JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 10)
Explanation
- Pitch of the screw gauge: 1 mm
- Number of divisions on the circular scale: 100 divisions
- Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.
- Measurement data:
- 4 linear scale divisions are visible.
- 60 divisions on the circular scale coincide with the reference line.
Step-by-Step Calculation
- Least Count of the screw gauge:
$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$
- Main Scale Reading (MSR):
The linear scale shows 4 divisions, so the main scale reading is:
$\text{MSR} = 4 \text{ mm}$
- Circular Scale Reading (CSR):
60 divisions coincide with the reference line, so the circular scale reading is:
$\text{CSR} = 60 \times \text{Least Count} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}$
- Zero Error:
The zero error is 5 divisions below the reference line, indicating a positive zero error:
$\text{Zero Error} = +5 \times \text{Least Count} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}$
- Total Reading without considering zero error:
$\text{Total Reading (without zero error)} = \text{MSR} + \text{CSR} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}$
- 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
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