If a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be 

: Initial resistance (RJ = R; Initial length is I and final length (12 ) = 0.5 I. Volume of a wire= I . A. Since the volume of the wire remains the same after recasting, therefore ${\ell }_1\cdot A_1={\ell }_2{A}_2$

or  $\frac{{\ell }_1}{{\ell }_2}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_2}\text{ or }\frac{\ell }{0.51}=\frac{{\mathrm{A}}_2}{{\mathrm{A}}_2}\text{ or }\frac{{\mathrm{A}}_2}{{\mathrm{A}}_2}=2.$

We also know that resistance of a wire (R)  

$\begin{array}{l}R=\rho \times \frac{\ell }{\mathrm{A}}.;\mathrm{R}\propto \frac{1}{\mathrm{A}}\\ \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\ell }_1}{{\ell }_2}\times \frac{{\mathrm{A}}_2}{{\mathrm{A}}_1}=\frac{\ell }{0.5\mathrm{l}}\times 2=4\end{array}$

Or $R_2=\frac{R_1}{4}=\frac{R}{4}$

 [ Alt: When wires are drawn from same volume but with different area of cross-section, then

$R\propto \frac{1}{\text{ (Area of cross }-{\text{ section) }}^2}$