Here,
I2 = 120cm = 1.20m
R1 = 5.0Ω
R2 = 2.5Ω
Now,
R1 = \(\frac{ρ\,\times\,I_1}{A}\)
And,
R2 = \(\frac{ρ\,\times\,I_2}{A}\)
(Here, resistivity ρ and the area A remain same as the wires are similar.)
∴ \(\frac{R_1}{R_2} = \frac{ρ\,\times\,\frac{I_1}{A}}{ρ\,\times\,\frac{I_2}{A}}\)
\(\frac{R_1}{R_2} = \frac{I_1}{I_2}=\frac{I_1}{120\,\times10^{-2}}\)
Now,
\( \frac{5.0}{2.5}=\frac{I_1}{120\,\times10^{-2}}\)
l1 = 2×120×10-2 m
l1 = 24×10-1 m
∴ l1 = 240 cm