91Ó°ÊÓ

• 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,

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

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, $\mathrm{ÒÏ}=\frac{\mathrm{m}}{\mathrm{V}}$ (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

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

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

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

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

From the given scale of volume, we get

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

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

From equation (ii), we get

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

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