Physics - Fluid Question with Solution | TestHub

It is given that

$A_1=10 {\mathrm{cm}}^2; A_2=5 {\mathrm{cm}}^2$

$∆P=3 \mathrm{N} {\mathrm{m}}^{-2}$

By continuity equation,

$$\begin{gathered} A_1{v}_1=A_2{v}_2 \\ \Rightarrow v_1=\frac{A_2{v}_2}{A_1} ...\left(i\right) \end{gathered}$$

By Bernoulli's equation,

$\frac{1}{2}\rho {v_2}^2-\frac{1}{2}\rho {v_1}^2=∆P ...\left(ii\right)$

Using equation (i) in (ii)

$$\begin{gathered} ∆P=\frac{\rho }{2}({v^2}_2-{\left(\frac{A_2{v}_2}{A_1}\right)}^2) \\ \Rightarrow ∆P=\frac{\rho {v^2}_2}{2}\left(1-{\left(\frac{A_2}{A_1}\right)}^2\right) \\ \Rightarrow 3=\frac{1.25\times {10}^3\times {v^2}_2}{2}\left(1-\frac{1}{4}\right) \\ \Rightarrow {v^2}_2=\frac{6\times 4}{1.25\times {10}^3\times 3} \\ \Rightarrow v_2=8\times {10}^{-2} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

So, the discharge rate is,

$$\begin{gathered} A_2{v}_2=x\times {10}^{-5} \\ \Rightarrow x=\frac{5\times {10}^{-4}\times 8\times {10}^{-2}}{{10}^{-5}}=4 \end{gathered}$$