1. Use same material and same mass condition

Since the two wires are made of the same material:

• resistivity $\rho$ is same
• density is same

Also, they have the same mass, so they have the same volume: $V_A = V_B$

For a wire, $V = \pi r^2 l$ Thus, $\pi r_A^2 l_A = \pi r_B^2 l_B$ $r_A^2 l_A = r_B^2 l_B$

Given: $r_A = 2\text{ mm}, \qquad r_B = 4\text{ mm}$ So, $l \propto \frac{1}{r^2}$ Hence, $\frac{l_A}{l_B} = \frac{r_B^2}{r_A^2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$ Therefore, $l_A = 4l_B$

2. Use resistance formula

Resistance of a wire is: $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$ So, $R \propto \frac{l}{r^2}$

Therefore, $\frac{R_A}{R_B} = \frac{l_A/r_A^2}{l_B/r_B^2}$ Substitute $l_A = 4l_B$: $\frac{R_A}{R_B} = \frac{4l_B/2^2}{l_B/4^2}$ $= \frac{4/4}{1/16} = 16$

So, $R_A = 16R_B$ Given, $R_B = 2\,\Omega$ Hence, $R_A = 16 \times 2 = 32\,\Omega$

3. Final Answer

$\boxed{32}$

4. Comparison with stored answer

Stored correct answer = $32$

My derived answer matches the stored answer.