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A steel rod at 25^(@)C is bolted at both ends and then cooled. By how many^(@)C should the rod be cooled so that it will rupture? Assume that till rupture, Hooke's law is obeyed. (alpha_("steel")=10xx10^(-6)//^(@)C,Y=2xx10^(11)N//m^(2)" and "sigma_(b)("breaking stress of steel rod")=4xx10^(8)N//m^(2)) |
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Answer» SOLUTION :`Deltal=alphalDeltaT` `(Deltal)/(l)=alphaDeltaT` `sigma_(b)=(T)/(A)=(yDeltal)/(l)=yalphaDeltaT=DeltaT=(sigma_(b))/(yalpha)=(4xx10^(8))/(2xx10^(11)xx10^(-5))=200^(@)C` |
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