A wire of length L and resistance R is reformed so that its length increases by 25% with no change in its volume. The resistance of the new wire is:

Correct option is B

Sol: The correct answer is (b) $$\frac{25}{16} R$$

Given:

Initial length of the wire = L

Initial resistance of the wire = R

Length of the wire increases by 25%, so new length = L+0.25L=1.25L

Concept:

Volume of a wire = Cross-sectional area (A) × Length (L)

Resistance of a wire R is given by:

R = $$\rho\frac{L}{A}$$

​​where $$\rho$$​ is the resistivity of the material.

Since the volume remains constant, the initial and final volumes are equal:

​A×L=A×L

where A and A are the initial and final cross-sectional areas:

​$$A= \frac{×}{1.25}$$ = $$\frac{A}{1.25}$$ = 0.8A

Solution:

New Resistance-

R= $$\rho\frac{L}{A}$$ = $$\rho\frac{1.25}{0.8}$$

R = $$\frac{1.25}{0.8}R$$

R = $$\frac{25}{16} R$$​​

The resistance of the new wire is $$\frac{25}{16} R$$​​