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Fundamental and Derived Units

18 questions

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Question 1Question →

The derived SI unit of force is the newton (N\text{N}), which can be expressed in terms of fundamental SI base units as kg⋅m⋅sx\text{kg}\cdot\text{m}\cdot\text{s}^{x}. What is the numerical value of the exponent xx?

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Answer: -2

Answer

The numerical value of the exponent xx for time in the base SI expression of the newton is −2-2.
Force is defined by Newton's second law as mass times acceleration (F=maF = ma). In SI fundamental units, mass is measured in kilograms (kg\text{kg}) and acceleration in meters per second squared (m⋅s−2\text{m}\cdot\text{s}^{-2}). Combining these yields 1 N=1 kg⋅m⋅s−2\text{1\ N} = \text{1\ kg}\cdot\text{m}\cdot\text{s}^{-2}. Therefore, the power of time in seconds (xx) is −2-2.

Step-by-Step Solution

1
Identify the relationship between force and fundamental quantities
Force is defined as mass multiplied by acceleration (F=maF = ma)
This relates force to fundamental quantities of mass, length, and time.
2
Substitute the SI base units of mass and acceleration
Mass unit is kg\text{kg}; acceleration unit is m⋅s−2\text{m}\cdot\text{s}^{-2}
Kilogram, meter, and second are standard fundamental SI units.
3
Determine the exponent of seconds
The combined base unit expression is kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}, giving x=−2x = -2
Equating the powers of seconds gives x=−2x = -2.

Key Concept

Expressing derived units in terms of fundamental SI base units
Question 2Question →

The derived SI unit of electric potential, the volt (V\text{V}), can be expressed in terms of fundamental SI base units as kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d. What is the value of the exponent cc?

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Answer: -3

Answer

The exponent cc corresponding to the base unit second (s\text{s}) is −3-3.
Electric potential difference (VV) is defined as work (WW) per unit charge (QQ). In terms of SI base units, work has units of kg⋅m2⋅s−2\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} and charge has units of A⋅s\text{A} \cdot \text{s}. Dividing work by charge gives kg⋅m2⋅s−2A⋅s=kg1⋅m2⋅s−3⋅A−1\frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}} = \text{kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}. Equating this to kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d yields c=−3c = -3.

Step-by-Step Solution

1
Relate electric potential to work and charge
1 V=1 J1 C\text{1 V} = \frac{\text{1 J}}{\text{1 C}}
Electric potential difference is defined as work done per unit electric charge.
2
Break down Joules into base SI units
1 J=1 N⋅m=(1 kg⋅m⋅s−2)⋅m=1 kg⋅m2⋅s−2\text{1 J} = \text{1 N} \cdot \text{m} = (\text{1 kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m} = \text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}
Work is force multiplied by displacement, and force is mass multiplied by acceleration.
3
Break down Coulombs into base SI units
1 C=1 A⋅s\text{1 C} = \text{1 A} \cdot \text{s}
Electric charge is electric current multiplied by time.
4
Substitute base unit expressions into the ratio for electric potential
1 V=1 kg⋅m2⋅s−21 A⋅s=1 kg1⋅m2⋅s−3⋅A−1\text{1 V} = \frac{\text{1 kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{1 A} \cdot \text{s}} = \text{1 kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}
Combining powers of base units yields the complete fundamental representation.
5
Determine the exponent value cc
c=−3c = -3
Comparing kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d to kg1⋅m2⋅s−3⋅A−1\text{kg}^1 \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1} shows that the exponent of s\text{s} is −3-3.

Key Concept

Expressing derived units in terms of fundamental SI base units
Question 3Question →

Match each derived SI unit listed on the left with its corresponding fundamental (base) SI unit representation on the right.

Click a left item, then click its matching right item

Items

Joule (J\text{J})
Pascal (Pa\text{Pa})
Watt (W\text{W})
Newton (N\text{N})

Matches

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Answer

Joule (J\text{J}) matches kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}; Pascal (Pa\text{Pa}) matches kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}; Watt (W\text{W}) matches kg⋅m2⋅s−3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}; Newton (N\text{N}) matches kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}.
Each derived unit is systematically expressed in terms of the fundamental SI units of mass (kg), length (m), and time (s) by substituting definitions: Newton is kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}, Joule is kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}, Pascal is kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, and Watt is kg⋅m2⋅s−3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}.

