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  • A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 degree C, if the original lengths are at 40.0 degree C,

Q. 11.10 A brass rod of length 50\; cm and diameter 3.0\; mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250^{\circ}C,  if the original lengths are at 40.0^{\circ}C\; ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear  expansion\; of\; brass=2.0\times 10^{-5}K^{-1},\; steel=1.2\times 10^{-5}K^{-1}).

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Length of the rods l = 50 cm

Co-efficient of linear expansion of brass, \\\alpha _{b}=2\times 10^{-5}K^{-1}

Co-efficient of linear expansion of steel, \\\alpha _{s}=1.2\times 10^{-5}K^{-1}

Initial Temperature T1 = 40.0 oC

Final Temperature T2 = 250 oC

Change in length of brass rod is

\\\Delta l_{t}=l_{t}\alpha_{b} \Delta T\\ \Delta l_{t}=50\times 2.0\times 10^{-5}\times (250-40)\\ \Delta l_{t}=0.21cm

Change in length of the steel rod is

\\\Delta l_{s}=l_{s}\alpha_{s} \Delta T\\ \Delta l_{s}=50\times 1.2\times 10^{-5}\times (250-40)\\ \Delta l_{s}=0.126cm

Change in length of the combined rod is

\\\Delta l=\Delta l_{s}+\Delta l_{b}\\ \Delta l=0.126+0.21\\ \Delta l=0.336cm

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