Question

A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ₹5 per 100 sq cm. [Use $\mathrm{\pi }$ = 3.14] [CBSE 2015]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cubical}\mathrm{block},a=10\mathrm{cm} \\ \mathrm{The}\mathrm{largest}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{hemisphere}=a=10\mathrm{cm} \\ \mathrm{Also},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hemisphere},r=\frac{10}{2}=5\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}=\mathrm{TSA}\mathrm{of}\mathrm{cube}+\mathrm{CSA}\mathrm{of}\mathrm{hemisphere}-\mathrm{Area}\mathrm{of}\mathrm{circle} \\ =6a^2+2\mathrm{\pi }{r}^2-\mathrm{\pi }{r}^2 \\ =6a^2+\mathrm{\pi }{r}^2 \\ =6\times 10\times 10+3.14\times 5\times 5 \\ =600+78.5 \\ =678.5{\mathrm{cm}}^2 \\ \mathrm{As},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{painting}\mathrm{the}\mathrm{solid}=₹5\mathrm{per}100{\mathrm{cm}}^2 \\ \mathrm{So},\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{painting}\mathrm{the}\mathrm{solid}=678.5\times \frac{5}{100}\approx ₹33.92 \end{gathered}$$

hence, the cost of painting the total surface area of the solid is ₹33.92.