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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Fluid Mechanics for Mechanical Engineering  >  Test: Fluid Properties - 2 - Mechanical Engineering MCQ

Test: Fluid Properties - 2 - Mechanical Engineering MCQ


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10 Questions MCQ Test Fluid Mechanics for Mechanical Engineering - Test: Fluid Properties - 2

Test: Fluid Properties - 2 for Mechanical Engineering 2024 is part of Fluid Mechanics for Mechanical Engineering preparation. The Test: Fluid Properties - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Fluid Properties - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Fluid Properties - 2 below.
Solutions of Test: Fluid Properties - 2 questions in English are available as part of our Fluid Mechanics for Mechanical Engineering for Mechanical Engineering & Test: Fluid Properties - 2 solutions in Hindi for Fluid Mechanics for Mechanical Engineering course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Fluid Properties - 2 | 10 questions in 20 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study Fluid Mechanics for Mechanical Engineering for Mechanical Engineering Exam | Download free PDF with solutions
Test: Fluid Properties - 2 - Question 1
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 Calculate the specific weight and weight of 20dm3 of petrol of specific gravity 0.6. 

Detailed Solution for Test: Fluid Properties - 2 - Question 1

► Specific weight = Density * Acceleration due to gravity
= 0.6*1000*9.81 = 5886N/m3
► Weight = Volume*Specific Weight
= 5886*0.02 = 117.2N.

Test: Fluid Properties - 2 - Question 2
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If 200m3 of fluid has a weight of 1060N measured on the planet having acceleration due to gravity 6.625m/s2, what will be it’s specific volume? 

Detailed Solution for Test: Fluid Properties - 2 - Question 2

► Specific Volume is defined as (Volume/Mass of the object) = 1/density 

► Mass of object = Weight of object/ acceleration due to gravity = 1060/6.625 = 160 kg

► Density= 160/200 =0.8

► Specific Volume = 1/Density = 1/0.8 = 1.25 m3/kg

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Test: Fluid Properties - 2 - Question 3
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 For an incompressible fluid does density vary with temperature and pressure?

Detailed Solution for Test: Fluid Properties - 2 - Question 3

For an incompressible fluid, the change in density is negligible. Thus it does not change with temperature and pressure.

Test: Fluid Properties - 2 - Question 4
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 Specific gravity is what kind of property?
Detailed Solution for Test: Fluid Properties - 2 - Question 4

It is independent of quantity of matter present.

Test: Fluid Properties - 2 - Question 5
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 If there is bucket full of oil and bucket full of water and you are asked to lift them, which one of the two will require more effort given that volume of buckets remains same?
Detailed Solution for Test: Fluid Properties - 2 - Question 5

Density of water is more that oil. Hence, its weight for same volume of oil will also be higher. Therefore, more effort will be required.

Test: Fluid Properties - 2 - Question 6
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If the fluid has specific weight of 10N/m3 for a volume of 100dm3 on a planet which is having acceleration due to gravity 20m/s2 , what will be its specific weight on a planet having acceleration due to gravity 4m/s2? 
Detailed Solution for Test: Fluid Properties - 2 - Question 6

For same volume, specific weight is directly proportional to acceleration due to gravity.
Specific Weight = 4*10/20 = 2.

Test: Fluid Properties - 2 - Question 7
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 Should specific weight of incompressible fluid only be taken at STP?
Detailed Solution for Test: Fluid Properties - 2 - Question 7
  • Specific weight is inversely proportional to volume.
  • For incompressible fluid , variation of volume with temperature and pressure is negligible for practical consideration.

Therefore, specific weight remains constant.

Test: Fluid Properties - 2 - Question 8
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An instrument with air as fluid was involved in some experiment (specific volume was the characteristic property utilized) which was conducted during day in desert. Due to some reason experiment couldn’t be conducted during day and had to be conducted during night. However there were considerable errors in obtained values. What might be the reason of these errors? 
Detailed Solution for Test: Fluid Properties - 2 - Question 8

In Desert areas, the temperature at night is considerably lower than at day. Due to this air contracts at night. Hence, its specific volume changes. As specific volume was characteristic property utilized, results obtained showed error due to change in specific volume.

Test: Fluid Properties - 2 - Question 9
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A stone weighed 177 N on earth. It was dropped in to oil of specific gravity 0.8 on a planet whose acceleration due to gravity is 5m/s2. It displaced oil having weight of 100N. What was the volume of oil displaced by the stone? 
Detailed Solution for Test: Fluid Properties - 2 - Question 9

Volume Displaced = Oil Displaced/(Specific Gravity * Water Density * Acceleration due to gravity) = 100/ (0.8*1000*5).

Test: Fluid Properties - 2 - Question 10
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An compressible fluid’s specific gravity was measured on earth, on a planet having acceleration due to gravity 5.5 times that of earth, and in space at STP. Where will it be having highest value?
Detailed Solution for Test: Fluid Properties - 2 - Question 10

Specific gravity is a characteristic property of fluid and is independent of external conditions.

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