We are given with

\(\displaystyle{P}_{{{1}}}={1.2}\times{10}^{{{4}}}{P}{a}\)

\(\displaystyle{P}_{{{2}}}={1.15}\times{10}^{{{4}}}{P}{a}\)

\(\displaystyle{v}_{{{1}}}={30}\) cm/s

=0.3 m/s

specific gravity RD = 1.06

\(\displaystyle{P}_{{{b}}}={R}{D}\times\) density of water

\(\displaystyle={1.06}\times{1000}{k}\frac{{g}}{{m}^{{{3}}}}\)

\(\displaystyle={1060}{k}\frac{{g}}{{m}^{{{3}}}}\)

as the heights are equal we can write

\(\displaystyle{h}_{{{1}}}={h}_{{{2}}}\)

and according to the bernoulis equation we get

\(\displaystyle{P}_{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={P}_{{{1}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={\left({P}_{{{1}}}-{P}_{{{2}}}\right)}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{v}_{{{2}}}=\sqrt{{{\left[{2}\frac{{{P}_{{{1}}}-{P}_{{{2}}}}}{{p}}{b}\right]}+{{v}_{{{1}}}^{{{2}}}}}}\)

=.......m/s

=.......cm/s

so according to the equation of continuity we get

\(\displaystyle{A}_{{{1}}}{v}_{{{1}}}={A}_{{{2}}}{v}_{{{2}}}\)

\(\displaystyle{\left(\frac{{A}_{{{2}}}}{{A}_{{{1}}}}\right)}=\)......%

\(\displaystyle{P}_{{{1}}}={1.2}\times{10}^{{{4}}}{P}{a}\)

\(\displaystyle{P}_{{{2}}}={1.15}\times{10}^{{{4}}}{P}{a}\)

\(\displaystyle{v}_{{{1}}}={30}\) cm/s

=0.3 m/s

specific gravity RD = 1.06

\(\displaystyle{P}_{{{b}}}={R}{D}\times\) density of water

\(\displaystyle={1.06}\times{1000}{k}\frac{{g}}{{m}^{{{3}}}}\)

\(\displaystyle={1060}{k}\frac{{g}}{{m}^{{{3}}}}\)

as the heights are equal we can write

\(\displaystyle{h}_{{{1}}}={h}_{{{2}}}\)

and according to the bernoulis equation we get

\(\displaystyle{P}_{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={P}_{{{1}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={\left({P}_{{{1}}}-{P}_{{{2}}}\right)}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}\)

\(\displaystyle{v}_{{{2}}}=\sqrt{{{\left[{2}\frac{{{P}_{{{1}}}-{P}_{{{2}}}}}{{p}}{b}\right]}+{{v}_{{{1}}}^{{{2}}}}}}\)

=.......m/s

=.......cm/s

so according to the equation of continuity we get

\(\displaystyle{A}_{{{1}}}{v}_{{{1}}}={A}_{{{2}}}{v}_{{{2}}}\)

\(\displaystyle{\left(\frac{{A}_{{{2}}}}{{A}_{{{1}}}}\right)}=\)......%