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. `160 Omega`
D. `120 Omega`

Explanation

To find the new resistance of the wire after its diameter is reduced to half of its original diameter, we can follow these steps:

### Step 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 \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

### Step 2: Determine the initial resistance
We are given that the initial resistance \( R \) is \( 10 \, \Omega \).

### Step 3: Analyze the change in diameter
If the original diameter of the wire is \( D \), then the new diameter after being drawn is:
\[ d = \frac{D}{2} \]

### Step 4: Calculate the initial and new areas
The cross-sectional area \( A \) of the wire is given by:
\[ A = \frac{\pi D^2}{4} \]
For the new diameter \( d \):
\[ A' = \frac{\pi (d)^2}{4} = \frac{\pi \left(\frac{D}{2}\right)^2}{4} = \frac{\pi D^2}{16} \]

### Step 5: Relate the volumes before and after drawing the wire
Since the volume of the wire remains constant, we have:
\[ V = A \cdot L = A' \cdot L' \]
Substituting the areas:
\[ \frac{\pi D^2}{4} \cdot L = \frac{\pi D^2}{16} \cdot L' \]
Cancelling \( \pi D^2 \) from both sides:
\[ \frac{L}{4} = \frac{L'}{16} \]
This simplifies to:
\[ L' = 4L \]

### Step 6: Substitute the new length and area back into the resistance formula
Now substituting \( L' \) and \( A' \) into the resistance formula for the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (4L)}{\left(\frac{\pi D^2}{16}\right)} = \frac{4 \rho L}{\frac{\pi D^2}{16}} = \frac{4 \cdot 16 \rho L}{\pi D^2} = 64 \cdot \frac{\rho L}{A} \]
Since \( R = \frac{\rho L}{A} = 10 \, \Omega \):
\[ R' = 64 \cdot 10 = 640 \, \Omega \]

### Step 7: Conclusion
Thus, the new resistance of the wire after it is drawn to half its original diameter is:
\[ \boxed{640 \, \Omega} \]