SSC CHSL 9 Jan 2017 Shift 3 Paper

The hemispherical bowl has three surfaces to calculate : the interior hemisphere $\left({\mathit{r}}_{\mathit{i}\mathit{n}\mathit{t}}=7\right)$ cm , the exterior hemisphere $\left({\mathit{r}}_{\mathit{e}\mathit{x}\mathit{t}}=7+1=8\right)$ cm and the annular ( ring shaped ) top edge $\left({\mathit{r}}_{\mathit{e}\mathit{x}\mathit{t}},{\mathit{r}}_{\mathit{i}\mathit{n}\mathit{t}}\right)$
Area of hemisphere = $2\mathit{\pi }{\mathit{r}}^2$ and area of annular = $\mathit{\pi }\left({{\mathit{r}}^2}_{\mathit{e}\mathit{x}\mathit{t}}-{{\mathit{r}}^2}_{\mathit{i}\mathit{n}\mathit{t}}\right)$
Total surface area of hemisphere is the sum of these 3 areas
= $\left[2\mathit{\pi }{\left(7\right)}^2\right]+\left[2\mathit{\pi }{\left(8\right)}^2\right]+\left[\mathit{\pi }\left(8^2-7^2\right)\right]$
= $\mathit{\pi }\left(98+128+64-49\right)=241\mathit{\pi }$
= $241\times \frac{22}{7}=757.43\mathit{c}{\mathit{m}}^2$