JEE PYQ: Units & Measurements - Question ID 4e54075614b5 (JEE Main 2019)

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

• Main (Linear) Scale: Fixed scale with markings in millimeters (mm).
• Circular (Vernier) Scale: Rotating scale with divisions that subdivide the smallest main-scale division.

The key concepts involved are:

1. Pitch ($p$): The distance moved by the screw per complete rotation of the circular scale. Here, $p = 0.5$ mm.
2. Least Count (LC): The smallest measurement possible with the screw gauge, given by: $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{p}{N}$ Here, $N = 100$, so $\text{LC} = \frac{0.5}{100} = 0.005$ mm.
3. Zero Error: If the zero of the circular scale does not align with the main scale’s zero when fully tightened, the instrument has a zero error. Here, the zero is 3 divisions below the mean line, which implies a positive zero error of: $\text{Zero Error} = + (3 \times \text{LC}) = + (3 \times 0.005) = +0.015 \text{ mm}.$ This means the instrument overestimates measurements by 0.015 mm, so we must subtract this from the observed reading.

Step 1: Calculate the observed reading.

The main scale reading is $5.5$ mm, and the circular scale reading is $48$ divisions. The total observed thickness ($T_{\text{obs}}$) is: $T_{\text{obs}} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{LC})$ $T_{\text{obs}} = 5.5 + (48 \times 0.005) = 5.5 + 0.24 = 5.74 \text{ mm}.$

Step 2: Correct for zero error.

The zero error is $+0.015$ mm (as derived above). To get the true thickness ($T_{\text{true}}$), subtract the zero error from the observed reading: $T_{\text{true}} = T_{\text{obs}} - \text{Zero Error}$ $T_{\text{true}} = 5.74 - 0.015 = 5.725 \text{ mm}.$

Step 3: Match with the given options.

The calculated thickness is $5.725$ mm, which corresponds to Option C.

1. Misinterpreting Zero Error:
   • Students often confuse whether to add or subtract the zero error. Remember: If the zero of the circular scale is below the mean line, the error is positive (instrument overestimates), so subtract it. If it were above, the error would be negative, and you’d add it.
2. Incorrect Least Count Calculation:
   • Some students divide the number of divisions by the pitch instead of the other way around. Always use: $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}}.$
3. Ignoring Units:
   • Ensure all values are in consistent units (here, mm). Mixing cm or meters will lead to incorrect results.
4. Circular Scale Reading:
   • Students sometimes miscount the circular scale divisions. Here, the reading is 48, not 48.0 or 4.8.

Exam Tip: Always verify the zero error first before calculating the final measurement. A quick sketch of the screw gauge’s scales can help avoid sign errors.