When researchers find a reasonably complete fossil of a dinosaur, they can determine the mass and weight of the living dinosaur with a scale model sculpted from plastic and based on the dimensions of the fossil bones. The scale of the model is 1/20; that is, lengths are 1/20actual length, areas are ${\left(1/20\right)}^{\mathbf{2}}$ actual areas, and volumes are ${\left(1/20\right)}^{\mathbf{3}}$actual volumes. First, the model is suspended from one arm of a balance and weights are added to the other arm until equilibrium is reached. Then the model is fully submerged in water and enough weights are removed from the second arm to re-establish equilibrium (Figure).For a model of a particular T.rexfossil,637.76 ghad to be removed to re-establish equilibrium. (a)What was the volume of the model? (b)What was the volume of the actual T.rex? (c) If the density of T.rexwas approximately the density of water, what was its mass?

• The scale of the model is 1/20
• The length of the model, $l_{model}=\frac{1}{20}{l}_{actual}$
• The area of the model,$A_{model}={\left(\frac{1}{20}\right)}^2{A}_{actual}$
• The volume of the model,$V_{model}={\left(\frac{1}{20}\right)}^3{V}_{actual}$
• The difference in mass,$∆m=637.76kg$
• The density of the T.rex,

• $$\begin{gathered} {\rho }_{actual}={\rho }_w \\ =1000\frac{kg}{m^3} \end{gathered}$$

We can use Archimedes’ principle to find the volume of the model. Then using the given scale of the model, we can find the volume of actual T.rex. From this volume, we can easily find the mass of T.rex using the density of the actual T.rex.

Formulae:

Force applied on body (or weight), $F_b=m_{f}g$ (i)

Density of a substance, $\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}$ (ii)

When the model is suspended in air, then the weight of the model is$F_g$

When the model is submerged in water, then the forces acting on it due to buoyant force can be given as the net force:$F_g-F_b$

Hence, the difference in weight is given as:

role="math" localid="1661156058315" $$\begin{gathered} ∆mg=F_g-\left(F_g-F_b\right) \\ ∆mg=F_b.........................................................................\left(\mathrm{iii}\right) \end{gathered}$$

Using equation (i) & (ii), we get

$F_b={\rho }_w{V}_{model}g..........................................................\left(\mathrm{iii}\right)$

From equations (iii) and (IV), we get

$$\begin{gathered} ∆mg={\rho }_w{V}_{model}g \\ ∆m={\rho }_w{V}_{model} \\ {V}_{model}=\frac{∆m}{{\rho }_w} \\ =\frac{637.76\times {10}^{-3}kg}{1000{\displaystyle \frac{kg}{m^3}}} \\ =6.378\times {10}^{-4}{m}^3 \end{gathered}$$

Hence, the volume of the model is$6.378\times {10}^{-4}{m}^3$

From the given scale of volume, we get

$$\begin{gathered} V_{model}={\left(\frac{1}{20}\right)}^2{V}_{actual} \\ {V}_{model}={\left(20\right)}^3{V}_{actual} \\ ={\left(20\right)}^3\times 6.378\times {10}^{-4}{m}^3 \\ =5.102m^3 \end{gathered}$$

Hence, the volume of actual T.rex is $5.102m^3$

From equation (ii), we get

$$\begin{gathered} m_{actual}={\rho }_{actual}{V}_{actual} \\ =1000\frac{kg}{m^3}\times 5.102m^3 \\ =5.102\times {10}^{3}kg \end{gathered}$$

Hence, the mass of actual T.rex is $5.102\times {10}^{3}kg$.