JEE Advanced 2026 Physics question
Two thin wires, Wire-1 of diameter 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 and there are divisions on the circular scale (CS). The smallest division on the linear scale (LS) is . The table shows the readings of LS and CS for the measurements. The value of (in µm) is:
| | Readings: LS (mm) | Readings: CS | |---|---|---| | Wire-1 | 0.5 | 42 | | Wire-2 | 1.5 | 95 |
Answer: 1915
Solution
**Least count.**
LC===0.005\ mm=5\ μm.
**Observed reading of Wire-1.**
LS+CS×LC=0.5+42(0.005)=0.5+0.210=0.710\ mm.
**Zero error.*
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