A cylindrical pencil of diameter 1.2 cm has one of its end sharpened into a conical shape of height 1.4 cm. The volume of the material removed is

The volume of the material removed = volume of cylinder - volume of cone

                                                    = $$\pi r^2 h  - \frac{1}{3}\pi r^2 h$$ 

                                                     =  $$ \frac{2}{3}\pi r^2 h$$      { $$ r = \frac{1.2}{2} = 0.6 , h =1.4$$ }

                                                      = $$ \frac{2}{3} \times \frac{22}{7}\times 0.6^2 \times1.4 = 2 \times 22 \times 0.2 \times 0.2 \times 0.6$$  = 1.056 $$cm^3$$