Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter are given. To obtain the value of , the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of (in ) is:
| Readings | ||
|---|---|---|
| LS (mm) | CS | |
| Wire-1 | 0.5 | 42 |
| Wire-2 | 1.5 | 95 |
Answer: 1915
Solution
Least count = mm.
Measured diameter of Wire-1: mm. True value is 0.65 mm, so positive zero error = mm.
Measured diameter of Wire-2: mm. True diameter: mm = 1915 m.
Measured diameter of Wire-1: mm. True value is 0.65 mm, so positive zero error = mm.
Measured diameter of Wire-2: mm. True diameter: mm = 1915 m.
JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer