A wire of resistance `5Omega` is uniformly stretched until its new length becomes 4 times the original length. Find its new resistance.

Answer

To find the new resistance of a wire that has been uniformly stretched, 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:
\[ R = \rho \frac{L}{A} \]
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.

### Step 2: Determine the change in dimensions
Given that the original length of the wire is \( L \) and it is stretched to a new length of \( 4L \), we can denote:
- Original length \( L_1 = L \)
- New length \( L_2 = 4L \)

### Step 3: Use the conservation of volume
The volume of the wire remains constant before and after stretching. The volume \( V \) is given by:
\[ V = A \cdot L \]
Thus, the original volume \( V_1 \) is:
\[ V_1 = A_1 \cdot L_1 = A \cdot L \]
After stretching, the new volume \( V_2 \) is:
\[ V_2 = A_2 \cdot L_2 = A_2 \cdot 4L \]
Setting \( V_1 = V_2 \):
\[ A \cdot L = A_2 \cdot 4L \]
From this, we can solve for the new area \( A_2 \):
\[ A_2 = \frac{A}{4} \]

### Step 4: Calculate the new resistance
Using the resistance formula with the new dimensions:
\[ R_2 = \rho \frac{L_2}{A_2} \]
Substituting \( L_2 \) and \( A_2 \):
\[ R_2 = \rho \frac{4L}{A/4} = \rho \frac{4L \cdot 4}{A} = \rho \frac{16L}{A} \]
Now, we can relate the new resistance \( R_2 \) to the original resistance \( R_1 \):
\[ R_2 = 16 \cdot R_1 \]
Given that the original resistance \( R_1 = 5 \, \Omega \):
\[ R_2 = 16 \cdot 5 \, \Omega = 80 \, \Omega \]

### Final Answer
The new resistance of the wire after being stretched is:
\[ \boxed{80 \, \Omega} \]