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**Quiz Description** :

**Name: **Dimensional Analysis of Physical Quantities mcq Test

**Subject: **Physics

**Topic: **Dimensional Analysis of Physical Quantities

**Questions: **15 Objective type

**Time Allowed: **10 minutes

**Important for: **11th & 12th school students, Engineering & Medical, M Sc / B Sc entrance exams etc.

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- Question 1 of 15
##### 1. Question

1 pointsA dimensionless quantity

CorrectWe know that angle is dimensionless quantity and it is measured in radians. Therefore a dimensionless quantity may have a unit.

IncorrectWe know that angle is dimensionless quantity and it is measured in radians. Therefore a dimensionless quantity may have a unit.

- Question 2 of 15
##### 2. Question

1 pointsWhich of the following physical quantity is dimensionless?

CorrectWe know that the angle, strain and specific gravity are the ratios of fundamental units. Therefore all of those are dimensionless quantities.

IncorrectWe know that the angle, strain and specific gravity are the ratios of fundamental units. Therefore all of those are dimensionless quantities.

- Question 3 of 15
##### 3. Question

1 pointsThe dimensional formula of angular velocity is

CorrectWe know that the angular velocity = [Angle/ Time]

Therefore dimensional formula of angular velocity = [Dimensions of angle / Dimensions of time]

= [M^{0}L^{0}T^{0}] / [M^{0}L^{0}T] = [M^{0}L^{0}T^{-1}]IncorrectWe know that the angular velocity = [Angle/ Time]

Therefore dimensional formula of angular velocity = [Dimensions of angle / Dimensions of time]

= [M^{0}L^{0}T^{0}] / [M^{0}L^{0}T] = [M^{0}L^{0}T^{-1}] - Question 4 of 15
##### 4. Question

1 pointsThe dimensions of impulse are

CorrectWe know that impulse = Force × Time.

Therefore dimensions of impulse = Dimensions of force × Dimensions of time

= [MLT^{-2}] [M^{0}L^{0}T] = [MLT^{-1}]IncorrectWe know that impulse = Force × Time.

Therefore dimensions of impulse = Dimensions of force × Dimensions of time

= [MLT^{-2}] [M^{0}L^{0}T] = [MLT^{-1}] - Question 5 of 15
##### 5. Question

1 pointsWhich of the following are the dimensions of coefficient of friction?

CorrectWe know that the coefficient of friction (

*µ*) = [Frictional force / Normal reaction].

Therefore dimensions of coefficient of friction = [Dimension of force / Dimension of reaction]

= [MLT^{-2}] / [MLT^{-2}] = [M^{0}LT^{0}]IncorrectWe know that the coefficient of friction (

*µ*) = [Frictional force / Normal reaction].

Therefore dimensions of coefficient of friction = [Dimension of force / Dimension of reaction]

= [MLT^{-2}] / [MLT^{-2}] = [M^{0}LT^{0}] - Question 6 of 15
##### 6. Question

1 pointsThe dimensions of energy are

CorrectWe know that the energy of a body is the capacity to do some work in joules.

Therefore dimensions of energy will be same as that of work/

Therefore dimensions of work = Dimensions of force × Dimensions of distance

= [MLT^{-2}] [M^{0}LT^{0}] = [ML^{2}T^{-2}]IncorrectWe know that the energy of a body is the capacity to do some work in joules.

Therefore dimensions of energy will be same as that of work/

Therefore dimensions of work = Dimensions of force × Dimensions of distance

= [MLT^{-2}] [M^{0}LT^{0}] = [ML^{2}T^{-2}] - Question 7 of 15
##### 7. Question

1 pointsWhich of the following physical quantity has the dimensions of [ML

^{2}T^{-3}]?CorrectWe know that power (

*P*) = [Work / Time]

Therefore dimensions of the power = [Dimensions of work / Dimensions of time] [ML^{2}T^{-2}] / [M^{0}L^{0}T]

= [ML^{2}T^{-3}]IncorrectWe know that power (

*P*) = [Work / Time]

Therefore dimensions of the power = [Dimensions of work / Dimensions of time] [ML^{2}T^{-2}] / [M^{0}L^{0}T]

= [ML^{2}T^{-3}] - Question 8 of 15
##### 8. Question

1 pointsDimensional formula for angular momentum is

CorrectWe know that angular momentum = Mass × Angular velocity × Radius.

Therefore dimensional formula for angular momentum

= Dimensions of mass × Dimensions of velocity × Dimensions of radius

= [ML^{0}T^{0}] [M^{0}LT^{-1}] [M^{0}LT^{0}] = [ML^{2}T^{-1}]IncorrectWe know that angular momentum = Mass × Angular velocity × Radius.

