Question

An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Answer 1

The resistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = ρ$\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π$0.25^{2}$
A1=π$0.50^{2}$

we get. R2 = 16R