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

1. Which among the following is not a fundamental dimension?

a) [L]

b) [M]

c) [T]

d) [kg]

View Answer

Explanation: It is essential to adopt a consistent dimensional quantity. Thus, we adopt a basic form to categorize dimension quantities. For this purpose, we adopt a comparison of the quantities in SI or MKS units.

2. The fundamental dimensional quantities are related by________

a) Avagadaro’s law

b) Newton’s second law

c) Newtons first law

d) Newton’s third law

View Answer

Explanation: Newton’s 2

^{nd}law is the most suitable one for determining the dimensional quantities. We know that, F=ma. Where F = Force , m = mass in kg, and a= acceleration in m/s

^{2}.

3. Force can be written as______

a) [M][L][T]^{-2}

b) [M][L][T]^{2}

c) [M][L][T]

d) [M][L][T]^{3}

View Answer

Explanation: Force can be written dimensionally by [F]= [M][L][T]

^{-2}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L][T]

^{-2}.

4. How can we write power using the MLT system?

a) [M][L][T]^{-2}

b) [M][L]^{2}[T]^{3}

c) [M][L][T]

d) [M][L][T]^{3}

View Answer

Explanation: Power can be written dimensionally by [M][L]2[T]

^{3}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L]

^{2}[T]

^{3}.

5. How can we write dynamic viscosity using the MLT system?

a) [M][L][T]^{-2}

b) [M][L]^{2}[T]^{3}

c) [M][L]^{-1}[T]^{-1}

d) [M][L][T]^{3}

View Answer

Explanation: Dynamic viscosity can be written dimensionally by [M][L]-1[T]

^{-1}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L]

^{-1}[T]

^{-1}.

6. How can we write kinematic viscosity using the MLT system?

a) [M][L][T]^{-2}

b) [M]^{0}[L]^{2}[T]^{-1}

c) [M][L]^{-1}[T]^{-1}

d) [M][L][T]^{3}

View Answer

Explanation: Kinematic viscosity can be written dimensionally by [M]0[L]2[T]

^{-1}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M]

^{0}[L]

^{2}[T]

^{-1}.

7. How can we write momentum using the MLT system?

a) [M][L][T]^{-2}

b) [M]^{0}[L]^{2}[T]^{-1}

c) [M][L][T]^{-1}

d) [M][L][T]^{3}

View Answer

Explanation: Momentum can be written dimensionally by [M][L][T]

^{-1}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L][T]

^{-1}.

8. How can we write specific weight using the FLT system?

a) [F]

b) [F][T]

c) [F][L][T]

d) [L]

View Answer

Explanation: Specific can be written dimensionally by [F]. This is by adopting the basic SI or MKS units (FLT system). Where, [F] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [F].

9. How can we write specific mass using the MLT system?

a) [M][L][T]^{-2}

b) [M]^{0}[L]^{2}[T]^{-1}

c) [M][L]^{-3}[T]^{0}

d) [M][L][T]^{3}

View Answer

Explanation: Specific mass can be written dimensionally by [M][L]-3[T]

^{0}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L]

^{-3}[T]

^{0}.

10. How can we write energy using the MLT system?

a) [M][L]^{2}[T]^{2}

b) [M]^{0}[L]^{2}[T]^{-1}

c) [M][L]^{-3}[T]^{0}

d) [M][L][T]^{3}

View Answer

Explanation: Energy or work can be written dimensionally by [M][L]2[T]

^{2}. This is by adopting the basic SI or MKS units. Where, [M] is the mass, [L] is the length and [T] is the time. Thus, the correct option is [M][L]

^{2}[T]

^{2}.

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