Question

When a wire is placed between studs of a screw gauge then no main scale division is completed and 34 circular scale division coincides with the reference line. If pitch is 0.5 mm and there is 50 divisions on head scale. Then diameter of wire is –

• A. 0.34 mm
• B. 0.17 mm
• C. 3.4 mm
• D. 1.7 mm

Answer 1

Answer: (A)

0.34 mm

Answer 2

LC = $\frac { 0.5 } { 50 }$ = 0.01 mm

MSR = 0 ; CSR = 34 × 0.01 = 0.34 mm

\ Diameter = MSR + CSR = 0.34 mm