A) \[{{\alpha }_{1}}+{{\alpha }_{2}}\]
B) \[2{{\alpha }_{1}}+{{\alpha }_{2}}\]
C) \[{{\alpha }_{1}}+2{{\alpha }_{2}}\]
D) None of these
Correct Answer: C
Solution :
\[V={{V}_{0}}(1+\gamma \Delta \theta )\] \[{{L}^{3}}={{L}_{0}}(1+{{\alpha }_{1}}\Delta \theta )L_{0}^{2}{{(1+{{\alpha }_{2}}\Delta \theta )}^{2}}\]\[=L_{0}^{3}(1+{{\alpha }_{1}}\Delta \theta )\,{{(1+{{\alpha }_{2}}\Delta \theta )}^{2}}\] Since \[L_{0}^{3}={{V}_{0}}\] and \[{{L}^{3}}=V\] Hence \[1+\gamma \Delta \theta =(1+{{\alpha }_{1}}\Delta \theta )\,{{(1+{{\alpha }_{2}}\Delta \theta )}^{2}}\] \[\tilde{=}\,(1+{{\alpha }_{1}}\Delta \theta )\,(1+2{{\alpha }_{2}}\Delta \theta )\] \[\tilde{=}\,(1+{{\alpha }_{1}}\Delta \theta +2{{\alpha }_{2}}\Delta \theta )\] Þ g = a1 + 2a2
You need to login to perform this action.
You will be redirected in
3 sec
