Indian Institute of Foreign Trade 2009 Solved Paper

Top face will look like the figure shown below.Curved surface area of the cylinder$=2\times \mathit{\pi }\times \mathit{r}\times \mathit{h}$
To calculate height of the cylinder in terms of 'r', we can see that Δ ADC is similar to Δ EFC
Therefore,
$\Rightarrow \frac{10}{4}=\frac{\mathit{h}}{4-\mathit{r}}$
$\Rightarrow \mathit{h}=\frac{5}{2}\left(4-\mathit{r}\right)$
Therefore, the curved surface area of the cylinder $=2\times \mathit{\pi }\times \mathit{r}\times \frac{5}{2}\left(4-\mathit{r}\right)=5\mathit{\pi }\mathit{r}\left(4-\mathit{r}\right)$
Hence, option B is the correct answer.