SSC CHSL 15 Nov 2015 Shift 1 Paper

Radius of hemispherical bowl = 3.5 cm
Curved surface area of the hemisphere = $2\mathit{\pi }{\mathit{r}}^2$
= $2\times \frac{22}{7}\times {\left(3.5\right)}^2$
= $\frac{44}{7}\times 12.25=77\mathit{c}{\mathit{m}}^2$
As bowl is to be painted inside and outside, thus total surface to be painted = $2\times 77=154\mathit{c}{\mathit{m}}^2$
Now,cost of painting $10\mathit{c}{\mathit{m}}^2=\mathit{R}\mathit{s}.5$
∴ ost of painting 154 $\mathit{c}{\mathit{m}}^2=\frac{5}{10}\times 154=\mathit{R}\mathit{s}.77$