Question
Explain the terms: Young's modulus of elasticity and elastic fatigue.
✓
Answer
We have $\text{Y}=\frac{\text{F}.\text{l}}{\text{a}\Delta\text{l}}$
For same force $\Delta\text{l}_{\text{rubber}}$ is more than of $\Delta\text{l}_{\text{steel}}.$
$\therefore\text{Y}_\text{s}>\text{Y}_\text{r}$
For same force $\Delta\text{l}_{\text{rubber}}$ is more than of $\Delta\text{l}_{\text{steel}}.$
$\therefore\text{Y}_\text{s}>\text{Y}_\text{r}$
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