Ends of two wires A and B, having resistivity`rho_(A)=3xx10^(5)Omegam` and `rho_(B)=6xx10^(-5) omega m` of same cross section ara are pointed together to form a single wire. If the resistance of the joined wire does not change with temperature, then find the ratio of their lengths given that temperatue coefficient of resistitivy of wires A and B are `alpha=4xx10^(-5)//^(@)C` and `alpha=-4xx10^(6)//^(@)C` . Assume that mechanical dimensions do not change with temperature.

Answer

To solve the problem, we need to find the ratio of the lengths of two wires A and B, given their resistivities and temperature coefficients of resistivity. Let's break down the solution step by step.

### Step 1: Understand the relationship between resistance, resistivity, and length
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: Write the expressions for the resistances of wires A and B
For wire A:
\[ R_A = \frac{\rho_A L_A}{A} \]
For wire B:
\[ R_B = \frac{\rho_B L_B}{A} \]

### Step 3: Combine the resistances for the joined wire
When wires A and B are joined together, they form a series combination. The total resistance \( R \) of the combined wire is:
\[ R = R_A + R_B = \frac{\rho_A L_A}{A} + \frac{\rho_B L_B}{A} \]

### Step 4: Consider the effect of temperature on resistivity
The resistivity of a material changes with temperature according to the formula:
\[ \rho = \rho_0 (1 + \alpha \Delta T) \]
where:
- \( \rho_0 \) is the resistivity at a reference temperature,
- \( \alpha \) is the temperature coefficient of resistivity,
- \( \Delta T \) is the change in temperature.

### Step 5: Set up the equation for the condition that resistance does not change with temperature
Since the resistance of the joined wire does not change with temperature, we can write:
\[ \rho_A L_A (1 + \alpha_A \Delta T) + \rho_B L_B (1 + \alpha_B \Delta T) = 0 \]
This implies:
\[ \rho_A L_A \alpha_A + \rho_B L_B \alpha_B = 0 \]

### Step 6: Rearrange the equation to find the ratio of lengths
From the equation:
\[ \rho_A L_A \alpha_A + \rho_B L_B \alpha_B = 0 \]
we can express the ratio of lengths:
\[ \frac{L_A}{L_B} = -\frac{\rho_B \alpha_B}{\rho_A \alpha_A} \]

### Step 7: Substitute the given values
Given:
- \( \rho_A = 3 \times 10^{-5} \, \Omega \cdot m \)
- \( \rho_B = 6 \times 10^{-5} \, \Omega \cdot m \)
- \( \alpha_A = 4 \times 10^{-5} \, \text{per} \, ^\circ C \)
- \( \alpha_B = -4 \times 10^{-6} \, \text{per} \, ^\circ C \)

Substituting these values into the ratio:
\[ \frac{L_A}{L_B} = -\frac{(6 \times 10^{-5})(-4 \times 10^{-6})}{(3 \times 10^{-5})(4 \times 10^{-5})} \]

### Step 8: Simplify the expression
Calculating the ratio:
\[ \frac{L_A}{L_B} = \frac{6 \times 10^{-5} \times 4 \times 10^{-6}}{3 \times 10^{-5} \times 4 \times 10^{-5}} = \frac{6}{3} \cdot \frac{10^{-5}}{10^{-5}} = 2 \]

### Final Answer
Thus, the ratio of the lengths of wires A and B is:
\[ \frac{L_A}{L_B} = 2 \]