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

Solution :
Volume of material remains same in stretching.
As volume remains same, $${A_1}{l_1} = {A_2}{l_2}$$
Now, given $${l_2} = n{l_1}$$
∴ New area $${A_2} = \frac{{{A_1}{l_1}}}{{{l_2}}} = \frac{{{A_1}}}{n}$$
Resistance of wire after stretching
$$\eqalign{ & {R_2} = \rho \frac{{{l_2}}}{{{A_2}}} = \rho \cdot \frac{{n{l_1}}}{{\frac{{{A_1}}}{n}}} \cr & = \left( {\rho \frac{{{l_1}}}{{{A_1}}}} \right) \cdot {n^2} = {n^2} \cdot R\,\,\left[ {\because R = \left( {\rho \frac{{{l_1}}}{{{A_1}}}} \right)} \right] \cr} $$