In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46ᵗʰ division the circular scale coincide with the reference line. The diameter of the wire is ______________ × 10⁻² mm.

Solution

Physics sir · 4 steps

Find the least count first, then read the scale and subtract the positive zero error.

1. Least count. One full turn moves the screw by the pitch, 0.5 mm, and the circular scale has 100 divisions. So the least count is 0.5/100 = 0.005 mm.
2. Zero error. With the studs touching, the circular zero sits 6 divisions below the reference line. So the gauge already reads +6 divisions with nothing between the studs. The zero error is positive: +6 × 0.005 = +0.03 mm.
3. Observed reading. Each linear division equals the pitch, 0.5 mm, so 4 divisions give 2 mm. The circular scale adds 46 × 0.005 = 0.23 mm, so the observed reading is 2.23 mm.

   d = observed reading - zero error = 2.23 - 0.03 = 2.20 mm

4. So. The diameter is 2.20 mm = 220 × 10⁻² mm, so the answer is 220.

The common slip: 226. Adding the zero error instead of subtracting it gives 2.23 + 0.03 = 2.26 mm, which is 226, not 220.

The idea to keep. If the circular zero is below the reference line, the zero error is positive. If it is above, the zero error is negative. Always subtract the zero error, with its sign, from the observed reading.

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