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 resistivity, resistance, and length
The resistance \( R \) of a wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: 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 3: Set up the equation for the combined resistance
When wires A and B are joined end to end, their resistances add up:
\[ R_{total} = R_A + R_B = \frac{\rho_A L_A}{A} + \frac{\rho_B L_B}{A} \]
This can be simplified to:
\[ R_{total} = \frac{1}{A} (\rho_A L_A + \rho_B L_B) \]

### Step 4: Substitute the temperature-dependent resistivity
Substituting the resistivity expressions into the resistance equation gives:
\[ R_{total} = \frac{1}{A} \left( \rho_A L_A (1 + \alpha_A \Delta T) + \rho_B L_B (1 + \alpha_B \Delta T) \right) \]
Expanding this, we have:
\[ R_{total} = \frac{1}{A} \left( \rho_A L_A + \rho_B L_B + \rho_A L_A \alpha_A \Delta T + \rho_B L_B \alpha_B \Delta T \right) \]

### Step 5: Set the temperature coefficient term to zero
Since the resistance does not change with temperature, the coefficient of \( \Delta T \) must equal zero:
\[ \rho_A L_A \alpha_A + \rho_B L_B \alpha_B = 0 \]

### Step 6: Rearrange to find the ratio of lengths
Rearranging the equation gives:
\[ \rho_A L_A \alpha_A = -\rho_B L_B \alpha_B \]
Taking the ratio of lengths \( \frac{L_A}{L_B} \):
\[ \frac{L_A}{L_B} = -\frac{\rho_B \alpha_B}{\rho_A \alpha_A} \]

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

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 numerator:
\[ 6 \times 10^{-5} \times -4 \times 10^{-6} = 24 \times 10^{-11} \]
Calculating the denominator:
\[ 3 \times 10^5 \times 4 \times 10^{-5} = 12 \]
Thus:
\[ \frac{L_A}{L_B} = \frac{24 \times 10^{-11}}{12} = 2 \times 10^{-11} \]

### Final Step: Calculate the numerical value
Simplifying gives:
\[ \frac{L_A}{L_B} = 0.2 \]

### Conclusion
The ratio of the lengths of wires A and B is:
\[ \frac{L_A}{L_B} = 0.2 \]