A copper wire has a diameter 0.5 mm and resistivity of 1.6×10-8Ωm. The length of the wire to make its resistance 10Ω is [[1]]m.

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A copper wire has a diameter $0.5 mm$ and resistivity of $1.6 \times 1 0^{- 8} Ωm .$ The length of the wire to make its resistance $10 Ω$ is [[1]] $m .$

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Detailed Solution

A copper wire has a diameter

0.5 mm

and resistivity of

1.6 \times 1 0^{- 8} Ωm .

The length of the wire to make its resistance

10 Ω

122.7 m

.
The resistance of a wire is:

R = \frac{ρl}{A}

………………. (i)
Where,

ρ, l,

and A are the resistivity of the copper, length, and the area of cross-section of the wire.
Therefore, the length of the wire can be found as:

l = \frac{RA}{ρ}

…………….. (ii)
We know that

A = π r^{2}

[ where r = radius of area of cross section of wire) …………. (iii)
also r =

\frac{d}{2}

Put value of r in equation (ii) we get,

A = \frac{π d^{2}}{4}

………….. (iv)
Put value of equation (iv) in equation (ii) we get,

= > l = \frac{R (π d^{2} / 4)}{ρ}

(where d is the diameter of the area of cross-section)

= > l = \frac{(10) (π {(0.5 \times 1 0^{- 3})}^{2} / 4)}{1.6 \times 1 0^{- 8}}

= > l = \frac{10 \times 22 \times {(0.5 \times 1 0^{- 3})}^{2})}{4 \times 7 \times 1.6 \times 1 0^{- 8}}

where

(π = \frac{22}{7})

= > l = \frac{10 \times 22 \times 0.25 \times 1 0^{- 6}}{4 \times 7 \times 1.6 \times 1 0^{- 8}}

= > l = \frac{55 \times 1 0^{- 6}}{44.8 \times 1 0^{- 8}}

= > l = 1.22 \times 1 0^{- 6 + 8}

= > l = 1.22 \times 1 0^{2}

= > l = 122.7 m

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