Question:
An open-ended U-tube of uniform cross-sectional area contains water
\((density\) \(10^{3} \,kg\, m ^{-3} )\). Initially the water level stands at \(0.29 \,m\) from the bottom in each arm Kerosene oil (a water-immiscible liquid) of density \(800 \,kg\, m ^{-3}\) is added to the left arm until its length is \(0.1 \,m\), as shown in the schematic figure below .The ratio \(\left(\frac{ h _{1}}{ h _{2}}\right)\) of the heights of the liquid in the two arms is
An open-ended U-tube of uniform cross-sectional area contains water
\((density\) \(10^{3} \,kg\, m ^{-3} )\). Initially the water level stands at \(0.29 \,m\) from the bottom in each arm Kerosene oil (a water-immiscible liquid) of density \(800 \,kg\, m ^{-3}\) is added to the left arm until its length is \(0.1 \,m\), as shown in the schematic figure below .The ratio \(\left(\frac{ h _{1}}{ h _{2}}\right)\) of the heights of the liquid in the two arms is
\((density\) \(10^{3} \,kg\, m ^{-3} )\). Initially the water level stands at \(0.29 \,m\) from the bottom in each arm Kerosene oil (a water-immiscible liquid) of density \(800 \,kg\, m ^{-3}\) is added to the left arm until its length is \(0.1 \,m\), as shown in the schematic figure below .The ratio \(\left(\frac{ h _{1}}{ h _{2}}\right)\) of the heights of the liquid in the two arms is
Updated On: Jul 13, 2024
- \(\frac{15}{14}\)
- \(\frac{35}{33}\)
- \(\frac{7}{6}\)
- \(\frac{5}{4}\)
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The Correct Option is B
Solution and Explanation
The Correct Option is (B): \(\frac{35}{33}\)
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Concepts Used:
Hydrostatics
A branch of physics that deals with the characteristics of fluids at rest, particularly with the pressure in a fluid or exerted by a fluid (gas or liquid) on an immersed body is called Hydrostatics.
Hydrostatic Pressure:
Hydrostatic pressure is proportional to the depth measured from the surface as the weight of the fluid increases when a force is applied in a downward direction.
Formula for Hydrostatic Pressure:
