Answer: Option C
$$\eqalign{
& {\text{Let the length of train be x m}} \cr
& {\text{When a train crosses a light }} \cr
& {\text{post in 9 second the distance covered}} \cr
& {\text{ = length of train }} \cr
& \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr
& {\text{Distance covered in crossing a}} \cr
& {\text{700 meter platfrom in 30 seconds}} \cr
& {\text{ = Length of platfrom + length of train}} \cr
& {\text{Speed of train = }}\frac{{x + 700}}{30} \cr
& \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr
& \Rightarrow \frac{x}{3} = \frac{{x + 700}}{{10}} \cr
& \Rightarrow 10x = 3x + 2100 \cr
& \Rightarrow 10x - 3x = 2100 \cr
& \Rightarrow 7x = 2100 \cr
& \Rightarrow x = \frac{{2100}}{7} = 300{\text{m}} \cr
& {\text{When the length of the platform be 800m,}} \cr
& {\text{then time T be taken by train to cross 800m}} \cr
& {\text{long platform}} \cr
& \frac{x}{9} = \frac{{x + 800}}{T} \cr
& \Rightarrow Tx = 9x + 7200 \cr
& \Rightarrow 300T = 2700 + 7200 \cr
& \Rightarrow 300T = 9900 \cr
& \Rightarrow T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr} $$