Question

A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are ${\gamma }_1$ and ${\gamma }_2$ respectively. If the temperature 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.

• A. $\frac{1}{({\gamma }_2-{\gamma }_1)\Delta t}$
• B. $\frac{1}{({\gamma }_1-{\gamma }_2)\Delta t}$
• C. $({\gamma }_1-{\gamma }_2)\Delta t$
• D. $({\gamma }_2-{\gamma }_1)\Delta t$

Answer 1

Answer: (D)

$({\gamma }_2-{\gamma }_1)\Delta t$

Weight of metal piece $w_m=mg$
When immersed in liquid, let fraction f be submerged in the liquid.
$\therefore f{V}_{0m}{\rho }_{0l}g=mg$ ......(1)
When the metal and the liquid are heated, let f' be the fraction which is submerged in the liquid.
$\therefore f^{\prime }{V}_{0m}(1+{\gamma }_m\Delta T){\rho }_{0l}(1-{\gamma }_l\Delta T)g=mg$ .....(2)
Dividing (1) by (2),
$\frac{f}{f^{\prime }(1+{\gamma }_m\Delta T)(1-{\gamma }_l\Delta T)}=1$
$\Rightarrow \frac{f^{\prime }}{f}=(1-{\gamma }_m\Delta T)(1+{\gamma }_l\Delta T)$
$\Rightarrow \frac{f^{\prime }}{f}-1=({\gamma }_l-{\gamma }_m)\Delta T$
$\frac{f^{\prime }-f}{f}=({\gamma }_l-{\gamma }_m)\Delta T=({\gamma }_2-{\gamma }_1)\Delta T$