Step-by-Step Solution

1
Express Newton (N) in fundamental units
Force=mass×acceleration⇒1 N=1 kg⋅m⋅s−2\text{Force} = \text{mass} \times \text{acceleration} \Rightarrow 1\text{ N} = 1\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}
Force is defined as mass multiplied by acceleration.
2
Express Joule (J) in fundamental units
Work=Force×distance⇒1 J=(kg⋅m⋅s−2)×m=1 kg⋅m2⋅s−2\text{Work} = \text{Force} \times \text{distance} \Rightarrow 1\text{ J} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-2}
Work done is force multiplied by displacement in the direction of the force.
3
Express Pascal (Pa) in fundamental units
Pressure=ForceArea⇒1 Pa=kg⋅m⋅s−2m2=1 kg⋅m−1⋅s−2\text{Pressure} = \frac{\text{Force}}{\text{Area}} \Rightarrow 1\text{ Pa} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Pressure is defined as force applied perpendicular to a surface per unit area.
4
Express Watt (W) in fundamental units
Power=Worktime⇒1 W=kg⋅m2⋅s−2s=1 kg⋅m2⋅s−3\text{Power} = \frac{\text{Work}}{\text{time}} \Rightarrow 1\text{ W} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{s}} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-3}
Power is the rate at which work is done or energy is transferred.

Key Concept

Expressing derived SI units in terms of base SI fundamental units
Question 4Question →

Which of the following correctly expresses the derived SI unit of power, the watt (W\text{W}), in terms of fundamental SI base units?

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Answer: kg⋅m2⋅s−3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

Answer

The watt (W\text{W}) expressed in fundamental SI base units is kg⋅m2⋅s−3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}.
Power is defined as energy transferred or work done per unit time (P=Wt\text{P} = \frac{W}{t}). Work is force times distance (N⋅m=kg⋅m2⋅s−2\text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}). Dividing work by time (s\text{s}) yields kg⋅m2⋅s−3\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}.

Step-by-Step Solution

1
Recall the defining formula for power
Power=WorkTime=Force×DisplacementTime\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{\text{Force} \times \text{Displacement}}{\text{Time}}
Relating the derived quantity to simpler mechanical quantities.
2
Break down force into fundamental base units using Newton's second law (F=maF = ma)
Unit of Force (Newton)=kg⋅m⋅s−2\text{Unit of Force (Newton)} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}
Mass is measured in kilograms (kg\text{kg}), acceleration in meters per second squared (m⋅s−2\text{m}\cdot\text{s}^{-2}).
3
Express work in fundamental base units
Unit of Work (Joule)=(kg⋅m⋅s−2)×m=kg⋅m2⋅s−2\text{Unit of Work (Joule)} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}
Work is force multiplied by distance.
4
Divide the unit of work by the unit of time (seconds)
Unit of Power (Watt)=kg⋅m2⋅s−2s=kg⋅m2⋅s−3\text{Unit of Power (Watt)} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}
Dividing by s\text{s} decreases the exponent of seconds by 1.

Key Concept

Expressing derived SI units in terms of fundamental SI base units
Estimated Time:1m 0s
Question 5Question →

The derived SI unit of dynamic viscosity, the pascal-second (Pa⋅s\text{Pa}\cdot\text{s}), can be expressed in fundamental SI base units as kga⋅mb⋅sc\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c. Determine the numerical value of the exponent of length, bb.

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Answer: -1

Answer

The numerical value of the exponent of length bb is −1-1.
Dynamic viscosity is measured in pascal-seconds (Pa⋅s\text{Pa}\cdot\text{s}). Substituting 1 Pa=1 N/m2=1 kg⋅m−1⋅s−21\text{ Pa} = 1\text{ N/m}^2 = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2} into the formula gives 1 Pa⋅s=1 kg1⋅m−1⋅s−11\text{ Pa}\cdot\text{s} = 1\text{ kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}. Comparing this with kga⋅mb⋅sc\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c, the exponent of length (meters) is b=−1b = -1.

Step-by-Step Solution

1
Express the newton in base SI units using F=maF = ma.
N=kg⋅m⋅s−2\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}.
Force is the product of mass and acceleration.
2
Derive the SI base unit expression for pressure (pascal, Pa\text{Pa}).
Pa=Nm2=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2\text{Pa} = \frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}.
Pressure is defined as force per unit area.
3
Multiply the base unit expression of pressure by seconds.
Pa⋅s=(kg⋅m−1⋅s−2)⋅s1=kg1⋅m−1⋅s−1\text{Pa}\cdot\text{s} = (\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}) \cdot \text{s}^1 = \text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}.
Dynamic viscosity is measured in pascal-seconds.
4
Extract the exponent of length (bb) corresponding to the meter unit.
b=−1b = -1.
The power of meters (m\text{m}) in kg1⋅m−1⋅s−1\text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1} is −1-1.

