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JEE Advanced 2021 · Paper 1, Physics question 1

The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

Two Vernier calipers readings (main scale in 0.1 cm divisions above, 10-division Vernier scale below). Left, jaws closed: main scale 0 to 1; the Vernier zero sits slightly to the left of the main-scale zero, and Vernier division 6 lines up with the main-scale 0.5 mark (negative zero error). Right, sphere between the jaws: main scale 3 to 4; the Vernier zero sits just past the 3.1 mark, and Vernier division 1 lines up with the 3.2 mark.
Choose one option
Answer and worked solution

Official answer: C

Least count. 10 Vernier divisions equal 9 main-scale divisions, so
LC=1 MSD−1 VSD=0.1−0.09=0.01 cm.LC=1\text{ MSD}-1\text{ VSD}=0.1-0.09=0.01\text{ cm}.

Zero error (left figure). The Vernier zero sits to the left of the main-scale zero, and Vernier division 6 lines up with the 0.5 cm mark. So the Vernier zero is at 0.5−6(0.09)=−0.040.5-6(0.09)=-0.04 cm, a zero error of −0.04-0.04 cm.

Reading with the sphere (right figure). Vernier division 1 lines up with the 3.2 cm mark, so the Vernier zero is at 3.2−0.09=3.113.2-0.09=3.11 cm.

Correct diameter.
3.11−(−0.04)=3.15 cm.3.11-(-0.04)=3.15\text{ cm}.
(C)

Topics:Vernier callipers and screw gaugeLeast count
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