JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 10)

There are 100 divisions on the circular scale of a screw gauge of pitch $$1 \mathrm{~mm}$$. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :
4.65 mm
4.60 mm
4.55 mm
3.35 mm

Explanation

  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

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