The densities of wood and benzene at $0^{\circ} \mathrm{C}$ are $880 \mathrm{~kg} \mathrm{~m}^{-3}$ and $900…
The densities of wood and benzene at $0^{\circ} \mathrm{C}$ are $880 \mathrm{~kg} \mathrm{~m}^{-3}$ and $900 \mathrm{~kg} \mathrm{~m}^{-3}$, respectively. The coefficient of volume expansion is $1.2 \times 10^{-3} \mathrm{C}^{-10}$ for wood and $1.5 \times 10^{-30} \mathrm{C}^{-1}$ for benzene. Then the temperature at which a piece of wood just sinks in benzene is
$88^{\circ} \mathrm{C}$
$90^{\circ} \mathrm{C}$
$83.3^{\circ} \mathrm{C}$
$90.3^{\circ} \mathrm{C}$
Solution
Given, density of wood at $0^{\circ} \mathrm{C}, \rho_w=880 \mathrm{Kg} \mathrm{m}^{-3}$ density of benzene at $0^{\circ} \mathrm{C}, \rho_b=900 \mathrm{~kg} \mathrm{~m}^{-3}$ coefficient of volume expansion of wood,
$
\gamma_w=1.2 \times 10^{-3 \circ} \mathrm{C}^{-1}
$
coefficient of volume expansion of benzene,
$
\gamma_b=1.5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}
$
and initial temperature, $T_1=0^{\circ} \mathrm{C}$
Let $T_2$ be the temperature at which pieces of wood will just sink in benzene and $\Delta T=T_2-T_1$
The piece of wood begins to sink when it weight is equal to the weight of benzene displaced.
$
\text { Mass }=\text { Volume } \times \text { Density }
$
Therefore, $V \rho_w g=V \rho_b g$
$
\begin{aligned}
& \therefore \quad \frac{\rho_w}{1+\gamma_w \Delta T}=\frac{\rho_b}{1+\gamma_b \Delta T} \\
& \frac{880}{1+1.2 \times 10^{-3} \Delta T}=\frac{900}{1+1.5 \times 10^{-3} \Delta \mathrm{T}} \\
& 880+880 \times 1.5 \times 10^{-3} \Delta T=900+900 \times 1.2 \times 10^{-3} \Delta T \\
&(1320-1080) \times 10^{-3} \Delta T=20 \\
& \Delta T=\frac{20}{240 \times 10^{-3}} \\
& \Delta T=83.3^{\circ} \mathrm{C} \\
& T_2-T_1=83.3^{\circ} \mathrm{C}
\end{aligned}
$
Hence, $\quad T_2=83.3^{\circ} \mathrm{C}$