Question

A piece of metal weighs 46g in air. When immersed in a liquid of specific gravity $1.24$ at ${27^\circ }$ it weighs $30g$. When the temperature of the liquid is raised to ${42^\circ }$ in a liquid of specific gravity $1.20$, it weighs $30.5g$. The coefficient of linear expansion of metal is
A.$2.23 \times {10^{ - 5}}{/^\circ }C$
B.$6.7 \times {10^{ - 5}}{/^\circ }C$
C.$4.46 \times {10^{ - 5}}{/^\circ }C$
D.none of these

Answer 1

We will use the principle of buoyancy to find the resultant mass of the metal in both the cases when the liquid is at ${27^ \circ }C$ as well as when the liquid is at ${42^ \circ }C$. Then we will get two different values of the volume of the liquid. These two volumes are related to each other with the relation $V' = V(1 + \gamma \Delta t)$.

Complete answer:
We know that the net weight (apparent mass) of the piece of metal is found by subtracting the mass of the liquid displaced from the mass of the metal. Thus
${M_{apparent}} = {M_{metal}} - {M_{displaced\,liquid}}$.
Now the mass of the displaced liquid is density times the volume of the liquid displaced. Thus
${M_{displaced\,liquid}} = \rho \times Vol.$,
where $\rho$ is the density of the liquid displaced.
Thus applying this formula in the first case, we get
$30\,g = 46\,g - V(1.24\,g/c{m^3})$.
$\Rightarrow V = \dfrac{{46\,g - 30\,g}}{{1.24\,g/c{m^3}}}$
Thus, we get $V = 12.903\,c{m^3}$.
Applying this same formula in the second case when the temperature of the liquid has been increased to the new temperature, we get
$30.5\,g = 46\,g - V'(1.20\,g/c{m^3})$.
$\Rightarrow V' = \dfrac{{46\,g - 30.5\,g}}{{1.20\,g/c{m^3}}}$
Thus we get $V' = 12.917\,c{m^3}$.
Now the relation between the two volumes is given with the help of the coefficient of thermal expansion as $V' = V(1 + \gamma \Delta t)$.
Here $\gamma$ is the coefficient of volumetric or cubical expansion given as $\gamma = 3\alpha$, where $\alpha$ is the coefficient of linear expansion.
Also $\Delta t$ is the change in temperature which is
$\Delta t = {42^ \circ }C - {27^ \circ }C = {15^ \circ }C$.
Thus we have
$V' = V(1 + 3\alpha \Delta t)$, i.e.
$V' = V(1 + 3\alpha ({15^ \circ }C))$
On substituting the values of $V$ and $V'$ we get,
$12.917\,c{m^3} = 12.903\,c{m^3}(1 + 3\alpha ({15^ \circ }C))$
$\Rightarrow 1 + 3\alpha ({15^ \circ }C) = \dfrac{{12.917\,c{m^3}}}{{12.903\,c{m^3}}}$
Upon further evaluating the equation we get,
$\alpha ({45^ \circ }C) = \dfrac{{12.917}}{{12.903}} - 1 = 108.502 \times {10^{ - 5}}$
$\Rightarrow \alpha = 2.41 \times {10^{ - 5}}{/^ \circ }C$.
Considering the options given in the above question, we can see that no option matches the answer found.

Thus the correct answer is option (D).

Note:
We should note that the relation $\gamma = 3\alpha$ is approximated and is valid for extremely small values of the coefficient of expansion which is generally the case. Also, the specific gravity is the same as the density of a material in $g/c{m^3}$.