Question

When a liquid in a glass vessel is heated, its apparent expansion is $10.30\times {10}^{-4}{/}^o$ C. Same liquid when heated in a metallic vessel, its apparent expansion is $10.06\times {10}^{-4}{/}^o$ C. The coefficient of linear expansion of metal is :

(${\alpha }_{glass}=9\times {10}^{-6}{/}^o$ C)

• A. $51\times {10}^{-6}{/}^{o}C$
• B. $43\times {10}^{-6}{/}^{o}C$
• C. $25\times {10}^{-6}{/}^{o}C$
• D. $17\times {10}^{-6}{/}^{o}C$

Answer 1

The correct option is D $17\times {10}^{-6}{/}^{o}C$

${\gamma }_r={\gamma }_{a1}+{\gamma }_{glass}={\gamma }_{a2}+{\gamma }_{metal}$ ......(1)
where ${\gamma }_{a1}$ is the apparent expansion coefficients when the the liquid is kept in the glass container and ${\gamma }_{a2}$is the apparent expansion coefficients when the the liquid is kept in the metal container
Given,
${\gamma }_{a1}=10.30\times {10}^{-4}$
${\gamma }_{a2}=10.06\times {10}^{-4}$
${\alpha }_{glass}=9\times {10}^{-6}$
${\gamma }_{glass}=3{\alpha }_{glass}=27\times {10}^{-6}$
${\gamma }_{metal}=3{\alpha }_{metal}$

Substituting the values in (1)
$10.30\times {10}^{-4}+27\times {10}^{-6}=10.06\times {10}^{-4}+3{\alpha }_{metal}$
$\Rightarrow 24\times {10}^{-6}+27\times {10}^{-6}=3{\alpha }_{metal}$
$\Rightarrow {\alpha }_{metal}=\left(\frac{51}{3}\right){10}^{-6}=17\times {10}^{-6}$