A liquid of density 750kg m ^ - 3 flows smoothly through a...

From the equation of continuity, ${v}_{2} = 2{v}_{1}$.

Applying the Bernoulli's Theorem,

{P}_{1} + \dfrac{ρ{{v}_{1}}^{2}}{2} = {P}_{2} + \dfrac{ρ4{\left({v}_{1}\right)}^{2}}{2} ⇒ {P}_{1} - {P}_{2} = \dfrac{3ρ{{v}_{1}}^{2}}{2}$</p><p>$⇒ 4500 = \dfrac{3 \times 750 \times {{v}_{1}}^{2}}{2} ⇒ {v}_{1} = 2m{s}^{- 1}$</p><p>Now volume flow rate,$Q = A{v}_{1} = 1 . 2 \times {10}^{- 2} \times 2 = 24 \times {10}^{- 3}{m}^{3}{s}^{- 1}