Question

Using clay, a student made a right circular cone of height $48$ cm and base radius $12$ cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.

Answer 1

Hint:
Using clay, a student made a right circular cone of height $48$ cm and base radius $12$ cm. Another student reshapes it in the form of a sphere. So first find the volume of the cone and since he reshapes it to the sphere, the volume of sphere and volume of the cone will be equal and simplify it, you will get the radius.

Complete step by step solution:
Using clay, a student made a right circular cone of height $48$ cm and base radius $12$ cm. Another student reshapes it in the form of a sphere. So first let us find the volume of the cone.
We know, volume of cone $=\dfrac{1}{3}\pi {{r}^{2}}h$ ………. (1)
Here, $h=48$cm and $r=12$cm.
So substituting above values in (1) we get,
volume of cone $=\dfrac{1}{3}\pi \times {{(12)}^{2}}\times 48$
volume of cone $=\pi \times {{(12)}^{2}}\times 16$ ………… (2)
Now, volume of sphere $=\dfrac{4}{3}\pi {{R}^{3}}$ ………….. (3)
So, since the cone is reshaped to sphere both the volumes will be the same.
volume of sphere$=$volume of cone $=\dfrac{4}{3}\pi {{R}^{3}}=\pi \times {{(12)}^{2}}\times 16$
now simplifying we get,
$\dfrac{4}{3}\pi {{R}^{3}}=\pi \times {{(12)}^{2}}\times 16$
$\Rightarrow{{R}^{3}}=\dfrac{3}{4}\times {{(12)}^{2}}\times 16$
$\Rightarrow{{R}^{3}}=3\times {{(12)}^{2}}\times 4$
$\Rightarrow{{R}^{3}}=12\times {{(12)}^{2}}$
$\Rightarrow{{R}^{3}}={{(12)}^{3}}$
So taking cube root we get,
$R=12$cm

Therefore, we get radius of sphere as $12$cm.

Additional information:
A cone is a three-dimensional shape in geometry that narrows smoothly from a flat base (usually circular base) to a point (which forms an axis to the centre of base) called the apex or vertex. We can also define the cone as a pyramid which has a circular cross-section, unlike a pyramid which has a triangular cross-section. These cones are also stated as a circular cone
A sphere is an object that is an absolutely round geometrical shape in three-dimensional space. Like a circle in 2D space, a sphere is a three-dimensional shape and it is mathematically defined as a set of points from the given point called “centre” with an equal distance called radius “r” in the three-dimensional space of Euclidean space. The diameter “d’ is twice the radius. The pair of points that connect the opposite sides of a sphere is called “antipodes”. The sphere is sometimes interchangeably called “ball”.

Note:
The only point which is important in the above problem is that as the cone is reshaped to the sphere, so the volume of sphere and volume of a cone is equal. From this assumption, the whole problem can be solved.
Also, volume of cone $=\dfrac{1}{3}\pi {{r}^{2}}h$ and volume of sphere $=\dfrac{4}{3}\pi {{R}^{3}}$.