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 their temperature is increased by $\Delta T$, the fraction of the volume of metal submerged in mercury changes by a factor

• A. $\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$
• B. $\left(\frac{1+{\gamma }_2\Delta T}{1-{\gamma }_1\Delta T}\right)$
• C. $\left(\frac{1-{\gamma }_2\Delta T}{1+\gamma \Delta T}\right)$
• D. $\frac{{\gamma }_2}{{\gamma }_1}$

Answer 1

The correct option is A

$\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$

Let density of the metal and mercury be $d_1$ and $d_2$ respectively.
The fraction of volume of metal submerged initially, $f=\frac{d_1}{d_2}$
Density of metal after increase of temperature by $\Delta T$, $d_1^{\prime }=\frac{d_1}{1+{\gamma }_1\Delta T}$
Density of mercury after increase of temperature by $\Delta T$, $d_2^{\prime }=\frac{d_2}{1+{\gamma }_2\Delta T}$
So, fraction of volume of metal submerged finally, $f^{\prime }=\frac{d_1^{\prime }}{d_2^{\prime }}=\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)\left(\frac{d_1}{d_2}\right)=\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)f$
$\therefore \frac{f^{\prime }}{f}=\left(\frac{1+{\gamma }_2\Delta T}{1+{\gamma }_1\Delta T}\right)$and the correct choice is (a).