A) \[\gamma = 3\alpha \]
B) \[\alpha = 3\gamma \]
C) \[\beta = 3\alpha \]
D) \[\gamma = 3\beta \]
Correct Answer: A
Solution :
[a] \[\operatorname{V}+\Delta V={{(L+\Delta L)}^{3}}={{(L+\alpha L\Delta T)}^{3}}\] \[={{L}^{3}}+(1+3\,\alpha \,\Delta T+3\,{{\alpha }^{2}}\Delta {{T}^{2}}+{{\alpha }^{3}}\Delta {{T}^{3}})\] \[\Rightarrow {{\alpha }^{2}}\]and \[{{\alpha }^{3}}\]terms are neglected. \[\therefore V(l+\gamma \,\Delta T)= V(1+3\,\alpha \,\Delta T)\] \[\operatorname{l}+\gamma \Delta T=l+3\alpha \Delta T\therefore \gamma =3\alpha .\].
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