An aluminium rod with Young’s modulus Y = 7 . 0 × 10 ^ 10 N m...

The data given is

$Y = 7 \times {10}^{10}N{m}^{- 2}\text{and}\dfrac{∆ l}{l} = \dfrac{0 . 04}{100}$

Young's modulus is given by the ratio of stress and strain,

$Y = \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}} = \dfrac{Fl}{A ∆ l}$

The energy stored is given as

$E = \left(\dfrac{YA}{2l}\right) ∆ {l}^{2} ⇒ E = \left(\dfrac{YA}{2}\right){\left(\dfrac{∆ l}{l}\right)}^{2} \times l . . . ( i )$

The energy per unit volume can be written from equation (i)

$\dfrac{E}{V} = \dfrac{Y}{2} \times {\left(\dfrac{∆ l}{l}\right)}^{2} . . . ( ii )$

Substituting the values in equation (ii)

$\dfrac{E}{V} = \dfrac{1}{2} \times 7 \times {10}^{10} \times \dfrac{0 . 04 \times 0 . 04}{{10}^{4}} = 56 \times {10}^{2}$