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A piece of metal has a weight of $49 \mathrm{gm}$ in air and $39 \mathrm{gm}$ in a liquid of density $1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3 \mathrm{kept}$ at $32^{\circ} \mathrm{C}$. When the temperature of the liquid is raised to $45^{\circ} \mathrm{C}$ the metal piece has a weight of $40 \mathrm{gm}$. If the density of the liquid at $42{ }^{\circ} \mathrm{C}$ is $1.0 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$, then the coefficient of linear expansion of the metal is
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The correct answer is:
$\frac{8}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Given weight of metal in air $\mathrm{M}=49 \mathrm{gm}$
Density $\mathrm{d}_1=1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
At temperature, $\mathrm{T}_1=32^{\circ} \mathrm{C}$
Weight of metal in liquid $\mathrm{m}_1=39 \mathrm{gm}$ $=39 \times 10^{-3} \mathrm{~kg}$
At temperature, $\mathrm{T}_2=42^{\circ} \mathrm{C}$
Weight of metal in liquid $\mathrm{m}=40 \mathrm{gm}=40 \times 10^{-3} \mathrm{~kg}$
Density, $\mathrm{d}_2=1.0 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
$\mathrm{W}=\mathrm{Mg}-$ Buoyancy Force
$\mathrm{mg}=\mathrm{Mg}-\mathrm{d}_1 \mathrm{vg}$
$\mathrm{V}_1=\frac{\mathrm{M}-\mathrm{m}_1}{\mathrm{~d}_1}=\frac{(49-39) \times 10^{-3}}{1.2 \times 10^3}$
$=8.33 \times 10^{-6} \mathrm{~m}^3$
$\mathrm{V}_2=\frac{\mathrm{M}-\mathrm{m}_2}{\mathrm{~d}_2}=\frac{(49-40) \times 10^{-3}}{10^3}=9 \times 10^{-6} \mathrm{~m}^3$
$\Delta \mathrm{V}=3 \alpha \mathrm{V}_1 \Delta \mathrm{T}$
$(9-8.33) \times 10^{-6}=3 \times \alpha \times 8.33 \times 10^{-6} \times 10$
The coefficient of linear expansion
$$
\alpha=8 / 3 \times 10^{-3} /{ }^{\circ} \mathrm{C}
$$
Density $\mathrm{d}_1=1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
At temperature, $\mathrm{T}_1=32^{\circ} \mathrm{C}$
Weight of metal in liquid $\mathrm{m}_1=39 \mathrm{gm}$ $=39 \times 10^{-3} \mathrm{~kg}$
At temperature, $\mathrm{T}_2=42^{\circ} \mathrm{C}$
Weight of metal in liquid $\mathrm{m}=40 \mathrm{gm}=40 \times 10^{-3} \mathrm{~kg}$
Density, $\mathrm{d}_2=1.0 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
$\mathrm{W}=\mathrm{Mg}-$ Buoyancy Force
$\mathrm{mg}=\mathrm{Mg}-\mathrm{d}_1 \mathrm{vg}$
$\mathrm{V}_1=\frac{\mathrm{M}-\mathrm{m}_1}{\mathrm{~d}_1}=\frac{(49-39) \times 10^{-3}}{1.2 \times 10^3}$
$=8.33 \times 10^{-6} \mathrm{~m}^3$
$\mathrm{V}_2=\frac{\mathrm{M}-\mathrm{m}_2}{\mathrm{~d}_2}=\frac{(49-40) \times 10^{-3}}{10^3}=9 \times 10^{-6} \mathrm{~m}^3$
$\Delta \mathrm{V}=3 \alpha \mathrm{V}_1 \Delta \mathrm{T}$
$(9-8.33) \times 10^{-6}=3 \times \alpha \times 8.33 \times 10^{-6} \times 10$
The coefficient of linear expansion
$$
\alpha=8 / 3 \times 10^{-3} /{ }^{\circ} \mathrm{C}
$$
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