Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ______________.
[Figure shows position of reference 'O' when jaws of screw gauge are closed]
Given pitch = 0.1 cm.

Solution

Physics sir · 3 steps

Find each gauge's zero error with its sign, then ask what each gauge must read for a true 0.322 cm.

1. Least count. Least count = (0.1 cm)/100 = 0.001 cm, so one circular division is 0.001 cm.
2. Zero errors. In gauge A, with the jaws closed, the reference line is at division 5, just past zero. That is a positive zero error of +5 × 0.001 = +0.005 cm. In gauge B the line is at division 92, which is 8 divisions short of zero. That is a negative zero error of -8 × 0.001 = -0.008 cm.

   observed reading = true value + zero error

   Gauge	Zero error	Observed reading	Main scale and circular scale
   A	+0.005 cm	0.322 + 0.005 = 0.327 cm	main scale 0.3 cm, circular scale 27
   B	-0.008 cm	0.322 - 0.008 = 0.314 cm	main scale 0.3 cm, circular scale 14

3. So. Both gauges show the same main scale reading of 0.3 cm. The circular scale readings differ by 27 - 14 = 13, so the answer is 13.

The common slip: 3. Giving both zero errors the same sign, for example taking B's error as +0.008 cm, gives circular readings of 27 and 30 and a difference of 3.

The idea to keep. Zero error takes its sign from where the reference line sits when the jaws are closed: past zero is positive, short of zero (like 92 of 100) is negative. The observed reading is the true value plus the zero error.

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