The radius of a wire having resistance R is made half and its length reduced to one-fourth of its original length. The new resistance of the wire will be

R

The correct answer is Option (2) → R

Resistance of a wire is given by:

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

Let the original radius be $r$ and length $L$, so cross-sectional area $A = \pi r^2$.

New radius: $r' = \frac{r}{2} \Rightarrow A' = \pi (r/2)^2 = \frac{\pi r^2}{4}$

New length: $L' = \frac{L}{4}$

New resistance:

$R' = \rho \frac{L'}{A'} = \rho \frac{L/4}{(\pi r^2)/4} = \rho \frac{L/4}{A/4} = \rho \frac{L}{A} = R$

New resistance R' = R