A) \[90.0096\,cm\]
B) \[80.0272\,cm\]
C) \[1\,cm\]
D) \[25.2\,cm\]
Correct Answer: B
Solution :
From relation \[{{l}_{t}}=10\,(1+\alpha t)\] Here, \[\alpha \] steel \[=11\times {{10}^{-6}}pe{{r}^{o}}C.\] Putting the value in equation (i) we get \[{{l}_{t}}=1\times [(1+11\times {{10}^{-6}})\times ({{40}^{o}}-{{20}^{o}})]\] \[\Rightarrow \]\[{{l}_{i}}=1.00022\,cm\] Now, length of copper rod at \[{{40}^{o}}C\] \[{{i}_{t}}={{i}_{0}}.\,\,(1+\alpha t)\] Given a copper \[=17\times {{10}^{-6}}({{40}^{o}}-{{20}^{o}})\] \[\Rightarrow \] \[{{i}_{1}}=80[1+17+{{10}^{-6}}(40-{{20}^{o}})]\] \[\Rightarrow \] \[{{i}_{l}}=80.0272cm.\] Hence, number of cms observed on the scale \[\frac{{{i}_{t}}}{{{l}_{t}}}=\frac{80.272}{1.00022}=80.0096.\]You need to login to perform this action.
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