Question

When a block of iron floats in mercury at $0^{o}C$, fraction $k_1$ of its volume is submerged, while at the temperature ${60}^{o}C$, 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 $k_1/k_2$ can be expressed as

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

Let V₀ is the volume of iron block at T=0 and V is the volume at T = 60

So,

$v=v_0+v_{0}T{\gamma }_{Fe}$

$=V_0(1+60{\gamma }_{Fe}).........\left(1\right)$

Let V₀’ and V’ be the volume of iron submerged at T=0 and T=60

$V^{\prime }=V_0(1+60{\gamma }_{Hg})$ ……….(2)

Dividing equation (1) by (2) and solving, we get

$\frac{V}{V^{\prime }}\times \frac{{V_0}^{{}^{\prime }}}{V_0}=\frac{1+60{\gamma }_{Fe}}{1+60{\gamma }_{Hg}}$

$\because k_1=\frac{{V_0}^{{}^{\prime }}}{V_0}and{k}_2=\frac{V^{{}^{\prime }}}{V}$

$\therefore \frac{k_1}{k_2}=\frac{1+60{\gamma }_{Fe}}{1+60{\gamma }_{Hg}}$