Question

if the resistance of a wire of length 1.20 m diameter 0.4 mm is found to be 2.5. What is the specific resistance of the material of the wire?

Answer 1

Answer:

26.17 *  10⁻⁸ Ωm

Step-by-step explanation:

if the resistance of a wire of length 1.20 m diameter 0.4 mm is found to be 2.5. What is the specific resistance of the material of the wire?

R = ρ L/A

ρ = specific resistance

R = 2.5 Ω

L = 1.2 m

A = πR² = π (D/2)²  

D = 0.4 mm  = 0.4/1000 m = 4 * 10⁻⁴ m

A = π (2 * 10⁻⁴)² = 4π * 10⁻⁸ m²

2.5 = ρ * 1.2 / 4π * 10⁻⁸

=> ρ  = 10π * 10⁻⁸/1.2

π = 3.14

=> ρ  = 10 * 3.14 * 10⁻⁸/1.2

=> ρ  = 26.17 *  10⁻⁸ Ωm

Answer 2

l = 1.2m

l = 1.2mD = 0.4 mm

l = 1.2mD = 0.4 mmR = 0.4/2

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8= 12.5 * 10 power -8

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8= 12.5 * 10 power -8R= 2.5 ohm

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8= 12.5 * 10 power -8R= 2.5 ohm rho = RA/ l

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8= 12.5 * 10 power -8R= 2.5 ohm rho = RA/ l= 2.5 * 12.5*10 power -8/1.2

l = 1.2mD = 0.4 mmR = 0.4/2R = 0.2 mm so,R = 0.2 / 1000 m= 0.2 * 10 power -4Area of cross section = πr2 = 22/7 *0.2*0.3*10 power - 4= 88/7 * 10 power -8= 12.5 * 10 power -8R= 2.5 ohm rho = RA/ l= 2.5 * 12.5*10 power -8/1.2= 26.01 * 10 power -8

step by step explaination: