A uniform metallic wire carries a current 2A , when 3 . 4V battery...

Given here, $I = 2A, V = 3 . 4V, m = 8 . 92 \times {10}^{- 3}kg,$

$d = 8 . 92 \times {10}^{3}kg{m}^{- 3}andρ = 1 . 7 \times {10}^{- 8}Ω - m$

From Ohm's law, resistance is $R = \dfrac{V}{I} = \dfrac{3 . 4}{2} = 1 . 7Ω$

As resistance of wire is given by $R = \dfrac{ρl}{A}$, where, $A$ is area and $l$ is length of wire.

$⇒ 1 . 7 = \dfrac{ρl}{A}$ $⇒ l = \dfrac{1 . 7A}{ρ}$

Now, volume$= \dfrac{m}{d} = \dfrac{8 . 92 \times {10}^{- 3}}{8 . 92 \times {10}^{3}} = {10}^{- 6}{m}^{3}$

Thus,

${l}^{2} = \dfrac{1 . 7A}{ρ} \times l = \dfrac{1 . 7\left(\dfrac{\text{volume}}{l}\right)}{ρ} \times l = \dfrac{1 . 7}{ρ}\left({10}^{- 6}\right) = \dfrac{1 . 7}{1 . 7 \times {10}^{- 8}} \times {10}^{- 6}$

Hence, $l = 10m$