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 ${\gamma _s} = 0.3 \times {10^{ - 5}}{C^{ - 1}}$ and of the liquid is ${\gamma _l} = 8 \times {10^{ - 5}}{C^{ - 1}}$.
A. 99.99
B. 88.88
C. 77.77
D. 66.66

Answer 1

The body being completely immersed in the liquid when its density equals the density of the liquid. Recall the formula for volume expansion and thus express the density of the solid and liquid at $0^\circ C$. The percentage faction of the solid immersed in the liquid is the ratio of density of solid and density of liquid at $0^\circ C$.

Formula used:
${\rho _f} = {\rho _i}\left( {1 + \gamma \Delta T} \right)$
Here, ${\rho _f}$ is the final density, ${\rho _i}$ is the initial density, $\gamma$ is the coefficient of expansion and $\Delta T$ is the change in temperature.

Complete step by step answer:
We have given that, at $t = 50^\circ C$, the body is completely immersed in the liquid. We know that the body is completely immersed in the liquid when its density equals the density of the liquid. Therefore,
$t = 50^\circ C$, ${\rho _l} = {\rho _s}$.
Now, at ${t_0} = 0^\circ C$, we can express the fraction of the body immersed in the liquid as,$\eta = \left( {\dfrac{{{\rho _s}}}{{{\rho _l}}}} \right) \times 100$ …… (1)
Since the temperature of the liquid is lowered in the second case, we can express the density of the liquid at $0^\circ C$ as follows,
${\rho _{o,l}} = {\rho _{50,l}}\left( {1 + {\gamma _l}\Delta T} \right)$ …… (2)
Here, ${\rho _{o,l}}$ is the density of liquid at $0^\circ C$, ${\rho _{50,l}}$ is the density of liquid at $50^\circ C$, ${\gamma _l}$ is the cubical expansion coefficient of liquid and $\Delta T$ is the change in temperature.

We can also express the density of the solid body at $0^\circ C$as follows,
${\rho _{o,s}} = {\rho _{50,s}}\left( {1 + {\gamma _s}\Delta T} \right)$ …… (3)
Here, ${\rho _{o,s}}$ is the density of solid at $0^\circ C$, ${\rho _{50,s}}$ is the density of solid at $50^\circ C$, ${\gamma _s}$ is the cubical expansion coefficient of solid and $\Delta T$ is the change in temperature.
Substituting equation (2) and (3) in equation (1), we get,
$\eta = \left( {\dfrac{{{\rho _{50,s}}\left( {1 + {\gamma _s}\Delta T} \right)}}{{{\rho _{50,l}}\left( {1 + {\gamma _l}\Delta T} \right)}}} \right) \times 100$

We have determined that the density of solid and liquid at $t = 50^\circ C$ is the same. Therefore, the above equation becomes,
$\eta = \dfrac{{\left( {1 + {\gamma _s}\Delta T} \right)}}{{\left( {1 + {\gamma _l}\Delta T} \right)}} \times 100$
Substituting ${\gamma _s} = 0.3 \times {10^{ - 5}}{C^{ - 1}}$, ${\gamma _l} = 8 \times {10^{ - 5}}{C^{ - 1}}$ and $\Delta T = 50^\circ C$ in the above equation, we get,
$\eta = \dfrac{{\left( {1 + \left( {0.3 \times {{10}^{ - 5}}} \right)\left( {50} \right)} \right)}}{{\left( {1 + \left( {8 \times {{10}^{ - 5}}} \right)\left( {50} \right)} \right)}} \times 100$
$\Rightarrow \eta = \dfrac{{\left( {1.00015} \right)}}{{\left( {1.004} \right)}} \times 100$
$\Rightarrow \eta = 99.6\%$
$\therefore \eta \approx 99.99\%$

So, the correct answer is option A.

Note: Students must note that the density of solids and liquids is greater at low temperatures and therefore, using this fact, we have taken the positive sign for $\Delta T$ even if the final temperature is less than the initial temperature. The crucial step in this solution is to find out why the solid is completely being immersed in the liquid at $50^\circ C$. The principle of floatation tells us that the solid with greater density than the liquid sinks in the liquid.