8.Mechanical Properties of Solids
medium

The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress ?

Option A
Option B
Option C
Option D

Solution

Young's modulus $=\frac{\text { Tensile stress }}{\text { Longitudinal strain }}$

For the same longitudinal strain, Young's modulus $\mathrm{Y}$ is proportional to tensile stress

$\therefore \quad \frac{\mathrm{Y}_{\text {steel }}}{\mathrm{Y}_{\text {rubber }}}=\frac{(\text { Stress })_{\text {steel }}}{(\text { Stress })_{\text {rubber }}}$

but $\mathrm{Y}_{\text {steel }}>\mathrm{Y}_{\text {rubber }}$

$\therefore \frac{\mathrm{Y}_{\text {steel }}}{\mathrm{Y}_{\text {rubber }}}>1$ $\therefore \frac{(\text { Stress })_{\text {steel }}}{(\text { Stress })_{\text {rubber }}}>1$ $\therefore$ (Stress) $_{\text {steel }}>$ (Stress) $_{\text {rubber }}$

Standard 11
Physics

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