Glycerine of density \(1.25 \times 10^3\,kg\,m ^{-3}\) is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is \(10\,cm ^2\) and \(5\,cm ^2\) and pressure drop across its length is \(3\,Nm ^{-2}\). The rate of flow of glycerine through the pipe is \(x \times 10^{-5} m ^3 s ^{-1}\). The value of \(x\) is \(..............\).

By continuity \(eq ^{ n }\)

\(A _1 v _1= A _2 v _2\)

\(\therefore \quad v_1=\frac{A_2}{A_1} v_2-(1)\)

By Bernoulli's equation.

\(P _1+\frac{1}{2} \rho v _1{ }^2= P _2+\frac{1}{2} \rho v _2{ }^2\)

\(P _1- P _2=\frac{1}{2} \rho\left( v _2^2- v _1^2\right)\)

\(\Delta P =\frac{1}{2} \rho\left( v _2^2-\frac{ A _2^2}{ A _1^2} v _2^2\right)\)

\(\Delta P =\frac{1}{2} \rho\left[1-\left(\frac{ A _2}{ A _1}\right)^2\right] v _2^2\)

\(3=\frac{1}{2} \times 1.25 \times 10^3\left[1-\left(\frac{5}{10}\right)^2\right] v _2^2\)

\(3=\frac{1}{2} \times 1.25 \times 10^3\left[1-\frac{1}{4}\right] v _2^2\)

\(3=\frac{1}{2} \times 1.25 \times 10^3 \times \frac{3}{4} v_2^2\)

So discharge rate \(= A _2 V _2\)

\(=5 \times 10^{-4} \times 8 \times 10^{-2}\)

\(=4 \times 10^{-5}\,m ^3 / s\)

Correct ans is \(x=4\)