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InterviewSolution
| 1. |
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 ? |
| Answer» Solution :`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 FORA = `(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` | |