Question

A mettalic cone having radius of base 7cm n height 28cm is melted and recast into a spehrical ball find the radius of thr spehrical ball

Answer 1

The radius of the spherical ball is approximately 5.809 cm.

Given : A mettalic cone having radius of base 7cm and height 28cm is melted and recast into a spehrical ball.

To find : The radius of the spherical ball.

Solution :

We can simply solve this mathematical problem by using the following mathematical process. (our goal is to calculate the radius of the spherical ball)

First of all, we have to calculate the volume of the cone.

So, the volume of the cone is :

= (1/3) × π × (radius)² × height

= [(1/3) × π × (7)² ×28] cm³

Let, the radius of the sphere = r cm

So, the volume of the sphere will be :

= (4/3) × π × (radius)³

= [(4/3) × π × (r)³] cm³

As, the spherical ball is formed from melting the cone, so both of them will have equal volumes.

So,

[(1/3) × π × (7)² ×28] = [(4/3) × π × (r)³]

(7)² × 28 = 4 × (r)³

(49 × 28)/7 = r³

r³ = 196

r = 5.809 cm (approx.)

Hence, the radius of the spherical ball is approximately 5.809 cm.

Answer 2

Step 1: Given data

radius of base of cone, $r=7cm$

height of cone, $h=28cm$

after recasting the cone into a spherical ball,

radius of spherical ball, $R=?$

Step 2: Using the formula

volume of cone $=\frac{1}{3}\pi r^{2} h$

volume of sphere $=\frac{4}{3} \pi r^{3}$

Step 3: Calculating the radius of spherical ball

Since, we recast the cone into a spherical ball

volume of cone $=$ volume of spherical ball

$\frac{1}{3}\pi r^{2} h=\frac{4}{3}\pi R^{3}$

$\frac{1}{3}\times \frac{22}{7}\times 7^{2} \times 28=\frac{4}{3}\times\frac{22}{7}\times R^{3}$

$R^{3} =7^{3}$

$R=7cm$

Hence, the radius of the spherical ball is $7cm$.

#SPJ2