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The specific resistance of a wire of length $1.1 m, 0.4 mm$ diameter and having a total resistance of $4.2 Ω$ is ____ $\times 10^{- 8} Ωm$ .

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Detailed Solution

The specific resistance of a wire of length

1.1 m, 0.4 mm

diameter and having a total resistance of

4.2 Ω

48 \times 10^{- 8} Ωm

.
Given,
Resistance,

R = 4.2 Ω

Length,

l = 1.1 m

Diameter,

d = 0.4 mm

Since diameter is double the radius, the radius is given by

r = \frac{d}{2}

On substituting the value, we get the radius as

\Rightarrow r = \frac{0.4}{2}

\Rightarrow r = 0.2 mm

\Rightarrow r = 0.2 \times 10^{- 3} m

Also, the area of cross-section

(A)

is given by

A = π r^{2}

On substituting the given values, we get the cross-sectional area as

\Rightarrow A = 3.14 \times {(0.2 \times 10^{- 3})}^{2}

\Rightarrow A = 0.1256 \times 10^{- 6} m^{2}

Now, it is known that the specific resistance or resistivity of a substance is numerically equal to the resistance when the substance is

1 m

long and has a cross-sectional area of

1 m^{2}

. Its SI unit is

ohm - metre (Ωm)

. It depends on the nature of the substance.
So, specific resistance

(ρ)

is given by

ρ = \frac{RA}{l}

On substituting the given values, we get the specific resistance as

\Rightarrow ρ = \frac{4.2 \times 0.1256 \times 10^{- 6}}{1.1}

\Rightarrow ρ = 48 \times 10^{- 8} Ωm

Hence, the specific resistance is

48 \times 10^{- 8} Ωm

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