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

ρ=mV\rho = \dfrac{m}{V}

Where :

  • ρ\rho is the density of the coffee .
  • mm is the mass of the coffee .
  • VV is the volume of the coffee .
  • rr is the radius of the cup .
  • AA is the area of the cup .
  • dd is the depth of the coffee .

Givens\textbf{Givens}: d=0.075 md = 0.075 \ \text{m} , m=0.375 kgm = 0.375\ \text{kg} and ρ=1000 kg/m3\rho = 1000\ \mathrm{kg/m^3} . Plugging\textbf{Plugging} known information to get :

V=mρAd=mρπr2d=mρr2=mπdρr=mπdρ=0.375π×0.075×1000=0.0399\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*}

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

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