$A$ wire $2 \,m$ in length suspended vertically stretches by $10 \,mm$ when a mass of $10 \,kg$ is attached to the lower end. The elastic potential energy gained by the wire is ...... $J$ (take $g=10 \,m/s^2$)

What is the work done in stretching a uniform metal wire of length $2 \ m$ to $2.004 \ m$ with an area of cross-section $10^{-6} \ m^2$ (in $J$)? [Young's modulus of the wire = $2 \times 10^{11} \ N/m^2$]

$A$ brass rod of cross-sectional area $1 \, cm^2$ and length $0.2 \, m$ is compressed lengthwise by a weight of $5 \, kg$. If Young's modulus of elasticity of brass is $1 \times 10^{11} \, N/m^2$ and $g = 10 \, m/s^2$,then the increase in the energy of the rod will be:

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: $Y_{cu} = 1 \times 10^{11}\,N/m^2$,$Y_{steel} = 2 \times 10^{11}\,N/m^2$):

The work done on a wire of volume $2 \,cm^3$ is $16 \times 10^2 \,J$. If the Young's modulus of the material of the wire is $4 \times 10^{12} \,N/m^2$, then the strain produced in the wire is:

$A$ wire of length $25 \ cm$ and radius $2 \ mm$ is fixed at one end. If a torque is applied at the other end to produce an angular displacement of $45^o$,calculate the work done in $J$. (Given: $\eta = 8 \times 10^{10} \ N/m^2$)