A rod of length 1000, mm and coefficient of l | vedclass

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(A) The free expansion of the rod due to temperature increase is given by $\Delta \ell_{free} = \ell \alpha \Delta T$.Given $\ell = 1000\, mm = 1\, m$

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(A) The free expansion of the rod due to temperature increase is given by $\Delta \ell_{free} = \ell \alpha \Delta T$.Given $\ell = 1000\, mm = 1\, m$,$\alpha = 10^{-4} / ^\circ C$,and $\Delta T = 20^\circ C$.$\Delta \ell_{free} = 1 	imes 10^{-4} 	imes 20 = 20 	imes 10^{-4}\, m = 2\, mm$.The gap between the rod and the walls is $1001\, mm - 1000\, mm = 1\, mm$.Since the rod expands by $2\, mm$ but only has $1\, mm$ of free space,it will be compressed by $\Delta \ell_{comp} = 2\, mm - 1\, mm = 1\, mm = 10^{-3}\, m$.The stress $\sigma$ developed in the rod is given by $\sigma = Y 	imes 	ext{strain} = Y 	imes \frac{\Delta \ell_{comp}}{\ell}$.$\sigma = 10^{11} 	imes \frac{10^{-3}}{1} = 10^8\, N/m^2$.Since $1\, MPa = 10^6\, N/m^2$,we have $\sigma = 100\, MPa$.

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