Question

From a solid right circular cylinder of height h and base radius r, a conical cavity of the same height and base is scooped out. Then the ratio of the volume of the cone and the remaining solid is __________.

Answer 1

It is given that, the radius and height of the solid right circular cylinder is r and h, respectively.

Now,

Radius of the conical cavity scooped out = Radius of the solid right circular cylinder = r

Height of the conical cavity scooped out = Height of the solid right circular cylinder = h

∴ Volume of the cone, V₁ = $\frac{1}{3}\mathrm{\pi }{r}^{2}h$

Also,

Volume of the remaining solid, V₂

= Volume of the cylinder − Volume of the conical cavity scooped out

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

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

$\therefore \frac{V_1}{V_2}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{2}h}{{\displaystyle \frac{2}{3}}\mathrm{\pi }{r}^{2}h}=\frac{1}{2}$

⇒ Volume of the cone, V₁ : Volume of the remaining solid, V₂ = 1 : 2

Thus, the the ratio of the volume of the cone and the remaining solid is 1 : 2.

From a solid right circular cylinder of height h and base radius r, a conical cavity of the same height and base is scooped out. Then the ratio of the volume of the cone and the remaining solid is ___1 : 2___.