Q124 / 200 · NEET 2024 · Physics

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:

  • Correct answer
    A
    $\frac{1}{100(N+1)}$
  • B
    100 N
  • C
    $10(N+1)$
  • D
    $\frac{1}{10 N}$

Solution

$V . C=M S D-V S D$....(1) given : $(N+1) V S D=N$ MSD $\operatorname{VSD}=\left(\frac{N}{N+1}\right) \mathrm{MSD}$....(2) From (1) and (2) $\begin{aligned} & V \cdot C=(\mathrm{MSD})-\frac{N}{N+1}(\mathrm{MSD}) \\ & =\operatorname{MSD}\left(1-\frac{N}{N+1}\right)=\frac{\mathrm{MSD}}{N+1} \\ & =\frac{0.01}{N+1}=\frac{1}{100(N+1)} \end{aligned}$

Hence, the correct answer is option: 'A'.

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