Properties Of Solids And Liquids Question 405
Question: A bar of iron is 10 cm at $ 20{}^\circ C $ . At $ 19{}^\circ C $ it will be ($ \alpha $ of iron$ =11\times {{10}^{-6}}/{}^\circ C $ )
Options:
A) $ 11\times {{10}^{-6}} $ cm longer
B) $ 11\times {{10}^{-6}} $ cm shorter
C) $ 11\times {{10}^{-5}} $ cm shorter
D) $ 11\times {{10}^{-5}} $ cm longer
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ L=L _{0}( 1+\alpha \Delta \theta )\Rightarrow \frac{L _{1}}{L _{2}}=\frac{1+\alpha {{( \Delta \theta )} _{1}}}{1+\alpha {{( \Delta \theta )} _{2}}} $
$ \Rightarrow \frac{10}{L _{2}}=\frac{1+11\times {{10}^{-6}}\times 20}{1+11\times {{10}^{-6}}\times 19}\Rightarrow L _{2}=9.99989 $
$ \Rightarrow $ Length is shorten by $ 10-9.99989=0.00011=11\times {{10}^{-5}}cm. $
BETA
