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The basic information, about a given screw gauge, is as follows

(i) Number of divisions on the circular scale = 50 

(ii) Distance moved, by the moving jaw, in 5 complete rotations = 5 mm 

(iii) Position of the zero of the circular scale, when the two jaws just touch each other; as shown in the figure When a wire is just held between the jaws of the screw gauge 

(a) Reading of the main scale = 6.2 mm 

(b) No. of circular scale division coinciding with the ‘reference line’ = 38 

The radius, of this wire, equals

(1) 0.326 cm

(2) 0.652cm

(3) 0.664 cm

(4) 0.332 cm

1 Answer

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Best answer

(1) 0.326 cm

Pitch = 5/5 mm  = 1mm = 0.1 cm

∴ Least count = 0.1/100 = 0.001 cm

Zero error = + 6 x 0.001 cm = + 0.006 cm 

Total observed reading = (0.62 + 38 x 0.001) cm

= 0.658 cm 

∴ Corrected reading = (0.658 – 0.006) cm 

= 0.652 cm 

Radius of wire = 0.326 cm

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