Chapter 13, Problem 70A

(a)

To determine

Volume of the statuette.

Answer to Problem 70A

The volume of statuette is $6.33\times {10}^{-5}{m}^3$ .

Explanation of Solution

Given:

Force of statuette when suspended in air: $F_g=\text{11}\text{.81}\text{N}$

Force of statuette when completely submerged in water: $F_g=\text{11}\text{.19}\text{N}$

The density of the fluid, ${\rho }_{water}=1.00\times {10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

One force is the upward force of the liquid on the object. There is also the downward force of the object’s weight. The difference between the buoyant force (or upward force) and the object’s weight whether an object floats or sinks.

The upward force or buoyant force can be calculated by using a formula,

  $F_{buoyant}\text{ =}\rho Vg$

Where, $\rho$ is the density of the fluid, $g$ is an acceleration of gravity, and $V$ is a volume of the fluid.

Calculation:

  $F_{buoyant}={\rho }_{water}Vg=F_g-F_{net}$ .

Thus, the volume of the statuette suspended in air is calculated by:

  $V=\frac{F_g-F_{net}}{{\rho }_{water}g}$

Where, $F_g$ is the air weight of the camera, $F_{net}$ is the net force, $\rho$ is the density of the fluid, $g$ is an acceleration of gravity, and $V$ is a volume of the fluid.

Apply the formula and calculate the volume of the statuette suspended in air is:

  $V=\frac{\text{11}\text{.81}\text{N-11}\text{.19}\text{N}}{\text{(1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{)(9}\text{.8}\text{N/kg)}}$

  $V=\text{6}{\text{.33×10}}^{\text{-5}}{\text{m}}^{\text{3}}$

Conclusion: Hence, Volume of the statuette is $6.33\times {10}^{-5}{m}^3$ .

(b)

To determine

Material of the statuette.

Answer to Problem 70A

As the density of statuette is $19\times {10}^{3}kg/m^3$ , thus, it is made of gold.

Explanation of Solution

Given:

The density of gold, ${\rho }_{go\mathrm{ld}}=\text{19}{\text{.3×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}$

The density of gold − plated aluminum ${\rho }_{{}_{aluminum}}=2.7\times {10}^3{\text{kg/m}}^{\text{3}}$

The volume of the statuette is $6.33\times {10}^{-5}{m}^3$

Force of statuette when suspended in the air: $F_g=\text{11}\text{.81}\text{N}$

Formula used:

Density:

  $\rho =\frac{m}{V}$

Where, $m$ is mass and $V$ is volume.

Calculation:

  $\rho =\frac{F_g}{V_g}$

Here, $F_g$ is the air weight of the camera and $V_g$ is the volume.

  $\rho =\frac{11.81\text{N}}{\left(\text{6}{\text{.33×10}}^{\text{-5}}{\text{m}}^{\text{3}}\right)\left(\text{9}\text{.8}\text{N/kg}\right)}$

  $\rho =19\times {10}^3{\text{kg/m}}^{\text{3}}$

Conclusion: The statuette is made of gold and its density is $19\times {10}^{3}kg/m^3$ .