Question

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Example 13 : Hanumappa and his wife Gangamma are making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm (see Fig. 13.22). If each $c{m}^3$ of molasses has about 1.2 g, find the mass of the molasses that can be poured into each mould. (Take $\mathrm{\pi }=\frac{22}{7}$)

Answer 1

Dear Student,

As the mould is in the shape of a frustum, so the volume of molasses is equal to volume of frustum.
Volume of frustum is given by V = $$\begin{gathered} \frac{1}{3}\mathrm{\pi h}({{\mathrm{r}}^2}_1+{{\mathrm{r}}^2}_2+{\mathrm{r}}_1{\mathrm{r}}_2),\mathrm{here}\mathrm{h}\mathrm{is}\mathrm{the}\mathrm{heght}\mathrm{of}\mathrm{the}\mathrm{frustum},{\mathrm{r}}_1\mathrm{is}\mathrm{the}\mathrm{smaller}\mathrm{radius}, \\ {\mathrm{r}}_2\mathrm{is}\mathrm{the}\mathrm{bigger}\mathrm{radiu}s. \end{gathered}$$
Hence the height is given as 14cm, r₂ = 35 cm and r₁ = 30 cm
$$\begin{gathered} \mathrm{V}=\frac{1}{3}\times \frac{22}{7}\times 14\left({30}^2+{35}^2+30\times 35\right)=46566.66{\mathrm{cm}}^3 \\ \mathrm{Since}1{\mathrm{cm}}^3=1.2\mathrm{gram} \\ \mathrm{So},46566.66{\mathrm{cm}}^3=46566.66\times 1.2=55879.99\approx 55880\mathrm{g} \end{gathered}$$

So the mass of molasses is 55880 g

Regards