Question Number 209853 by peter frank last updated on 23/Jul/24
Commented by mr W last updated on 23/Jul/24
$${the}\:{iron}\:{rod}\:{must}\:{be}\:{of}\:{a}\:{length}\:{of} \\ $$$$\mathrm{44482}\:{cm}! \\ $$
Answered by mr W last updated on 23/Jul/24
$${L}_{{B}} ={length}\:{of}\:{brass}\:{bar}\:{at}\:\mathrm{10}°{C} \\ $$$${L}_{{I}} ={length}\:{of}\:{iron}\:{bar}\:{at}\:\mathrm{10}°{C} \\ $$$${L}_{{I}} −{L}_{{B}} =\mathrm{14}\:\:\:\:\:…\left({i}\right) \\ $$$${at}\:\mathrm{100}°{C}: \\ $$$${length}\:{of}\:{brass}\:{bar}:\: \\ $$$${L}_{{B}\mathrm{100}} ={L}_{{B}} +{L}_{{B}} ×\left(\mathrm{100}−\mathrm{10}\right)×\mathrm{19}×\mathrm{10}^{−\mathrm{6}} \\ $$$${length}\:{of}\:{iron}\:{bar}:\: \\ $$$${L}_{{I}\mathrm{100}} ={L}_{{I}} +{L}_{{I}} ×\left(\mathrm{100}−\mathrm{10}\right)×\mathrm{12}×\mathrm{10}^{−\mathrm{6}} \\ $$$${L}_{{B}\mathrm{100}} −{L}_{{I}\mathrm{100}} =\mathrm{14} \\ $$$${L}_{{B}} +{L}_{{B}} ×\left(\mathrm{100}−\mathrm{10}\right)×\mathrm{19}×\mathrm{10}^{−\mathrm{6}} −{L}_{{I}} −{L}_{{I}} ×\left(\mathrm{100}−\mathrm{10}\right)×\mathrm{12}×\mathrm{10}^{−\mathrm{6}} =\mathrm{14}\:\:\:…\left({ii}\right) \\ $$$${L}_{{B}} ×\mathrm{90}×\mathrm{19}−{L}_{{I}} ×\mathrm{90}×\mathrm{12}=\mathrm{28}×\mathrm{10}^{\mathrm{6}} \\ $$$$\Rightarrow\left({L}_{{I}} −\mathrm{14}\right)×\mathrm{90}×\mathrm{19}−{L}_{{I}} ×\mathrm{90}×\mathrm{12}=\mathrm{28}×\mathrm{10}^{\mathrm{6}} \\ $$$$\Rightarrow{L}_{{I}} \approx\mathrm{44482}\:{cm},\:{L}_{{B}} \approx\mathrm{44468}\:{cm} \\ $$$${check}\:{at}\:\mathrm{100}°{C}: \\ $$$${iron}\:{bar}:\:\mathrm{44482}+\mathrm{44482}×\mathrm{90}×\mathrm{12}×\mathrm{10}^{−\mathrm{6}} =\mathrm{44530}\:{cm} \\ $$$${brass}\:{bar}:\:\mathrm{44468}+\mathrm{44468}×\mathrm{90}×\mathrm{19}×\mathrm{10}^{−\mathrm{6}} =\mathrm{44544}\:{cm} \\ $$
Commented by mr W last updated on 23/Jul/24
Commented by mr W last updated on 23/Jul/24
$${at}\:\mathrm{10}°{C}\:{the}\:{difference}\:{in}\:{their} \\ $$$${lengthes}\:{is}\:\mathrm{14}\:{cm}. \\ $$$${at}\:\mathrm{100}°{C}\:{the}\:{difference}\:{in}\:{their} \\ $$$${lengthes}\:{should}\:{also}\:{be}\:\mathrm{14}\:{cm}. \\ $$$${this}\:{is}\:{only}\:{possible}\:{when}\:{at}\:\mathrm{10}°{C} \\ $$$${the}\:{brass}\:{rod}\:{is}\:{shorter}\:{than}\:{the} \\ $$$${iron}\:{rod}\:{and}\:{at}\:\mathrm{100}°{C}\:{the}\:{brass}\:{rod} \\ $$$${is}\:{longer}\:{than}\:{the}\:{iron}\:{rod}.\:{but}\:{their} \\ $$$${difference}\:{is}\:{in}\:{both}\:{cases}\:\mathrm{14}\:{cm}. \\ $$
Answered by Spillover last updated on 23/Jul/24
![l_B =length of brass l_I =length of iron α_B =19×10^(−6) /k α_I =12×10^(−6) /k l_B −l_I =14cm or l_I − l_B =14cm α_B =((△l_B )/(l_B △θ)) l_B =((△l_B )/(α_B △θ)) α_I =((△l_I )/(l_I △θ)) l_I =((△l_I )/(α_I △θ)) l_I −l_B =((△l_I )/(α_I △θ)) −((△l_B )/(α_B △θ)) 14=((△l_I )/(α_I △θ)) −((△l_B )/(α_B △θ)) △θ=100−10=90°C 14=((△l_I )/(12×10^(−6) /k×90))−((△l_B )/(19×10^(−6) /k×90)) 14=925.93△l_I −584.8△l_B N.B for the difference to remain the same △l_I must be equal to △l_B [△l_I =△l_B ] 14=925.93△l_I −584.8△l_B 14=925.93△l_I −584.8△l_I 14=341.13△l_I △l_I =0.041cm α_I =((△l_I )/(l_I △θ)) l_I =((△l_I )/(α_I △θ)) =((0.041)/(12×10^(−6) ×90))=37.96 l_I =37.96≈38 Length of the iron must be 38cm](https://www.tinkutara.com/question/Q209867.png)
