Chapter 13, Problem 86GP

(a)

To determine

To Find: The floating position of cube in the mercury.

Answer to Problem 86GP

The cube will float lower in mercury when the temperature in increased up to $2{\text{5}}^{\text{o}}\text{C}$ .

Explanation of Solution

Given:

Initial temperature, $T_0=0°\text{C}$ .

Final temperature, $T_f=25°\text{C}$ .

Formula used:

Buoyant force:

  $F_b\text{ = }{V}_{\text{submerged}}\times {\rho }_{Hg}\times g=F_g$

Here, $V_{\text{submerged}}$ is the volume of the submerged portion of iron,${\rho }_{Hg}$ is the density of the mercury.

  $g$ is the acceleration due to gravity.

Calculation:

Write the expression for buoyant force.

  $\mathrm{Buoyant}force{F}_b=Gravityforce{F}_g$ .

  $\begin{array}{l}{V}_{\text{submerged}}\times {\rho }_{Hg}\times g\text{ = }{V}_{Fe}\times {\rho }_{Fe}\times g\\ V_{S_{Fe}}\times {\rho }_{Hg}\times g=V_{Fe}\times {\rho }_{Fe}\times g\\ \frac{V_{S_{Fe}}}{V_{Fe}}=\frac{{\rho }_{Fe}}{{\rho }_{Hg}}\end{array}$

It can be safely assumed that volume of submerged portion of Iron increases slightly and therefore the block will float lower in mercury.

Conclusion:

Thus, the cube will float lower in mercury when the temperature in increased up to $25°\text{C}$ .

(b)

To determine

To Find: The percent fractionof the change in the volume.

Answer to Problem 86GP

The percent fractionof the change in the volume submerged is $0.36\text{ \%}$ .

Explanation of Solution

Given:

Initial temperature, $T_0=0°\text{C}$ .

Final temperature, $T_f=25°\text{C}$ .

Thermal expansion of mercury, ${\beta }_{Hg}=180\times {10}^{-6}{\text{ °C}}^{\text{-1}}$ .

Thermal expansion of iron, ${\beta }_{Fe}=35\times {10}^{-6}{\text{ °C}}^{\text{-1}}$ .

Formula used:

The percentage fraction of the change in volume submerged is determined by the following formula,

  $\begin{array}{l}\Delta V=\frac{\frac{V_{Hg}}{V_{Fe}}-\frac{V_{0_{Hg}}}{V_{0_{Fe}}}}{\frac{V_{0_{Hg}}}{V_{0_{Fe}}}}\\ V_{Hg}=V_{0_{Hg}}\left(1+{\beta }_{Hg}\times \Delta T\right)\end{array}$

Here, $V_{Hg}$ represents final Volume of mercury.

  $V_{0_{Fe}}$ represents the initial volume of iron.

  $V_{Fe}$ represents the final volume of iron.

  $V_{0_{Hg}}$ represents the initial volume of mercury.

Calculation:

Substitute the value of $V_{Hg}$ in the above equation,

  $\begin{array}{l}\Delta V\text{= }\frac{\frac{V_{0_{Hg}}\left(1+{\beta }_{Hg}\times \Delta T\right)}{V_{0_{Fe}}\left(1+{\beta }_{Fe}\times \Delta T\right)}-\frac{V_{0_{Hg}}}{V_{0_{Fe}}}}{\frac{V_{0_{Hg}}}{V_{0_{Fe}}}}\\ \text{= }\frac{1+{\beta }_{Hg}\times \Delta T}{1+{\beta }_{Fe}\times \Delta T}-1\\ \text{= }\frac{1+(180\times {10}^{-6}\times 25)}{1+(35\times {10}^{-6}\times 25)}-1\\ =3.6\times {10}^{-3}=0.0036=0.36\text{\%}\end{array}$

Conclusion:

Thus, the percent fractionof the change in the volume submerged is $0.36\%$ .