Question

A wire of length 100 cm and radius 0.1m has a resistance of 100Ω. Find resistivity of the material.
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Answer 1

Explanation:

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⟹ R = $p \frac{ l}{a}$

⟹R = $r = \: P \frac{r\pi {r}^{2} }{l}$

⟹$P= \frac{100 \times 3.14(10 - 4) {}^{2} }{1}$

⟹$P= 3.14 \times 10 {}^{ - 6} Ω - m$

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Answer 2

Given : A wire of length 100 cm and radius 0.1m has a resistance of 100Ω.

Length of wire , L = 100 cm = 100/100 = 1 m

Radius ,r = 0.1 m

Resistance,R = 100Ω

To Find : Resistivity of the material.

 

Solution:

Resistivity = ρ

Resistance of the wire , R = ρ L / A

ρ = A R / L

ρ = πr² R / L

ρ = 3.14 × (0.1)² × 100 / 1

ρ = 3.14 ×0 .01 × 100  

ρ = 314 × 0.01

ρ = 3.14 Ω-m

Hence, the resistivity of the material is 3.14 Ω-m

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