Q. A piece of metal weighs $45\ g$ in air and $25\ g$ in a liquid of density $1.5 \times 10^{3}kg - m^{- 3}$ kept at $30℃$. When the temperature of the liquid is raised to $40℃$, the metal piece is weighs $27\ g.$ The density of liquid at $40℃$, is $1.25 \times 10^{3}kg - m^{- 3}.$ The coefficient of linear expansion of metal is
A) $1.3 \times 10^{- 3}/℃$
B) $5.2 \times 10^{- 3}/℃$
C) $2.6 \times 10^{- 3}/℃$
D) $0.26 \times 10^{- 3}/℃$
Explanation
(c) The volume of the metal at $30℃$ is
$V_{30} = \frac{loss\ of\ weight}{Specific\ gravity\ \times g} = \frac{(45 - 25)g}{1.5 \times g} = 13.33\ cm^{3}$
Similarly, Volume of metal at $40℃$ is
$V_{40} = \frac{(45 - 27)g}{1.25 \times g} = 14.40cm^{3}$
Now, $V_{40} = V_{30}\lbrack 1 + \gamma(t_{2} - t_{1})\rbrack$
$\Rightarrow \gamma = \frac{V_{40} - V_{30}}{V_{30}(t_{2} - t_{1})} = \frac{14.40 - 13.33}{13.33(40 - 30)} = 8.03 \times 10^{- 3}/℃$
$\therefore$ Coefficient of linear expansion of the metal is
$\alpha = \frac{\gamma}{3} = \frac{8.03 \times 10^{- 3}}{3} = 2.6 \times 10^{- 3}/℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.