Question

Resistance of a metal wire of length 1 m is 26 W at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?

Answer 1

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Resistance , R = 26Ω

Resistance , R = 26ΩLength of wire , L = 1m

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³m

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/A

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/AHere, R is the resistance, ρ is the resistivity , L is the length of wire and A is the base area of it .

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/AHere, R is the resistance, ρ is the resistivity , L is the length of wire and A is the base area of it .So , ρ = RA/L = 26Ω × 7.06 × 10⁻⁸m²/1m

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/AHere, R is the resistance, ρ is the resistivity , L is the length of wire and A is the base area of it .So , ρ = RA/L = 26Ω × 7.06 × 10⁻⁸m²/1mρ = 1.8356 × 10⁻⁶ Ωm

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/AHere, R is the resistance, ρ is the resistivity , L is the length of wire and A is the base area of it .So , ρ = RA/L = 26Ω × 7.06 × 10⁻⁸m²/1mρ = 1.8356 × 10⁻⁶ ΩmHence, resistivity is 1.8356 × 10⁻⁶ Ωm at 20°C

Resistance , R = 26ΩLength of wire , L = 1mdiameter of circular part of wire ,d = 0.3mm = 0.3 × 10⁻³mso, base area of wire , A = πr² = πd²/4 [ ∵d = 2r]= 3.14 × (0.3 × 10⁻³)²/4m²= 3.14 × 9 × 10⁻⁸/4 m²= 28.26/4 × 10⁻⁸ m² = 7.06 × 10⁻⁸ m²Now, use formula,R = ρL/AHere, R is the resistance, ρ is the resistivity , L is the length of wire and A is the base area of it .So , ρ = RA/L = 26Ω × 7.06 × 10⁻⁸m²/1mρ = 1.8356 × 10⁻⁶ ΩmHence, resistivity is 1.8356 × 10⁻⁶ Ωm at 20°C4.6

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Answer 2

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