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JEE Main 2026 · 2 Apr · Shift 2, Physics question 32

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is ________ mm.

Choose one option
Answer and worked solution

Official answer: A

Least count. 0.1100=0.001\frac{0.1}{100}=0.001 mm. The zero error is +5×0.001=+0.005+5\times0.001=+0.005 mm.

Reading. 5+50×0.001=5.0505+50\times0.001=5.050 mm.

Corrected diameter. 5.050−0.005=5.0455.050-0.005=5.045 mm.

(A)

Topics:Vernier callipers and screw gaugeLeast count
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