In a vernier callipers the main scale and the vernier scale are made up different materials. When the room temperature increases by `DeltaT^(@)C`, it is found the reading of the instrument remains the same. Earlier it was observed that the front edge of the wooden rod placed for measurement crossed the `N^(th)` main scale division and N + 2 MSD coincided with the 2nd VSD. Initially, 10 VSD coincided with 9MSD. If coefficient of linear expansion of the main scale is `alpha_(1)` and that of the vernier scale is `alpha_(2)` then what is the value of `alpha_(1)//alpha_(2)` ? (Ignore the expansion of the rod on heating)

Options

A. `1.8//N`
B. `1.8//(N+3.8)`
C. `1.8//(N-2)`
D. `1.8//N+2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a vernier caliper with a main scale (MS) and a vernier scale (VS) made of different materials. The reading remains unchanged when the temperature increases by ΔT.

### Step 2: Identify the initial conditions
Initially, the front edge of a wooden rod crosses the N-th main scale division, and the (N + 2)-th main scale division coincides with the 2nd vernier scale division.

### Step 3: Establish the relationship between the scales
From the problem, we know:
- 10 VSD coincides with 9 MSD.
- Let the length of one main scale division (MSD) be \( x \).
- Let the length of one vernier scale division (VSD) be \( y \).

From the information given:
\[ 10y = 9x \implies x = \frac{10}{9}y \]

### Step 4: Calculate the change in length
When the temperature increases by ΔT, the change in length of the main scale and the vernier scale can be expressed as:
- Change in length of main scale (for N + 2 divisions):
\[ \Delta L_{MS} = (N + 2)x \alpha_1 \Delta T \]
- Change in length of vernier scale (for 2 divisions):
\[ \Delta L_{VS} = 2y \alpha_2 \Delta T \]

### Step 5: Set the changes equal
Since the reading remains the same, we set the changes equal:
\[ (N + 2)x \alpha_1 \Delta T = 2y \alpha_2 \Delta T \]
Cancelling ΔT from both sides gives:
\[ (N + 2)x \alpha_1 = 2y \alpha_2 \]

### Step 6: Substitute for \( x \)
Substituting \( x = \frac{10}{9}y \) into the equation:
\[ (N + 2) \left(\frac{10}{9}y\right) \alpha_1 = 2y \alpha_2 \]
Cancelling \( y \) (assuming \( y \neq 0 \)):
\[ (N + 2) \frac{10}{9} \alpha_1 = 2 \alpha_2 \]

### Step 7: Solve for the ratio \( \frac{\alpha_1}{\alpha_2} \)
Rearranging gives:
\[ \frac{\alpha_1}{\alpha_2} = \frac{2 \cdot 9}{10(N + 2)} = \frac{18}{10(N + 2)} = \frac{9}{5(N + 2)} \]

### Final Result
Thus, the value of \( \frac{\alpha_1}{\alpha_2} \) is:
\[ \frac{\alpha_1}{\alpha_2} = \frac{9}{5(N + 2)} \]