Key Concept

Deriving base SI units for derived physical quantities
Estimated Time:1m 30s
Question 6Question →

The derived SI unit of electrical resistance is the ohm (Ω\Omega). Which of the following correctly expresses the ohm strictly in terms of fundamental SI base units?

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Answer: kg⋅m2⋅s−3⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}

Answer

kg⋅m2⋅s−3⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}
Resistance is defined by Ohm's Law as voltage divided by current (R=V/IR = V / I). Voltage is work done per unit charge (V=W/QV = W / Q), and charge is current multiplied by time (Q=I⋅tQ = I \cdot t). Substituting the unit of work (kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}) gives R=kg⋅m2⋅s−2A2⋅s=kg⋅m2⋅s−3⋅A−2R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}.

Step-by-Step Solution

1
Relate resistance to potential difference and current using Ohm's Law.
R=VIR = \frac{V}{I}
Resistance is defined as the ratio of potential difference to electric current.
2
Express electric potential difference (VV) in terms of work (WW) and charge (QQ), where Q=I⋅tQ = I \cdot t.
V=WI⋅t  ⟹  R=WI2⋅tV = \frac{W}{I \cdot t} \implies R = \frac{W}{I^2 \cdot t}
Electric potential difference is work done per unit charge, and electric charge is current multiplied by time.
3
Break down work (W=Force×distanceW = \text{Force} \times \text{distance}) into base SI units.
Unit of Work = kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}
Force is mass times acceleration (kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}), so work is kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}.
4
Substitute all base units into the resistance formula.
Unit of R=kg⋅m2⋅s−2A2⋅s=kg⋅m2⋅s−3⋅A−2R = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}^2 \cdot \text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}
Combining powers of base units yields the complete fundamental SI base unit expression for the ohm.

Key Concept

Derivation of SI base units from defining formulas of physical quantities
Question 7Question →

Match each physical quantity listed on the left with its corresponding expression in fundamental SI base units on the right.

Click a left item, then click its matching right item

Items

Pressure
Surface Tension
Specific Heat Capacity
Electric Capacitance

Matches

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Answer

Pressure matches with kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, Surface Tension matches with kg⋅s−2\text{kg}\cdot\text{s}^{-2}, Specific Heat Capacity matches with m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}, and Electric Capacitance matches with kg−1⋅m−2⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2.
Each quantity on the left is matched correctly to its fundamental SI base unit equivalent derived directly from its defining physical formula.

Step-by-Step Solution

1
Derive base SI units for Pressure
P=FA=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2P = \frac{F}{A} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Pressure is force per unit area.
2
Derive base SI units for Surface Tension
γ=FL=kg⋅m⋅s−2m=kg⋅s−2\gamma = \frac{F}{L} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}} = \text{kg}\cdot\text{s}^{-2}
Surface tension is force per unit length.
3
Derive base SI units for Specific Heat Capacity
c=QmΔT=kg⋅m2⋅s−2kg⋅K=m2⋅s−2⋅K−1c = \frac{Q}{m\Delta T} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}
Specific heat capacity is thermal energy per unit mass per kelvin.
4
Derive base SI units for Electric Capacitance
C=QV=A⋅skg⋅m2⋅s−3⋅A−1=kg−1⋅m−2⋅s4⋅A2C = \frac{Q}{V} = \frac{\text{A}\cdot\text{s}}{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}} = \text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^4\cdot\text{A}^2
Capacitance is electric charge divided by electric potential difference.

Key Concept

Expressing derived SI units in terms of fundamental base SI units
Question 8Question →

In the International System of Units (SI), physical quantities are categorized as either fundamental or derived. Which of the following is a fundamental SI base unit?

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Answer: Kelvin

Answer

Kelvin is the fundamental SI base unit.
The kelvin is the standard SI base unit for measuring thermodynamic temperature and is one of the seven defined fundamental units.

Step-by-Step Solution

1
Identify the seven SI fundamental (base) units
The seven fundamental SI base units are the meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).
Fundamental units are basic units that are independent of one another and cannot be expressed in terms of other units.
2
Compare the given options against the set of fundamental units
Among Joule, Kelvin, Pascal, and Watt, only Kelvin is in the list of fundamental SI base units.
Joule, Pascal, and Watt are all derived units defined by combinations of base units.

