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Question Number 45127 by Necxx last updated on 09/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18
$$\left.\mathrm{23}\right)\bigtriangleup{L}_{{Al}} ={L}\propto_{{Al}} \bigtriangleup{T} \\ $$$$\bigtriangleup{L}_{{steel}} ={L}\propto_{{steel}} \bigtriangleup{T} \\ $$$${reauired}\:{space}=\frac{\mathrm{1}}{\mathrm{2}}{L}\bigtriangleup{T}\left(\propto_{{Al}} +\propto_{{steel}} \right) \\ $$$${because}\:{alternate}\:{aluminium}\:{and}\:{steel}\:{are}\:{elongated} \\ $$$${both}\:{side}... \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{44}×\left(\mathrm{1}.\mathrm{2}×\mathrm{10}^{−\mathrm{5}} +\mathrm{2}.\mathrm{4}×\mathrm{10}^{−\mathrm{5}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{44}×\mathrm{3}.\mathrm{6}×\mathrm{10}^{−\mathrm{5}} \\ $$$$=\mathrm{0}.\mathrm{95}{cm} \\ $$
Commented by Necxx last updated on 09/Oct/18
$${hmmmm}_{.._{} } {Its}\:{really}\:{clear}_{} \\ $$