The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the of wire has a length of 1m at `10^@C.` Now the end P is maintained at `10^@C,` while the ends S is heated and maintained at `400^@C.` The system is thermally insultated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is `1.2xx10^-5K^-1,` the change in length of the wire PQ is

Options

A. `0.78"mm"`
B. `0.90"mm"`
C. `1.56"mm"`
D. `2.34"mm"`

Explanation

To solve the problem, we need to determine the change in length of wire PQ when one end is kept at 10°C and the other end is connected to wire RS, which is heated to 400°C.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two wires: PQ and RS, each initially 1 meter long at 10°C.
- End P of wire PQ is kept at 10°C, and end S of wire RS is heated to 400°C.
- The thermal conductivity of wire PQ (K1) is twice that of wire RS (K2): \( K1 = 2K2 \).
- The coefficient of linear thermal expansion (α) for wire PQ is given as \( 1.2 \times 10^{-5} \, \text{K}^{-1} \).

2. **Finding the Junction Temperature (T)**:
- The rate of heat flow through both wires must be equal since they are in thermal contact:
\[ \frac{K2A(400 - T)}{L} = \frac{K1A(T - 10)}{L} \]
- Canceling \( A \) and \( L \) from both sides, we get:
\[ K2(400 - T) = K1(T - 10) \]
- Substituting \( K1 = 2K2 \):
\[ K2(400 - T) = 2K2(T - 10) \]
- Dividing through by \( K2 \) (assuming \( K2 \neq 0 \)):
\[ 400 - T = 2(T - 10) \]
- Simplifying this equation:
\[ 400 - T = 2T - 20 \]
\[ 400 + 20 = 3T \]
\[ T = \frac{420}{3} = 140 \, \text{°C} \]

3. **Calculating the Change in Length of Wire PQ**:
- The temperature difference for wire PQ is \( 140 - 10 = 130 \, \text{°C} \).
- The change in length \( \Delta L \) due to thermal expansion is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
- Substituting the values:
\[ \Delta L = 1 \cdot (1.2 \times 10^{-5}) \cdot 130 \]
- Calculating the change in length:
\[ \Delta L = 1.2 \times 130 \times 10^{-5} = 156 \times 10^{-5} \, \text{m} = 1.56 \times 10^{-3} \, \text{m} = 0.156 \, \text{mm} \]

4. **Final Result**:
- The change in length of wire PQ is approximately \( 0.156 \, \text{mm} \).