A given copper wire of resistance R is stretched to reduce its diameter to half its previous value. What sould be its new resistance ?

Answer

To find the new resistance of a copper wire when it is stretched to reduce its diameter to half its previous value, we can follow these steps:

### Step 1: Understand the relationship between resistance, length, and cross-sectional area.
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
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 dimensions of the wire.
Let the initial diameter of the wire be \( D \) and the initial length be \( L_1 \). The initial cross-sectional area \( A_1 \) can be calculated as:
\[ A_1 = \frac{\pi D^2}{4} \]
The initial resistance is given as \( R = R_1 \).

### Step 3: Analyze the changes after stretching the wire.
When the wire is stretched to reduce its diameter to half, the new diameter \( D_2 \) becomes:
\[ D_2 = \frac{D}{2} \]
The new cross-sectional area \( A_2 \) becomes:
\[ A_2 = \frac{\pi (D_2)^2}{4} = \frac{\pi \left(\frac{D}{2}\right)^2}{4} = \frac{\pi D^2}{16} \]

### Step 4: Relate the lengths before and after stretching.
Since the volume of the wire remains constant during stretching, we have:
\[ A_1 L_1 = A_2 L_2 \]
Substituting the areas:
\[ \frac{\pi D^2}{4} L_1 = \frac{\pi D^2}{16} L_2 \]
Cancelling \( \pi D^2 \) from both sides gives:
\[ 4 L_1 = L_2 \quad \Rightarrow \quad L_2 = 4 L_1 \]

### Step 5: Calculate the new resistance.
Now we can find the new resistance \( R_2 \):
\[ R_2 = \rho \frac{L_2}{A_2} \]
Substituting \( L_2 \) and \( A_2 \):
\[ R_2 = \rho \frac{4 L_1}{\frac{\pi D^2}{16}} = \rho \frac{4 L_1 \cdot 16}{\pi D^2} = 64 \left(\rho \frac{L_1}{A_1}\right) \]
Since \( R_1 = \rho \frac{L_1}{A_1} \), we have:
\[ R_2 = 64 R_1 \]

### Step 6: Substitute the initial resistance.
Given that the initial resistance \( R_1 = R \), we find:
\[ R_2 = 64 R \]

### Final Answer:
The new resistance after stretching the wire is:
\[ \boxed{64R} \]