1. Find the least count of the Vernier calipers

Given:

• $10$ Vernier scale divisions $= 9$ main scale divisions
• $1$ main scale division $(1\,\text{MSD}) = 1\,\text{mm}$

So, $10\,\text{VSD} = 9\,\text{mm}$ $1\,\text{VSD} = \frac{9}{10}\,\text{mm} = 0.9\,\text{mm}$

Least count, $\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1.0 - 0.9 = 0.1\,\text{mm}$

2. Determine the zero error

It is given that when jaws are closed, the zero of Vernier scale is shifted to the left of the zero of main scale. This means the instrument has a negative zero error.

Also, the $4^{\text{th}}$ Vernier division coincides with a main scale division.

For a negative zero error, the zero error magnitude is: $\text{Zero error} = -(10 - 4)\times \text{LC} = -6\times 0.1 = -0.6\,\text{mm}$

So, $\text{Zero correction} = +0.6\,\text{mm}$

3. Find the observed reading of the spherical body

While measuring the diameter:

• Vernier zero lies between $30$ and $31$ main scale divisions
• Therefore, main scale reading: $\text{MSR} = 30\,\text{mm}$
• $6^{\text{th}}$ Vernier division coincides

Vernier reading: $\text{VR} = 6\times \text{LC} = 6\times 0.1 = 0.6\,\text{mm}$

Observed reading: $\text{Observed reading} = \text{MSR} + \text{VR} = 30 + 0.6 = 30.6\,\text{mm}$

4. Apply zero correction

$\text{True reading} = \text{Observed reading} + 0.6\,\text{mm}$ $\text{True reading} = 30.6 + 0.6 = 31.2\,\text{mm}$

Convert into cm: $31.2\,\text{mm} = 3.12\,\text{cm}$

5. Match with the nearest option

The exact value obtained is $3.12\,\text{cm}$, but this is not present in the options. The nearest option is: $\boxed{3.10\,\text{cm}}$

So the intended answer is Option C.