The temperature coefficient of resistivity of material is 0.0004/k. When the temperature of the material is increased by `50^(@)C`, its resistivity increases by `2xx10^(-8)` ohm-m. The initial resistivity of the material in ohm-m is 

Options

A. `50xx10^(-8)`
B. `90xx10^(-8)`
C. `100xx10^(-8)`
D. `200xx10^(-8)`

Answer

To solve the problem, we need to find the initial resistivity of the material given the temperature coefficient of resistivity and the change in resistivity due to a temperature increase.

### Step-by-Step Solution:

1. **Identify the given values:**
- Temperature coefficient of resistivity (α) = 0.0004 K⁻¹
- Change in temperature (Δθ) = 50 °C
- Change in resistivity (Δρ) = 2 × 10⁻⁸ ohm-m

2. **Use the formula for change in resistivity:**
The change in resistivity due to a change in temperature can be expressed as:
\[ \Delta \rho = \rho_1 \cdot \alpha \cdot \Delta \theta \]
where:
- Δρ = ρ₂ - ρ₁ (the increase in resistivity)
- ρ₁ = initial resistivity
- ρ₂ = final resistivity

3. **Rearranging the equation:**
From the equation above, we can express the initial resistivity (ρ₁) as:
\[ \rho_1 = \frac{\Delta \rho}{\alpha \cdot \Delta \theta} \]

4. **Substituting the known values:**
Now, substitute the values into the equation:
- Δρ = 2 × 10⁻⁸ ohm-m
- α = 0.0004 K⁻¹
- Δθ = 50 °C

Thus:
\[ \rho_1 = \frac{2 \times 10^{-8}}{0.0004 \cdot 50} \]

5. **Calculating the denominator:**
Calculate α × Δθ:
\[ 0.0004 \cdot 50 = 0.02 \]

6. **Calculating the initial resistivity:**
Now substitute back into the equation:
\[ \rho_1 = \frac{2 \times 10^{-8}}{0.02} = 1 \times 10^{-6} \text{ ohm-m} \]

7. **Final result:**
To express this in the required format:
\[ \rho_1 = 100 \times 10^{-8} \text{ ohm-m} \]

### Conclusion:
The initial resistivity of the material is:
\[ \rho_1 = 100 \times 10^{-8} \text{ ohm-m} \]