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Chapter:

buoyancy-and-floatation

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

L/D<9/16 L/D<3/4 L/D<2/3 None of the mentioned

L/D<3/4

By principle of buoyancy,
Weight of cylinder = Weight of water displaced
h=2L/3
AG=L/2
AB=L/3
BG=AG-AB=L/6
GM= Metacentric height=I/? ? BG=3D2/32L-L/6
For stable equilibrium, GM should be positive
GM>0
i.e L/D<3/4.

Hint Answer

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

True False

True

The height of manometric fluid in a U-tube manometer in test column would fall if re is a positive gauge pressure. The height would increase if re is a negative gauge pressure. It is possible to measure negative gauge pressures with a U-tube manometer. However, negative pressure cannot fall below -1 Bar.

Hint Answer

3. A uniform body of size 4 m long * 2.5 m wide * 1.5 m deep floats in water. What is weight of body if depth of immersion is 1 m ?

147.1 kN 294.3 kN 73.5 kN 588.6 kN

147.1 kN

Weight of Body = Weight of water displaced
= ?*g*Volume of displaced water=9.81*1000*4*2.5*1.5=147.1kN.

Hint Answer

4. Proper explanation for metacentre is:

Point at which line of action of force meets the normal axis of body when it is given angular displacement Intersection of line passing through new centre of buoyancy and centre of gravity. point about which body starts oscillating when it is given small angular displacement All of the mentioned

All of the mentioned

All of above explanation are apt.

Hint Answer

5. In stable equilibrium for completely submerged bodies what is relation between forces?

Buoyancy force= Weight of body,the centre of buoyancy is below the centre of gravity. Buoyancy force=Weight of body, the centre of buoyancy is above the centre of gravity. Buoyancy force < Weight of body None of the mentioned

Buoyancy force=Weight of body, the centre of buoyancy is above the centre of gravity.

Fb=W and centre of buoyancy is above centre of gravity.

Hint Answer

6. In a U-tube mercury manometer, one end is exposed to atmosphere and or end is connected to a pressurized gas. The gauge pressure of gas is found to be 40 kPa. Now, we change manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m³, ρwater = 1000 kg/m³).

1260% 92.64 % Remains unchanged (0%) 13.6%

1260%

Since gauge pressure remains same ?*(h₂ ? h₁) = constant. The height difference in mercury manometer is 0.30 m and that in a water manometer is 4.08 m. Percent change is thus, 1260%. Be careful about denominator used for computing percent change.

Hint Answer

7. A body of dimensions 2.7 m * 3.8 m * 2.5 m, weighs 2500 N in water.Find its weight in air.

254.12 kN 508.25 kN 101.65 kN 127.06 kN

254.12 kN

Weight of stone in air = Weight of water displaced+Weight of stone in water
= 9.81*1000*2.7*3.8*2.5+2500=254.12 kN.

Hint Answer

8. For a completely immersed body, metacentric height is always zero.

True False

False

The metacentric height may or may not be zero as metacentre will not always coincide with centre of gravity.

Hint Answer

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

With depth only In the horizontal direction only Both with depth and along horizontal direction Neither with depth nor along horizontal direction

With depth only

According to Pascal?s law, local pressure of a fluid is same in all directions. Hence, pressure won?t vary along x and y direction. The local pressure will increase with an increase in depth due to extra weight of water column above that point.

Hint Answer

10. As a balloon rises in air its volume increases, at end it acquires a stable height and cannot rise any furr.

True False

True

As balloon rises in air, pressure acting on it reduces and refore its volume increases. Also, a rising balloon ceases rising when it and displaced air are equal in weight.

Hint Answer

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All Chapters

View all Chapter and number of question available From each chapter from Fluid-Mechanics-Mcqs

Properties of Fluids

Properties of Fluids

Pressure and Its Measurement

Pressure and Its Measurement

Hydrostatic Forces on Surfaces

Hydrostatic Forces on Surfaces

Buoyancy and Floatation

Buoyancy and Floatation

Kinematics of Flow and Ideal Flow

Kinematics of Flow and Ideal Flow

Fluid Dynamics

Orifices, Mouthpieces, Notches and Weirs

Orifices, Mouthpieces, Notches and Weirs

Viscous Flow & Turbulent Flow

Viscous Flow & Turbulent Flow

Flow Through Pipes

Flow Through Pipes

Indicator Diagrams

Indicator Diagrams

Dimensional and Model Analysis

Dimensional and Model Analysis

Boundary Layer Flow & Forces on Sub-merged Bodies

Boundary Layer Flow & Forces on Sub-merged Bodies

Compressible Flow

Compressible Flow

Flow in Open Channels

Flow in Open Channels

Non Uniform Flow in Channels

Non Uniform Flow in Channels

Impact of Jets and Jet Propulsion

Impact of Jets and Jet Propulsion

Topics

This Chapter buoyancy-and-floatation consists of the following topics

Buoyancy and Floatation

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

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

A uniform body of size 4 m long * 2.5 m wide * 1.5 m deep floats in water. What is weight of body if depth of immersion is 1 m ?

Proper explanation for metacentre is:

In stable equilibrium for completely submerged bodies what is relation between forces?

In a U-tube mercury manometer, one end is exposed to atmosphere and or end is connected to a pressurized gas. The gauge pressure of gas is found to be 40 kPa. Now, we change manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m³, ρwater = 1000 kg/m³).

A body of dimensions 2.7 m * 3.8 m * 2.5 m, weighs 2500 N in water.Find its weight in air.

For a completely immersed body, metacentric height is always zero.

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

As a balloon rises in air its volume increases, at end it acquires a stable height and cannot rise any furr.

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