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 $..............$.
JEE MAIN 2023, Diffcult
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$\Delta P = P _1- P _2=3 N / m ^2 \text { (given) }$
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 $x=4$
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