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:
\[ R = \frac{\rho L}{A} \]
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.

### Step 2: Identify the initial conditions
Given:
- Original resistance \( R = 5 \, \Omega \)
- Original length \( L \)
- New length \( L' = 4L \) (the wire is stretched to 4 times its original length)

### Step 3: Apply the volume conservation principle
When the wire is stretched, its volume remains constant. Therefore:
\[ A \cdot L = A' \cdot L' \]
Substituting \( L' = 4L \):
\[ A \cdot L = A' \cdot (4L) \]
Cancelling \( L \) from both sides:
\[ A = 4A' \]
This implies:
\[ A' = \frac{A}{4} \]
(The new cross-sectional area \( A' \) is one-fourth of the original area \( A \)).

### Step 4: Write the expression for new resistance
Using the new length and new area in the resistance formula:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' = 4L \) and \( A' = \frac{A}{4} \):
\[ R' = \frac{\rho (4L)}{\frac{A}{4}} = \frac{4\rho L \cdot 4}{A} = \frac{16\rho L}{A} \]

### Step 5: Relate new resistance to original resistance
From the original resistance \( R = \frac{\rho L}{A} = 5 \, \Omega \), we can substitute:
\[ R' = 16 \cdot \frac{\rho L}{A} = 16R \]
Thus:
\[ R' = 16 \cdot 5 \, \Omega = 80 \, \Omega \]

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

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