A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10-8 Ohm m|10 cbse electricity|Ex 6|

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Resistivity (ρ) = 1.6 × 10^-8 Ω m

Resistance (R) = 10 Ω

Diameter (d) = 0.5 mm

d = 5 × 10⁻⁴ m

Hence, we will get radius

Radius (r) = 0.25 *10^--3 m

r = 0.25 × 10⁻³ m

r = 2.5 × 10⁻⁴ m

We need to find the area of cross-section

A = πr2

A = (22/7)(2.5 × 10⁻⁴)2

A = (22/7)(6.25×10⁻⁸)

A = 1.964 × 10-7 m^2

Formula
We know that

R = ρ (L) / (A)

L = (R × A) / ρ

Substituting the values in the above equation we get

L = (10 × 1.964 × 10⁻⁷) / 1.6 × 10⁻⁸ m

L = 1.964×10-6 /1.6 × 10-8

L = 122.6 m

If the diameter of the wire is doubled, the new diameter = 2 × 0.5 = 1mm = 0.001m
Let new resistance be Rʹ

R = ρ (L) / (A)

R’ = ρ (L) / (4A)

R’ = ρ (L) X 1/(4A)

Hence, if diameter doubles, resistance becomes 1/4 times.

Answer
Therefore, the length of the wire is 122.6 m and the new resistance becomes 1/4 times.