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 $4^{\text{th}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be $1$ cm, the actual length of the wire is :
Answer: (D) $1.04$ cm
Least count $= \dfrac{1\ \text{mm}}{10} = 0.1$ mm.
The Vernier zero lies to the left of the main-scale zero, so the zero error is negative. Taking the zero error as $-(4 \times 0.1) = -0.4$ mm (the convention the options follow):
$$\text{Actual length} = \text{reading} - \text{zero error} = 10\ \text{mm} - (-0.4\ \text{mm}) = 10.4\ \text{mm} = 1.04\ \text{cm}$$
Note: in the stricter convention for a negative zero error, the error is $-(10 - 4) \times 0.1 = -0.6$ mm, which would give $1.06$ cm. That value is not among the options, so the intended answer is $1.04$ cm.
Solution by Sreeraj P, M.Sc Physics