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JEE Main 2022 (26 Jun, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (26 Jun, Shift 1)

Q33JEE Main 2022 · 26 JUN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _________ $\times$ 10 $-$ 2 mm.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

1 MSD $= \dfrac{1\text{ cm}}{20} = 0.5$ mm
$10\text{ VSD} = 9\text{ MSD} \Rightarrow$ VC $= \dfrac{1\text{ MSD}}{10} = 0.05$ mm
$= 5\times10^{-2}$ mm, answer 5

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