The strain energy stored in a body of volume $V$ due to shear strain $\phi$ is (shear modulus is $\eta$ )

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 of the wire and the work done in displacing it through $1.5\, mm$ by the weight is

An Indian rubber cord $L$ metre long and area of cross-section $A$ $metr{e^2}$ is suspended vertically. Density of rubber is $D$ $kg/metr{e^3}$ and Young's modulus of rubber is $E$ $newton/metr{e^2}$. If the wire extends by $l$ metre under its own weight, then extension $l$ is

The work per unit volume to stretch the length by $1\%$ of a wire with cross sectional area of $1\,m{m^2}$ will be. $[Y = 9 \times {10^{11}}\,N/{m^2}]$

The work done per unit volume to stretch the length of area of cross-section $2 \,mm ^2$ by $2 \%$ will be ……. $MJ / m ^3$ $\left[Y=8 \times 10^{10} \,N / m ^2\right]$