Question

A piece of metal weighs $46\text{g}$ in air. When immersed in a liquid of specific gravity $1.24$ at ${27}^{\circ }\text{C}$, it weighs $30\text{g}$. When the temperature of liquid is raised to ${42}^{\circ }\text{C}$, the metal piece weighs $30.5\text{g}$. Specific gravity of liquid at ${42}^{\circ }\text{C}$ is $1.20$. Calculate the coefficient of linear expansion of metal

• A. $2.23\times {10}^{-5}{}^o{\text{C}}^{-1}$
• B. $6.7\times {10}^{-5}{}^o{\text{C}}^{-1}$
• C. $4.46\times {10}^{-5}{}^o{\text{C}}^{-1}$
• D. none of these

Answer 1

The correct option is A $2.23\times {10}^{-5}{}^o{\text{C}}^{-1}$

Let volume of metal piece be $V_1$ at $t_1^{\circ }\text{C}(={27}^{\circ }\text{C})$ and $V_2$ at $t_2^{\circ }\text{C}(={42}^{\circ }\text{C})$.
Given weight of metal piece in liquid at ${27}^{\circ }\text{C}=30\text{g}$.
As weight of metal piece in air $=46\text{g}$, hence loss of weight of metal piece in liquid $=46-30=16\text{g}=$ weight of liquid displaced = volume of liquid $\times$ density.
$\therefore 16=V_1\times 1.24$ or $V_1=\frac{16}{1.24}{\text{cm}}^3$
Similarlry, $V_2=\frac{46-30.5}{1.20}=\frac{15.5}{1.20}{\text{cm}}^3$
Now, $V_{42}=V_{27}(1+\gamma \Delta \theta )$
or $V_2=V_1(1+\gamma \Delta \theta )=V_1(1+\gamma \times 15)$
$\therefore 1+15\gamma =\frac{V_2}{V_1}=\frac{15.5/1.20}{16/1.24}=1.0010$
$\therefore \gamma =6.7\times {10}^{-5}{}^o{\text{C}}^{-1}$
$\therefore \alpha =\frac{\gamma }{3}=2.23\times {10}^{-5}{}^o{\text{C}}^{-1}$
Hence, the correct answer is option (a).