Question

A wire of resistance \(5\,\Omega\) is drawn out so that its length is increased to twice its original length, its new resistance is:

• A. \(45\,\Omega\)
• B. \(54\,\Omega\)
• C. \(20\,\Omega\)
• D. \(5\,\Omega\)

Hint:

If length doubles \(\Rightarrow\) resistance becomes 4 times (volume constant).

Answer 1

The Correct Option is C

To solve this problem, we need to understand how resistance changes when the dimensions of the wire are altered. The resistance \( R \) of a wire is given by the formula:

\(R = \rho \frac{L}{A}\)

where:

• \(\rho\) is the resistivity of the material.
• \(L\) is the length of the wire.
• \(A\) is the cross-sectional area of the wire.

In this case, the original resistance \( R_1 \) is \( 5\,\Omega \). The original length of the wire is \( L \) and the cross-sectional area is \( A \). We need to determine the new resistance \( R_2 \) when the length of the wire is doubled, i.e., the new length is \( 2L \).

Since the volume of the wire remains the same before and after stretching, we have:

\(L \times A = 2L \times A_{\text{new}}\)

Solving for the new cross-sectional area \( A_{\text{new}} \):

\(A_{\text{new}} = \frac{A}{2}\)

The new resistance \( R_2 \) is given by:

\(R_2 = \rho \frac{2L}{A_{\text{new}}}\)

Substitute \( A_{\text{new}} = \frac{A}{2} \) into the equation:

\(R_2 = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A}\)

Which implies:

\(R_2 = 4 \times R_1\)

Substituting the initial resistance \( R_1 = 5\,\Omega \):

\(R_2 = 4 \times 5\,\Omega = 20\,\Omega\)

Therefore, the new resistance of the wire is \( 20\,\Omega \).

Conclusion: The correct answer is \( 20\,\Omega \).