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?

Two wires of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are stretched by forces $F_A$ and $F_B$ respectively to produce the same extension,then the ratio $\frac{F_A}{F_B}$ is:

$A$ block of weight $100 \ N$ is suspended by copper and steel wires of same cross-sectional area $0.5 \ cm^2$ and lengths $\sqrt{3} \ m$ and $1 \ m$,respectively. Their other ends are fixed on a ceiling as shown in the figure. The angles subtended by the copper and steel wires with the ceiling are $30^{\circ}$ and $60^{\circ}$,respectively. If the elongation in the copper wire is $\Delta \ell_C$ and the elongation in the steel wire is $\Delta \ell_S$,then the ratio $\frac{\Delta \ell_C}{\Delta \ell_S}$ is. . . . . .
[Young's modulus for copper and steel are $1 \times 10^{11} \ N/m^2$ and $2 \times 10^{11} \ N/m^2$ respectively]

$A$ uniform cylindrical rod of length $L$ and radius $r$ is made from a material whose Young's modulus of elasticity equals $Y$. When this rod is heated by temperature $T$ and simultaneously subjected to a net longitudinal compressional force $F$,its length remains unchanged. The coefficient of volume expansion of the material of the rod is (nearly) equal to:

$A$ rod is fixed between two points at $20\,^oC$. The coefficient of linear expansion of the material of the rod is $1.1 \times 10^{-5}/\,^oC$ and Young's modulus is $1.2 \times 10^{11}\,N/m^2$. Find the stress developed in the rod if the temperature of the rod becomes $10\,^oC$.

$A$ steel wire of length $20 \text{ cm}$ and area of cross-section $1 \text{ mm}^2$ is tied rigidly at both the ends. When the temperature of the wire is changed from $40^{\circ} \text{C}$ to $20^{\circ} \text{C}$, find the change in its tension. Given, the coefficient of linear expansion for steel is $\alpha = 1.1 \times 10^{-5} {}^{\circ} \text{C}^{-1}$ and Young's modulus of steel is $Y = 2.0 \times 10^{11} \text{ N/m}^2$. (in $\text{ N}$)