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JEE Main 2024 (30 Jan, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2024 (30 Jan, Shift 2)

Q26JEE Main 2024 · 30 JAN · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

If 50 Vernier divisions are equal to 49 main scale divisions of a traveling microscope and one smallest reading of main scale is $0.5\text{ mm}$ , the Vernier constant of traveling microscope is

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Answer: (a)

$50$ VSD $= 49$ MSD, so VC $= \dfrac{1}{50}\text{ MSD}$
VC $= \dfrac{0.5\text{ mm}}{50} = 0.01\text{ mm}$
So the answer is (a) $0.01\text{ mm}$ .

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