Question

The resistance of a wire is $'R'$ ohm. If it is melted and stretched to $'n'$ times its original length, its new resistance will be :

• A. $\frac{R}{n}$
• B. $n^2 R $
• C. $\frac{R}{n^2} $
• D. $nR $

Answer 1

The Correct Option is B

The resistance of a wire is a measure of how much it opposes the flow of electric current. When a wire is melted and stretched, its physical dimensions change, which affects its resistance.

The resistance R of a wire is directly proportional to its length L and inversely proportional to its cross-sectional area A, and is given by the formula:

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

Where:

• \rho is the resistivity of the material.

When the wire is melted and stretched to n times its original length, its length becomes nL. Since the volume of the wire remains constant during the stretching process, we have:

L \cdot A = nL \cdot A'

Where A' is the new cross-sectional area. Solving for A', we get:

A' = \frac{A}{n}

Substituting the new length and area into the resistance formula, the new resistance R' is:

R' = \rho \frac{nL}{A'}

Substitute A' in the above equation:

R' = \rho \frac{nL}{\frac{A}{n}} = \rho \frac{n^2 L}{A}

This shows that the new resistance R' = n^2 R, where R is the original resistance. Therefore, the correct answer is n^2 R.

Hence, the correct option is:

• n^2 R