A 1m long wire of diameter `0.31mm` has a resistance of `4.2 Omega`. If it is replaced by another wire of same material of length `1.5m` and diameter `0.155 mm`, then the resistance of wire is

Options

A. `25.2 Omega`
B. `0.6 Omega`
C. `26.7 Omega`
D. `0.8 Omega`

Answer

To find the resistance of the new wire, we can use the relationship between resistance, length, and diameter of the wire. 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 of the wire.

The cross-sectional area \( A \) for a wire with diameter \( d \) is given by:

\[ A = \frac{\pi d^2}{4} \]

### Step 1: Calculate the resistivity \( \rho \) of the first wire
Given:
- Length of the first wire \( L_1 = 1 \, \text{m} \)
- Diameter of the first wire \( d_1 = 0.31 \, \text{mm} = 0.31 \times 10^{-3} \, \text{m} \)
- Resistance of the first wire \( R_1 = 4.2 \, \Omega \)

First, we calculate the area \( A_1 \):

\[ A_1 = \frac{\pi d_1^2}{4} = \frac{\pi (0.31 \times 10^{-3})^2}{4} \]

Calculating \( A_1 \):

\[ A_1 = \frac{\pi (0.0961 \times 10^{-6})}{4} \approx 7.57 \times 10^{-8} \, \text{m}^2 \]

Now, substituting into the resistance formula to find \( \rho \):

\[ R_1 = \frac{\rho L_1}{A_1} \implies \rho = R_1 \cdot \frac{A_1}{L_1} \]

Substituting the values:

\[ \rho = 4.2 \cdot \frac{7.57 \times 10^{-8}}{1} \approx 3.18 \times 10^{-7} \, \Omega \cdot \text{m} \]

### Step 2: Calculate the resistance of the second wire
Given:
- Length of the second wire \( L_2 = 1.5 \, \text{m} \)
- Diameter of the second wire \( d_2 = 0.155 \, \text{mm} = 0.155 \times 10^{-3} \, \text{m} \)

Calculate the area \( A_2 \):

\[ A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (0.155 \times 10^{-3})^2}{4} \]

Calculating \( A_2 \):

\[ A_2 = \frac{\pi (0.024025 \times 10^{-6})}{4} \approx 1.89 \times 10^{-8} \, \text{m}^2 \]

Now, using the resistivity \( \rho \) calculated earlier, we find the resistance \( R_2 \):

\[ R_2 = \frac{\rho L_2}{A_2} = \frac{3.18 \times 10^{-7} \cdot 1.5}{1.89 \times 10^{-8}} \]

Calculating \( R_2 \):

\[ R_2 \approx \frac{4.77 \times 10^{-7}}{1.89 \times 10^{-8}} \approx 25.2 \, \Omega \]

### Final Answer
The resistance of the new wire is approximately \( 25.2 \, \Omega \).