A) \[{{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}\]
B) \[{{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}\]
C) \[\alpha _{1}^{2}{{l}_{1}}=\alpha _{2}^{2}{{l}_{2}}\]
D) \[{{\alpha }_{1}}{{l}_{1}}={{\alpha }_{2}}{{l}_{2}}\]
Correct Answer: D
Solution :
\[{{L}_{2}}={{l}_{2}}(1+{{\alpha }_{2}}\Delta \theta )\] and \[{{L}_{1}}={{l}_{1}}(1+{{\alpha }_{1}}\Delta \theta )\] \[\Rightarrow ({{L}_{2}}-{{L}_{1}})=({{l}_{2}}-{{l}_{1}})+\Delta \theta ({{l}_{2}}{{\alpha }_{2}}-{{l}_{1}}{{\alpha }_{1}})\] Now \[({{L}_{2}}-{{L}_{1}})=({{l}_{2}}-{{l}_{1}})\] so, \[{{l}_{2}}{{\alpha }_{2}}-{{l}_{1}}{{\alpha }_{1}}=0\]You need to login to perform this action.
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