$A$ copper wire has a diameter of $0.5\, mm$ and a resistivity of $1.6 \times 10^{-8}\, \Omega \, m$. What will be the length of this wire to make its resistance $10\, \Omega$? How much does the resistance change if the diameter is doubled?

(A) The resistance $(R)$ of a copper wire of length $l$ and cross-sectional area $A$ is given by the formula:
$R = \rho \frac{l}{A}$
Given:
Resistivity of copper,$\rho = 1.6 \times 10^{-8}\, \Omega \, m$
Diameter,$d = 0.5\, mm = 5 \times 10^{-4}\, m$
Resistance,$R = 10\, \Omega$
Area of cross-section,$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5 \times 10^{-4}}{2}\right)^2 = \pi \times 6.25 \times 10^{-8}\, m^2$
Calculating length $(l)$:
$l = \frac{R \times A}{\rho} = \frac{10 \times 3.14 \times 6.25 \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{196.25}{1.6} \approx 122.7\, m$
If the diameter is doubled,the new diameter $d' = 2d = 1.0\, mm = 10^{-3}\, m$.
Since $R \propto \frac{1}{d^2}$,if the diameter is doubled,the area $A$ becomes $4$ times larger.
Therefore,the new resistance $R' = \frac{R}{4} = \frac{10}{4} = 2.5\, \Omega$.