Physics Elasticity Question 136
Question: The coefficient of linear expansion of brass and steel are $ {{\alpha } _{1}} $ and $ {{\alpha } _{2}} $ . If we take a brass rod of length $ l _{1} $ and steel rod of length $ l _{2} $ at 0°C, their difference in length $ (l _{2}-l _{1}) $ will remain the same at a temperature if
[EAMCET (Med.) 1995]
Options:
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} $
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Answer:
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 $
BETA
