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 :

Solution

Physics sir · 4 steps

Read the gauge as usual, then remove the 0.05 mm it was already showing with the jaws shut.

1. Least count. One full turn moves the screw by the pitch, 1 mm, across 100 circular divisions. So the least count is 1/100 = 0.01 mm.
2. Zero error. With nothing between the jaws, the circular zero lies 5 divisions below the reference line. That means the gauge already reads 5 divisions too much, a positive zero error of +5 × 0.01 = +0.05 mm.
3. Observed reading. Main scale shows 4 mm and circular scale shows 60 × 0.01 = 0.60 mm. The observed reading is 4.60 mm.

   d = 4.60 mm - (+0.05 mm) = 4.55 mm

4. So. The true diameter of the wire is 4.55 mm, option (c).

The idea to keep. True reading = observed reading minus zero error, with the sign included. Circular zero below the reference line gives a positive zero error. Circular zero above the line gives a negative one.

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