Question

When a block of iron floats in mercury at $0^{\circ }C$, a fraction $k_1$ of its volume submerged, while at the temperature ${60}^{\circ }C$, a fraction $k_2$ is keen to be submerged the coefficient of volume expansion of iron of $g_{Fe}$ and that of mercury is $g_{Hg}$, then the ratio $\frac{k_1}{k_2}$ can be expressed as

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Let $V_o$ and $V$ be the volume of the iron block at $T=0^{o}C$ and ${60}^{o}C$ respectively, then we can write

$V=V_o(1+60g_{Fe})$......................(I)

Let $V_o$' and $V$′ be the volume of the iron block submerged in the mercury at $T=0^{o}C$ and ${60}^{o}C$ respectively, then we can write

$V\text{'}=V_o^{\prime }(1+{60}_{Hg})$.........(II)

Dividing (I) by (II), we get
$\frac{V}{V^{\prime }}=\frac{V_o}{V_o^{\prime }}=\frac{V_o(1+60g_{Fe})}{V_o^{\prime }(1+{60}_{Hg})}$

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