JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05$ mm, then the least count of the vernier callipers is _____ mm.

A
0.002
B
0.05
C
0.02
D
0.005

Solution

Least count = 1 MSD − 1 VSD.

$1 VSD = \frac{48}{50} MSD$ , so $1 MSD - 1 VSD = (1 - \frac{48}{50}) MSD = \frac{2}{50} MSD$ .

$LC = \frac{2}{50} \times 0.05 mm = 0.002 mm$ .

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