Question

a wire is 1 mtr long 2 mm in diameter and has a resistance of 10 ohm calculate the resistivity of its material​

Answer 1

Answer:

Here , it is Given that:

l=1m ,d=0.2mm=>r=0.1mm and R=10 ohm

by using the formula,

R= p౹/ A

= p౹/ πr²

therefore, p= Rπr²/l

= 10× 3.14×( 0.1× 10 -³)²

= 10× 3.14× 0.1×0.1× 10 to the power -6

= 0.314× 10 to the power-6

= 3.14× 10 to the power-7 ohm- m

hence, the resistivity of the conductor is 3.14× 10 to the power-7 ohm- m

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Answer 2

Answer:

here , it is Given that:

l=1m ,d=0.2mm=>r=0.1mm and R=10 ohm

by using the formula,

R= p౹/ A

= p౹/ πr²

therefore, p= Rπr²/l

= 10× 3.14×( 0.1× 10 -³)²

= 10× 3.14× 0.1×0.1× 10 to the power -6

= 0.314× 10 to the power-6

= 3.14× 10 to the power-7 ohm- m

hence, the resistivity of the conductor is 3.14× 10 to the power-7 ohm- m

Given :-

l = 0.1 m = 1000 mm

d = 0.2 mm

r = 0.1 mm

R = 10 Ω

To Find :-

Resistivity of its material.

Solution :-

ρ = R × A/l

Cross sectional area = πr²

⇒ A = π × (0.1)²

⇒ A = 0.0314 mm²

⇒ ρ = 0.000314 Ω - mm

⇒ ρ≃ 3.14 × 10⁻⁷  Ω m

Hence, the resistivity of its material is 3.14 × 10⁻⁷  Ω m.