The elastic potential energy stored in a copper rod of length $1 \,m$ and area of cross-section $1 \,mm^2$ when stretched by $1 \,mm$ is (Young's modulus of copper $= 1.2 \times 10^{11} \,N/m^2$).

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

Two wires of the same diameter and the same material have lengths $l$ and $2l$. If the same force $F$ is applied to each,what is the ratio of the work done in stretching the two wires?

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 = $2 \times 10^{11} \ N/m^2$]

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

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 $:1$)