Explanation

(a) $\mathrm{\Delta}L = \frac{FL}{AY}$
Because, wires of the same material are stretched by the same load. So, F and Y will be constant.
$\therefore\ \ \ \ \ \ \ \ \ $ $\mathrm{\Delta}L \propto \frac{L}{{\pi r}^{2}}$
$\mathrm{\Delta}L_{1} = \frac{100}{\pi \times (1 \times 10^{- 3})^{2}} = \frac{100}{\pi \times 10^{- 6}}$
$= \frac{100}{\pi} \times 10^{- 6}$
$\therefore\mathrm{\Delta}L_{2} = \frac{200}{\pi \times (3 \times 10^{- 3})^{2}} = \frac{200}{\pi \times 9 \times 10^{- 6}}$
$= \frac{22.2}{\pi} \times 10^{6}$
$\therefore\mathrm{\Delta}L_{3} = \frac{300}{\pi \times (3 \times 10^{- 3})^{2}} = \frac{100}{\pi \times {9 \times 10}^{- 6}}$
$= \frac{33.3}{\pi} \times 10^{6}$
$\therefore\mathrm{\Delta}L_{4} = \frac{400}{\pi \times (4 \times 10^{- 3})^{2}} = \frac{400}{\pi \times {16 \times 10}^{- 6}}$
$= \frac{25}{\pi} \times 10^{6}$
We can see that, L=100 cm and r = 1mm will elongate most.