Question

What is the new resistance of a wire of resistance 10 Ω is stretched to thrice its original length?

Answer 1

To determine the new resistance of a wire when it is stretched to three times its original length, we can use the relationship between resistance, length, and cross-sectional area of the wire. Here's a step-by-step explanation:

1. **Initial Resistance and Properties:**
   - Let the original length of the wire be \( L \).
   - Let the original cross-sectional area of the wire be \( A \).
   - The initial resistance \( R_0 \) of the wire is given by the formula:
     \[
     R_0 = \rho \frac{L}{A}
     \]
     where \( \rho \) is the resistivity of the material of the wire.

2. **Stretching the Wire:**
   - When the wire is stretched to three times its original length, the new length \( L' \) becomes:
     \[
     L' = 3L
     \]
   - The volume of the wire remains constant during stretching, so the product of the length and the cross-sectional area remains the same. Therefore:
     \[
     L \times A = L' \times A'
     \]
     where \( A' \) is the new cross-sectional area after stretching. Substituting \( L' = 3L \) into the equation gives:
     \[
     L \times A = 3L \times A'
     \]
     Solving for \( A' \):
     \[
     A' = \frac{A}{3}
     \]

3. **New Resistance Calculation:**
   - The resistance of the wire after stretching is given by:
     \[
     R' = \rho \frac{L'}{A'}
     \]
   - Substitute \( L' = 3L \) and \( A' = \frac{A}{3} \) into the formula:
     \[
     R' = \rho \frac{3L}{\frac{A}{3}} = \rho \frac{3L \times 3}{A} = 9 \rho \frac{L}{A}
     \]
     Thus:
     \[
     R' = 9 R_0
     \]
   - Given that the initial resistance \( R_0 \) is 10 Ω, the new resistance \( R' \) is:
     \[
     R' = 9 \times 10\, \Omega = 90\, \Omega
     \]

So, when the wire is stretched to three times its original length, its resistance increases to 90 Ω.

Answer 2

To determine the new resistance of a wire when it is stretched to three times its original length, we need to consider how stretching the wire affects its resistance. The resistance of a wire is given by:

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

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

When the wire is stretched, its length increases while its volume remains constant (assuming the stretching is done uniformly).

Let's denote:
- The original length of the wire as \( L_0 \),
- The original cross-sectional area as \( A_0 \),
- The original resistance as \( R_0 \).

Given that:
\[ R_0 = \rho \frac{L_0}{A_0} \]

When the wire is stretched to three times its original length:
- The new length, \( L_{\text{new}} \), is \( 3L_0 \).

The volume of the wire is conserved, so:
\[ L_0 \cdot A_0 = L_{\text{new}} \cdot A_{\text{new}} \]

Substitute \( L_{\text{new}} = 3L_0 \):
\[ L_0 \cdot A_0 = 3L_0 \cdot A_{\text{new}} \]

Solving for \( A_{\text{new}} \):
\[ A_{\text{new}} = \frac{A_0}{3} \]

The new resistance, \( R_{\text{new}} \), is:
\[ R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}} \]

Substitute \( L_{\text{new}} = 3L_0 \) and \( A_{\text{new}} = \frac{A_0}{3} \):
\[ R_{\text{new}} = \rho \frac{3L_0}{\frac{A_0}{3}} \]
\[ R_{\text{new}} = \rho \frac{3L_0 \cdot 3}{A_0} \]
\[ R_{\text{new}} = 9 \rho \frac{L_0}{A_0} \]
\[ R_{\text{new}} = 9 R_0 \]

Since the original resistance \( R_0 \) is 10 Ω:
\[ R_{\text{new}} = 9 \times 10 \, \Omega \]
\[ R_{\text{new}} = 90 \, \Omega \]

Therefore, the new resistance of the wire, after being stretched to three times its original length, is 90 Ω.