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JEE Main 2009 (AIEEE) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2009 (AIEEE)

Q64JEE Main 2009 · AIEEEMedium

Least Count, Vernier Callipers & Screw Gauge

In an experiment the angles are required to be measured using an instrument, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a degree(= $0.5^\circ$ ), then the least count of the instrument is:

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Answer: (a)

$30$ VSD $= 29$ MSD, so LC $= \dfrac{1}{30}\text{ MSD}$
LC $= \dfrac{0.5^\circ}{30} = \dfrac{1}{60}^\circ = 1'$
So the answer is (a) one minute.

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