Question

A copper wire has a diameter of 0.5mm . Its resistivity is 1.6×10^-8 ohm .

1. What will be the length of this wire to make its resistance 10 ohm .

2. How much does the resistance change when the diameter is doubled .​

Answer 1

Answer:

The length of the wire is 122.7 m and the new resistance becomes 1/4 times.

Explanation:

Given,

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 mm

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 m2

We have to find the length of the wire.

Let the length of the wire be L

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 = 122.72 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.

hopefully its helped u dear :)