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 ?
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) $_{\text {rubber }}$
If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modules of the material of the wire will :
In which case there is maximum extension in the wire, if same force is applied on each wire
A uniform plank of Young’s modulus $Y $ is moved over a smooth horizontal surface by a constant horizontal force $F.$ The area of cross section of the plank is $A.$ The compressive strain on the plank in the direction of the force is
A load of $2 \,kg$ produces an extension of $1 \,mm$ in a wire of $3 \,m$ in length and $1 \,mm$ in diameter. The Young's modulus of wire will be .......... $Nm ^{-2}$
A copper wire $(Y = 1 \times 10^{11}\, N/m^2)$ of length $6\, m$ and a steel wire $(Y = 2 \times 10^{11}\, N/m^2)$ of length $4\, m$ each of cross section $10^{-5}\, m^2$ are fastened end to end and stretched by a tension of $100\, N$. The elongation produced in the copper wire is ......... $mm$