Description

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

QUESTION: 1

Which one of the following is the unit of specific weight?

Solution:

Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,

γ = w / v

Thus, unit of is N / m^{3}.

QUESTION: 2

The specific gravity of a liquid has

Solution:

Explanation: The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it unitless.

QUESTION: 3

The specific volume of a liquid is the reciprocal of

Solution:

Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).

v = v⁄m = 1 / m⁄v = 1⁄p

where p is the mass density.

QUESTION: 4

** Match the following physical quantities in Group 1 with their dimensions in Group 2.**

**1. Mass Density (p) A. [M ^{0} L^{ 0 }T ^{0 }]**

2. Specific Gravity B. [M L^{– 1 }T^{ – 1 }]

3. Specific Volume (v) C. [M^{1 }L^{ -3 }T ^{0}^{ }]

4. Specific Weight (γ) D. [M^{-1} L^{3} T ^{0 }]

5. Dynamic viscosity (μ) E. [M L^{– 2 }T ^{– 2 }]

Solution:

Explanation:

-Mass Density(p) is defined as the mass(m) per unit volume(V ), i.e.,

[p] = [m]/[v] = [m] /[L^{3}] = [ML^{-3}]

-The specific gravity of a liquid is the ratio of two similar quantities (densities) which makes it dimensionless.

-Specific volume(v) is defined as the volume(V ) per unit mass(m). Thus,

[v] = [V]/[m] = [L^{3}]/[M] = [M^{-1}L^{3}].

-Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,

QUESTION: 5

**Which of the following forces generally act on fluid while considering fluid dynamics?**

**1. Viscous force
2. Pressure force
3. Gravity force
4. Turbulent force
5. Compressibility force**

Solution:

QUESTION: 6

**Which property of the fluid offers resistance to deformation under the action of shear force?**

Solution:

QUESTION: 7

**The study of force which produces motion in a fluid is called as**

Solution:

Explanation: Fluid dynamics is the study of forces responsible for the motion of fluids.

QUESTION: 8

**The specific weight of the fluid depends upon**

Solution:

QUESTION: 9

Two fluids 1 and 2 have mass densities of p1 and p2 respectively. If p1 > p2, which one of the following expressions will represent the relation between their specific volumes v1 and v2?

Solution:

Explanation: Specific volume(v) is defined as the volume(V ) per unit mass(m).

v = v⁄m = 1 / m⁄v = 1⁄p

where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2

will be represented by v1 < v2.

QUESTION: 10

A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific weight of the liquid will be

Solution:

Explanation: Specific weight(γ) is defined as the weight(w) per unit volume(V ), i.e.,

γ = w⁄V

Thus, γ = 6.5 ⁄10^{-3} N ⁄ m^{3} = 6.5 kN/m^{3}.

### Fluid Properties

Doc | 4 Pages

### Fluid Properties

Video | 27:12 min

### Fluid Properties - Fluid Mechanics, GATE

Doc | 14 Pages

### Fluid | Module 1 | Properties of Fluid | Part 1 (Lecture 2)

Video | 32:54 min

- Test: Fluid Properties - 1
Test | 10 questions | 20 min

- Fluid Properties - 1
Test | 10 questions | 30 min

- Test: Fluid Properties - 2
Test | 10 questions | 20 min

- Test: Fluid Mechanics- 1
Test | 25 questions | 50 min

- Fluid Properties - 2
Test | 10 questions | 30 min