Question

Short answer

(a) The iron cube will float lower in the mercury, and the fraction of change of the volume submerged will be (b) \(0.36\% \).

Step-by-step solution

Given data

The initial temperature is \(T = 0^\circ {\rm{C}}\).

The final temperature is \(T' = 25^\circ {\rm{C}}\).

Understanding densities of material

In this problem, first, determine the densities of iron and mercury and then compare them to evaluate whether the iron cube will float lower or higher in the mercury.

Determine whether the iron cube will float higher or lower in the mercury

When iron and mercury are heated, their densities decrease because of the volume expansion. The density of mercury reduces more upon heating as compared to the density of the iron cube. The total effect is that the densities get closer together. Hence, relatively, more mercury will have to be displaced to hold up the iron. Therefore, the iron will float lower in the mercury.

Calculation of the fraction of volume submerged

The relation to find the fraction of volume submerged can be written as:

\(N = \left( {\frac{{1 + \beta \Delta T}}{{1 + \beta '\Delta T}}} \right) - 1\)

Here,\(\beta \)and\(\beta '\)are the volume expansions of mercury and iron, respectively and\(\Delta T\)is the change in temperature.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}N &= \left( {\frac{{1 + \left( {180 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right)\left( {{\rm{25^\circ C}} - 0{\rm{^\circ C}}} \right)}}{{1 + \left( {35 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right)\left( {{\rm{25^\circ C}} - 0{\rm{^\circ C}}} \right)}}} \right) - 1\\N &= \frac{{1.0045}}{{1.0008}} - 1\\N &= 3.69 \times {10^{ - 3}}\\\% N &= 0.36\% \end{aligned}\)

Thus, \(0.36\% \) is the required percent.