Question

A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is
${\text{A}}{\text{.}}\dfrac{1}{1}$
${\text{B}}{\text{.}}\dfrac{1}{2}$
${\text{C}}{\text{.}}\dfrac{2}{1}$
${\text{D}}{\text{.}}\dfrac{2}{3}$

Answer 1

Hint: Volume of right circular cone is $\dfrac{1}{3}\pi {r^2}h$ and volume of sphere is $\dfrac{4}{3}\pi {r^3}$, taking 'r' same in both the formulas as its given.

Complete step-by-step solution-

Also its given:
Volume of cone $ = \dfrac{1}{2} \times $ Volume of sphere
Putting value of both formulas:
$
  \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{2} \times \dfrac{4}{3}\pi {r^3} \\
    \\
$
after solving this equation we get:
$ \Rightarrow h = 2r$
Hence the ratio of the altitude of the cone to the radius of its base is 2:1.
Correct option is C.

Note: It is given in question that the volume of cone is one-half of the volume of sphere therefore putting the volume formula both sides and keeping the radius r as common between the two which is given we simplify the equation and got the result as $ \Rightarrow h = 2r$, which signifies that the altitude of the cone is twice of the radius of its base.