There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :
- 4.65 mm
- 4.60 mm
- 4.55 mm
- 3.35 mm
Answer
(3) 4.55 mm
Solution
Least count = pitch ÷ circular divisions = 1 mm100 = 0.01 mm
zero of circular scale below the reference line ⇒ positive zero error = +5 0.01 = +0.05 mm
Observed reading = 4 1 + 60 0.01 = 4.60 mm
Hence diameter = observed reading − zero error = 4.60 - 0.05 = 4.55 mm
Answer: (3)