The work done in increasing the length of a $1 \ m$ long wire of cross-section area $1 \ mm^2$ by $1 \ mm$ will be ....... $J$ $(Y = 2 \times 10^{11} \ Nm^{-2})$

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

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

Two wires having the same length and material are stretched by the same force. Their diameters are in the ratio $1:3$. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter),when stretched,is (in $:1$)

Wires $A$ and $B$ are made from the same material. $A$ has twice the diameter and three times the length of $B$. If the elastic limits are not reached,when each is stretched by the same tension,the ratio of energy stored in $A$ to that in $B$ is

$A$ wire of Young's modulus $1.6 \times 10^{12} \,N/m^{2}$ is stretched by a force so as to produce a strain of $2 \times 10^{-4}$. The energy density of the wire is