Question

A cylindrical pencil is sharpened to produce a perfect cone at one end with no loss in the total length of the pencil.
a) If the diameter of the pencil is 1 cm and the length of the conical portion is 2 cm, calculate the amount of shavings up to 2 decimals places.
b) If the diameter of the graphite used is 2 mm and the length of the pencil is 10 cm, find the volume of wood used in the pencil before sharpening. (Take pi=3.14)

Answer 1

Answer :

Given : A cylindrical pencil is sharpened to produce a perfect cone at one end with no loss in the total length of the pencil.

a ) Amount of shaving in pencil = Volume of cylindrical part with height 2 cm - Volume of conical portion of pencil

We know Volume of cylinder = ${\mathrm{\pi r}}^2\mathrm{h}$
Here we take Diameter = 1 cm , So radius r = 0.5 cm
And
height h = 2 cm
So,
Volume of cylindrical part with height 2 cm = $\frac{22}{7}\times 0.5\times 0.5\times 2=\frac{11}{7}=1.57c{m}^3$
And
We know Volume of Cone = $\frac{{\mathrm{\pi r}}^2\mathrm{h}}{3}$
So,
Volume of conical portion of pencil = $\frac{\frac{22}{7}\times 0.5\times 0.5\times 2}{3}=\frac{11}{21}=0.52c{m}^3$
So,
Volume of shaving = 1.57 - 0.52 = 1.05 cm³ ( Ans )

b ) Volume of wood used in the pencil before sharpening = Volume of whole pencil - Volume of graphite used .

We know Volume of cylinder = ${\mathrm{\pi r}}^2\mathrm{h}$
Here we take Diameter = 1 cm , So radius r = 0.5 cm = 5 mm ( As we know 1 cm = 10 mm )
And
height h = 10 cm = 100 mm
So,
Volume of whole pencil = $3.14\times 5\times 5\times 100=7850m{m}^3$
And
Diameter of graphite used = 2 mm , So Radius of graphite used r = 1 mm

Volume of graphite used = $3.14\times 1\times 1\times 100=314m{m}^3$
So,
Volume of wood used in the pencil before sharpening = 7850 - 314 = 7536 mm³ = 7.536 cm³ ( Ans )