$A$ block of mass $M$ is suspended from a wire of length $L$,area of cross-section $A$ and Young's modulus $Y$. The elastic potential energy stored in the wire is

The force constant of a wire is $k$ and that of another wire is $2k$. When both the wires are stretched through the same distance,then the work done is:

The elastic potential energy stored in a steel wire of length $20 \, m$ stretched by $2 \, cm$ is $80 \, J$. The cross-sectional area of the wire is $......... \, mm^2$ (Given,$Y = 2.0 \times 10^{11} \, N/m^2$).

$A$ 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 $\times 10^{-10} \, \text{m}$)

$A$ metal rod of area of cross-section $3 \,cm^2$ is stretched along its length by applying a force of $9 \times 10^4 \,N$. If the Young's modulus of the material of the rod is $2 \times 10^{11} \,Nm^{-2}$, the energy stored per unit volume in the stretched rod is:

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 $:1$)