Key Concept

Fundamental SI Base Units
Estimated Time:35s
Question 9Question →

Match each physical quantity on the left with its corresponding SI unit expressed in terms of fundamental (base) units on the right.

Click a left item, then click its matching right item

Items

Frequency
Electric charge
Mass density
Acceleration

Matches

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Answer

Frequency matches s−1\text{s}^{-1}, Electric charge matches A⋅s\text{A}\cdot\text{s}, Mass density matches kg⋅m−3\text{kg}\cdot\text{m}^{-3}, and Acceleration matches m⋅s−2\text{m}\cdot\text{s}^{-2}.
Frequency (f=1/Tf = 1/T) is expressed in reciprocal seconds (s−1\text{s}^{-1}). Electric charge (Q=I⋅tQ = I \cdot t) is current times time, giving A⋅s\text{A}\cdot\text{s}. Mass density (ρ=m/V\rho = m/V) is mass per volume, giving kg⋅m−3\text{kg}\cdot\text{m}^{-3}. Acceleration (a=Δv/Δta = \Delta v / \Delta t) is rate of velocity change, giving m⋅s−2\text{m}\cdot\text{s}^{-2}.

Step-by-Step Solution

1
Identify the defining formula for each physical quantity
Frequency f=1Tf = \frac{1}{T}, Charge Q=I⋅tQ = I \cdot t, Density ρ=mV\rho = \frac{m}{V}, Acceleration a=ΔvΔta = \frac{\Delta v}{\Delta t}.
Relating derived quantities to their defining equations allows reduction into fundamental quantities.
2
Substitute the SI base units for mass (kg\text{kg}), length (m\text{m}), time (s\text{s}), and current (A\text{A})
Frequency: s−1\text{s}^{-1}; Charge: A⋅s\text{A}\cdot\text{s}; Density: kgm3=kg⋅m−3\frac{\text{kg}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-3}; Acceleration: m/ss=m⋅s−2\frac{\text{m/s}}{\text{s}} = \text{m}\cdot\text{s}^{-2}.
This expresses each derived unit strictly in terms of fundamental SI units.
3
Match each physical quantity to its calculated base unit representation
Frequency →s−1\rightarrow \text{s}^{-1}, Electric charge →A⋅s\rightarrow \text{A}\cdot\text{s}, Mass density →kg⋅m−3\rightarrow \text{kg}\cdot\text{m}^{-3}, Acceleration →m⋅s−2\rightarrow \text{m}\cdot\text{s}^{-2}.
Completes the pairing verification.

Key Concept

Expressing derived physical quantities in terms of SI fundamental (base) units
Estimated Time:45s
Question 10Question →

Match each physical quantity listed on the left with its corresponding SI unit expressed strictly in terms of fundamental (base) units on the right.

Click a left item, then click its matching right item

Items

Electric permittivity of free space (ε0\varepsilon_0)
Magnetic permeability of free space (μ0\mu_0)
Thermal conductivity (kk)
Specific heat capacity (cc)

Matches

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Answer

Electric permittivity of free space maps to kg−1⋅m−3⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2, Magnetic permeability of free space maps to kg⋅m⋅s−2⋅A−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}, Thermal conductivity maps to kg⋅m⋅s−3⋅K−1\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}, and Specific heat capacity maps to m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Each physical quantity is matched to its exact fundamental unit expression obtained by substituting basic formulas into SI base units (kg\text{kg}, m\text{m}, s\text{s}, A\text{A}, K\text{K}). Electric permittivity resolves to kg−1⋅m−3⋅s4⋅A2\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2, magnetic permeability resolves to kg⋅m⋅s−2⋅A−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}, thermal conductivity resolves to kg⋅m⋅s−3⋅K−1\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}, and specific heat capacity resolves to m2⋅s−2⋅K−1\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.

