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JEE Main 2022 · 26 Jul · Shift 2, Physics question 32

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th^{\text{th}} Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th^{\text{th}} Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :

Choose one option
Answer and worked solution

Official answer: C

Least count 1−0.9=0.11-0.9=0.1 mm.

Zero error. The vernier zero is left of the main zero (negative error); NTA's key takes it as −4×0.1=−0.4-4\times0.1=-0.4 mm.

Reading 30+6×0.1=30.630+6\times0.1=30.6 mm; corrected 30.6+0.4=31.030.6+0.4=31.0 mm =3.10=3.10 cm.

(Locating the shifted zero exactly, with the 4th vernier line on a main line, gives −0.6-0.6 mm and 3.12 cm, which is not offered.)

(C)

Topics:Vernier callipers and screw gaugeLeast count
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