Question

a hemispherical bowl is made of 0.2 cm thick steel . the inner diameter of bowl is 8 cm. find the outer curved surface area of the bowl. also find the cost of polishing its outer surface area at the rate of rs. 2 per cm sqaure.?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{inner}\mathrm{diameter}\mathrm{of}\mathrm{hemispherical}\mathrm{bowl}=8\mathrm{cm} \\ \mathrm{Then}\mathrm{its}\mathrm{inner}\mathrm{radius}r,=4\mathrm{cm} \\ \mathrm{Thickness}\mathrm{of}\mathrm{steel}=0.2\mathrm{cm} \\ \mathrm{Hence},\mathrm{outer}\mathrm{radius}\mathrm{of}\mathrm{bowl}\mathrm{R},=\left(4+0.2\right)\mathrm{cm}=4.2\mathrm{cm} \\ \mathrm{Hence}\mathrm{the}\mathrm{outer}\mathrm{curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bowl}=2\mathrm{\pi }{\mathrm{R}}^2 \\ =\left(2\times \frac{22}{7}\times 4.2\times 4.2\right){\mathrm{cm}}^2 \\ =\left(\frac{44}{7}\times \frac{42}{10}\times \frac{42}{10}\right){\mathrm{cm}}^2 \\ =\left({\displaystyle \frac{44\times 6\times 42}{100}}\right){\mathrm{cm}}^2 \\ =110.88{\mathrm{cm}}^2 \\ \mathrm{Hence}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{polishing}\mathrm{its}\mathrm{outer}\mathrm{surface}\mathrm{area}=\mathrm{Rs}\left(2\times 110.88\right) \\ =\mathrm{Rs}221.76 \end{gathered}$$