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Class 10 > Learn Science > A wire is 1m long, 0.2mm in diameter and has resistance...
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A wire is 1m long, 0.2mm in diameter and has resistance of 10?. Calculate its resistivity.

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Resistivity (ρ) = (R.A)/l Where, R= resistance of wire A= Cross sectional area of wire l = length of wire Given: R= 10Ω , Diammeter of wire (D) = 0.2 mm ,Length of wire = 1m Radius of wire (r) = D/2 = 0.2/2 = 0.1mm Cross sectional area of wire A=πr^2= 3.14 X〖(0.1x〖10〗^(-3))〗^2...
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Resistivity (ρ) = (R.A)/l Where, R= resistance of wire A= Cross sectional area of wire l = length of wire Given: R= 10Ω , Diammeter of wire (D) = 0.2 mm ,Length of wire = 1m Radius of wire (r) = D/2 = 0.2/2 = 0.1mm Cross sectional area of wire A=πr^2= 3.14 X〖(0.1x〖10〗^(-3))〗^2 = 3.14 X 〖10〗^(-8) Resistivity (ρ) = (R.A)/l = (10x3.14 X 〖10〗^(-8))/1 = 3.14 X 〖10〗^(-7) Ω.m read less
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ρ=R A/l ρ=4*10^-5 Ω.m
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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...
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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 read less
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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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Resistance = R = 10 Ω Resistivity = ρ = RA / l = 10 × 0.00000314 / 1 = 0.0000314 Ω m
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