The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

The main scale is calibrated in $$0.1\ \text{cm}$$ steps.
Ten Vernier‐scale (V-scale) divisions coincide with nine main-scale (M-scale) divisions, so

Total length of $$10$$ V-scale divisions $$= 9 \times 0.1\ \text{cm}=0.9\ \text{cm}$$.
Therefore one V-scale division $$= \dfrac{0.9\ \text{cm}}{10}=0.09\ \text{cm}$$.

The least count (LC) is the difference between one M-scale division and one V-scale division:

$$LC = 0.10\ \text{cm}-0.09\ \text{cm}=0.01\ \text{cm}$$.

Zero error from the left-hand figure (jaws closed)
With the jaws in contact, the 0-mark of the V-scale is not aligned with the 0-mark of the M-scale. It lies to the right of the M-scale zero. The 3ᵗʰ Vernier division (counting from the zero) exactly coincides with a main-scale division.

Observed zero reading $$= n \times LC = 3 \times 0.01\ \text{cm}=+0.03\ \text{cm}$$.
Because the zero of the V-scale is to the right of the M-scale zero, the zero error is positive $$+0.03\ \text{cm}$$.
Zero correction (to be applied to every measurement) $$= -0.03\ \text{cm}$$.

Reading with the sphere between the jaws (right-hand figure)

1. Main-scale reading (MSR): the V-scale zero lies between the 3.1 cm and 3.2 cm marks, so $$MSR = 3.1\ \text{cm}$$.

2. Vernier coincidence: the 8ᵗʰ V-scale division coincides with a main-scale division, so additional reading $$= n \times LC = 8 \times 0.01\ \text{cm}=0.08\ \text{cm}$$.

3. Observed reading (before correction):
$$\text{Observed length}=MSR + \text{Vernier reading} = 3.1\ \text{cm}+0.08\ \text{cm}=3.18\ \text{cm}$$.

4. Apply the zero correction:
$$\text{True length}=3.18\ \text{cm}+(-0.03\ \text{cm})=3.15\ \text{cm}$$.

Hence the correct diameter of the sphere is $$3.15\ \text{cm}$$.

Option C which is: 3.15 cm