Question

A student is trying to measure the diameter of a lead shot using a screw gauge. The pitch of the screw gauge is 1 mm and has 50 divisions on its circular scale. When the two jaws of the screw gauge are in contact with each other, the zero of the circular scale lies 6 division below the line of graduation. When the lead shot is placed between the jaws, 3 linear scale divisions are clearly visible while 31^st division on the circular scale coincides with the reference line. The diameter of the lead shot is:

• A. 3.62 mm
• B. 3.74 mm
• C. 3.55 mm
• D. 3.50 mm

Answer 1

The correct option is D

3.50 mm

We know that, $\text{Least Count}=\frac{\text{Pitch}}{\text{No.of divisions on the head scale}}=\frac{1mm}{50}=0.02mm$
The zero error is positive, given by Zero error $=6\times \text{Least count}=6\times 0.02mm=0.12mm$
First part of the measurement, main scale reading = 3 mm
Head scale division that coincides with the pitch scale axis, circular scale division = 31
Second part of the measurement, circular scale division x least count = $31\times 0.02mm=0.62mm$
Measurement with zero error = Main scale reading + (Circular scale division $\times$ Least count) = 3 + 0.62 = 3.62 mm.
Zero error = + 0.12 mm [Positive because zero is below the datum line]
So, corrected reading = 3.62 - 0.12 = 3.50 mm.