A) \[\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{1}}}{{\text{Y}}_{\text{1}}}}{{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\]
B) \[\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{{{\text{L}}_{1}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{1}}}{{\text{Y}}_{\text{1}}}}{{{\text{L}}_{2}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{2}}}{{\text{Y}}_{\text{2}}}}\]
C) \[\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{{{\text{L}}_{2}}{{\text{ }\!\!\alpha\!\!\text{ }}_{2}}{{\text{Y}}_{2}}}{{{\text{L}}_{1}}{{\text{ }\!\!\alpha\!\!\text{ }}_{1}}{{\text{Y}}_{1}}}\]
D) \[\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{{{\text{ }\!\!\alpha\!\!\text{ }}_{2}}{{\text{Y}}_{2}}}{{{\text{ }\!\!\alpha\!\!\text{ }}_{1}}{{\text{Y}}_{1}}}\]
Correct Answer: D
Solution :
\[Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T/A}{\Delta \ell /\ell }\] |
\[\text{T=}\frac{\text{Y}\text{. }\!\!\Delta\!\!\text{ l}}{\text{l}}\text{A=Y}\text{.A }\!\!\alpha\!\!\text{ }\!\!\Delta\!\!\text{ T}\] |
In both the rods tension will be same so |
\[{{\text{T}}_{\text{1}}}\text{ = }{{\text{T}}_{\text{2}}}\] |
Hence \[{{\text{Y}}_{\text{1}}}{{\text{A}}_{\text{1}}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{1}}}\text{ }\!\!\Delta\!\!\text{ T = }{{\text{Y}}_{\text{2}}}{{\text{A}}_{\text{2}}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{2}}}\text{ }\!\!\Delta\!\!\text{ T}\] |
\[\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{{{\text{Y}}_{\text{2}}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{2}}}}{{{\text{Y}}_{\text{1}}}{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{1}}}}\] |
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