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 × 10³ kg/m³ and resistivity is 1.7 × 10⁻⁸ Ω-m. The length of wire is :

Solution

Physics sir · 4 steps

Mass and density give the volume. Then write area as volume over length, and R depends on l².

1. Resistance from Ohm's law. R = V/I = 3.4/2 = 1.7 Ω.
2. Volume from mass. Volume = m/d = (8.92 × 10⁻³)/(8.92 × 10³) = 1 × 10⁻⁶ m³. The wire is uniform, so volume = Al, which gives A = Volume/l.

   R = (ρ l)/A = (ρ l²)/Volume ⇒ l² = (R × Volume)/ρ

3. Put in numbers. l² = (1.7 × 1 × 10⁻⁶)/(1.7 × 10⁻⁸) = 100 m², so l = √100 = 10 m.
4. So. The length of the wire is 10 m, which is option (d).

The idea to keep. When a problem gives mass and density but not area, find the volume first and replace A with Volume/l. Resistance then varies as l², so take a square root at the end.

Answer from the NTA final answer key. Solution by PrepWiser.