Step-by-Step Solution

1
Derive the base SI units for Electric permittivity of free space (ε0\varepsilon_0).
From Coulomb's Law, F=q1q24πε0r2  ⟹  ε0=q1q24πFr2F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2} \implies \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}. Expressing terms in SI fundamental units: charge q=A⋅sq = \text{A}\cdot\text{s}, force F=kg⋅m⋅s−2F = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}, distance r=mr = \text{m}. Thus, [ε0]=(A⋅s)2(kg⋅m⋅s−2)⋅m2=kg−1⋅m−3⋅s4⋅A2[\varepsilon_0] = \frac{(\text{A}\cdot\text{s})^2}{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2} = \text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2.
Relating derived electromagnetic quantities to fundamental SI units via governing physical equations.
2
Derive the base SI units for Magnetic permeability of free space (μ0\mu_0).
From the force per unit length between parallel current-carrying conductors, FL=μ0I1I22πr  ⟹  μ0=2πFrI1I2L\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \implies \mu_0 = \frac{2\pi F r}{I_1 I_2 L}. Units: [μ0]=(kg⋅m⋅s−2)⋅mA2⋅m=kg⋅m⋅s−2⋅A−2[\mu_0] = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}}{\text{A}^2\cdot\text{m}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}.
Using Ampère's force law to solve for magnetic permeability in base units.
3
Derive the base SI units for Thermal conductivity (kk).
From Fourier's Law of Heat Conduction, Qt=kAΔTΔx  ⟹  k=Q⋅ΔxtAΔT\frac{Q}{t} = k A \frac{\Delta T}{\Delta x} \implies k = \frac{Q \cdot \Delta x}{t A \Delta T}. Heat energy Q=kg⋅m2⋅s−2Q = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}, time t=st = \text{s}, area A=m2A = \text{m}^2, thickness Δx=m\Delta x = \text{m}, temperature difference ΔT=K\Delta T = \text{K}. Thus, [k]=(kg⋅m2⋅s−2)⋅ms⋅m2⋅K=kg⋅m⋅s−3⋅K−1[k] = \frac{(\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2})\cdot\text{m}}{\text{s}\cdot\text{m}^2\cdot\text{K}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}.
Connecting thermal conduction equations to fundamental mechanics and thermodynamic units.
4
Derive the base SI units for Specific heat capacity (cc).
From Q=mcΔT  ⟹  c=QmΔTQ = m c \Delta T \implies c = \frac{Q}{m \Delta T}. Units: [c]=kg⋅m2⋅s−2kg⋅K=m2⋅s−2⋅K−1[c] = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}.
Applying the defining equation of heat capacity to reduce the unit to fundamental base units.

Key Concept

Derivation of complex derived SI units from fundamental base units using fundamental physical laws.
Question 11Question →

Match each physical quantity to its correct SI unit.

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Items

Electric current
Force
Thermodynamic temperature
Energy

Matches

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Answer

Electric current matches Ampere (A); Force matches Newton (N); Thermodynamic temperature matches Kelvin (K); Energy matches Joule (J).
Electric current and thermodynamic temperature are fundamental quantities measured in amperes and kelvins respectively. Force and energy are derived quantities whose units (newton and joule) are combinations of base SI units.

Step-by-Step Solution

1
Identify fundamental physical quantities and their base SI units.
Electric current is measured in amperes (A), and thermodynamic temperature is measured in kelvins (K). Both are fundamental SI units.
Fundamental units are basic units that are defined independently of other quantities.
2
Identify derived physical quantities and their derived SI units.
Force is measured in newtons (N), and energy is measured in joules (J). Both are derived SI units.
Derived units are obtained by combining fundamental SI base units according to physical equations.

Key Concept

Classification of fundamental and derived SI units
Estimated Time:45s
Question 12Question →

According to Newton's law of universal gravitation, the gravitational force FF between two masses m1m_1 and m2m_2 separated by a distance rr is given by F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}. Which of the following correctly expresses the derived SI unit of the universal gravitational constant, GG, in terms of fundamental (base) units?

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Answer: kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

Answer

The SI unit of the universal gravitational constant GG expressed in fundamental base units is kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}.
The expression kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2} is correct because rearranging F=Gm1m2r2F = \frac{G m_1 m_2}{r^2} yields G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Replacing FF with its base equivalent kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}, rr with m\text{m}, and m1,m2m_1, m_2 with kg\text{kg} gives (kg⋅m⋅s−2)⋅m2kg2=kg−1⋅m3⋅s−2\frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2}{\text{kg}^2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}.

