Q. When a liquid in a glass vessel is heated, is apparent expansion is$10.30 \times 10^{- 4}℃^{- 1}$. When the same liquid is heated in a metal vessel, its apparent expansion is $10.06 \times 10^{- 4}℃^{- 1}.$ If the coefficient of linear expansion of glass = $9 \times 10^{- 6}℃^{- 1},$ what is the coefficient of linear expansion of metal?
A) $51\ \times 10^{- 6}℃^{- 1}$
B) $17\ \times 10^{- 6}℃^{- 1}$
C) $25\ \times 10^{- 6}℃^{- 1}$
D) $43\ \times 10^{- 6}℃^{- 1}$
Explanation
(b) Here, $\gamma_{ag} = 10.30\ \times 10^{- 4}℃^{- 1}$
$\gamma_{am} = 10.06 \times 10^{- 4}\ ℃^{- 1}$
$\alpha_{a} = 9 \times 10^{- 6}℃^{- 1},\ \alpha_{m} = \ ?$
Now , $\gamma_{r} = \ \gamma_{ag} + g_{glass} = \ \gamma_{am} + g_{m}$
$\therefore 10.30 \times 10^{- 4} + 3 \times 9 \times 10^{- 6} = 10.06 \times 10^{- 4} + g_{m}\ \ \ \ \ \ \lbrack\because g_{g} = 3 \times \alpha_{a}\rbrack$
$\therefore\ g_{m} = (10.30 + 0.27 - 10.06)10^{- 4} = 0.51 \times 10^{- 4}$
$\alpha_{m} = \frac{1}{3}g_{m} = \ \frac{0.50 \times 10^{- 4}}{3}$
= 0.17$\times 10^{- 4} = 17 \times 10^{- 6}℃^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.