A copper wire has diameter 0.5mm and resistivity of $1.6 \times 1 0^{- 8} Ω m$ . 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?

Question

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Answer 1

Given, diameter, $d = 0.5 m m$
resistivity, $ρ = 1.6 \times 1 0^{- 8} Ω m$
Resistance, $R = 10 Ω$
Let the length of wire be $l$ .

$A = π d^{2} / 4$
$R = ρ l / A$

$= \frac{ρ l}{π d ^{2} / 4}$

$⟹ l = \frac{R π d ^{2}}{4 ρ}$
$l = \frac{1 0 \times 3 . 1 4 \times ( 0 . 5 \times 1 0 ^{- 3} ) ^{2}}{4 \times 1 . 6 \times 1 0 ^{- 8}}$
$l = 122.7 m$

$R \propto 1 / d^{2}$
If the diameter is doubled, resistance will be one-fourth.
Hence, new resistance $= 2.5 Ω$

Answer 2

Given, diameter,

d = 0.5 m m

resistivity,

ρ = 1.6 \times 1 0^{- 8} Ω m

Resistance,

R = 10 Ω

Let the length of wire be

l

.

A = π d^{2} / 4

R = ρ l / A

= \frac{ρ l}{π d ^{2} / 4}

⟹ l = \frac{R π d ^{2}}{4 ρ}

l = \frac{1 0 \times 3 . 1 4 \times ( 0 . 5 \times 1 0 ^{- 3} ) ^{2}}{4 \times 1 . 6 \times 1 0 ^{- 8}}

l = 122.7 m

R \propto 1 / d^{2}

If the diameter is doubled, resistance will be one-fourth.

Hence, new resistance

= 2.5 Ω