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Question

# Suppose you have a coffee mug with a circular cross-section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm? Assume coffee has the same density as water.

Solution

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

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

Where :

• $\rho$ is the density of the coffee .
• $m$ is the mass of the coffee .
• $V$ is the volume of the coffee .
• $r$ is the radius of the cup .
• $A$ is the area of the cup .
• $d$ is the depth of the coffee .

$\textbf{Givens}$: $d = 0.075 \ \text{m}$ , $m = 0.375\ \text{kg}$ and $\rho = 1000\ \mathrm{kg/m^3}$ . $\textbf{Plugging}$ known information to get :

\begin{align*} V& = \dfrac{m}{\rho} \\ A \cdot d & = \dfrac{m}{\rho}\\ \pi r^2 \cdot d & = \dfrac{m}{\rho} \\ r^2 & = \dfrac{m}{ \pi d \rho } \\ r& = \sqrt{ \dfrac{m}{ \pi d \rho } } \\ &=\sqrt{ \dfrac{0.375 }{ \pi \times 0.075 \times 1000 } } \\ &=0.0399 \end{align*}

$\boxed{ r = 0.0399\ \text{m} }$

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