Step-by-Step Solution

1
Rearrange the gravitational formula to solve for GG
G=F⋅r2m1⋅m2G = \frac{F \cdot r^2}{m_1 \cdot m_2}
To express the unit of GG, we need it in terms of quantities with known units.
2
Substitute the SI unit for force in fundamental base units
\text{Unit of } F = \text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}
Force is mass times acceleration (F=maF = ma), so its base unit is kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}.
3
Substitute all base units into the formula for GG
\text{Unit of } G = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{m}^2}{\text{kg} \cdot \text{kg}} = \frac{\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}}{\text{kg}^2}
Combining length terms (m⋅m2=m3\text{m} \cdot \text{m}^2 = \text{m}^3) and mass terms in the denominator.
4
Simplify the mass exponent using laws of indices
\text{Unit of } G = \text{kg}^{1 - 2}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}
Dividing by kg2\text{kg}^2 subtracts 2 from the mass exponent.

Key Concept

Derivation of Derived Units in Base SI Units
Estimated Time:1m 15s
Question 13Question →

Pair each physical quantity in List I with its equivalent SI unit expressed purely in terms of fundamental base units in List II.

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Items

Electric Charge
Linear Momentum
Young's Modulus
Magnetic Flux Density

Matches

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Answer

Electric Charge matches A⋅s\text{A}\cdot\text{s}; Linear Momentum matches kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}; Young's Modulus matches kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}; Magnetic Flux Density matches kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.
Each physical quantity is resolved to fundamental SI base units using defining equations: Electric charge (Q=ItQ=It) yields A⋅s\text{A}\cdot\text{s}, linear momentum (p=mvp=mv) yields kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}, Young's modulus (stress/strain) yields kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}, and magnetic flux density (B=FILB=\frac{F}{IL}) yields kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.

Step-by-Step Solution

1
Derive base units for Electric Charge
A⋅s\text{A}\cdot\text{s}
From Q=ItQ = I t, electric current has the fundamental unit ampere (A) and time has the fundamental unit second (s).
2
Derive base units for Linear Momentum
kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}
From p=mvp = m v, mass is in kilograms (kg) and velocity is in meters per second (m/s).
3
Derive base units for Young's Modulus
kg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
Tensile strain is dimensionless. Tensile stress is force per area: Nm2=kg⋅m⋅s−2m2=kg⋅m−1⋅s−2\frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}.
4
Derive base units for Magnetic Flux Density
kg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}
From magnetic force F=ILBF = I L B, solving for B=FILB = \frac{F}{I L} gives kg⋅m⋅s−2A⋅m=kg⋅s−2⋅A−1\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{A}\cdot\text{m}} = \text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.

Key Concept

Expressing derived physical quantities in terms of fundamental SI base units
Question 14Question →

Match each derived physical quantity in Column I with its equivalent SI unit expressed in terms of fundamental (base) units in Column II.

Click a left item, then click its matching right item

Items

Gravitational Potential
Electrical Conductance
Self-Inductance
Impulse

Matches

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Answer

Gravitational Potential matches m2⋅s−2\text{m}^2\cdot\text{s}^{-2}; Electrical Conductance matches kg−1⋅m−2⋅s3⋅A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2; Self-Inductance matches kg⋅m2⋅s−2⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}; Impulse matches kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}.
Each derived unit is systematically expressed in terms of the fundamental SI base units (kilogram, meter, second, ampere) by substituting defining formulas. Gravitational potential is energy per unit mass (J⋅kg−1=m2⋅s−2\text{J}\cdot\text{kg}^{-1} = \text{m}^2\cdot\text{s}^{-2}). Electrical conductance is reciprocal resistance (Siemens =A⋅V−1=kg−1⋅m−2⋅s3⋅A2= \text{A}\cdot\text{V}^{-1} = \text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2). Self-Inductance is electromotive force per rate of change of current (Henry =V⋅s⋅A−1=kg⋅m2⋅s−2⋅A−2= \text{V}\cdot\text{s}\cdot\text{A}^{-1} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}). Impulse is force times time (N⋅s=kg⋅m⋅s−1\text{N}\cdot\text{s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}).

Step-by-Step Solution

1
Derive base units for Gravitational Potential
m2⋅s−2\text{m}^2\cdot\text{s}^{-2}
Gravitational potential is potential energy per unit mass (V=Epm=kg⋅m2⋅s−2kgV = \frac{E_p}{m} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}}).
2
Derive base units for Electrical Conductance
kg−1⋅m−2⋅s3⋅A2\text{kg}^{-1}\cdot\text{m}^{-2}\cdot\text{s}^3\cdot\text{A}^2
Conductance is current divided by voltage (G=IV=Akg⋅m2⋅s−3⋅A−1G = \frac{I}{V} = \frac{\text{A}}{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}}).
3
Derive base units for Self-Inductance
kg⋅m2⋅s−2⋅A−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\cdot\text{A}^{-2}
Inductance is voltage multiplied by time divided by current (L=V⋅tI=kg⋅m2⋅s−3⋅A−1⋅sAL = \frac{V \cdot t}{I} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1} \cdot \text{s}}{\text{A}}).
4
Derive base units for Impulse
kg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}
Impulse is force multiplied by time (I=F⋅t=(kg⋅m⋅s−2)⋅sI = F \cdot t = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{s}).

