Fluid Mechanics Questions and Answers – Bouyancy

This set of Fluid Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Bouyancy”.

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.
a) 0.95
b) 0.85
c) 1.05
d) 1.65
View Answer

Answer: b
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.

2. A stone weighs 450 N in air and 200 N in water. Compute the volume of stone.
a) .025 m3
b) .05 m3
c) .075 m3
d) None of the mentioned
View Answer

Answer: a
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.

3. A stone weighs 650 N in air and 275 N in water. Compute its specific gravity.
a) 1.73
b) 2.45
c) 3.46
d) 0.865
View Answer

Answer: a
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.
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4. A body of dimensions 2.7 m * 3.8 m * 2.5 m, weighs 2500 N in water.Find its weight in air.
a) 254.12 kN
b) 508.25 kN
c) 101.65 kN
d) 127.06 kN
View Answer

Answer: a
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.

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.
a) 6040 kg/m3
b) 12080 kg/m3
c) 24160 kg/m3
d) 3020 kg/m3
View Answer

Answer: a
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.
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6. What is the principal cause of action of buoyant force on a body submerged partially or fully in fluid?
a) Displacement of fluid due to submerged body
b) Development of force due to dynamic action
c) Internal shear forces mitigating external forces
d) None of the mentioned
View Answer

Answer: a
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.

7. How can relatively denser object be made to float on the less dense fluid?
a) By altering the shape.
b) By altering the forces acting on the object
c) By altering the shear forces acting on the object
d) None of the mentioned
View Answer

Answer: a
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.
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8. What happens to the buoyant force acting on the airship as it rises in the air?
a) Buoyant force increases
b) Buoyant force decreases
c) Buoyant force remains constant
d) Buoyant force first increases then shows decrease
View Answer

Answer: b
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.

9. As a balloon rises in the air its volume increases, at the end it acquires a stable height and cannot rise any further.
a) True
b) False
View Answer

Answer: a
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.
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10. Submarines use principle of ‘neutral buoyancy’ to go into the water.
a) True
b) False
View Answer

Answer: a
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.

Sanfoundry Global Education & Learning Series – Fluid Mechanics.

To practice all areas of Fluid Mechanics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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