A uniform wire having a resistance of 100 is melted and recast into...

Let the original resistance be $R_1 = 100 \Omega$, original length be $L_1$, and original area of cross-section be $A_1$.

The resistance of the original wire is given by $R_1 = \rho \dfrac{L_1}{A_1}$.

When the wire is melted and recast, its volume remains constant. Thus, $V = L_1 A_1 = L_2 A_2$.

Given that the new length is $L_2 = 2 L_1$.

The new area of cross-section is $A_2 = \dfrac{V}{L_2} = \dfrac{L_1 A_1}{2 L_1} = \dfrac{A_1}{2}$.

The resistance of the new wire is $R_2 = \rho \dfrac{L_2}{A_2} = \rho \dfrac{2 L_1}{\dfrac{A_1}{2}} = 4 \rho \dfrac{L_1}{A_1}$.

Therefore, $R_2 = 4 R_1 = 4 \times 100 = 400 \Omega$.