In a vernier callipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is: (a) 1/10N (b) (1)/(100(N+1)) (c) 100N (d) 10(N + 1)

Answer: (b). (N+1) VSD = N MSD ⇒ 1 VSD = N/N+1 MSD. VC = 1 MSD - 1 VSD = 1/N+1 MSD = (0.1 mm)/(N+1) = (0.01)/(N+1) cm So the answer is (b) (1)/(100(N+1)) cm.

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Q15NEET 2024Easy

Least Count, Vernier Callipers & Screw Gauge

In a vernier callipers, $(N + 1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1\text{ MSD}$ represents $0.1\text{ mm}$ , the vernier constant (in $\text{cm}$ ) is:

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Answer: (b)

$(N+1)\text{ VSD} = N\text{ MSD} \Rightarrow 1\text{ VSD} = \dfrac{N}{N+1}$ MSD.
VC $= 1\text{ MSD} - 1\text{ VSD} = \dfrac{1}{N+1}$ MSD $= \dfrac{0.1\text{ mm}}{N+1} = \dfrac{0.01}{N+1}$ cm
So the answer is (b) $\dfrac{1}{100(N+1)}$ cm.

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