Glycerine of density \(1.25 \times 10^3 \mathrm{~kg} / \mathrm{m}^{3^4}\) is flowing in conical shaped horizontal pipe. Crosssectional area of the pipe at its both ends is \(10 \mathrm{~cm}^2\) and \(5 \mathrm{~cm}^2\) respectively. Pressure difference at both the ends is \(3 \mathrm{~N} / \mathrm{m}^2\). Rate of flow of liquid in the pipe is

Using Bernoulli's equation,
\(\begin{aligned}
& P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2+\rho g h_2 \\
& P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\
& P_1-P_2=\frac{1}{2}\left(\rho v_2^2-\rho v_1^2\right) \\
& v_2^2-v_1^2=\frac{2 \times 3}{1.25 \times 10^3}=4.8 \times 10^{-3}
...(i)\end{aligned}\)
From equation of continuity,
\(\begin{aligned}
& \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 \\
& \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{5}{10}=0.5 \\
& \mathrm{v}_1=0.5 \mathrm{v}_2
..(ii)\end{aligned}\)
Substituting (ii) into (i),
\(\begin{aligned} 0.75 \mathrm{v}_2^2=4.8 \times 10^{-3} \Rightarrow \mathrm{v}_2 & =0.08 \mathrm{~m} / \mathrm{s} \\ \text { Rate of flow of glycerine } & =A_2 \mathrm{v}_2 \\ & =5 \times 10^{-4} \times 0.08 \\ & =4 \times 10^{-5} \mathrm{~m}^3 / \mathrm{s}\end{aligned}\)