Question

A tightly coiled spring having 75 turns, each 3.5 cm in diameter, is made of an insulated metal wire of cross section area 5 * 10^(-6) m2. The resistance across it is 1.74 ohm. Find the resistivity...​

Answer 1

Explanation:

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Answer 2

Solution: The resistance R of metallic conductor with length L, cross-sectional area A and resistivity ρ can

be calculated as

A

L

R    . In our case L  n   D ,

4

2Solution: The resistance R of metallic conductor with length L, cross-sectional area A and resistivity ρ can

be calculated as

A

L

R    . In our case L  n   D ,

4

2

d

A







, where n and D are the number and

diameter (m) of the spring coils; d is the diameter of the metal wire, m.

Then,   ohmm.

 

 









 

 











6

2 2 3 2

1.75 10

4 75 0.035

1.74 (3.25 10 )

4 4n D

R d

n D

R d

L

R A







Answer: 1.75·10-6

ohm·m. hope it is clear

d

A







, where n and D are the number and

diameter (m) of the spring coils; d is the diameter of the metal wire, m.

Then,   ohmm.

 

 









 

 











6

2 2 3 2

1.75 10

4 75 0.035

1.74 (3.25 10 )

4 4n D

R d

n D

R d

L

R A







Answer: 1.75·10-6

ohm·m.