An aluminium rod `(Y = 7xx10^9 N/m^2)` has a breaking strain of `0.2%`. What is minimum cross sectional area to support a load of `10^4` N ?

Explanation

`Y = (M g L)/(A Delta l)` there for `A = (M g L)/(Y Delta l)` But `(Deltal)/L` = `0.2 xx 10^-2` = Breaking strain. There for A = `(M g L)/(Y Delta l)` = `(10^4)/ (7 xx 10^9 xx 2 xx 10^-3)` `10^-2/ 14` = `7.1 xx 10^-4m^2`