Engineering Questions with Answers - Multiple Choice Questions
MCQs on Dimensional Homogenity
1 - Question
Which among the following is not a fundamental dimension?
a) [L]
b) [M]
c) [T]
d) [kg]
View Answer
d
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 - Question
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
b
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 - Question
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
a
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 - Question
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
b
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 - Question
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
c
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 - Question
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
b
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 - Question
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
c
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 - Question
How can we write specific weight using the FLT system?
a) [F]
b) [F][T]
c) [F][L][T]
d) [L]
View Answer
a
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 - Question
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
c
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 - Question
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
a
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.