Key Concept

Break down derived physical quantities into fundamental SI quantities (mass in kg, length in m, time in s, electric current in A) using defining physical formulas.
Estimated Time:1m 30s
Question 15Question →

Which of the following combinations of fundamental SI base units is equivalent to the volt (V\text{V}), the SI unit of electric potential difference?

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Answer: kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}

Answer

kg⋅m2⋅s−3⋅A−1\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}
Electric potential difference (VV) is defined as work done (WW) per unit charge (QQ), giving V=WQV = \frac{W}{Q}. In fundamental SI units, work is kg⋅m2⋅s−2\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2} and electric charge is A⋅s\text{A}\cdot\text{s}. Dividing work by charge yields kg⋅m2⋅s−2A⋅s=kg⋅m2⋅s−3⋅A−1\frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.

Step-by-Step Solution

1
State the defining physical equation for electric potential difference.
V=WQV = \frac{W}{Q}, where VV is electric potential, WW is work done in joules, and QQ is electric charge in coulombs.
Electric potential difference is defined as work done per unit electric charge.
2
Express work (WW) in terms of fundamental SI base units.
W=Force×distance=(kg⋅m⋅s−2)×m=kg⋅m2⋅s−2W = \text{Force} \times \text{distance} = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}.
Work is the product of force and displacement, where force is mass times acceleration.
3
Express electric charge (QQ) in terms of fundamental SI base units.
Q=I×t=A⋅sQ = I \times t = \text{A}\cdot\text{s}.
Electric charge is defined as electric current multiplied by time.
4
Divide the base units of work by the base units of electric charge.
V=kg⋅m2⋅s−2A⋅s=kg⋅m2⋅s−3⋅A−1V = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}.
Simplifying the exponents gives the equivalent combination of fundamental SI base units.

Key Concept

Expressing derived SI units in terms of fundamental (base) SI units.
Question 16Question →

The derived SI unit of electric capacitance, the farad (F\text{F}), can be expressed in terms of fundamental SI base units as kga⋅mb⋅sc⋅Ad\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d. What is the numerical value of the sum of the exponents a+b+c+da + b + c + d?

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Answer: 3

Answer

The numerical value of the sum of the exponents a+b+c+da + b + c + d is 3.
The farad (F\text{F}), when resolved into fundamental SI base units, is kg−1⋅m−2⋅s4⋅A2\text{kg}^{-1} \cdot \text{m}^{-2} \cdot \text{s}^4 \cdot \text{A}^2. Adding the exponents yields a=−1a = -1, b=−2b = -2, c=4c = 4, and d=2d = 2, giving a total sum of (−1)+(−2)+4+2=3(-1) + (-2) + 4 + 2 = 3.

Step-by-Step Solution

1
Relate electric capacitance to fundamental physical quantities
Capacitance is defined as C=QVC = \frac{Q}{V}, where QQ is electric charge and VV is electric potential difference.
This fundamental relation connects capacitance to charge and energy per unit charge.
2
Express charge and potential difference in terms of SI base units
Electric charge QQ has base units A⋅s\text{A} \cdot \text{s}. Potential difference V=WorkQV = \frac{\text{Work}}{Q} has base units kg⋅m2⋅s−2A⋅s=kg⋅m2⋅s−3⋅A−1\frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}.
Work is force times distance (kg⋅m⋅s−2×m=kg⋅m2⋅s−2\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \times \text{m} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}) and electric current is a fundamental base quantity.
3
Determine the base unit representation of the farad
The farad is expressed as A⋅skg⋅m2⋅s−3⋅A−1=kg−1⋅m−2⋅s4⋅A2\frac{\text{A} \cdot \text{s}}{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}} = \text{kg}^{-1} \cdot \text{m}^{-2} \cdot \text{s}^4 \cdot \text{A}^2.
Dividing the unit of charge by the unit of potential difference yields the derived base unit expression.
4
Calculate the sum of the exponents
a+b+c+d=(−1)+(−2)+4+2=3a + b + c + d = (-1) + (-2) + 4 + 2 = 3.
Summing the powers corresponding to kilograms, meters, seconds, and amperes gives the final scalar value.

