A wire `50 cm` long and `0.12mm` diameter has a resistance of `4.0 Omega` find the resistance of another wire of the same material whose length is `1.5m` and diameter is `0.15 mm`

Explanation

To find the resistance of the second 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:
- \( R \) = resistance,
- \( \rho \) = resistivity of the material,
- \( L \) = length of the wire,
- \( A \) = cross-sectional area of the wire.

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

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

### Step 1: Calculate the resistance of the first wire

Given:
- Length \( L_1 = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Diameter \( d_1 = 0.12 \, \text{mm} = 0.12 \times 10^{-3} \, \text{m} \)
- Resistance \( R_1 = 4.0 \, \Omega \)

First, calculate the cross-sectional area \( A_1 \):

\[ A_1 = \frac{\pi (d_1)^2}{4} = \frac{\pi (0.12 \times 10^{-3})^2}{4} \]

Calculating \( A_1 \):

\[ A_1 = \frac{\pi (0.12^2 \times 10^{-6})}{4} = \frac{\pi (0.0144 \times 10^{-6})}{4} = \frac{0.04524 \times 10^{-6}}{4} \approx 0.01131 \times 10^{-6} \, \text{m}^2 \]

Now, we can express the resistance in terms of resistivity:

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

Rearranging gives us:

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

### Step 2: Calculate the resistivity \( \rho \)

Substituting the known values:

\[ \rho = 4.0 \cdot \frac{0.01131 \times 10^{-6}}{0.5} \]

Calculating \( \rho \):

\[ \rho = 4.0 \cdot 0.02262 \times 10^{-6} \approx 9.048 \times 10^{-6} \, \Omega \cdot \text{m} \]

### Step 3: Calculate the resistance of the second wire

Given:
- Length \( L_2 = 1.5 \, \text{m} \)
- Diameter \( d_2 = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \)

Calculate the cross-sectional area \( A_2 \):

\[ A_2 = \frac{\pi (d_2)^2}{4} = \frac{\pi (0.15 \times 10^{-3})^2}{4} \]

Calculating \( A_2 \):

\[ A_2 = \frac{\pi (0.15^2 \times 10^{-6})}{4} = \frac{\pi (0.0225 \times 10^{-6})}{4} = \frac{0.0706858 \times 10^{-6}}{4} \approx 0.0176715 \times 10^{-6} \, \text{m}^2 \]

Now, we can find the resistance \( R_2 \):

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

Substituting the known values:

\[ R_2 = \frac{9.048 \times 10^{-6} \cdot 1.5}{0.0176715 \times 10^{-6}} \]

Calculating \( R_2 \):

\[ R_2 \approx \frac{13.572 \times 10^{-6}}{0.0176715 \times 10^{-6}} \approx 1.92 \, \Omega \]

### Final Answer:
The resistance of the second wire is approximately \( R_2 \approx 1.92 \, \Omega \).

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