Question

A cuboidal tank of length, breadth and height $10\mathrm{m},5\mathrm{m}$ and $5\mathrm{m}$ is filled with $900\mathrm{Kg}$oil. Find the volume of the cuboidal tank and the density of the liquid in Kg m^-3.

Answer 1

Step 1:

Given data:

The length of the cuboidal tank (l) = $10\mathrm{m}$

The breadth of the cuboidal tank (b) = $5\mathrm{m}$

The height of the cuboidal tank (h)= $5\mathrm{m}$

Mass of the oil in the tank (m)= $900\mathrm{Kg}$

Step 2:

Formula used:

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{a}\mathrm{cuboid}\left(\mathrm{V}\right)=\mathrm{length}\left(\mathrm{l}\right)\times \mathrm{breadth}\left(\mathrm{b}\right)\times \mathrm{height}\left(\mathrm{h}\right) \\ \mathrm{And}, \\ \mathrm{Density}\mathrm{of}\mathrm{a}\mathrm{substance}\left(\mathrm{\rho }\right)=\frac{\mathrm{Mass}\left(\mathrm{m}\right)\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Volume}\left(\mathrm{V}\right)\mathrm{of}\mathrm{the}\mathrm{substance}} \end{gathered}$$

Step 3:

Calculation of Volume of the tank:

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{cuboid}=\mathrm{length}\left(\mathrm{l}\right)\times \mathrm{breadth}\left(\mathrm{b}\right)\times \mathrm{height}\left(\mathrm{h}\right) \\ \mathrm{Hence}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cuboidal}\mathrm{tank}=10\mathrm{m}\times 5\mathrm{m}\times 5\mathrm{m} \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cuboidal}\mathrm{tank}=50\times 5 \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cuboidal}\mathrm{tank}=250{\mathrm{m}}^3 \\ \mathrm{Thus},\mathrm{thye}\mathrm{volume}\left(\mathrm{V}\right)\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{is}250{\mathrm{m}}^3. \end{gathered}$$

Step 4:

Calculation of density of the oil in the tank:

$$\begin{gathered} \mathrm{Density}\mathrm{of}\mathrm{oil}\mathrm{in}\mathrm{the}\mathrm{tank}=\frac{900\mathrm{Kg}}{250{\mathrm{m}}^3} \\ \mathrm{Density}\mathrm{of}\mathrm{oil}\mathrm{in}\mathrm{the}\mathrm{tank}=3.6\mathrm{Kg}/{\mathrm{m}}^3 \\ \mathrm{Thus}, \\ \mathrm{The}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{oil}\mathrm{in}\mathrm{the}\mathrm{tank}\mathrm{is}3.6\mathrm{Kg}/{\mathrm{m}}^3 \end{gathered}$$

Thus,

$$\begin{gathered} {\text{The volume of the cuboidal tank is 250 m}}^3\mathrm{and}, \\ \mathrm{The}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{liquid}\left(\mathrm{oil}\right)\mathrm{in}\mathrm{the}\mathrm{tank}\mathrm{is}3.6\mathrm{Kg}/{\mathrm{m}}^{3.} \end{gathered}$$