# Fluid Mechanics Questions and Answers – Dimensional Homogenity

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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]

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________
b) Newton’s second law
c) Newtons first law
d) Newton’s third law

Explanation: Newton’s 2nd 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/s2.

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

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.
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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

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

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

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

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]

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

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

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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