Question

two spheres have volume in the ratio 64:729 .if 160 ml paint is required for painting the surface area of the smaller sphere , how much paint is required to paint the the larger one?

Answer 1

Dear student

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{larger}\mathrm{sphere}\mathrm{be}\mathrm{R}\mathrm{and}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{sphere}\mathrm{be}\mathrm{r}. \\ \mathrm{Given}:\mathrm{Ratio}\mathrm{of}\mathrm{volumes}=64:729 \\ \Rightarrow \frac{{\displaystyle \frac{4}{3}{\mathrm{\pi r}}^3}}{{\displaystyle \frac{4}{3}{\mathrm{\pi R}}^3}}=\frac{64}{729} \\ \Rightarrow {\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^3=\frac{64}{729} \\ \Rightarrow \frac{\mathrm{r}}{\mathrm{R}}=\frac{4}{9} \\ \mathrm{So}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{smaller}\mathrm{sphere}=4{\mathrm{\pi r}}^2=4\times \frac{22}{7}\times 4^2=\frac{1408}{7}=201.14\mathrm{unit}\mathrm{sq}. \\ \mathrm{Also},160\mathrm{ml}\mathrm{is}\mathrm{required}\mathrm{for}\mathrm{painting}\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{smaller}\mathrm{sphere} \\ \mathrm{ie}160\mathrm{ml}\mathrm{is}\mathrm{required}\mathrm{for}\mathrm{painting}\frac{1408}{7}\mathrm{sq}\mathrm{units} \\ \mathrm{so}1\mathrm{sq}\mathrm{unit}=160÷\frac{1408}{7}=160\times \frac{7}{1408}=0.795\mathrm{ml} \\ \mathrm{So}\mathrm{paint}\mathrm{required}\mathrm{to}\mathrm{paint}\mathrm{the}\mathrm{larger}\mathrm{one}=4{\mathrm{\pi R}}^2\times 0.795=4\times \frac{22}{7}\times 9^2\times 0.795=809.5\mathrm{ml} \end{gathered}$$
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