Question:
A tank contains 140 cm height of water at the bottom and an oil of density \(900\,kg\,m^{-3}\) to a height of 200 cm above water. If the liquids are immiscible, then the initial velocity of efflux of water through a small opening at the bottom of the tank is:
A tank contains 140 cm height of water at the bottom and an oil of density \(900\,kg\,m^{-3}\) to a height of 200 cm above water. If the liquids are immiscible, then the initial velocity of efflux of water through a small opening at the bottom of the tank is:
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For layered liquids:
\[
h_{eq}
=
\sum \frac{\rho_i}{\rho_w}h_i.
\]
Convert every liquid column into equivalent water head.
Updated On: Jun 18, 2026
- \(12\,ms^{-1}\)
- \(6\,ms^{-1}\)
- \(4\,ms^{-1}\)
- \(8\,ms^{-1}\)
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The Correct Option is D
Solution and Explanation
Concept:
The pressure at the opening is due to both the water column and the oil column.
Equivalent water head is
\[
h_{eq}
=
h_w+\frac{\rho_o}{\rho_w}h_o.
\]
Efflux velocity:
\[
v=\sqrt{2gh_{eq}}.
\]
Step 1: Compute equivalent water head.
\[ h_w=1.4m, \qquad h_o=2m. \] \[ \rho_o=900, \qquad \rho_w=1000. \] \[ h_{eq} = 1.4+\frac{900}{1000}(2). \] \[ = 1.4+1.8. \] \[ =3.2m. \]
Step 2: Apply Torricelli's theorem.
\[ v=\sqrt{2gh}. \] \[ =\sqrt{2\times10\times3.2}. \] \[ =\sqrt{64}. \] \[ =8\,ms^{-1}. \] Hence \[ \boxed{8\,ms^{-1}}. \]
Step 1: Compute equivalent water head.
\[ h_w=1.4m, \qquad h_o=2m. \] \[ \rho_o=900, \qquad \rho_w=1000. \] \[ h_{eq} = 1.4+\frac{900}{1000}(2). \] \[ = 1.4+1.8. \] \[ =3.2m. \]
Step 2: Apply Torricelli's theorem.
\[ v=\sqrt{2gh}. \] \[ =\sqrt{2\times10\times3.2}. \] \[ =\sqrt{64}. \] \[ =8\,ms^{-1}. \] Hence \[ \boxed{8\,ms^{-1}}. \]
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