Within the elastic limit, stress of brass is `3.5xx10^(10)N//m^(2)`. Find the maximum load that can be applied to a brass wire of 0.75 mm diameter without exceeding the elastic limit.

Explanation

`"Stress"=(F)/(A)`
For elastic limit,
`"Stress"=3.5xx10^(10)N//m^(2)` (given)
Thus `F=(pir^2)xx"stress"`
`=3.14xx((0.75)/(2)xx10^(-3))^(2)xx3.5xx10^(10)`
`=3.14xx((0.75)/(2))^(2)xx3.5xx10^(4)`
`=1.55xx10^(4)N`