Question

The pitch of a screw gauge is 0.5 mm and it has 50 divisions on circular scale. When the studs are in contact, 46^th division of circular scale coincides with reference line. When a wire is kept between the studs the linear scale reads 2 division and 20^th division of circular coincides with the reference line. The correct statement is –

• A. Least count is 0.01 cm
• B. Zero correction is – 0.04 mm
• C. Radius is 0.62 mm
• D. All of the above

Answer 1

Answer: (C)

Radius is 0.62 mm

Answer 2

e = – 4 × 0.01 mm

c = 4 × 0.01 mm

2r = 2 × 0.5 + 20 × 0.01 + 0.04

2r = 1.24

r = 0.62 mm