Examkin

Question

Difficulty: MediumFundamental and Derived Units

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

  • Electric ChargeA⋅s\text{A}\cdot\text{s}
  • Linear Momentumkg⋅m⋅s−1\text{kg}\cdot\text{m}\cdot\text{s}^{-1}
  • Young's Moduluskg⋅m−1⋅s−2\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}
  • Magnetic Flux Densitykg⋅s−2⋅A−1\text{kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}

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
Rate this question

Topics

Fundamental and Derived Units

More questions from this topic