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A copper wire has diameter $0.5 mm$ and resistivity ρ = $1.6 \times 10^{- 8} ohm .$ What will be the length of its wire to make its resistance 10 ohm. In what ratio will the resistance change if diameter gets doubled?

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100 m

1 : 4

122.66 m

1 : 2

122.66 m

1 : 4

None of these

answer is C.

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Detailed Solution

A copper wire has diameter

0.5 mm

and resistivity ρ =

1.6 \times 10^{- 8} ohm .

The length of its wire should be

122.66 m

to make its resistance

10 ohm

. If diameter gets doubled, the resistance becomes

1 : 4 .

Radius = \frac{diameter}{2}

= \frac{0.5}{2}

= 0.25 mm

= 0.25 \times 10^{- 3} m

Resistivity ρ =

1.6 \times 10^{- 8} oh m^{- 1}

and

R = 10 ohm

We know that, resistance R =

\frac{ρl}{A}

10 = \frac{1.6 \times 10^{- 8} \times l}{π {(0.25 \times 10^{- 3})}^{2}}

Length l = 122.66 m

As we know, diameter

d = 2 r

. So,

r = d / 2

. Then,

{A = π d}^{2} / 4

If we double the diameter, then resistance will become one fourth.
Substituting this in the resistance formula, we get the ratio of resistances as

1 : 4 .

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