Question

A solid body floats in a liquid at a temperature $t={50}^{\circ }C$ being completely submerged in it. What percentage of the volume of the body is submerged in the liquid after it is cooled to $t_0=0^{\circ }C$, if the coefficient of cubic expansion for the solid is ${\gamma }_s=0.3\times {10}^{-5}{}^0{C}^{-1}$ and of the liquid ${\gamma }_l=8\times {10}^{-5}{}^{\circ }{C}^{-1}$.

• A. 99.61
• B. 88.88
• C. 77.77
• D. 66.66

Answer 1

The correct option is A 99.61

Let the volume of the body be $V$ at $0^{\circ }\text{C}$ and its mass be $m$.

Let the density of the liquid be $\rho$ at $0^{\circ }\text{C}$.

Thus, volume of the body at ${50}^{\circ }\text{C}$ is $V_1=V(1+{\gamma }_s\Delta T)$

$\therefore$ $V_1=V(1+0.3\times {10}^{-5}\times 50)=1.00015V$

Density of liquid at ${50}^{\circ }\text{C}$ is ${\rho }_1=\rho (1-{\gamma }_l\Delta T)$

$\therefore$ ${\rho }_1=\rho (1-8\times {10}^{-5}\times 50)=0.996\rho$

Thus, by Archimedes principle, $mg={\rho }_1{V}_{1}g$

Or $mg=\left(0.996\rho \right)g\left(1.00015V\right)\Rightarrow m=0.99615\rho V$

AT $0^{\circ }\text{C}$, let the volume of solid immersed be $V_2$.

Thus, $\rho g{V}_2=mg$ (by Archimedes principle)

Or $\rho g{V}_2=0.99615\rho Vg$

$\Rightarrow V_2=0.99615V$

Hence, $99.6\%$ of the solid is immersed in the liquid.