A material '$$B$$' has twice the specific resistance of '$$A$$'. A circular wire made of '$$B$$' has twice the diameter of a wire made of '$$A$$'. Then for the two wires to have the same resistance, the ratio $$\ell_A/\ell_B$$ of their respective lengths must be

The specific resistance (resistivity) of material $$B$$ is twice that of material $$A$$:

$$\rho_B = 2\rho_A$$

The diameter of wire $$B$$ is twice the diameter of wire $$A$$, which means its radius is also twice as large:

$$r_B = 2r_A$$

The problem states that both wires have the same resistance:

$$R_A = R_B$$

The formula for the resistance of a cylindrical wire is:

$$R = \rho \frac{\ell}{A} = \rho \frac{\ell}{\pi r^2}$$

Substitute the resistance formula for both wires $$A$$ and $$B$$:

$$\rho_A \frac{\ell_A}{\pi r_A^2} = \rho_B \frac{\ell_B}{\pi r_B^2}$$

Now, substitute the given relations $$\rho_B = 2\rho_A$$ and $$r_B = 2r_A$$ into the right side of the equation:

$$\rho_A \frac{\ell_A}{\pi r_A^2} = (2\rho_A) \frac{\ell_B}{\pi (2r_A)^2}$$

Square the radius term in the denominator on the right side:

$$\rho_A \frac{\ell_A}{\pi r_A^2} = 2\rho_A \frac{\ell_B}{4\pi r_A^2}$$

Simplify the fraction on the right side:

$$\rho_A \frac{\ell_A}{\pi r_A^2} = \rho_A \frac{\ell_B}{2\pi r_A^2}$$

Cancel the common terms ($$\rho_A$$, $$\pi$$, and $$r_A^2$$) from both sides of the equation:

$$\ell_A = \frac{\ell_B}{2}$$

Rearrange to find the ratio of their lengths:

$$\frac{\ell_A}{\ell_B} = \frac{1}{2}$$