A) \[9\times {{10}^{-6}}/{}^\circ C\]
B) \[6\times {{10}^{-6}}/{}^\circ C\]
C) \[36\times {{10}^{-6}}/{}^\circ C\]
D) \[27\times {{10}^{-6}}/{}^\circ C\]
Correct Answer: A
Solution :
\[{{\gamma }_{real}}={{\gamma }_{app.}}+{{\gamma }_{vessel}}\] |
So \[{{\left( {{\text{ }\!\!\gamma\!\!\text{ }}_{\text{app}\text{.}}}\text{+ }{{\text{ }\!\!\gamma\!\!\text{ }}_{\text{vessel}}} \right)}_{\text{glass}}}\text{=}{{\left( {{\text{ }\!\!\gamma\!\!\text{ }}_{\text{app}\text{.}}}\text{+ }{{\text{ }\!\!\gamma\!\!\text{ }}_{\text{vessel}}} \right)}_{\text{steel}}}\] |
\[\Rightarrow \,\text{153 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{+}{{\left( {{\text{ }\!\!\gamma\!\!\text{ }}_{\text{vessel}}} \right)}_{\text{glass}}}\text{=}{{\left( \text{144 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{+}{{\text{ }\!\!\gamma\!\!\text{ }}_{\text{vessel}}} \right)}_{\text{steel}}}\] |
\[Further,{{\left( {{\gamma }_{vessel}} \right)}_{steel}}=3\alpha =3\times \left( 12\times {{10}^{-6}} \right)=36\times {{10}^{-6}}/{}^\circ C\]\[\Rightarrow \,153\times {{10}^{-6}}+{{\left( {{\gamma }_{veseel}} \right)}_{glass}}={{\left( 144\times {{10}^{-6}}+{{\gamma }_{vessel}} \right)}_{steel}}\]\[\Rightarrow {{\left( {{\gamma }_{vessel}} \right)}_{glass}}=3\alpha =27\times {{10}^{-6}}+36\times {{10}^{-6}}\] |
\[\Rightarrow \alpha =9\times {{10}^{-6}}/{}^\circ C\] |
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