A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is (4/3)πa³. Is the following statement true or false and justify your answer
Solution:
Given, a solid ball is exactly fitted inside the cubical box side of side a.
We have to determine if the volume of the ball is (4/3)πa³.
Since the solid ball exactly fits inside the cubical box the diameter of the ball will be equal to the distance between the opposite faces of the cube.
So, diameter of ball = a
Radius = a/2
We know that, Volume of the sphere = (4/3)πr³
= (4/3)π(a/2)³
= (4/3)π(a³/8)
= (1/3)π(a³/2)
= πa³/6
Therefore, the volume of the ball is πa³/6.
✦ Try This: A solid ball is exactly fitted inside the cubical box of side a. What is the volume of remaining space inside the cubical box?
Given, a solid ball is exactly fitted inside the cubical box side of side a.
We have to determine if the volume of the ball is (4/3)πa³.
Since the solid ball exactly fits inside the cubical box the diameter of the ball will be equal to the distance between the opposite faces of the cube.
So, diameter of ball = a
Radius = a/2
We know that, Volume of the sphere = (4/3)πr³
= (4/3)π(a/2)³
= (4/3)π(a³/8)
= (1/3)π(a³/2)
= πa³/6
= (22/7)(a³/6)
= (11/7)(a³/3)
= 11a³/21
Volume of cube = a³
Volume of remaining space = volume of cube - volume spherical ball
= a³ - 11a³/21
= (21a³ - 11a³)/21
= 10a³/21
Therefore, the volume of the remaining space in the box is 10a³/21.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 13
NCERT Exemplar Class 10 Maths Exercise 12.2 Problem 4
A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is (4/3)πa³. Is the following statement true or false and justify your answer
Summary:
The statement “A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is (4/3)πa³.” is false
☛ Related Questions:
- The volume of the frustum of a cone is πh/3[r₁² + r₂² - r₁r₂], where h is vertical height of the fru . . . .
- The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as sho . . . .
- The curved surface area of a frustum of a cone is πl(r₁ + r₂), where l = √h² + (r₁ + r₂)², r₁ and r₂ . . . .
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