The tensile stress \sigma on a solid wire subjected to a stretching force F is defined as … — JAMB UTME

The tensile stress $\sigma$ on a solid wire subjected to a stretching force $F$ is defined as force per unit cross-sectional area, while the fractional change in length is the tensile strain $\epsilon$ . If Young's modulus $Y$ of the material is given by $Y = \frac{\sigma}{\epsilon}$ , which of the following is the SI unit of Young's modulus expressed in fundamental SI base units?

$\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$ Answer
B
$\text{kg}\cdot\text{m}\cdot\text{s}^{-2}$
C
$\text{kg}\cdot\text{m}^{-2}\cdot\text{s}^{-2}$
D
$\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}$

Answer

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Young's modulus is calculated as tensile stress divided by tensile strain. Since tensile strain is the ratio of change in length to original length, it has no units. Therefore, the SI unit of Young's modulus is identical to that of stress. Stress is force divided by area:

\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Step-by-Step Solution

Determine the dimensions of the force component

Force

F = ma \implies [F] = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Mass has fundamental unit

\text{kg}

and acceleration has derived unit

\text{m}\cdot\text{s}^{-2}

Determine the SI base units of tensile stress

\sigma

[\sigma] = \frac{[F]}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Stress is defined as force divided by cross-sectional area (

A = \text{m}^2

Evaluate the unit of Young's modulus

Y

[Y] = \frac{[\sigma]}{[\epsilon]} = \frac{\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}}{1} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Strain

\epsilon = \frac{\Delta L}{L}

is the ratio of two lengths and is dimensionless.

Key Concept

Derivation of SI base units for mechanical moduli from physical definitions

Estimated Time:1m 15s

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