Physics - Heat & Thermo Question with Solution | TestHub

Volume of the metal at $30^{\circ }\mathrm{C}$, $$\begin{array}{rl}{V}_{30}& =\frac{\text{ loss of weight }}{\text{ specific gravity }\times g}\\ & =\frac{(45-25)g}{1.5\times g}=13.33{\mathrm{cm}}^3\end{array}$$ Similarly, volume of metal at $40^{\circ }\mathrm{C}$ $$V_{40}=\frac{(45-27)g}{1.25\times g}=14.40{\mathrm{cm}}^3$$ Now, $V_{40}=V_{30}\left[1+\gamma \left(t_2-t_1\right)\right]$ $$\text{ or }\begin{array}{rl}& \gamma =\frac{V_{40}-V_{30}}{V_{30}\left(t_2-t_1\right)}=\frac{14.40-13.33}{13.33(40-30)}\\ & =8.03\times 10^{-3}/{}^{\circ }\mathrm{C}\end{array}$$ $\therefore$ Coefficient of linear expansion of the metal $$\alpha =\frac{\gamma }{3}=\frac{8.03\times 10^{-3}}{3}\approx 2.6\times 10^{-3}/{}^{\circ }\mathrm{C}$$