Examkin

Question

Difficulty: MediumFundamental and Derived Units

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.

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
Rate this question

Topics

Fundamental and Derived Units

More questions from this topic