Question

a solid ball is exactly fitted inside the cubical box of side a, then volume of the ball is

Answer 1

Given:

Side of the cubical box=a

To find:

The volume of the solid ball

Solution:

The volume of the solid ball is 11$a^{3}$/21.

We can find the volume by following the steps given below-

We know that the side of the cubical box is equal to the diameter of the solid ball.

The side of the cube=a

So, the diameter of the ball=a

The radius of the solid ball=diameter/2=a/2

Now, the volume of a solid ball=4/3π$r^{3}$, where r is the radius of the ball.

On putting r=a/2, we get

The volume of the solid ball=4/3π(a/2$)^{3}$

=4/3π×$a^{3}$/8

=π$a^{3}$/3×2

=π$a^{3}$/6

Substituting π with 22/7,

=22/7×$a^{3}$/6

=11$a^{3}$/21

Therefore, the volume of the solid ball is 11$a^{3}$/21.

Answer 2

Answer:

volume of the ball = $\frac{11}{3} a^{3}$ cube units

Step-by-step explanation:

Given that a solid ball is exactly fitted inside the cubical box

                                                     side of the cubical box = a

here we need to find volume of the ball which is in spherical shape

the ball is excatly fitted in cubical box

then diameter of the ball will be the side of the cubical box

⇒ diameter d = 2r  = a  ( r is radius )

from above data radius of the ball (r) = $\frac{d}{2}$ = $\frac{a}{2}$    

volume of the  sphere is given by   = $\frac{4}{3} \pi r^{3}$

                          volume of the ball  = $\frac{4}{3} (\frac{22}{7} )(\frac{a}{2})^{3}$

                                                          = $(\frac{4}{3} )(\frac{22}{7})( \frac{a}{2} )(\frac{a}{2})(\frac{a}{2})$  

                                                           = $\frac{11}{3} a^{3}$ cube units