# A piece of metal floats on mercury .the coefficient of expansion of metal and mercury are gamma_1 and gamma_2 ,respectively .if the temperature of both metal and mercury increased by an amount Delta T . See description ?

## by what factor does the fraction of the volume of the metal submerged in mercury changes .

Jun 18, 2017

This is what I get.

#### Explanation:

Let density of metal and mercury be ${d}_{1} \mathmr{and} {d}_{2}$ respectively.

Fraction of volume of metal submerged initially $= {d}_{1} / {d}_{2}$ .....(1)

We know that Volume of metal will change after increase of temperature $\Delta T$ by a factor $\left(1 + {\gamma}_{1} \Delta T\right)$
Similarly Volume of mercury will change after increase of temperature $\Delta T$ by a factor $\left(1 + {\gamma}_{2} \Delta T\right)$

(It is assumed that $\gamma$ is coefficient of volume expansion.)

Density of metal after increase of temperature $\Delta T = {d}_{1} / \left(1 + {\gamma}_{1} \Delta T\right)$
Density of mercury after increase of temperature $\Delta T = {d}_{2} / \left(1 + {\gamma}_{1} \Delta T\right)$

Fraction of volume of metal submerged after increase of temperature $= \frac{{d}_{1} / \left(1 + {\gamma}_{1} \Delta T\right)}{{d}_{2} / \left(1 + {\gamma}_{2} \Delta T\right)}$
$= {d}_{1} / {d}_{2} \left(\frac{1 + {\gamma}_{2} \Delta T}{1 + {\gamma}_{1} \Delta T}\right)$ .....(2)

Change in fraction of volume of metal submerged $= {d}_{1} / {d}_{2} \left(\frac{1 + {\gamma}_{2} \Delta T}{1 + {\gamma}_{1} \Delta T}\right) - {d}_{1} / {d}_{2}$
$= {d}_{1} / {d}_{2} \left(\frac{1 + {\gamma}_{2} \Delta T}{1 + {\gamma}_{1} \Delta T} - 1\right)$

$= {d}_{1} / {d}_{2} \frac{\left(1 + {\gamma}_{2} \Delta T\right) - \left(1 + {\gamma}_{1} \Delta T\right)}{\left(1 + {\gamma}_{1} \Delta T\right)}$
$= {d}_{1} / {d}_{2} \frac{\left({\gamma}_{2} - {\gamma}_{1}\right) \Delta T}{1 + {\gamma}_{1} \Delta T}$

Factor of the fraction of the volume of the metal submerged in mercury changed after increase of volume$= \frac{{d}_{1} / {d}_{2} \frac{\left({\gamma}_{2} - {\gamma}_{1}\right) \Delta T}{1 + {\gamma}_{1} \Delta T}}{{d}_{1} / {d}_{2}}$
$= \frac{\left({\gamma}_{2} - {\gamma}_{1}\right) \Delta T}{1 + {\gamma}_{1} \Delta T}$