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| 1. |
A metallic rod breaks when strain produced in the rod is 0.2%. The young's modulus of the material of the rod is 7 xx 10^9 N/m^2. What slould be it's area of cross section to support a load of 10^4 N. |
| Answer» Solution :Breaking stress = Y xx breaking STRAIN = `7 xx 10^9 xx 2/1000 = 14 xx 10^6 N/m^2` `there for` but breaking stress F/A `there for` `F/ "breaking stress" = `10^4/(14 xx 10^6) = 7.14 xx 10^-4 m^2` | |