated as: $\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$

When measuring an object:

• The main scale reading (MSR) is the reading on the linear scale just before the zero of the circular scale.
• The circular scale reading (CSR) is the division on the circular scale that aligns with the reference line.
• The zero error occurs if the zero of the circular scale does not coincide with the reference line when the jaws are closed. It can be positive or negative.

The actual measurement is: $\text{Actual Reading} = \text{MSR} + (\text{CSR} \times \text{LC}) - \text{Zero Error}$

The radius of the wire is half of its diameter.

Step 1: Determine the Least Count (LC)

Step 2: Analyze the Zero Error

When nothing is placed between the jaws, the zero of the circular scale lies 8 divisions below the reference line. This means the instrument has a positive zero error of: $\text{Zero Error} = +8 \times \text{LC} = +8 \times 0.01 \text{ mm} = +0.08 \text{ mm}$ This error must be subtracted from the observed reading to get the correct measurement.

Step 3: Record the Measurement with the Wire

When the wire is placed: - The first division of the linear scale is clearly visible → Main Scale Reading (MSR) = $1 \text{ mm}$ - The 72^nd division on the circular scale coincides with the reference line → Circular Scale Reading (CSR) = $72$

The observed diameter is: $\text{Observed Reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 1 \text{ mm} + (72 \times 0.01 \text{ mm}) = 1 \text{ mm} + 0.72 \text{ mm} = 1.72 \text{ mm}$

Step 4: Apply Zero Error Correction

The actual diameter of the wire is: $\text{Actual Diameter} = \text{Observed Reading} - \text{Zero Error} = 1.72 \text{ mm} - 0.08 \text{ mm} = 1.64 \text{ mm}$

Step 5: Calculate the Radius

The radius is half the diameter: $\text{Radius} = \frac{\text{Actual Diameter}}{2} = \frac{1.64 \text{ mm}}{2} = 0.82 \text{ mm}$

Conclusion: The radius of the wire is $0.82 \text{ mm}$, which corresponds to option C.

• Misinterpreting Zero Error Direction: Students often confuse whether the zero error is positive or negative. If the zero of the circular scale is below the reference line, the error is positive and must be subtracted. If it were above, the error would be negative and added.
• Ignoring Zero Error: Many students forget to correct for zero error, leading to incorrect diameter calculations. Always check for zero error before finalizing the measurement.
• Confusing Diameter with Radius: The question asks for the radius, not the diameter. Ensure you divide the final diameter by 2.
• Incorrect Least Count Calculation: Some students mistakenly take the least count as 0.1 mm instead of 0.01 mm. Always verify the number of divisions on the circular scale.

Exam Tip: Always write down the least count, zero error, and observed reading clearly. Double-check the direction of zero error and apply the correction properly. This systematic approach minimizes errors in screw gauge problems.