Steel and copper wires of the same length are stretched by the same weight one after the other. The Young's modulus of steel and copper are $2 \times 10^{11} \, N/m^2$ and $1.2 \times 10^{11} \, N/m^2$,respectively. What is the ratio of the increase in their lengths?

Two wires of diameter $0.25 \; cm,$ one made of steel and the other made of brass,are loaded as shown in the figure. The unloaded length of the steel wire is $1.5 \; m$ and that of the brass wire is $1.0 \; m.$ Compute the elongations of the steel and the brass wires. (Given: Young's modulus of steel $Y_s = 2.0 \times 10^{11} \; Pa,$ Young's modulus of brass $Y_b = 0.91 \times 10^{11} \; Pa$)

$A$ uniform steel rod of mass $1.8 \,kg$ and length $0.8 \,m$ is hung from a nail with the help of two steel wires,each of area of cross-section $0.01 \,mm^2$ and unstretched length $0.5 \,m$,as shown in the figure. The centre of mass of the rod lies vertically below the nail. The increase in the distance between the centre of mass of the rod and the nail due to stretching of the wires as the rod hangs is . . . . . . $mm$. (Young's modulus of steel $= 2 \times 10^{11} \,N/m^2$ and acceleration due to gravity $= 10 \,m/s^2$)

$A$ rigid massless rod of length $6L$ is suspended horizontally by means of two elastic rods $PQ$ and $RS$ as shown in the figure. Their area of cross-section,Young's modulus,and lengths are mentioned in the figure. Find the deflection of end $S$ in the equilibrium state. The free end of the rigid rod is pushed down by a constant force $F$. $A$ is the area of cross-section,$Y$ is Young's modulus of elasticity.

$A$ force of $10^3 \ N$ stretches the length of a hanging wire by $1 \ mm$. The force required to stretch a wire of the same material and length,but having four times the diameter,by $1 \ mm$ is:

If Young's modulus of iron is $2 \times 10^{11} \ N/m^2$ and the interatomic spacing between two molecules is $3 \times 10^{-10} \ m$,the interatomic force constant is ......... $N/m$.