Question

When a block of iron floats in mercury at $0^{\circ }\text{C}$, fraction $K_1$ of its volume is submerged, while at the temperature ${60}^{\circ }\text{C}$, a fraction $K_2$ is seen to be submerged. If the coefficient of volume expansion of iron is ${\gamma }_{Fe}$ and that of mercury is ${\gamma }_{Hg}$, then the ratio of $\frac{K_1}{K_2}$ can be expressed as

• A. $\frac{1+60{\gamma }_{Hg}}{1+60{\gamma }_{Fe}}$
• B. $\frac{1-60{\gamma }_{Hg}}{1+60{\gamma }_{Fe}}$
• C. $\frac{1+60{\gamma }_{Fe}}{1+60{\gamma }_{Hg}}$
• D. $\frac{1-60{\gamma }_{Hg}}{1+60{\gamma }_{Fe}}$

Answer 1

The correct option is C $\frac{1+60{\gamma }_{Fe}}{1+60{\gamma }_{Hg}}$

Let density of the block be ${\rho }_{Fe}$ and density of the mercury be ${\rho }_{Hg}$.

Let total volume of the block be $V$ and volume submerged be $V_{sub}$.

From equilibrium,
Weight of the block = upthrust
$\Rightarrow V{\rho }_{Fe}g=V_{sub}{\rho }_{Hg}g$
$\Rightarrow \frac{V_{sub}}{V}=\frac{{\rho }_{Fe}}{{\rho }_{Hg}}$ ...(i)

Now, according to question
$K_1=\frac{{\rho }_{Fe}}{{\rho }_{Hg}}$ (at $0^{\circ }\text{C})$ ...(ii)
$K_2=\frac{{\rho }_{Fe}}{{\rho }_{Hg}}$ (at ${60}^{\circ }\text{C})$ ...(iii)
In case of expansion,
$\frac{{\rho }_{{\theta }^{\circ }\text{C}}}{{\rho }_{0^{\circ }\text{C}}}=\frac{1}{1+\gamma \theta }$ ....(iv)

From the equations (ii), (iii) and (iv),
$\frac{K_1}{K_2}={\left(\frac{{\rho }_{Fe}}{{\rho }_{Hg}}\right)}_{0^{\circ }\text{C}}\times {\left(\frac{{\rho }_{Hg}}{{\rho }_{Fe}}\right)}_{{60}^{\circ }\text{C}}=\frac{(1+60{\gamma }_{Fe})}{(1+60{\gamma }_{Hg})}$
Hence option (c) is correct.