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  • Glycerin of density <img alt="1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.25%20%5Ctimes%2010%5E3%20%5Cmathrm%7B%7Ekg%7D%20%5Cmathrm%7B%7Em

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

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\Delta P=P_1-P_2=3 \mathrm{~N} / \mathrm{m}^2 (given)
By continuity equation
\mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2
$$ \therefore \mathrm{v}_1=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \mathrm{v}_2\, \, \, \, \, \, .....(1)
By bernoulli's equation
$$ \begin{aligned} & \mathrm{P}_1+\frac{1}{2} \rho \mathrm{v}_1^2=\mathrm{P}_2+\frac{1}{2} \rho \mathrm{v}_2^2 \\ & \mathrm{P}_1-\mathrm{P}_2=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\mathrm{v}_1^2\right) \\ & \Delta \mathrm{P}=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\frac{\mathrm{A}_2^2}{\mathrm{~A}_1^2} \mathrm{v}_2^2\right) \\ & \Delta \mathrm{P}=\frac{1}{2} \rho\left[1-\left(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right)^2\right] \mathrm{v}_2^2 \\ & 3=\frac{1}{2} \times 1.25 \times 10^3\left[1-\left(\frac{5}{10}\right)^2\right) \mathrm{v}_2^2 \\ & 3=\frac{1}{2} \times 1.25 \times 10^3\left[1-\frac{1}{4}\right] \mathrm{v}_2^2 \\ & 3=\frac{1}{2} \times 1.25 \times 10^3 \times \frac{3}{4} \mathrm{v}_2^2 \end{aligned}

\begin{aligned} & \therefore \mathrm{v}_2=8 \times 10^{-2} \mathrm{~m} / \mathrm{s} \\ & \text { So discharge rate }=\mathrm{A}_2 \mathrm{v}_2 \\ & =5 \times 10^{-4} \times 8 \times 10^{-2} \\ & =4 \times 10^{-5} \mathrm{~m}^3 / \mathrm{s} \end{aligned}

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Anam Khan

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