Question

A cylindrical block of wood floats vertically with $80\%$ of its volume immersed in a liquid at $0^{\circ }\text{C}$. When the temperature of the liquid is raised to ${62.5}^{\circ }\text{C}$, the block just sinks in the liquid. The coefficient of cubical expansion of the liquid is

• A. $1\times {10}^{-3}{\text{K}}^{-1}$
• B. $2\times {10}^{-3}{\text{K}}^{-1}$
• C. $3\times {10}^{-3}{\text{K}}^{-1}$
• D. $4\times {10}^{-3}{\text{K}}^{-1}$

Answer 1

The correct option is D $4\times {10}^{-3}{\text{K}}^{-1}$

Given:
Percentage of volume of cylindrical block immersed in liquid $=80\%$
Initial temperature of liquid, $T_1=0^{\circ }\text{C}$
Final temperature of liquid, $T_2={62.5}^{\circ }\text{C}$
To find:
Coefficient of cubical expansion of liquid, $\gamma =?$
We know that,
weight of cylinder $=$ weight of the liquid displaced
$\Rightarrow Ah{\rho }_b=0.8\times A\times h\times {\rho }_0$


$\Rightarrow {\rho }_0=\frac{{\rho }_b}{0.8}=\frac{5{\rho }_b}{4}$ ........(1)
Here, ${\rho }_0$ is the density of the liquid at $0^{\circ }\text{C}$ and ${\rho }_b$ is the density of the block.
Now, due to thermal expansion, density as function of temperature is given by $\rho =\frac{{\rho }_0}{1+\gamma \Delta T}$ ......(2)
$\rho =\frac{\frac{5{\rho }_b}{4}}{1+\gamma \Delta T}$
[from (1) and (2) ]
For the block to sink in the liquid at ${62.5}^{\circ }\text{C}$, $\rho ={\rho }_b$
${\rho }_b=\frac{\frac{5{\rho }_b}{4}}{1+\gamma \Delta T}$
$\Rightarrow 1+\gamma \Delta T=\frac{5}{4}$
$\Rightarrow 1+\gamma (T_2-T_1)=\frac{5}{4}$
$\Rightarrow 1+\gamma (62.5-0)=\frac{5}{4}$
$\Rightarrow \gamma =4\times {10}^{-3}{{}^{\circ }\text{C}}^{-1}$ or ${\text{K}}^{-1}$