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# MCQs on Dimensional Homogenity

Which among the following is not a fundamental dimension?

a) [L]

b) [M]

c) [T]

d) [kg]

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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.

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

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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

Force can be written as______

a) [M][L][T]-2

b) [M][L][T]2

c) [M][L][T]

d) [M][L][T]3

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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.

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

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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.

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

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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.

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

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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.

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

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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.

How can we write specific weight using the FLT system?

a) [F]

b) [F][T]

c) [F][L][T]

d) [L]

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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].

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

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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.

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

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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.