Glycerin of density 1 . 25 × 10 ^ 3 kg m ^ – 3 is flowing through...

It is given that

${A}_{1} = 10{cm}^{2}; {A}_{2} = 5{cm}^{2}$

$∆ P = 3N{m}^{- 2}$

By continuity equation,

${A}_{1}{v}_{1} = {A}_{2}{v}_{2} ⇒ {v}_{1} = \dfrac{{A}_{2}{v}_{2}}{{A}_{1}} . . . ( i )$

By Bernoulli's equation,

$\dfrac{1}{2}ρ{{v}_{2}}^{2} - \dfrac{1}{2}ρ{{v}_{1}}^{2} = ∆ P . . . ( ii )$

Using equation (i) in (ii)

$∆ P = \dfrac{ρ}{2} ( {v}^{2}2 - {\left(\dfrac{{A}_{2}{v}_{2}}{{A}_{1}}\right)}^{2} ) ⇒ ∆ P = \dfrac{ρ{v}^{2}2}{2}\left(1 - {\left(\dfrac{{A}_{2}}{{A}_{1}}\right)}^{2}\right) ⇒ 3 = \dfrac{1 . 25 \times {10}^{3} \times {v}^{2}2}{2}\left(1 - \dfrac{1}{4}\right) ⇒ {v}^{2}2 = \dfrac{6 \times 4}{1 . 25 \times {10}^{3} \times 3} ⇒ {v}_{2} = 8 \times {10}^{- 2}m{s}^{- 1}$

So, the discharge rate is,

${A}_{2}{v}_{2} = x \times {10}^{- 5} ⇒ x = \dfrac{5 \times {10}^{- 4} \times 8 \times {10}^{- 2}}{{10}^{- 5}} = 4$