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. \(nR\)
• B. \(\frac Rn\)
• C. \(n^2R\)
• D. \(\frac {R}{n^2}\)

Answer 1

The Correct Option is C

To find the new resistance of the wire after it is melted and stretched to ‘n’ times its original length, we start with the basics of resistance.

The resistance \( R \) of a wire is given by the formula:

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

where:

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

Initially, the resistance is R. When the wire is melted and stretched, its volume remains constant. Therefore, the product of its original length and cross-sectional area is equal to the new product after being stretched:

L_1 \cdot A_1 = L_2 \cdot A_2

where:

• L_1 and A_1 are the original length and area.
• L_2 = nL_1 is the new length.
• A_2 is the new cross-sectional area.

From volume conservation, we get:

A_2 = \frac{A_1}{n}

Let's calculate the new resistance R_2:

R_2 = \rho \frac{L_2}{A_2} = \rho \frac{nL_1}{A_1/n} = \rho \frac{n^2L_1}{A_1} = n^2R

Therefore, the new resistance of the wire when it is elongated to 'n' times its original length is:

• \(\boxed{n^2R}\)

Thus, the correct answer is:

• n^2R