The least count of the main scale of a vernier calipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7$$^{th}$$ division of the vernier scale coincides with a division of the main scale and the zero of vernier scale is lying right side of the zero of the main scale. When this vernier is used to measure the length of the cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4$$^{th}$$ VSD coincides with the main scale division. The length of the cylinder is (VSD is vernier scale division):

We are told that the least count of the main scale of the vernier calipers is $$1\ \text{mm}$$. Converting into centimetres first, we note that $$1\ \text{mm}=0.1\ \text{cm}$$, so each main-scale division (MSD) is $$0.1\ \text{cm}$$.

Next, the statement “10 vernier scale divisions (VSD) coincide with 9 main scale divisions” gives us the actual size of one vernier division. If 10 VSD equal 9 MSD, then one VSD is

$$1\ \text{VSD}=\frac{9\ \text{MSD}}{10}= \frac{9\times 0.1\ \text{cm}}{10}=0.09\ \text{cm}.$$

The least count (LC) of a vernier is defined by the formula

$$\text{Least Count}=1\ \text{MSD}-1\ \text{VSD}.$$

Substituting the values just obtained, we have

$$\text{LC}=0.1\ \text{cm}-0.09\ \text{cm}=0.01\ \text{cm}.$$

This $$0.01\ \text{cm}$$ (or $$0.1\ \text{mm}$$) is the resolution with which measurements can be refined by the vernier scale.

Now we examine the zero error. When the jaws touch each other, the seventh vernier division (i.e., $$n=7$$) lines up with a main-scale division and, importantly, the zero of the vernier lies to the right of the zero of the main scale. Whenever the vernier zero is to the right, the instrument is giving a reading even though the true length is zero; hence the zero error is positive.

The magnitude of that error equals the number of the coinciding division multiplied by the least count:

$$\text{Zero error}=n\times\text{LC}=7\times 0.01\ \text{cm}=0.07\ \text{cm}.$$

Because the error is positive, the zero correction (what we must algebraically add to any observed reading) is the negative of this value, namely $$-0.07\ \text{cm}.$$

We now proceed to the actual measurement of the cylinder. The problem states that the zero of the vernier scale lies between $$3.1\ \text{cm}$$ and $$3.2\ \text{cm}.$$ Therefore the main-scale reading (MSR)—the value just to the left of the vernier zero—is

$$\text{MSR}=3.1\ \text{cm}.$$

Further, the 4$$^{\text{th}}$$ vernier division coincides with a main-scale mark, so

$$\text{Vernier scale reading (VSR)}=4\times\text{LC}=4\times 0.01\ \text{cm}=0.04\ \text{cm}.$$

Adding the two parts together gives the observed reading (OR):

$$\text{OR}=\text{MSR}+\text{VSR}=3.1\ \text{cm}+0.04\ \text{cm}=3.14\ \text{cm}.$$

Finally, we must apply the zero correction. Since the correction is $$-0.07\ \text{cm},$$ we obtain the true length (TL) as

$$\text{TL}=\text{OR}+\text{correction}=3.14\ \text{cm}+(-0.07\ \text{cm})=3.07\ \text{cm}.$$

Hence, the correct answer is Option C.