1. Least count of each screw gauge

Given:

• Pitch $=0.1\text{ cm}$
• Number of circular scale divisions $=100$

So the least count is

2. Actual radius of the wire

This corresponds to:

• Main scale reading $=0.3\text{ cm}$
• Remaining reading $=0.022\text{ cm}$

Hence the correct circular scale contribution should be

So, for an error-free screw gauge, the circular scale reading would be $22$ divisions.

3. Zero error from the figures

From the given closed-jaw positions of the two screw gauges:

• For student A's screw gauge, the reference line passes through division $8$ when closed. This means the zero of circular scale is 8 divisions below the reference line, so the instrument has positive zero error of

  $$+8\times 0.001=+0.008\text{ cm}$$

• For student B's screw gauge, the reference line passes through division $5$ from the other side, i.e. the zero is 5 divisions above the reference line. So this is a negative zero error of magnitude

  $$-5\times 0.001=-0.005\text{ cm}$$

4. Observed final circular scale readings

For a screw gauge:

• positive zero error makes the observed reading larger by that many divisions,
• negative zero error makes the observed reading smaller accordingly.

Thus,

For student A:

For student B:

5. Absolute difference

Hence, the required absolute difference is