question_answer 9)

is cooled to a temperature of \[-{{39}^{o}}C\], what is the tension developed in the wire, if the diameter is 2.0 mm? Coefficient of linear expansion of brass \[=2.0\times {{10}^{-5}}\,{{C}^{-1}},\] and Young's modulus of brass \[{{\text{l}}_{\text{1}}}\text{=l;}{{\text{A}}_{\text{1}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\]

Answer:

Here, \[{{\text{Y}}_{\text{1}}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\text{N}{{\text{m}}^{\text{-2}}}\] \[\text{B,}{{\text{l}}_{\text{2}}}\text{=l;}{{\text{A}}_{\text{2}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\]\[{{\text{Y}}_{\text{2}}}\text{=7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}\text{N}{{\text{m}}^{\text{-2}}}\]\[{{\text{F}}_{\text{1}}}\text{ and }{{\text{F}}_{\text{2}}}\] \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{A}}_{\text{1}}}}\text{=}\frac{{{\text{F}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}}\]\[\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{\text{1}}{\text{2}}\]\[{{F}_{1}}x={{F}_{2}}\left( 1\cdot 05-x \right)\] \[\frac{1\cdot 05-x}{x}=\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{\text{1}}{\text{2}}\] From \[2\cdot 10-2x=x\] \[x=0\cdot \text{70 m=70 cm}\] Also, \[{{\text{F}}_{\text{1}}}\text{ and }{{\text{F}}_{\text{2}}}\] \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{A}}_{\text{1}}}{{\text{Y}}_{\text{1}}}}\text{=}\frac{{{\text{F}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\]or \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{{{\text{A}}_{\text{1}}}{{\text{Y}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\text{=}\] \[\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}}{\text{7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}}\text{=}\frac{\text{10}}{\text{7}}\] \[{{F}_{1}}x={{F}_{2}}\left( 1\cdot 05-x \right)\] Negative sign indicates that the force is inwards due to contraction of the wire.