Question

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 $72^{\text {nd }}$ division on circular scale coincides with the reference line. The radius of the wire is

• A. $1.64\, mm$
• B. $0.82\, mm$
• C. $1.80\, mm$
• D. $0.90 \,mm$

Answer 1

Answer: (B)

$0.82\, mm$

Answer 2

Least count $=\frac{1 \,mm }{100}=0.01 \,mm$ zero error $=+8 \times LC =+0.08 \,mm$ True reading (Diameter) $=(1\, mm +72 \times LC )-\text { (Zero error) }$ $=(1 mm +72 \times 0.01 mm ) -0.08 \,mm$ $=1.72 \,mm -0.08 \,mm$ therefore radius $=\frac{1.64}{2}$ $=0.82\,mm$