## Related questions with answers

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³ of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes' principle.)

Solution

VerifiedFrom $\textbf{ the definition of the average density}$ we know that :Density The average density of a substance or object is defined as its mass per unit volume

$\rho = \dfrac{m}{V}$

, $\textbf{Archimedes’ Principle}$ The buoyant force on an object equals the weight of the fluid it displaces.

$F_{B} = w_{fl}$

Where:

- $F_{B}$ is the buoyant force .
- $w_\text{fl}$ is the weight of the displaced water .
- $\rho_\text{rock}$ is the average density of the rock .
- $\rho_\text{fl}$ is the average density of the water.
- $m_\text{rock}$ is the mass of the rock.
- $m_\text{fl}$ is the mass of the water.
- $V_\text{rock}$ is the volume of the rock

$\textbf{Givens}$: $\rho_\text{fl} = 1 \mathrm{g/cm^3}$ , $V =89 \mathrm{cm^3}$ .

$\textbf{Plugging}$ known information to get :

$\begin{align*} F_\text{B}& = w_\text{fl} \\ m_\text{rock} g &= m_\text{fl} g\\ m_\text{rock} &= \rho_\text{fl} V_\text{fl} \\ \rho_\text{rock} &= \dfrac{\rho_\text{fl} V_\text{fl}}{ V_\text{rock}} \\ &= \dfrac{1 \mathrm{g/cm^3} \times 89 \mathrm{cm^3} }{ V_\text{rock}} \\ &= 89 \end{align*}$

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