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 formula for resistance, which is given by:

\[ R = \frac{\rho L}{A} \]

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.

### Step 1: Calculate the cross-sectional area of the original wire

The diameter of the original wire \( d_1 = 0.31 \, \text{mm} = 0.31 \times 10^{-3} \, \text{m} \).

The radius \( r_1 \) is half of the diameter:

\[ r_1 = \frac{d_1}{2} = \frac{0.31 \times 10^{-3}}{2} = 0.155 \times 10^{-3} \, \text{m} \]

Now, calculate the area \( A_1 \):

\[ A_1 = \pi r_1^2 = \pi \left(0.155 \times 10^{-3}\right)^2 \]

Calculating this gives:

\[ A_1 = \pi \times (0.155 \times 10^{-3})^2 \approx 7.55 \times 10^{-8} \, \text{m}^2 \]

### Step 2: Calculate the resistance of the original wire

Given that the resistance \( R_1 = 4.2 \, \Omega \) and the length \( L_1 = 1 \, \text{m} \), we can express the resistivity \( \rho \):

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

Rearranging for \( \rho \):

\[ \rho = R_1 \cdot \frac{A_1}{L_1} = 4.2 \cdot \frac{7.55 \times 10^{-8}}{1} \]

Calculating this gives:

\[ \rho \approx 3.17 \times 10^{-7} \, \Omega \cdot \text{m} \]

### Step 3: Calculate the cross-sectional area of the new wire

The diameter of the new wire \( d_2 = 0.155 \, \text{mm} = 0.155 \times 10^{-3} \, \text{m} \).

The radius \( r_2 \):

\[ r_2 = \frac{d_2}{2} = \frac{0.155 \times 10^{-3}}{2} = 0.0775 \times 10^{-3} \, \text{m} \]

Now, calculate the area \( A_2 \):

\[ A_2 = \pi r_2^2 = \pi \left(0.0775 \times 10^{-3}\right)^2 \]

Calculating this gives:

\[ A_2 \approx 1.89 \times 10^{-8} \, \text{m}^2 \]

### Step 4: Calculate the resistance of the new wire

The length of the new wire \( L_2 = 1.5 \, \text{m} \). Now we can find the resistance \( R_2 \):

\[ R_2 = \frac{\rho L_2}{A_2} \]

Substituting the values:

\[ R_2 = \frac{3.17 \times 10^{-7} \cdot 1.5}{1.89 \times 10^{-8}} \]

Calculating this gives:

\[ R_2 \approx 25.2 \, \Omega \]

### Final Answer

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

---