A cylindrical wire has a resistance of `18 Omega`. The resistance of another wire of the same material with the same cross section, but 1.5 times the length is :

Options

A. `18 Omega`
B. `24 Omega`
C. `12 Omega`
D. `9 Omega`

Answer

To solve the problem, we need to find the resistance of a cylindrical wire that has a length 1.5 times that of another wire, while keeping the cross-sectional area and material constant.

### Step-by-Step Solution:

1. **Understand the relationship between resistance, length, and cross-sectional area**:
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.

2. **Identify the known values**:
From the problem, we know:
- The resistance \( R = 18 \, \Omega \) for the original wire.
- The length of the new wire is \( 1.5 \) times the length of the original wire.

3. **Express the new length**:
Let the original length of the wire be \( L \). Then the new length \( L' \) is:
\[ L' = 1.5L \]

4. **Calculate the new resistance**:
Since the cross-sectional area \( A \) and the resistivity \( \rho \) remain constant, we can express the new resistance \( R' \) as:
\[ R' = \frac{\rho L'}{A} \]
Substituting \( L' \):
\[ R' = \frac{\rho (1.5L)}{A} = 1.5 \cdot \frac{\rho L}{A} \]
Since \( \frac{\rho L}{A} = R \):
\[ R' = 1.5R \]

5. **Substitute the known resistance**:
Now, substituting \( R = 18 \, \Omega \):
\[ R' = 1.5 \cdot 18 \, \Omega = 27 \, \Omega \]

### Final Answer:
The resistance of the new wire is \( 27 \, \Omega \).