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JEE Advanced 2026 · Paper 2, Physics question 10

Two thin wires, Wire-1 of diameter 0.6500.650 mm and Wire-2 of unknown diameter dd are given. To obtain the value of dd, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.50.5 mm and there are 100100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.50.5 mm. The table shows the readings of LS and CS for the measurements. The value of dd (in µm) is:

LS (mm) CS
Wire-1 0.5 42
Wire-2 1.5 95
Answer and worked solution

Official answer: 1915

Least count =0.5100=0.005=\dfrac{0.5}{100}=0.005 mm.

Wire-1. Reading =0.5+42×0.005=0.710=0.5+42\times0.005=0.710 mm. The true diameter is 0.6500.650 mm, so the zero error is +0.060+0.060 mm.

Wire-2. Reading =1.5+95×0.005=1.975=1.5+95\times0.005=1.975 mm. Corrected:
d=1.975−0.060=1.915 mm=1915 μm.d=1.975-0.060=1.915\text{ mm}=1915\ \mu\text{m}.

Topics:Vernier callipers and screw gaugeLeast count
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