Least count of a vernier caliper is 1/20N cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is: (a) ((2N-1)/(2N)) (b) ((2N-1)/(20N)) (c) (2N - 1) (d) (2N-1/2)

Answer: (d). LC = 1/20N cm = 1/2N mm; let N VSD = x MSD, so 1 VSD = x/N mm. LC = 1 - x/N = 1/2N x = N - 1/2 = 2N-1/2 So the answer is (d).

Least count of a vernier caliper is 1/20N cm. The value of one… | JEE Main 2024 (8 Apr, Shift 2)

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Least Count, Vernier Callipers & Screw Gauge

Least count of a vernier caliper is $\frac{1}{20N}\text{ cm}$ . The value of one division on the main scale is $1\text{ mm}$ . Then the number of divisions of main scale that coincide with $N$ divisions of vernier scale is:

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Answer: (d)

LC $= \dfrac{1}{20N}$ cm $= \dfrac{1}{2N}$ mm; let $N$ VSD $= x$ MSD, so 1 VSD $= \dfrac{x}{N}$ mm.
LC $= 1 - \dfrac{x}{N} = \dfrac{1}{2N}$
$x = N - \dfrac{1}{2} = \dfrac{2N-1}{2}$
So the answer is (d).

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