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
(A)3.07 cm
(B)3.11 cm
(C)3.15 cm✓ Correct
(D)3.17 cm
Answer: (C)
Solution
Given 10 VSD = 9 MSD, so 1 VSD = (9/10) MSD.
Least count = 1 MSD - 1 VSD = (1 - 9/10) MSD = 0.1 MSD = 0.1 × 0.1 cm = 0.01 cm.
From the left figure, the '0' of V.S. lies before '0' of M.S., so zero error = -[10 - 6] × LC = -4 × 0.01 cm = -0.04 cm.
From the right figure, reading = 3.1 cm + 1 × LC = 3.1 cm + 0.01 cm = 3.11 cm.
True diameter = Reading - Zero error = 3.11 - (-0.04) = 3.15 cm.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option
