Question
One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{\text{th}\text{~}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm, the actual length of the wire is :
Options
Solution
Sol. [Key Concept: Vernier calliper — least count and reading]
⇒ There is negative zero error
$${\begin{array}{r} \Rightarrow \text{~}\text{Zero error}\text{~} = - (10 - 4) \times \text{~}\text{L.C.}\text{~}\#(i) \end{array} } {From\ ...(i) }{Zero\ error\ = - (10 - 4) \times 0.01 = - 0.06\text{ }cm }$$⇒ Zero correction = - Zero error
$${\begin{matrix} & \ = - ( - 0.06) \\ & \ = 0.06\text{ }cm \end{matrix} } $$⇒ Correct reading $= R^{'}$
$${\begin{matrix} & R^{'} = R + \text{~}\text{Zero correction}\text{~} \\ & \ = 1 + 0.06 \\ & \ = 1.06\text{ }cm \end{matrix} } $$∴ No option is correct
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