QJEEM24Ph1646: JEE Main 2024 Physics - Class 11 JEEM24Ph1646 Solution: There are 100 divisions on the circular scale of a screw gauge…

1. Pitch of the screw gauge: 1 mm


2. Number of divisions on the circular scale: 100 divisions


3. Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.


4. Measurement data:

• 4 linear scale divisions are visible.


• 60 divisions on the circular scale coincide with the reference line.

Step-by-Step Calculation

1. Least Count of the screw gauge:

$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$

1. Main Scale Reading (MSR):

The linear scale shows 4 divisions, so the main scale reading is:

$\text{MSR} = 4 \text{ mm}$

1. Circular Scale Reading (CSR):

60 divisions coincide with the reference line, so the circular scale reading is:

$\text{CSR} = 60 \times \text{Least Count} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}$

1. Zero Error:

The zero error is 5 divisions below the reference line, indicating a positive zero error:

$\text{Zero Error} = +5 \times \text{Least Count} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}$

1. Total Reading without considering zero error:

$\text{Total Reading (without zero error)} = \text{MSR} + \text{CSR} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}$

1. Corrected Reading considering zero error:

Since the zero error is positive, we subtract it from the total reading:

$\text{Corrected Reading} = \text{Total Reading (without zero error)} - \text{Zero Error} = 4.60 \text{ mm} - 0.05 \text{ mm} = 4.55 \text{ mm}$

Conclusion

The diameter of the wire is:

Option C: 4.55 mm