A wire of lenth `L` is drawn such that its diameter is reduced to half of its original diamter. If the initial resistance of the wire were `10 Omega`, its new resistance would be

Options

A. `40 Omega`
B. `80 Omega`
C. `120 Omega`
D. `160 Omega`

Explanation

To solve the problem, we need to determine the new resistance of a wire after it has been drawn such that its diameter is reduced to half of its original diameter. We know the initial resistance of the wire is \(10 \, \Omega\).

### Step-by-step Solution:

1. **Understand the relationship between resistance, length, and 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 of the wire.

2. **Calculate the initial area**:
The cross-sectional area \(A\) of a wire with diameter \(d\) is given by:
\[ A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
Let the original diameter be \(d\). Therefore, the initial area \(A\) is:
\[ A = \frac{\pi d^2}{4} \]

3. **Determine the new dimensions after drawing the wire**:
When the diameter is reduced to half, the new diameter \(d' = \frac{d}{2}\). The new radius \(r' = \frac{d'}{2} = \frac{d}{4}\).

4. **Calculate the new area**:
The new cross-sectional area \(A'\) becomes:
\[ A' = \pi \left(\frac{d'}{2}\right)^2 = \pi \left(\frac{d/2}{2}\right)^2 = \pi \left(\frac{d}{4}\right)^2 = \frac{\pi d^2}{16} \]

5. **Use the volume conservation principle**:
The volume of the wire remains constant before and after drawing. Thus:
\[ A \cdot L = A' \cdot L' \]
Substituting the areas we found:
\[ \frac{\pi d^2}{4} \cdot L = \frac{\pi d^2}{16} \cdot L' \]
Simplifying this gives:
\[ L' = 4L \]

6. **Calculate the new resistance**:
The new resistance \(R'\) can be calculated using the resistance formula:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \(L' = 4L\) and \(A' = \frac{\pi d^2}{16}\):
\[ R' = \frac{\rho (4L)}{\frac{\pi d^2}{16}} = \frac{64 \rho L}{\pi d^2} \]
Since \(R = \frac{\rho L}{A} = 10 \, \Omega\), we can relate \(R'\) to \(R\):
\[ R' = 16R \]
Therefore:
\[ R' = 16 \times 10 \, \Omega = 160 \, \Omega \]

### Final Answer:
The new resistance of the wire is \(160 \, \Omega\).