JEE Advanced 2026 — Paper 2 — PHY-III solution

JEE Advanced2026Paper 2PHY-III

Two thin wires, Wire-1 of diameter $0.650$ mm and Wire-2 of unknown diameter $d$ are given. To obtain the value of $d$ , 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 $d$ (in $μ$ m) is:

	LS (mm)	CS
Wire-1	0.5	42
Wire-2	1.5	95

Your answer

Solution

Least count. $LC = \frac{0.5 mm}{100} = 0.005$ mm.

Wire-1 measured reading. $0.5 + 42 \times 0.005 = 0.71$ mm. True diameter is $0.650$ mm.

Zero error. $Δ_{0} = 0.71 - 0.65 = + 0.060$ mm (positive zero error).

Wire-2 measured reading. $1.5 + 95 \times 0.005 = 1.975$ mm.

Corrected diameter of Wire-2. $d = 1.975 - 0.060 = 1.915$ mm $= 1915 μ$ m.

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