An ideal fluid of density 800kg m ^ - 3 , flows smoothly through a...

From equation of continuity,

$a{v}_{1} = \dfrac{a}{2}{v}_{2} ⇒ {v}_{2} = 2{v}_{1}$

From Bernoulli's theorem,

${P}_{1} + \dfrac{1}{2}ρ{{v}_{1}}^{2} + ρg{h}_{1} = {P}_{2} + \dfrac{1}{2}ρ{{v}_{2}}^{2} + ρg{h}_{2} ⇒ {P}_{1} - {P}_{2} = ρ\left[g\left({h}_{2} - {h}_{1}\right) + \dfrac{1}{2}\left({{v}_{2}}^{2} - {{v}_{1}}^{2}\right)\right] ⇒ 4100 = 800\left[10\left(- 1\right) + \dfrac{1}{2}3{{v}_{1}}^{2}\right] ⇒ \dfrac{4100}{800} + 10 = \dfrac{3}{2}{{v}_{1}}^{2} ⇒ {v}_{1} = \sqrt{\dfrac{121}{4 \times 3}} = \dfrac{\sqrt{363}}{6}m{s}^{- 1}$

Therefore, $x = 363$.