Explanation

(b) Initial elastic potential energy
$U_{1} = \frac{1}{2}\ F\mathrm{\Delta}l = \frac{1}{2} = \frac{1}{2}\ \times (100 \times 1000) \times \left( 1.59 \times 10^{- 3} \right) = \ 79.5\ J$
Let $\mathrm{\Delta}l_{1},$ be the elongation in the rod when stretching force is increased by, 200N, Since, $\mathrm{\Delta}l = \frac{F}{\pi r^{2}}\ \times \frac{l}{Y}\ ;so,\ \mathrm{\Delta}l\ \propto F\ $
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\mathrm{\Delta}l_{1}}{\mathrm{\Delta}l} = \frac{F_{1}}{F} = \ \frac{100 + 200}{100} = 3\ $
Or $\mathrm{\Delta}l_{1} = 3\mathrm{\Delta}l = 3\ \times 1.59\ \times 10^{- 3}m = 4.77 \times 10^{- 3}\ m$
Final elastic potential energy is
$U_{1} = \frac{1}{2}F_{1}\mathrm{\Delta}l_{1} = \frac{1}{2} \times \left( 300 \times 10^{3} \right) \times \left( 4.77 \times 10^{- 3} \right) = 715.5\ J\ $
Increase in elastic potential energy
= 715.5-79.5 = 636.0 J