A uniform metallic wire carries a current 2 A , when 3 . 4 V battery is connected across it. The mass of uniform metallic wire is 8 . 92 × 10 - 3 kg , density is 8 . 92 × 10 3 kg m - 3 and resistivity is 1 . 7 × 10 - 8 Ω - m The length of wire is :

Step-by-step solution

Given here, I = 2   A ,   V = 3 . 4   V ,   m = 8 . 92 × 10 - 3   kg , d = 8 . 92 × 10 3   kg   m - 3   and   ρ = 1 . 7 × 10 - 8   Ω - m From Ohm's law, resistance is R = V I = 3 . 4 2 = 1 . 7   Ω As resistance of wire is given by R = ρ l A , where, A is area and l is length of wire. ⇒ 1 . 7 = ρ l A ⇒ l = 1 . 7 A ρ Now, volume = m d = 8 . 92 × 10 - 3 8 . 92 × 10 3 = 10 - 6   m 3 Thus, l