Applying continuity equation $=>\frac{{0\mathrm{.}25}^2}{4}{v}_1=\frac{{0\mathrm{.}5}^2}{4}{V}_2$

$=>V_1=4V_2$

putting in $(1)$

$65=\frac{0\mathrm{.}5v_1^2}{2g}+\frac{0\mathrm{.}011500v_1^2}{2g0\mathrm{.}25}+\frac{g}{16}\frac{v_1^2}{2g}+\frac{0\mathrm{.}0081500v_1^2}{162g0\mathrm{.}5}$

$+\frac{V_1^2}{16*2g}$

$=>V_1=4\mathrm{.}5126\frac{m}{sec.\text{R}\text{a}\text{t}\text{e}of\text{f}\text{l}\text{o}\text{w}Q}=A_1{V}_1$

$Q=\frac{d_1^2}{4}{V}_1=\frac{{0\mathrm{.}25}^2}{4}4\mathrm{.}5126=0\mathrm{.}22\frac{m^3}{sec.}$

CAN SOMEONE PLEASE EXPLAIN THESE STEPS IN MORE DETAIL