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) $_{\text {rubber }}$
A wire extends by $1 mm$ when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be ........
A steel rod of length $1\,m$ and area of cross section $1\,cm^2$ is heated from $0\,^oC$ to $200\,^oC$ without being allowed to extend or bend. Find the tension produced in the rod $(Y = 2.0 \times 10^{11}\,Nm^{-2}$, $\alpha = 10^{-5} C^{-1})$
Two wires each of radius $0.2\,cm$ and negligible mass, one made of steel and other made of brass are loaded as shown in the figure. The elongation of the steel wire is $.........\times 10^{-6}\,m$. [Young's modulus for steel $=2 \times 10^{11}\,Nm ^{-2}$ and $g =10\,ms ^{-2}$ ]
A rod of uniform cross-sectional area $A$ and length $L$ has a weight $W$. It is suspended vertically from a fixed support. If Young's modulus for rod is $Y$, then elongation produced in rod is ......
Three bars having length $l, 2l$ and $3l$ and area of cross-section $A, 2 A$ and $3 A$ are joined rigidly end to end. Compound rod is subjected to a stretching force $F$. The increase in length of rod is (Young's modulus of material is $Y$ and bars are massless)