$A$ copper wire of length $2.2 \; m$ and a steel wire of length $1.6 \; m$,both of diameter $3.0 \; mm$,are connected end to end. When stretched by a load,the net elongation is found to be $0.70 \; mm$. Obtain the load applied in $N$.

What should be the diameter of a copper wire $(Y=12 \times 10^{10} \text{ N/m}^2)$ of length $5 \text{ m}$ to produce the same elongation produced by a $5 \text{ m}$ long aluminium wire $(Y=7 \times 10^{10} \text{ N/m}^2)$ of diameter $3 \text{ mm}$ with the same $40 \text{ kg}$ mass (in $\text{ mm}$)?

$5 \,m$ long aluminium wire $(Y = 7 \times 10^{10} \,N/m^2)$ of diameter $3 \,mm$ supports a $40 \,kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} \,N/m^2)$ of the same length under the same weight, the diameter should be (in $mm$):

Which one is more elastic,steel or plastic? Why?

$A$ steel wire $1.5\,m$ long and of radius $1\,mm$ is attached with a load of $3\,kg$ at one end,while the other end is fixed. It is whirled in a vertical circle with a frequency of $2\,Hz$. Find the elongation of the wire when the weight is at the lowest position. (Given: $Y = 2 \times 10^{11}\,N/m^2$ and $g = 10\,m/s^2$)

$A$ 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.