A wire of resistance 9 $\mathrm{\Omega }$ is broken in two parts. The length ratio being 1 : 2. The two pieces are connected in parallel. The net resistance will be

We have,   $R=\rho \frac{l}{A}$

$R\propto l$

$\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{1}{2}$

$\Rightarrow R_2 = 2R_1$                 …(i)

$\because R_1+R_2 = 9$               …(ii)

From Eqs. (i) and (ii), we get

R₁ + 2R₁ = 9

$\Rightarrow R_1=3\mathrm{\Omega }$

$\therefore R_2=2R_1=2\times 3=6\mathrm{\Omega }$

Net resistance, ,$R_{\text{net }}=\frac{R_1{R}_2}{R_1+R_2}=\frac{3\times 6}{3+6}=2\Omega$