$${l}_{{B}} ={length}\:{of}\:{brass}\:\:\:\:\:\:\:\:\:\:\:{l}_{{I}} ={length}\:{of}\:{iron} \\ $$$$\alpha_{{B}} =\mathrm{19}×\mathrm{10}^{−\mathrm{6}} /{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{{I}} =\mathrm{12}×\mathrm{10}^{−\mathrm{6}} /{k} \\ $$$${l}_{{B}} −{l}_{{I}} =\mathrm{14}{cm}\:\:\:\:\:\:\:\:{or}\:\:\:\:{l}_{{I}} −\:{l}_{{B}} =\mathrm{14}{cm} \\ $$$$\alpha_{{B}} =\frac{\bigtriangleup{l}_{{B}} }{{l}_{{B}} \bigtriangleup\theta}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{l}_{{B}} =\frac{\bigtriangleup{l}_{{B}} }{\alpha_{{B}} \bigtriangleup\theta}\:\: \\ $$$$\alpha_{{I}} =\frac{\bigtriangleup{l}_{{I}} }{{l}_{{I}} \bigtriangleup\theta}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{l}_{{I}} =\frac{\bigtriangleup{l}_{{I}} }{\alpha_{{I}} \bigtriangleup\theta} \\ $$$${l}_{{I}} −{l}_{{B}} =\frac{\bigtriangleup{l}_{{I}} }{\alpha_{{I}} \bigtriangleup\theta}\:\:−\frac{\bigtriangleup{l}_{{B}} }{\alpha_{{B}} \bigtriangleup\theta} \\ $$$$\mathrm{14}=\frac{\bigtriangleup{l}_{{I}} }{\alpha_{{I}} \bigtriangleup\theta}\:\:−\frac{\bigtriangleup{l}_{{B}} }{\alpha_{{B}} \bigtriangleup\theta}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bigtriangleup\theta=\mathrm{100}−\mathrm{10}=\mathrm{90}°{C} \\ $$$$\mathrm{14}=\frac{\bigtriangleup{l}_{{I}} }{\mathrm{12}×\mathrm{10}^{−\mathrm{6}} /{k}×\mathrm{90}}−\frac{\bigtriangleup{l}_{{B}} }{\mathrm{19}×\mathrm{10}^{−\mathrm{6}} /{k}×\mathrm{90}}\: \\ $$$$\mathrm{14}=\mathrm{925}.\mathrm{93}\bigtriangleup{l}_{{I}} −\mathrm{584}.\mathrm{8}\bigtriangleup{l}_{{B}} \\ $$$${N}.{B} \\ $$$${for}\:{the}\:{difference}\:{to}\:{remain}\:{the}\:{same} \\ $$$$\bigtriangleup{l}_{{I}} \:{must}\:{be}\:{equal}\:{to}\:\bigtriangleup{l}_{{B}} \:\:\:\:\:\:\:\:\:\:\:\:\left[\bigtriangleup{l}_{{I}} \:=\bigtriangleup{l}_{{B}} \right] \\ $$$$\mathrm{14}=\mathrm{925}.\mathrm{93}\bigtriangleup{l}_{{I}} −\mathrm{584}.\mathrm{8}\bigtriangleup{l}_{{B}} \\ $$$$\mathrm{14}=\mathrm{925}.\mathrm{93}\bigtriangleup{l}_{{I}} −\mathrm{584}.\mathrm{8}\bigtriangleup{l}_{{I}} \\ $$$$\mathrm{14}=\mathrm{341}.\mathrm{13}\bigtriangleup{l}_{{I}} \\ $$$$\bigtriangleup{l}_{{I}} =\mathrm{0}.\mathrm{041}{cm} \\ $$$$\alpha_{{I}} =\frac{\bigtriangleup{l}_{{I}} }{{l}_{{I}} \bigtriangleup\theta}\:\:\:\:\:\: \\ $$$$\:{l}_{{I}} =\frac{\bigtriangleup{l}_{{I}} }{\alpha_{{I}} \bigtriangleup\theta}\:\:=\frac{\mathrm{0}.\mathrm{041}}{\mathrm{12}×\mathrm{10}^{−\mathrm{6}} ×\mathrm{90}}=\mathrm{37}.\mathrm{96} \\ $$$$\:{l}_{{I}} =\mathrm{37}.\mathrm{96}\approx\mathrm{38} \\ $$$${Length}\:{of}\:{the}\:{iron}\:{must}\:{be}\:\mathrm{38}{cm} \\ $$$$ \\ $$