A) \[\frac{a}{N+1}\]
B) \[\frac{a}{N-1}\]
C) \[\frac{2a}{N-1}\]
D) \[\frac{2a}{N+1}\]
Correct Answer: A
Solution :
Least count of vernier callipers = value of one division of main scale - value of one division of vernier scale Now, \[N\times a=(N+1)a'\] (a? = value of one division of vernier scale) \[a'=\frac{N}{N+1}a\] \[\therefore \] Least count \[=a-a'=\frac{a}{N+1}\]
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