The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while division on circular scale coincides with the reference line, The radius of the wire is :
- (A)0.82 mm✓ Correct
- (B)1.64 mm
- (C)0.90 mm
- (D)1.80 mm
Answer: (A)
Solution
Least count (LC) = pitch/divisions = 1 mm/100 = 0.01 mm.
Zero error = +8 divisions = +0.08 mm (positive because zero is below reference line).
Reading = Main scale reading + (Circular scale reading × LC) - Zero error
= 1 mm + (72 × 0.01 mm) - 0.08 mm
= 1 + 0.72 - 0.08 = 1.64 mm
This is the diameter. Radius = 1.64/2 = 0.82 mm.
Zero error = +8 divisions = +0.08 mm (positive because zero is below reference line).
Reading = Main scale reading + (Circular scale reading × LC) - Zero error
= 1 mm + (72 × 0.01 mm) - 0.08 mm
= 1 + 0.72 - 0.08 = 1.64 mm
This is the diameter. Radius = 1.64/2 = 0.82 mm.
JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option