Two wires 1 and 2, each of length 2L and L are made of the same material. The radius of wire 1 is r and of wire 2 is 2r. The current I flows through the composite wire (see Fig). Choose the correct statement from the following. 

 

$R=\frac{\rho \mathcal{l}}{\pi r^2}$.

Since the two wires are made of the same material, resistivity $\rho$ is the same for wires 1 and 2. 

$R_1=\frac{\rho \left(2L\right)}{\pi r^2} and R_2=\frac{\rho \left(L\right)}{\pi {\left(2r\right)}^2}$

$$\begin{gathered} \therefore \frac{R_1}{R_2}=8 \\ \Rightarrow R_1=8R_2 \end{gathered}$$

Since the current I is the same in wires 1 and 2 (as they are in series). 

$$\begin{gathered} {\mathrm{V}}_1={\mathrm{IR}}_1 \mathrm{and} {\mathrm{V}}_2={\mathrm{IR}}_2 \\ \therefore \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=8 \end{gathered}$$

Powers in wires are P ₁ = I² R₁ and I² R₂. So 

 $\frac{P_1}{P_2}=\frac{R_1}{R_2}=8.$

Current density $J=\frac{I}{{\mathrm{\pi r}}^2}.$

                        $J_1=\frac{I}{{\mathrm{\pi r}}^2}$

                      $J_2=\frac{I}{\pi {\left(2r\right)}^2}$

                     $\therefore \frac{J_1}{J_2}=4.$

Electric field E =$\frac{V}{\mathcal{l}}$

                 $E_1=\frac{V_1}{2L} and E_2=\frac{V_2}{L}$

$\therefore \frac{E_1}{E_2}=\frac{V_1}{2V_2}=\frac{1}{2}\times 8=4$.