The young's modulus of the material of a rod is `20xx10^(10)` pascal. When the longitudal strain is `0.04%`. The energy stored per unit volume is

Options

A. `16xx10^(3)J//m^(3)`
B. `8xx10^(3)J//m^(3)`
C. `16xx10^(2)J//m^(3)`
D. `0.8xx10^(3)J//m^(3)`

Answer

A

Explanation

`u=(1)/(2) y("strain")^(2)`
`=(1)/(2) xx 20 xx 10^(10) xx (4 xx 10^(-4))^(2)`
`=10^(11) xx 16xx10^(-8)`
`=16 xx10^(3)J//m^(3)`.