1 Answer
When a wire is stretched, its resistance increases because its length increases and its cross-sectional area decreases. The resistance of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
where:
Given:
From volume conservation:
\[
A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1 L_1}{3L_1} = \frac{A_1}{3}
\]
New resistance:
\[
R_2 = \rho \frac{L_2}{A_2} = \rho \frac{3L_1}{A_1/3} = \rho \frac{3L_1 \times 3}{A_1} = 9 \times R_1
\]
\[
R_2 = 9 \times 10Ω = 90Ω
\]
Final Answer:
The new resistance of the wire is 90Ω.
\[
R = \rho \frac{L}{A}
\]
where:
- \( R \) = Resistance of the wire
- \( \rho \) = Resistivity (remains constant)
- \( L \) = Length of the wire
- \( A \) = Cross-sectional area of the wire
Given:
- Initial resistance, \( R_1 = 10Ω \)
- Initial length, \( L_1 \)
- Final length, \( L_2 = 3L_1 \) (since the wire is stretched to thrice its original length)
- Volume remains constant: \( A_1 L_1 = A_2 L_2 \)
From volume conservation:
\[
A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1 L_1}{3L_1} = \frac{A_1}{3}
\]
New resistance:
\[
R_2 = \rho \frac{L_2}{A_2} = \rho \frac{3L_1}{A_1/3} = \rho \frac{3L_1 \times 3}{A_1} = 9 \times R_1
\]
\[
R_2 = 9 \times 10Ω = 90Ω
\]