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