$A$ uniform heavy rod of weight $10 \, N$,cross-sectional area $100 \, \text{cm}^2$ and length $20 \, \text{cm}$ is hanging from a fixed support. The Young's modulus of the material of the rod is $2 \times 10^{11} \, \text{N/m}^2$. Neglecting the lateral contraction,find the elongation of the rod due to its own weight. (In $\times 10^{-10} \, \text{m}$)

The work done per unit volume to stretch a wire by $1\%$ of its length,having a cross-sectional area of $1\,mm^2$,is: $[Y = 9 \times 10^{11}\,N/m^2]$

$A$ metal wire of length $L$ is suspended vertically from a rigid support. When a body of mass $M$ is attached to the lower end of the wire,the elongation in the wire is $l$. Consider the following statements:
$(I)$ The loss of gravitational potential energy of mass $M$ is $Mgl$.
$(II)$ The elastic potential energy stored in the wire is $Mgl$.
$(III)$ The elastic potential energy stored in the wire is $\frac{1}{2} Mgl$.
$(IV)$ The heat produced is $\frac{1}{2} Mgl$.
Which of the following statements are correct?

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \, N$,then the increase in length is $0.5 \, mm$. The ratio of the energy stored in the wire to the work done in displacing it through $1.5 \, mm$ by the weight is

An $8\,m$ long copper wire and $4\,m$ long steel wire,each of cross-section $0.5\,cm^2$,are fastened end to end and stretched by a $500\,N$ force. The elastic potential energy of the system is (Young's modulus: $Y_{cu} = 1 \times 10^{11}\,N/m^2$,$Y_{steel} = 2 \times 10^{11}\,N/m^2$):

Two wires of the same material and length but diameters in the ratio $1:2$ are stretched by the same force. The elastic potential energy per unit volume for the wires,when stretched by the same force,will be in the ratio: (in $:1$)