Step-by-Step Solution

1. Determine the Least Count (LC) of the Vernier Scale

The least count of a vernier scale is the difference between one main scale division (MSD) and one vernier scale division (VSD).

• Given, the least count of the main scale is $1.0 \, \text{mm}$. Therefore, $1 \, \text{MSD} = 1.0 \, \text{mm}$.
• It's also given that $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. $10 \, \text{VSD} = 9 \, \text{MSD}$ $1 \, \text{VSD} = \frac{9}{10} \, \text{MSD} = 0.9 \times 1.0 \, \text{mm} = 0.9 \, \text{mm}$
• Now, we can calculate the least count (LC): $\text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD} = 1.0 \, \text{mm} - 0.9 \, \text{mm} = 0.1 \, \text{mm}$

2. Calculate the Elongation of the Steel Wire

The elongation $(\Delta L)$ of the wire is caused by the additional load. We can find it using the formula for Young's modulus, $Y = \frac{\text{Stress}}{\text{Strain}}$.

$Y = \frac{F/A}{\Delta L/L} \implies \Delta L = \frac{F \cdot L}{A \cdot Y}$

Let's calculate the terms in this formula:

• Additional Force (F): The load is increased by a mass $\Delta M = 1.2 \, \text{kg}$. $F = \Delta M \cdot g = 1.2 \, \text{kg} \times 10 \, \text{m/s}^2 = 12 \, \text{N}$
• Length of the wire (L): The length of the wire under the initial load is given as $L = 1.0 \, \text{m}$. We use this as the reference length for calculating the additional elongation.
• Cross-sectional Area (A): The diameter is $d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$. The radius is $r = d/2 = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m}$. $A = \pi r^2$ Using the given value $\pi = 3.2$: $A = 3.2 \times (0.25 \times 10^{-3})^2 \, \text{m}^2 = 3.2 \times 0.0625 \times 10^{-6} \, \text{m}^2$ $A = 0.2 \times 10^{-6} \, \text{m}^2$
• Young's Modulus (Y): Given as $Y = 2 \times 10^{11} \, \text{N/m}^2$.

Now, substitute these values into the elongation formula:

$\Delta L = \frac{(12 \, \text{N}) \times (1.0 \, \text{m})}{(0.2 \times 10^{-6} \, \text{m}^2) \times (2 \times 10^{11} \, \text{N/m}^2)}$ $\Delta L = \frac{12}{0.4 \times 10^5} \, \text{m} = \frac{120}{4 \times 10^5} \, \text{m} = 30 \times 10^{-5} \, \text{m}$

To use this value with the vernier scale, we convert it to millimeters:

$\Delta L = 30 \times 10^{-5} \, \text{m} \times \frac{1000 \, \text{mm}}{1 \, \text{m}} = 30 \times 10^{-2} \, \text{mm} = 0.3 \, \text{mm}$

3. Determine the Coinciding Vernier Scale Division

The elongation of the wire, $\Delta L$, is the distance the vernier scale moves. This value is the reading on the vernier instrument.

• Reading = $0.3 \, \text{mm}$.

The formula for the reading of a vernier caliper is:

$\text{Reading} = \text{Main Scale Reading (MSR)} + (\text{Vernier Scale Coincidence (VSC)}) \times \text{LC}$

• The MSR is the reading on the main scale immediately to the left of the zero mark of the vernier scale. Since the total reading is $0.3 \, \text{mm}$, which is less than one main scale division ($1.0 \, \text{mm}$), the zero of the vernier scale lies between the $0 \, \text{mm}$ and $1 \, \text{mm}$ marks of the main scale. Thus, MSR = $0 \, \text{mm}$.

• Now we can find the VSC: $0.3 \, \text{mm} = 0 \, \text{mm} + (\text{VSC}) \times (0.1 \, \text{mm})$ $\text{VSC} = \frac{0.3 \, \text{mm}}{0.1 \, \text{mm}} = 3$

Therefore, the 3rd division of the vernier scale coincides with a division on the main scale.

The final answer is 3.