Question

How many time the resistance of a wire changes when it is uniformly stretched to thrice its length?

Answer 1

When a wire is uniformly stretched, its resistance changes due to the alteration of its length and cross-sectional area. To find how many times the resistance changes when a wire is stretched to **thrice its length**, we need to use the formula for the resistance of a wire:

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

Where:
- \(R\) is the resistance,
- \(\rho\) is the resistivity (a constant for the material of the wire),
- \(L\) is the length of the wire,
- \(A\) is the cross-sectional area of the wire.

Now, let's go through the process step by step:

### Step 1: Understanding how length affects resistance
Initially, let the original length of the wire be \(L\) and its resistance be \(R_1\). The formula for the original resistance is:

\[
R_1 = \rho \frac{L}{A}
\]

When the wire is stretched to thrice its original length, the new length \(L_2\) becomes:

\[
L_2 = 3L
\]

### Step 2: Effect of stretching on cross-sectional area
When the wire is stretched, its volume remains constant because the wire is uniformly stretched. The volume \(V\) of the wire is given by:

\[
V = A \times L
\]

Since the volume remains constant before and after stretching, we have:

\[
A \times L = A_2 \times L_2
\]

Substitute \(L_2 = 3L\) into the equation:

\[
A \times L = A_2 \times 3L
\]

Canceling \(L\) from both sides:

\[
A = 3A_2
\]

So, the new cross-sectional area \(A_2\) becomes:

\[
A_2 = \frac{A}{3}
\]

### Step 3: Calculating the new resistance
Now, we can find the new resistance \(R_2\) of the stretched wire. Using the formula for resistance:

\[
R_2 = \rho \frac{L_2}{A_2}
\]

Substitute \(L_2 = 3L\) and \(A_2 = \frac{A}{3}\) into the equation:

\[
R_2 = \rho \frac{3L}{\frac{A}{3}} = \rho \frac{3L \times 3}{A} = 9 \times \rho \frac{L}{A}
\]

From this, we can see that the new resistance \(R_2\) is 9 times the original resistance:

\[
R_2 = 9R_1
\]

### Conclusion:
When a wire is uniformly stretched to **thrice its original length**, its resistance increases by a factor of **9**. Therefore, the resistance changes **9 times**.

Answer 2

When a wire is stretched to three times its original length, its resistance changes due to two factors: length and cross-sectional area.

1. **Resistance and Length:** Resistance \( R \) is directly proportional to the length \( L \) of the wire. If the length is increased to \( 3L \), the resistance would initially increase proportionally.

2. **Resistance and Cross-Sectional Area:** However, when the wire is stretched, its cross-sectional area \( A \) decreases. For a uniform wire, the volume remains constant, so if the length increases to \( 3L \), the cross-sectional area decreases to \( \frac{A}{3} \).

The resistance \( R \) is also inversely proportional to the cross-sectional area. Thus, the decrease in area increases the resistance.

Combining these factors, the new resistance \( R' \) can be calculated as:

\[ R' = \rho \frac{3L}{A/3} = 9 \cdot \rho \frac{L}{A} \]

So, the resistance increases by a factor of 9.