Question

A piece of metal floats in mercury. The coefficients of volume expansion of the metal and mercury are ${\gamma }_1$ and ${\gamma }_2$ respectively. If the temperatures of both mercury and metal are increased by an amount $\Delta T$, the fraction of the volume of the metal submerged in mercury changes by the factor of $\frac{x+{\gamma }_2\Delta T}{x+{\gamma }_1\Delta T}$. Find $x$

Answer 1

The condition for floating is
Weight = Upthrust
$V{\rho }_{1}g=V_1{\rho }_{2}g$
where ${\rho }_1$ = density of metal, ${\rho }_2$ = density of mercury
$\therefore$ fraction of volume of metal submerged in mercury
$\frac{V_1}{V}=\frac{{\rho }_1}{{\rho }_2}=x$ (say)

Now, when the temperature is increased by $\Delta T$,
${\rho }_1=\frac{{\rho }_1}{1+{\gamma }_1\Delta T}$ ...(1)
and ${\rho }_2=\frac{{\rho }_2}{1+{\gamma }_2\Delta T}$ ...(2)

$\therefore$ new volume fraction,
$x^{\prime }=\frac{{\rho }_1}{{\rho }_2}=\left(\frac{{\rho }_1}{1+{\gamma }_1\Delta T}\right)\times \left(\frac{1+{\gamma }_2\Delta T}{{\rho }_2}\right)$
$=\frac{{\rho }_1}{{\rho }_2}\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$

$\Rightarrow x^{\prime }=x\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$
$\therefore \frac{x^{\prime }}{x}=\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}$