Math, asked by upendarbhatt480, 6 months ago

Calculate the resistivity of a wire of diameter 3.5 mm and length 15 cm. Resistance of the wire is 25 ohm. *​

Answers

Answered by jaya9561
0

Answer:

hi bro so sorry i don't know your answer

Step-by-step explanation:

copy from Google

Answered by Ankitachettri
0

Strategy

We can rearrange the equation

R

=

ρ

L

A

R=ρLA to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.

Solution

The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in

R

=

ρ

L

A

R=ρLA, is

A

=

ρ

L

R

A=ρLR.

Substituting the given values, and taking ρ from Table 1, yields

A

=

(

5.6

×

10

8

Ω

m

)

(

4.00

×

10

2

m

)

0.350

Ω

=

6.40

×

10

9

m

2

A=(5.6×10−8Ω⋅m)(4.00×10−2m)0.350Ω=6.40×10−9m2.

The area of a circle is related to its diameter D by

A

=

π

D

2

4

A=πD24.

Solving for the diameter D, and substituting the value found for A, gives

D

=

2

(

A

p

)

1

2

=

2

(

6.40

×

10

9

m

2

3.14

)

1

2

=

9.0

×

10

5

m

D=2(Ap)12=2(6.40×10−9m23.14)12=9.0×10−5m.

Discussion

The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.Strategy

We can rearrange the equation

R

=

ρ

L

A

R=ρLA to find the cross-sectional area A of the filament from the given information. Then its diameter can be found by assuming it has a circular cross-section.

Solution

The cross-sectional area, found by rearranging the expression for the resistance of a cylinder given in

R

=

ρ

L

A

R=ρLA, is

A

=

ρ

L

R

A=ρLR.

Substituting the given values, and taking ρ from Table 1, yields

A

=

(

5.6

×

10

8

Ω

m

)

(

4.00

×

10

2

m

)

0.350

Ω

=

6.40

×

10

9

m

2

A=(5.6×10−8Ω⋅m)(4.00×10−2m)0.350Ω=6.40×10−9m2.

The area of a circle is related to its diameter D by

A

=

π

D

2

4

A=πD24.

Solving for the diameter D, and substituting the value found for A, gives

D

=

2

(

A

p

)

1

2

=

2

(

6.40

×

10

9

m

2

3.14

)

1

2

=

9.0

×

10

5

m

D=2(Ap)12=2(6.40×10−9m23.14)12=9.0×10−5m.

Discussion

The diameter is just under a tenth of a millimeter. It is quoted to only two digits, because ρ is known to only two digits.

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