Question

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

• A. $3.316\times {10}^{-5}{/}^{\circ }\text{C}$
• B. $2.316\times {10}^{-5}{/}^{\circ }\text{C}$
• C. $4.316\times {10}^{-5}{/}^{\circ }\text{C}$
• D. None of these

Answer 1

The correct option is B $2.316\times {10}^{-5}{/}^{\circ }\text{C}$

Given,
Original weight of metal $\left(m_1\right)=46\text{gm}$
Specific gravity of liquid at ${27}^{\circ }\text{C}$ is 1.24
Specific gravity of liquid at ${42}^{\circ }\text{C}$ is 1.20
Weight of metal at ${27}^{\circ }\text{C}\left(m_2\right)=30\text{gm}$
Weight of metal at ${27}^{\circ }\text{C}\left(m_3\right)=30.5\text{gm}$
Let the density of metal at ${27}^{\circ }\text{C}$ be ${\rho }_1=1.24\times {\rho }_w$ and at ${42}^{\circ }\text{C}$ be ${\rho }_2=1.20\times {\rho }_w$
$\therefore$ Loss of weight at ${27}^{\circ }\text{C}$ is
$=46-30=16=V_1\times (1.24\times {\rho }_w)\times g....\left(1\right)$
Loss of weight at ${42}^{\circ }\text{C}$
$=46-30.5=15.5=V_2\times (1.20\times {\rho }_w)\times g....\left(2\right)$
Now dividing $\left(1\right)$ by $\left(2\right)$, we get $\frac{16}{15.5}=\frac{V_1}{V_2}\times \frac{1.24}{1.2}$
But from the definition of volume expansion of a solid, we can write that,$\frac{V_2}{V_1}=1+3\alpha (t_2-t_1)$
$\therefore 1+3\alpha (42-27)=\frac{15.5\times 1.24}{16\times 1.2}=1.001042$
$\Rightarrow 45\alpha =0.001042$
So, $\alpha =2.316\times {10}^{-5}{/}^{\circ }\text{C}$
Hence, option (b) is the correct answer.