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Quiz Description :
Name: System of Units – Mechanics Test
Subject: Physics Mechanics)
Topic: System of Units
Questions: 15 MCQ
Time Allowed: 15 Minutes
Important for: IIT JEE, UPSEE, WBJEE Engineering Entrance exams, AIPMT and other PMT entrance test, +2 school students and B. Sc. / M. Sc. College Students.
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 Question 1 of 15
1. Question
1 pointsWhich of the following is not a unit of time?
CorrectWe know that light year is the distance traveled by the light in one year. Therefore light year is not a unit of time.
IncorrectWe know that light year is the distance traveled by the light in one year. Therefore light year is not a unit of time.
UnattemptedWe know that light year is the distance traveled by the light in one year. Therefore light year is not a unit of time.
 Question 2 of 15
2. Question
1 points‘Parsec’ is the unit of
CorrectWe know that ‘Parsec (Pc)’ is an astronomical unit of distance.
Mathematically, 1 Parsec = 3.085677 × 10^{16} m = 3.26 light years.IncorrectWe know that ‘Parsec (Pc)’ is an astronomical unit of distance.
Mathematically, 1 Parsec = 3.085677 × 10^{16} m = 3.26 light years.UnattemptedWe know that ‘Parsec (Pc)’ is an astronomical unit of distance.
Mathematically, 1 Parsec = 3.085677 × 10^{16} m = 3.26 light years.  Question 3 of 15
3. Question
1 pointsThe S.I unit of force is
CorrectWe know that S.T. unit of force is Newton.
Mathematically, 1 Newton is a force which produces an acceleration of 1 ms2 to a body of mass 1 kg.IncorrectWe know that S.T. unit of force is Newton.
Mathematically, 1 Newton is a force which produces an acceleration of 1 ms2 to a body of mass 1 kg.UnattemptedWe know that S.T. unit of force is Newton.
Mathematically, 1 Newton is a force which produces an acceleration of 1 ms2 to a body of mass 1 kg.  Question 4 of 15
4. Question
1 pointsThe unit of power is
CorrectWe know that the unit of power is ‘Watt’. It is also written as W.
Mathematically, 1 W = 1 Js^{1}IncorrectWe know that the unit of power is ‘Watt’. It is also written as W.
Mathematically, 1 W = 1 Js^{1}UnattemptedWe know that the unit of power is ‘Watt’. It is also written as W.
Mathematically, 1 W = 1 Js^{1}  Question 5 of 15
5. Question
1 pointsThe unit of gravitational constant is
CorrectWe know that the gravitational constant = [Unit of force × (Unit of distance)^{2}] / (Unit of mass)^{2} = [Nm^{2}]/kg^{2} = Nm^{2}kg^{2}
IncorrectWe know that the gravitational constant = [Unit of force × (Unit of distance)^{2}] / (Unit of mass)^{2} = [Nm^{2}]/kg^{2} = Nm^{2}kg^{2}
UnattemptedWe know that the gravitational constant = [Unit of force × (Unit of distance)^{2}] / (Unit of mass)^{2} = [Nm^{2}]/kg^{2} = Nm^{2}kg^{2}
 Question 6 of 15
6. Question
1 pointsThe S.I. unit of gravitational potential is
CorrectWe know that the gravitational potential = Work / Mass
Therefore S.I. unit of gravitational potential = Unit of work (in joules) / Units of mass (in kg)
= J / kg = Jkg^{1}IncorrectWe know that the gravitational potential = Work / Mass
Therefore S.I. unit of gravitational potential = Unit of work (in joules) / Units of mass (in kg)
= J / kg = Jkg^{1}UnattemptedWe know that the gravitational potential = Work / Mass
Therefore S.I. unit of gravitational potential = Unit of work (in joules) / Units of mass (in kg)
= J / kg = Jkg^{1}  Question 7 of 15
7. Question
1 points‘Decibel’ is the unit of
CorrectWe know that decibel is a unit used to compare two power level usually applied to sound or electrical signals. Therefore it is used for measuring the loudness of sound.
IncorrectWe know that decibel is a unit used to compare two power level usually applied to sound or electrical signals. Therefore it is used for measuring the loudness of sound.
UnattemptedWe know that decibel is a unit used to compare two power level usually applied to sound or electrical signals. Therefore it is used for measuring the loudness of sound.
 Question 8 of 15
8. Question
1 pointsThe S.I. unit of temperature is
CorrectWe know that S.I. unit of temperature is Kelvin, written as ‘K’. One Kelvin is defined as the 1/100^{th} part of the difference in temperatures of the melting ice and the boiling water at normal atmospheric pressure.
IncorrectWe know that S.I. unit of temperature is Kelvin, written as ‘K’. One Kelvin is defined as the 1/100^{th} part of the difference in temperatures of the melting ice and the boiling water at normal atmospheric pressure.
UnattemptedWe know that S.I. unit of temperature is Kelvin, written as ‘K’. One Kelvin is defined as the 1/100^{th} part of the difference in temperatures of the melting ice and the boiling water at normal atmospheric pressure.
