The curved surface area of a right circular cone is 65 $$\pi$$ $$cm^{2}$$ and the radius of its base is 5 cm. What is 40% of the volume of the cone, in $$cm^{3}$$?

Given, 

Curved Surface Area (C.S.A) of cone = 65$$\pi\ $$'

We know, 

C.S.A of cone = $$\pi rl$$

$$\pi\times\ 5\times\ l=65\pi\ $$

$$l=13\ cm$$

$$\therefore\ l^2=h^2+r^2$$

$$\therefore\ 13^2=h^2+5^2$$

by solving, We get h = 12cm

Now, Volume of cone = $$\frac{1}{3}\pi\ r^2h$$

$$\therefore\ $$ 40% of volume of cone = $$\frac{40}{100}\times\ \frac{1}{3}\pi\ r^2h$$

$$\therefore\ \frac{4}{10}\times\ \frac{1}{3}\times\pi\ \times25\times\ 12\ $$

= 40$$\pi\ $$

Hence, Option B is correct.