How can relatively denser object be made to float on the less dense fluid?

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How can relatively denser object be made to float on the less dense fluid?

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Find the density of metallic body which floats at the interface of mercury of sp.gr 13.6 and water such that 40 % of its volume is sub-merged in mercury and 60% in water.

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Find the position of centre of buoyancy for a wooden block of width 3.5 m and depth 1 m, when it floats horizontally in water. The density of wooden block id 850 kg/m3 and its length 7.0 m.

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A wooden cylinder of sp.gr. = 0.6 and circular in cross-section is required to float in oil(sp.gr. = 0.90). Find the L/D ratio for the cylinder to float with its longitudinal axis vertical in oil, where L is the height of cylinder and D is its diameter.

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In a U-tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s^2).

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In a stationary fluid, how does the local pressure of the fluid vary?

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In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m^3, ρwater = 1000 kg/m^3).

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In case of spherically shaped bodies of uniform mass distribution and completely immersed in fluid and floating, the centre of buoyancy coincides with centre of gravity.

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The floating body is said to be in neutraL equilibrium if the metacentre is above the centre of gravity.

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In the manometer given above, 2 immiscible fluids mercury (ρ = 13600 kg/m^3) and water (ρ = 1000 kg/m^3) are used as manometric fluids. The water end is exposed to atmosphere (100 kPa) and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. Ignore the width of the manometer tube and the radius of curvature. The value of h is found to be 9.45 m. The height of the mercury column is given to be 75 cm. Find the gauge pressure of the gas. (g = 9.8 m/s^2)

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The floating body is said to be in unstable equilibrium if the metacentre is below the centre of gravity.

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A simple U-tube manometer can measure negative gauge pressures.

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The principle of floatation of bodies is based on the premise of

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Both ends of a U-tube manometer are exposed to the atmosphere. There exists a possibility that the height difference of the manometer is non-zero. True or False?

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A student wants to find the absolute pressure of water at a point below the surface of water. He has a barometer and a manometer pressure gauge. The barometer reads 1.3152 bar where as the manometer pressure gauge reads 0.3152 bar. What is the absolute pressure? (Assume that pressure at one end of the manometer is atmospheric.)

asked Jan 30, 2022 in Fluid Mechanics by Charlotte (118k points)

How can relatively denser object be made to float on the less dense fluid?

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