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$N$ divisions on the main scale of a vernier calliper coincide with $(N+1)$ divisions of the vernier scale. If each division of main scale is ' $a$ ' units, then the least count of the instrument is
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The correct answer is:
$\frac{a}{N+1}$
$\frac{a}{N+1}$
No of divisions on main scale $=N$
No of divisions on vernier scale $=N+1$ size of main scale division $=a$
Let size of vernier scale division be $b$ then we have
$$
a N=b(N+1) \Rightarrow \mathrm{b}=\frac{a N}{N+1}
$$
Least count is $\mathrm{a}-\mathrm{b}=\mathrm{a}-\frac{a N}{N+1}$ $=a\left[\frac{N+1-N}{N+1}\right]=\frac{a}{N+1}$
No of divisions on vernier scale $=N+1$ size of main scale division $=a$
Let size of vernier scale division be $b$ then we have
$$
a N=b(N+1) \Rightarrow \mathrm{b}=\frac{a N}{N+1}
$$
Least count is $\mathrm{a}-\mathrm{b}=\mathrm{a}-\frac{a N}{N+1}$ $=a\left[\frac{N+1-N}{N+1}\right]=\frac{a}{N+1}$
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