Question

if the resistance of a wire of length 120 cm and diameter 0.04 mm is 25 ohm then calculate the specific resistance of the material of the wire.

Answer 1

Hi there.

Given conditions ⇒

Length of the wire (l) = 120 cm.
Diameter of the cross section of the wire wire = 0.04 mm.
∴ Radius of the cross section of the wire(r) = 0.02 mm.
   = 0.002 cm.

Area of the cross-section of the wire = πr²
= 22/7 × (0.002)²
= 22/7 × 0.000004
= 12.57 × 10⁻⁶ cm²

Resistance of the wire (R) = 25 Ω

Using the Formula,

 Specific Resistance (ρ) = Ra/l
∴ ρ = (25 × 12.57 × 10⁻⁶) ÷ 120
⇒ ρ = (314.25 × 10⁻⁶)/120
∴ ρ = 2.61875 × 10⁻⁶
∴ ρ = 2.62 × 10⁻⁶ Ω-cm.


Hence, the Specific Resistance of the Resistivity of the wire is 2.62 × 10⁻⁶ Ω-cm.


Hope it helps.

Answer 2

Explanation:

Given conditions ⇒

Length of the wire (l) = 120 cm.

Diameter of the cross section of the wire wire = 0.04 mm.

∴ Radius of the cross section of the wire(r) = 0.02 mm.

  = 0.002 cm.

Area of the cross-section of the wire = πr²

= 22/7 × (0.002)²

= 22/7 × 0.000004

= 12.57 × 10⁻⁶ cm²

Resistance of the wire (R) = 25 Ω

Using the Formula,

Specific Resistance (ρ) = Ra/l

∴ ρ = (25 × 12.57 × 10⁻⁶) ÷ 120

⇒ ρ = (314.25 × 10⁻⁶)/120

∴ ρ = 2.61875 × 10⁻⁶

∴ ρ = 2.62 × 10⁻⁶ Ω-cm.

Hence, the Specific Resistance of the Resistivity of the wire is 2.62 × 10⁻⁶ Ω-cm.