Question:
In a slide callipers, $(m + 1)$ number of vernier divisions is equal to m number of smallest main scale divisions. If d unit is the magnitude of the smallest main scale division, then the magnitude of the vernier constant is
In a slide callipers, $(m + 1)$ number of vernier divisions is equal to m number of smallest main scale divisions. If d unit is the magnitude of the smallest main scale division, then the magnitude of the vernier constant is
Updated On: Apr 26, 2024
- $\frac{d}{\left(m+1\right)}$ uint
- $\frac{d}{m}$ uint
- $\frac{md}{\left(m+1\right)}$ uint
- $\frac{\left(m+1\right)d}{m}$ uint
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The Correct Option is A
Solution and Explanation
$(m+1)$ vernier division $=m$ main scale division One division on vernier scale $=\left(\frac{m}{m+1}\right)$ division on main scale Vernier constant $=\left(1-\frac{m}{m+1}\right) d=\frac{d}{m+1}$
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