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# Test: Buoyancy

## 10 Questions MCQ Test Fluid Mechanics | Test: Buoyancy

Description
This mock test of Test: Buoyancy for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 10 Multiple Choice Questions for Civil Engineering (CE) Test: Buoyancy (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Buoyancy quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Test: Buoyancy exercise for a better result in the exam. You can find other Test: Buoyancy extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1

### 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.

Solution:

Explanation: Weight of the block=ρ*g*Volume=850*9.81*7*3.5*1=204.29 kN
Volume of
water displaced= Weight of water displaced/weight density of water
= 20.825 m3.
h=20.825/3.5*7=0.85 m.

QUESTION: 2

### A stone weighs 450 N in air and 200 N in water. Compute the volume of stone.

Solution:

Explanation: Weight of water displaced=Weight of stone in air – Weight of stone in water
=250
Volume of water displaced=Volume of stone=250/9.81*1000=0.025 m3.

QUESTION: 3

### A stone weighs 650 N in air and 275 N in water. Compute its specific gravity.

Solution:

Explanation: Weight of water displaced=Weight of stone in air – Weight of stone in water
=375
Volume of water displaced=Volume of stone=375/9.81*1000=0.038 m3
Density of stone= mass/volume=650/9.81*0.038=1733 kg/m3
specific gravity= Density of stone/Density of water=1.73.

QUESTION: 4

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

Solution:

Explanation: 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.

QUESTION: 5

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.

Solution:

Explanation: Total Bouyant force=Force of bouyancy due to water+Force of bouyancy due to mercury
For equilibrium, Total bouyant force= Weiht of body
1000*9.81*0.6*V + 13.6*1000*9.81*0.4*V=ρ*g*V
ρ=6040 kg/m3.

QUESTION: 6

What is the principal cause of action of buoyant force on a body submerged partially or fully in fluid?

Solution:

Explanation: The principal cause of action of buoyant force on a body submerged partially or fully in fluid is the force equal in magnitude to the weight of the volume of displaced fluid.

QUESTION: 7

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

Solution:

Explanation: By changing the shape of an object it can be made to float on a fluid even if it is denser than that fluid. This principle is used in ship building.

QUESTION: 8

What happens to the buoyant force acting on the airship as it rises in the air?

Solution:

Explanation: Buoyant force acting on the airship decreases as it rises in the air as air at higher altitude becomes rarer and its density decreases.

QUESTION: 9

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

Solution:

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

QUESTION: 10

Submarines use principle of ‘neutral buoyancy’ to go into the water.

Solution:

Explanation: To dive, the submarine tanks are opened to allow air to exhaust, while the water flows in. When the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and hence will go down.