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  • Tanks A and B open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth ' h ' from the surface of the liqu

Tanks A and B open at the top contain two different liquids upto certain height in them. A hole is made to the wall of each tank at a depth ' h ' from the surface of the liquid. The area of the hole in A is twice that of in B. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is:

Option: 1

\frac{2}{1}


Option: 2

\frac{3}{2}


Option: 3

\frac{2}{3}


Option: 4

\frac{1}{2}


Answers (1)

best_answer

- According to the equation of continuity \mathrm{A_1 v_1=A_2 v_2}, we have

\mathrm{ \frac{v_1}{v_2}=\frac{A_2}{A_1}=\frac{2 A_1}{A_1}=2 \quad\left(\because A_2=2 A_1\right) }
Now mass flux = rate of mass of liquid flowing per unit area

\mathrm{ \begin{aligned} & =\frac{\text { mass }}{\text { area }} \div \text { time } \\\\ & =\frac{\text { mass of liquid }}{\text { area } \times \text { time }} \\\\ & =\frac{\text { mass }}{\text { volume }} \times \frac{\text { distance }}{\text { time }} \\\\ & =\text { density } \times \text { velocity of flow } \\ & =\rho v \end{aligned} }
Therefore, \mathrm{m_1=\rho_1 v_1 \, \, and \, \, m_2=\rho_2 v_2. \, \, Given \, \, m_1=m_2.} Hence

or     \mathrm{ \begin{aligned} \rho_1 v_1 & =\rho_2 v_2 \\\\ \frac{\rho_1}{\rho_2} & =\frac{v_2}{v_1}=\frac{1}{2} \quad\left(\because \frac{v_1}{v_2}=2\right) \end{aligned} }

Posted by

Divya Prakash Singh

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