Key Concept

Expressing derived SI units in terms of fundamental SI base units
Question 17Question →

The coefficient of dynamic viscosity η\eta of a fluid can be defined using Newton's formula for viscous flow, η=FAΔvΔx\eta = \frac{F}{A \frac{\Delta v}{\Delta x}}, where FF is the viscous drag force, AA is the contact surface area, and ΔvΔx\frac{\Delta v}{\Delta x} is the velocity gradient perpendicular to the flow. Which of the following combinations of fundamental SI base units is equivalent to the coefficient of dynamic viscosity?

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Answer: kg⋅m−1⋅s−1\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}

Answer

The coefficient of dynamic viscosity expressed in fundamental SI base units is kg⋅m−1⋅s−1\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}.
The unit of dynamic viscosity is derived from η=FA(Δv/Δx)\eta = \frac{F}{A (\Delta v / \Delta x)}. Substituting base units yields kg⋅m⋅s−2m2⋅s−1=kg⋅m−1⋅s−1\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2 \cdot \text{s}^{-1}} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}, which correctly represents the unit in base SI quantities.

Step-by-Step Solution

1
Identify the fundamental SI base units for each quantity in the formula
Force FF has units kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}; Area AA has units m2\text{m}^2; Velocity gradient ΔvΔx\frac{\Delta v}{\Delta x} has units m⋅s−1m=s−1\frac{\text{m}\cdot\text{s}^{-1}}{\text{m}} = \text{s}^{-1}.
Decomposing derived physical quantities into SI base units is essential for dimensional reduction.
2
Substitute these fundamental units into the given formula for η\eta
Unit of η=kg⋅m⋅s−2m2⋅s−1\text{Unit of } \eta = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2 \cdot \text{s}^{-1}}.
Direct algebraic substitution yields the compound unit expression.
3
Simplify the powers of mass, length, and time
kg1⋅m1−2⋅s−2−(−1)=kg⋅m−1⋅s−1\text{kg}^1 \cdot \text{m}^{1 - 2} \cdot \text{s}^{-2 - (-1)} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}.
Applying exponent rules simplifies the expression to base units.

Key Concept

Expressing derived units in terms of fundamental SI base units (kilogram, meter, second)
Question 18Question →

The force constant (stiffness) kk of a helical spring measures its resistance to elastic deformation and is defined by the relationship k=Fek = \frac{F}{e}, where FF represents the restoring force and ee represents the extension. Which of the following represents the SI unit of kk expressed strictly in terms of fundamental SI base units?

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Answer: kg⋅s−2\text{kg} \cdot \text{s}^{-2}

Answer

kg⋅s−2\text{kg} \cdot \text{s}^{-2}
The force constant kk is defined as force per unit extension (k=Fek = \frac{F}{e}). Since force has base units of kg⋅m⋅s−2\text{kg} \cdot \text{m} \cdot \text{s}^{-2} and extension has base units of m\text{m}, dividing force by extension yields kg⋅m⋅s−2m=kg⋅s−2\frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}} = \text{kg} \cdot \text{s}^{-2}.

Step-by-Step Solution

1
Identify the defining formula and constituent quantities.
The force constant formula is k=Fek = \frac{F}{e}, where force FF is measured in newtons (N\text{N}) and extension ee is measured in metres (m\text{m}).
Determining SI base units requires substituting base dimensions into the governing physical equation.
2
Express force in fundamental SI base units using Newton's second law (F=maF = ma).
1 N=1 kg⋅m⋅s−21\text{ N} = 1\text{ kg} \cdot \text{m} \cdot \text{s}^{-2}.
Mass has fundamental unit kg\text{kg}, and acceleration has fundamental unit m⋅s−2\text{m} \cdot \text{s}^{-2}.
3
Substitute the base unit expression of force into the formula for kk and simplify.
\text{SI unit of } k = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{m}} = \text{kg} \cdot \text{s}^{-2}.
The unit of length (m\text{m}) in the numerator cancels out with the unit of length in the denominator.

Key Concept

Deriving SI units of physical quantities from fundamental base units using defining equations.
Fundamental and Derived Units Practice Questions — JAMB UTME | Examkin