Therefore dimensional formula for angular momentum

= Dimensions of mass × Dimensions of velocity × Dimensions of radius

= [ML^{0}T^{0}] [M^{0}LT^{-1}] [M^{0}LT^{0}] = [ML^{2}T^{-1}] - Question 9 of 15
##### 9. Question

1 pointsThe dimensions of torque is

CorrectWe know that torque (τ) = Force × Distance

Therefore dimensions of torque

= Dimensions of force × Dimensions of distance

= [MLT^{-2}] [M^{0}LT^{0}] = [ML^{2}T^{-2}]IncorrectWe know that torque (τ) = Force × Distance

Therefore dimensions of torque

= Dimensions of force × Dimensions of distance

= [MLT^{-2}] [M^{0}LT^{0}] = [ML^{2}T^{-2}] - Question 10 of 15
##### 10. Question

1 pointsThe dimensions of universal gravitational constant are

CorrectWe know that gravitational constant (G)

= [Force × (Distance)^{2}] / (Mass)^{2}

= [Dimensions of Force × (Dimensions of Distance)^{2}] / (Dimensions of Mass)^{2}

= [MLT^{-2}] [M^{0}LT^{0}]^{2}/ [ML^{0}T^{0}]^{2}= [M^{-1}L^{3}T^{-2}]IncorrectWe know that gravitational constant (G)

= [Force × (Distance)^{2}] / (Mass)^{2}

= [Dimensions of Force × (Dimensions of Distance)^{2}] / (Dimensions of Mass)^{2}

= [MLT^{-2}] [M^{0}LT^{0}]^{2}/ [ML^{0}T^{0}]^{2}= [M^{-1}L^{3}T^{-2}] - Question 11 of 15
##### 11. Question

1 pointsWhat is the dimensional representation of Young’s modulus or modulus of elasticity?

CorrectWe know that Young’s modulus (

*Y*) = Stress / Strain.

Therefore dimensional representation of Young’s modulus (*Y*)

= Dimensions of stress / Dimensions of strain

= [ML^{-1}T^{-2}] / [M^{0}L^{0}T^{0}]

= [ML^{-1}T^{-2}]IncorrectWe know that Young’s modulus (

*Y*) = Stress / Strain.

Therefore dimensional representation of Young’s modulus (*Y*)

= Dimensions of stress / Dimensions of strain

= [ML^{-1}T^{-2}] / [M^{0}L^{0}T^{0}]

= [ML^{-1}T^{-2}] - Question 12 of 15
##### 12. Question

1 pointsStrain has dimensions equal to

CorrectWe know that strain = Change in length / Original length

since both the denominator and numerator have the same dimensions, i.e. [M^{0}L^{0}T^{0}], Therefore strain is a dimensionless quantity.IncorrectWe know that strain = Change in length / Original length

since both the denominator and numerator have the same dimensions, i.e. [M^{0}L^{0}T^{0}], Therefore strain is a dimensionless quantity. - Question 13 of 15
##### 13. Question

1 pointsWhich of the following is a dimensionless quantity?

CorrectWe know that strain = Change in length / Original length

Therefore dimensions of strain = Dimensions of change in length / Dimensions of original length

= [M^{0}LT^{0}] / [M^{0}LT^{0}] = [M^{0}L^{0}T^{0}]

Therefore strain is a dimensionless quantity.IncorrectWe know that strain = Change in length / Original length

Therefore dimensions of strain = Dimensions of change in length / Dimensions of original length

= [M^{0}LT^{0}] / [M^{0}LT^{0}] = [M^{0}L^{0}T^{0}]

Therefore strain is a dimensionless quantity. - Question 14 of 15
##### 14. Question

1 pointsDimensional formula for angle of contact is

CorrectWe know that angle of contact is an angle which is a dimensionless quantity. Therefore dimensional formula for angle of contact = [M

^{0}L^{0}T^{0}]IncorrectWe know that angle of contact is an angle which is a dimensionless quantity. Therefore dimensional formula for angle of contact = [M

^{0}L^{0}T^{0}] - Question 15 of 15
##### 15. Question

1 pointsThe dimensional formula of relative density is

CorrectWe know that relative density (

*ρ*) = Density of substance / Density of water at 4℃_{r}

Since both the denominator and numerator has the dimensions of density i.e., [ML^{-3}T^{0}]

therefore dimensional formula of relative density = [M^{0}L^{0}T^{0}]IncorrectWe know that relative density (

*ρ*) = Density of substance / Density of water at 4℃_{r}

Since both the denominator and numerator has the dimensions of density i.e., [ML^{-3}T^{0}]

therefore dimensional formula of relative density = [M^{0}L^{0}T^{0}]