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. `188 ohm`
B. `72 ohm`
C. `288 ohm`
D. `388 ohm`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Initial Resistance**:
The initial resistance \( R \) of the wire is given as 18 ohms. The resistance \( R \) can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.

2. **Calculating the Initial Area**:
The cross-sectional area \( A \) of the wire with radius \( r \) is given by:
\[ A = \pi r^2 \]

3. **Volume Conservation**:
When the wire is drawn, its volume remains constant. The initial volume \( V \) is:
\[ V = A \cdot L = \pi r^2 L \]
After drawing, the new radius becomes \( \frac{r}{2} \), so the new area \( A' \) is:
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]
Let \( L' \) be the new length of the wire. The volume after drawing is:
\[ V' = A' \cdot L' = \pi \frac{r^2}{4} \cdot L' \]
Setting the initial and final volumes equal gives:
\[ \pi r^2 L = \pi \frac{r^2}{4} L' \]
Simplifying this, we find:
\[ r^2 L = \frac{r^2}{4} L' \implies L' = 4L \]

4. **Calculating the New Resistance**:
The new resistance \( R' \) can be calculated using the new length \( L' \) and new area \( A' \):
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' = 4L \) and \( A' = \frac{A}{4} \) (since \( A = \pi r^2 \)):
\[ R' = \frac{\rho (4L)}{\frac{A}{4}} = \frac{16 \rho L}{A} \]
Since \( \frac{\rho L}{A} = R \) (the original resistance), we have:
\[ R' = 16R \]
Substituting \( R = 18 \, \text{ohms} \):
\[ R' = 16 \times 18 = 288 \, \text{ohms} \]

### Final Answer:
The new resistance of the wire after it has been drawn is **288 ohms**.