Chapter 10, Problem 77GP

To determine

To Find: The mass of the copper.

Answer to Problem 77GP

The mass of the copper if the top of the wood block is exactly at the surface of water is $0.3\text{ kg}$ .

Explanation of Solution

Given:

Mass of wood, $m_w=0.5\text{ kg}$

Density of wood, ${\rho }_w=0.6\times {10}^3{\text{ kg / m}}^{\text{3}}$

Density of water, ${\rho }_{\text{water}}={10}^3{\text{ kg / m}}^{\text{3}}$

Formula used:

Write the expression for the density.

  $\begin{array}{l}\rho =\frac{m}{V}\\ m=\rho \times V\end{array}$

Write the expression for the force balance

  $F_B=F_{{\text{g}}_{\text{copper}}}+F_{{\text{g}}_{\text{wood}}}$

Here, $F_B$ is the buoyant force,

  $F_{{\text{g}}_{\text{wood}}}$ is the gravity force on wood,

  $F_{\text{gcopper}}$ is the gravity force on copper.

Write the expression for force in terms of mass and gravity.

  $F_B=m_w\times g$

Here, $m_w$ is the mass of water displaced.

  $g$ is the acceleration due to gravity.

Calculation:

Substitute the given values in the above equation,

  $V_{\text{wood}}=\frac{0.5}{0.6\times {10}^3}=0.8\times {10}^{-3}{\text{ m}}^{\text{3}}$

Write the expression for the total buoyant force.

  $F_B=F_{{\text{g}}_{\text{copper}}}+F_{{\text{g}}_{\text{wood}}}$

Here, $F_{g\text{copper}}=m_{Cu}\times g\text{ and }{F}_B=m_w\times g$ .

Substituting the values,

  $\begin{array}{l}{F}_B=F_{{\text{g}}_{\text{copper}}}+F_{{\text{g}}_{\text{wood}}}\\ m_w\times g=\left(m_{Cu}+m_{\text{wood}}\right)\times g\\ m_w=m_{Cu}+m_{\text{wood}}\end{array}$

Write the expression for mass of water displaced.

  $\begin{array}{l}{m}_w=\rho \times V\\ {\rho }_w\times V_w=m_{Cu}+m_{\text{wood}}\\ {10}^3\times 0.8\times {10}^{-3}=m_{Cu}+0.5\\ m_{Cu}=0.8-0.5=0.3\text{ kg}\end{array}$

Conclusion:

Thus, mass of the copper if the top of the wood block is exactly at the surface of water is $0.3\text{ kg}$ .