Question

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10 −8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled?

Answer 1

here is your answer

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Answer 2

If the diameter doubled, the resistance changes about 2.5 ohm.

Explanation:

We know that area of the copper wire that is taken in this case will be $= \pi (\frac{d}{2})^2$.

Diameter given Is 0.5 mm which becomes 0.0005 m on converting into metre.

Resistance required is 10 ohms while the resistivity is $1.6 \times 10^{-8}$

Applying formula to find the length of wire R = ⲣl/A

Putting the given values in the formula for solving and calculating the answer we get length L is equal to 122.72 m

Now if the Diameter of the wire is doubled Then the resistance will be

$=1.6 \times 10^{-8} \times \frac{122.72}{(3.14 \times 0.001 \times 0.001)}= 2.5$ ohm .

To know more:

A copper wire 4 in long has diameter of 1 mm. if a load of 10 kg wt is attached at other end. What extension is produced, if Poisson's ratio is 0.26? How much lateral compression is produced in it ? (Y_{cu} = 12.5 x 10¹⁰ N / m²) (Ans : 3.992 mm, 2.595 x 10⁻⁴ mm)

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