Question

The pitch of a screw gauge is 1 mm and there are 50 divisions on its cap. When nothing is put in between the studs, 44th division of the circular scale coincides with the reference line. When a glass plate is placed between the studs, the main scale reads three divisions and the circular scale reads 26 divisions. Calculate the thickness of the plate –

• A. (a) 3.52 mm
• A. (b) 3.26 mm
• A. (c) 3.40 mm
• A. (d) 3.64 mm

Answer 1

(d)

Sol. LC = $\frac { 1 } { 50 }$ mm = 0.02 mm

error = – 6 × 0.02 = – 0.12

observed reading = 3 mm + 26 × (0.02) mm

= 3.52 mm

true reading = 3.52 – (– 0.12)

= 3.64 mm