 Question 9 of 15
9. Question
1 pointsThe unit of Young’s modulus is
CorrectWe know that Young’s modulus of elasticity = Stress / Strain
Therefore units of Young’s modulus = Unit of stress / Unit of strain = Unit of stress = Nm^{2}IncorrectWe know that Young’s modulus of elasticity = Stress / Strain
Therefore units of Young’s modulus = Unit of stress / Unit of strain = Unit of stress = Nm^{2}UnattemptedWe know that Young’s modulus of elasticity = Stress / Strain
Therefore units of Young’s modulus = Unit of stress / Unit of strain = Unit of stress = Nm^{2}  Question 10 of 15
10. Question
1 pointsThe unit of pressure, in the S.I. system is
CorrectWe know that the unit of pressure, in the S.I. system, is ‘Pascal’ written as Pa. Mathematically, 1 pascal = 1 Nm^{2}.
IncorrectWe know that the unit of pressure, in the S.I. system, is ‘Pascal’ written as Pa. Mathematically, 1 pascal = 1 Nm^{2}.
UnattemptedWe know that the unit of pressure, in the S.I. system, is ‘Pascal’ written as Pa. Mathematically, 1 pascal = 1 Nm^{2}.
 Question 11 of 15
11. Question
1 pointsS.I. unit of surface tension is
CorrectWe know that surface tension = Force / length.
Therefore S.I. unit of surface tension = Unit of force (in newton) / Unit of length (in metre)
= N/m = Nm^{1}IncorrectWe know that surface tension = Force / length.
Therefore S.I. unit of surface tension = Unit of force (in newton) / Unit of length (in metre)
= N/m = Nm^{1}UnattemptedWe know that surface tension = Force / length.
Therefore S.I. unit of surface tension = Unit of force (in newton) / Unit of length (in metre)
= N/m = Nm^{1}  Question 12 of 15
12. Question
1 pointsUnit of Stefan’s constant is
CorrectWe know that Stefan’s constant (σ)
= Energy radiated per second per unit area / (Temperature in Kelvin)^{4}
Therefore unit of Stefan’s constant
= Unit of energy radiated per second per unit area / (Unit of temperature)^{4}
= Js^{1}m^{2}/K^{4} = Wm^{2}/K^{4} = Wm^{2}K^{4}IncorrectWe know that Stefan’s constant (σ)
= Energy radiated per second per unit area / (Temperature in Kelvin)^{4}
Therefore unit of Stefan’s constant
= Unit of energy radiated per second per unit area / (Unit of temperature)^{4}
= Js^{1}m^{2}/K^{4} = Wm^{2}/K^{4} = Wm^{2}K^{4}UnattemptedWe know that Stefan’s constant (σ)
= Energy radiated per second per unit area / (Temperature in Kelvin)^{4}
Therefore unit of Stefan’s constant
= Unit of energy radiated per second per unit area / (Unit of temperature)^{4}
= Js^{1}m^{2}/K^{4} = Wm^{2}/K^{4} = Wm^{2}K^{4}  Question 13 of 15
13. Question
1 pointsS.I. unit of electric permittivity of free space is
CorrectWe know from the Coulomb’s law of electrostatics that electric permittivity of free space (ε_{0})
= q_{1}q_{2}/4πFr^{2}
Therefore S.I. unit of electric permittivity of free space
= (Unit of charge)^{2} / [Unit of force × (Unit of distance)^{2}]IncorrectWe know from the Coulomb’s law of electrostatics that electric permittivity of free space (ε_{0})
= q_{1}q_{2}/4πFr^{2}
Therefore S.I. unit of electric permittivity of free space
= (Unit of charge)^{2} / [Unit of force × (Unit of distance)^{2}]UnattemptedWe know from the Coulomb’s law of electrostatics that electric permittivity of free space (ε_{0})
= q_{1}q_{2}/4πFr^{2}
Therefore S.I. unit of electric permittivity of free space
= (Unit of charge)^{2} / [Unit of force × (Unit of distance)^{2}]  Question 14 of 15
14. Question
1 pointsThe unit of electric field strength is
CorrectWe know that electric field strength = Force / Charge
Therefore the unit of electric field strength
= Unit of force (in newton) / Unit of charge (in coulomb)
= N / C = NC^{1}IncorrectWe know that electric field strength = Force / Charge
Therefore the unit of electric field strength
= Unit of force (in newton) / Unit of charge (in coulomb)
= N / C = NC^{1}UnattemptedWe know that electric field strength = Force / Charge
Therefore the unit of electric field strength
= Unit of force (in newton) / Unit of charge (in coulomb)
= N / C = NC^{1}  Question 15 of 15
15. Question
1 pointsThe unit of electric flux is
CorrectWe know that electric flux (ϕ_{E}) = Electric field × Area
Therefore unit of electric flux= Unit of electric field × Unit of area = NC^{1}×m^{2} = Nm^{2}C^{1}IncorrectWe know that electric flux (ϕ_{E}) = Electric field × Area
Therefore unit of electric flux= Unit of electric field × Unit of area = NC^{1}×m^{2} = Nm^{2}C^{1}UnattemptedWe know that electric flux (ϕ_{E}) = Electric field × Area
Therefore unit of electric flux= Unit of electric field × Unit of area = NC^{1}×m^{2} = Nm^{2}C^{1}