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

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 heated to 400°C. Let's break down the solution step by step.

### Step 1: Understanding the Setup
We have two wires, PQ and RS, each of length 1 meter at an initial temperature of 10°C. The end P of wire PQ is kept at 10°C, and the end S of wire RS is heated to 400°C. The thermal conductivity of wire PQ is twice that of wire RS.

### Step 2: Establishing the Heat Flow
Since the system is thermally insulated, the heat flow through both wires must be equal at the junction where they are soldered. We can express this as:
\[ \frac{K_1 A (T_S - T)}{L} = \frac{K_2 A (T - T_P)}{L} \]
where:
- \( K_1 \) is the thermal conductivity of wire PQ,
- \( K_2 \) is the thermal conductivity of wire RS,
- \( A \) is the cross-sectional area (assumed equal for both wires),
- \( T_S = 400°C \) (temperature at end S),
- \( T_P = 10°C \) (temperature at end P),
- \( T \) is the temperature at the junction.

### Step 3: Relating Thermal Conductivities
Given that \( K_1 = 2K_2 \), we can substitute this into our equation:
\[ \frac{2K_2 A (400 - T)}{L} = \frac{K_2 A (T - 10)}{L} \]
Cancelling \( K_2 \), \( A \), and \( L \) from both sides gives:
\[ 2(400 - T) = T - 10 \]

### Step 4: Solving for Junction Temperature \( T \)
Expanding and rearranging the equation:
\[ 800 - 2T = T - 10 \\ 800 + 10 = 3T \\ 810 = 3T \\ T = 270°C \]

### Step 5: Temperature Gradient in Wire PQ
The temperature gradient in wire PQ can be expressed as:
\[ \frac{dT}{dx} = \frac{T - T_P}{L} = \frac{270 - 10}{1} = 260°C/m \]

### Step 6: Change in Length of Wire PQ
The change in length \( \Delta L \) of wire PQ due to thermal expansion can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where:
- \( L = 1 \, \text{m} \),
- \( \alpha = 1.2 \times 10^{-5} \, \text{K}^{-1} \),
- \( \Delta T = T - T_P = 270 - 10 = 260°C \).

Substituting the values:
\[ \Delta L = 1 \cdot (1.2 \times 10^{-5}) \cdot 260 = 3.12 \times 10^{-3} \, \text{m} = 0.00312 \, \text{m} = 3.12 \, \text{mm} \]

### Final Answer
The change in length of wire PQ is \( 3.12 \, \text{mm} \).

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