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 72$$^{nd}$$ division on circular scale coincides with the reference line. The radius of the wire is

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale, so the least count is $$\text{LC} = \frac{1}{100} = 0.01$$ mm.

When nothing is placed between the jaws, the zero of the circular scale lies 8 divisions below the reference line. This means the zero error is positive: $$\text{Zero error} = +8 \times 0.01 = +0.08$$ mm.

When the wire is placed between the jaws, the first linear scale division is clearly visible, giving a main scale reading of 1 mm. The 72nd circular scale division coincides with the reference line, giving a circular scale reading of $$72 \times 0.01 = 0.72$$ mm.

The observed reading is $$1 + 0.72 = 1.72$$ mm.

The corrected reading (diameter of the wire) is $$\text{Observed reading} - \text{Zero error} = 1.72 - 0.08 = 1.64$$ mm.

The radius of the wire is $$\frac{1.64}{2} = 0.82$$ mm.

The correct answer is Option (2): 0.82 mm.