Properties Of Solids And Liquids Question 470
A bar of iron is 10 cm at 20°C. At 19°C it will be (a of iron = 11 ´ 10^-6/°C) [EAMCET 1997]
Options:
A) 11 × 10⁻⁶ cm longer
B) 11 × 10⁻⁶ cm shorter
C) 11 × 10⁻⁵ cm shorter
D) 11 × 10⁻⁵ cm longer
Show Answer
Answer:
Correct Answer: C
Solution:
$ L=L _{0}(1+\alpha \Delta T ) $
therefore $ \frac{L _{1}}{L _{2}}=\frac{1+\alpha {{(\Delta \theta )} _{1}}}{1+\alpha {{(\Delta \theta )} _{2}}} $
therefore $ \frac{10}{L _{2}}=\frac{1+11\times {{10}^{-6}}\times 20}{1+11\times {{10}^{-6}}\times 19} $
therefore $ L _{2}=9.99989 $
therefore Length is shortened by $ 10-9.99989=0.00011=11\times {{10}^{-4}}cm $
BETA
