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

Solution :
Let initial resistance of wire be $$R,$$ its initial length is $${l_1}$$ and final length is $${l_2}.$$ According to problem $${l_2} = 0.5\,{l_1},$$   volume of the wire is $$= lA.$$  Since, the volume of wire remains the same after recasting, therefore
$$\eqalign{ & {l_1}{A_1} = {l_2}{A_2} \cr & \therefore \frac{{{l_1}}}{{{l_2}}} = \frac{{{A_2}}}{{{A_1}}}\,\,{\text{or}}\,\,\frac{{{l_1}}}{{0.5\,{l_1}}} = \frac{{{A_2}}}{{{A_1}}} \cr & \therefore \frac{{{A_2}}}{{{A_1}}} = 2 \cr} $$
As resistance of wire, $$R \propto \frac{l}{A}$$
$$\eqalign{ & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} \times \frac{{{A_2}}}{{{A_1}}} = \frac{{{l_1}}}{{0.5\,{l_1}}} \times 2 = 4 \cr & {\text{or}}\,\,{R_2} = \frac{{{R_1}}}{4} = \frac{R}{4} \cr} $$
NOTE
As we know the resistance $$R$$ of wire of length $$l$$ and area of cross-section, $$A$$ is given by $$R = \int {\frac{l}{A}} $$