A wire of resistance 18 ohm is drawn until its radius reduce `(1)/(2)`th of its original radius then resistance of the wire is

Options

A. `188Omega`
B. `72Omega`
C. `288Omega`
D. `388Omega`

Explanation

To find the new resistance of a wire after it has been drawn to reduce its radius to half of its original value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between resistance, resistivity, length, and area**:
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

2. **Calculate the original area**:
The area \( A \) of the wire is given by:
\[ A = \pi R^2 \]
where \( R \) is the radius of the wire.

3. **Determine the new radius**:
If the radius is reduced to half, then the new radius \( R' \) is:
\[ R' = \frac{R}{2} \]

4. **Calculate the new area**:
The new cross-sectional area \( A' \) after the radius is reduced is:
\[ A' = \pi (R')^2 = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} = \frac{A}{4} \]

5. **Volume conservation**:
Since the volume of the wire remains constant during stretching, we have:
\[ A \cdot L = A' \cdot L' \]
Substituting \( A' = \frac{A}{4} \):
\[ A \cdot L = \frac{A}{4} \cdot L' \]
This simplifies to:
\[ L' = 4L \]
Thus, the new length \( L' \) is four times the original length.

6. **Calculate the new resistance**:
Substitute \( L' \) and \( A' \) into the resistance formula:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (4L)}{\frac{A}{4}} = \frac{16 \rho L}{A} \]
Since \( R = \frac{\rho L}{A} \), we can express \( R' \) in terms of \( R \):
\[ R' = 16R \]

7. **Substituting the original resistance**:
Given that the original resistance \( R = 18 \, \Omega \):
\[ R' = 16 \times 18 = 288 \, \Omega \]

### Final Answer:
The new resistance of the wire after it has been drawn is \( 288 \, \Omega \).

---