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 72ⁿᵈ division on circular scale coincides with the reference line. The radius of the wire is :

Solution

Physics sir · 4 steps

Find the zero error first, subtract it from the reading, and only then halve to get the radius.

1. Least count. Least count = pitch/(circular divisions) = (1 mm)/100 = 0.01 mm.
2. Zero error. With the jaws closed, the circular zero sits 8 divisions below the reference line. So the line already reads 8 divisions when it should read zero. The gauge over-reads, giving a positive zero error of 8 × 0.01 = +0.08 mm.
3. Observed reading. Main scale shows 1 mm and circular scale shows 72 divisions, so the observed reading is 1 + 72 × 0.01 = 1.72 mm.

   d = 1.72 - 0.08 = 1.64 mm

4. So. The gauge measures the diameter, so the radius is 1.64/2 = 0.82 mm, option (c).

The idea to keep. Circular zero below the reference line means positive zero error, so subtract it. A screw gauge always gives the diameter, so halve it if the question asks for the radius.

Answer from the NTA final answer key. Solution by PrepWiser.