A) \[R={{R}_{0}}(6T+2)\]
B) \[R=2{{R}_{0}}(3+2T)\]
C) \[R={{R}_{0}}(1+{{T}^{2}}+{{T}^{3}})\]
D) \[R={{R}_{0}}(1-T+{{T}^{2}}+{{T}^{3}})\]
Correct Answer: C
Solution :
[c] \[\alpha (T)=\frac{1}{{{R}_{0}}}\frac{dR}{dT}\]or \[(3{{T}^{2}}+2T)=\frac{1}{{{R}_{0}}}\frac{dR}{dT}\] Or \[dR={{R}_{0}}(3{{T}^{2}}+2T)dT\] Or \[\int\limits_{{{R}_{0}}}^{R}{dR={{R}_{0}}\left[ 3\int\limits_{0}^{T}{{{T}^{2}}dT+2\int\limits_{0}^{T}{TdT}} \right]}\] Or \[R={{R}_{0}}[1+{{T}^{2}}+{{T}^{3}}]\]
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