A hemisphere and a cone both have same diameter. These two metal solids are joined by putting their bases together. The height of the cone is equal to the diameter of the sphere. This solid is melted and recast into a sphere of a diameter equal to one third of the diameter of the hemisphere. [2] (a) if radius of the hemisphere is r , find the volume of the combined solid. (b) find the number of spheres.
Answers
Answer:
a = (4/3)πr³
b= 27
Explanation:
Diameter of Cone and Hemisphere = 2r
Therefore, radius of Cone and Hemisphere = 2r/2 =r
Height of cone = 2r
Volume of Hemisphere is given by (2/3)πr³
and Volume of cone = (1/3)πr²(2r) = (2/3)πr³
hence, total Volume will be = (2/3)πr³ + (2/3)πr³ = (4/3)πr³
Given the diameter of new Sphere = 2r/3
then the radius of new Sphere = r/3
Hence, the volume of new Sphere = (4/3) π(r/3)³
= (4/3)πr³ / 27
on observing we can say
Volume of new Sphere = (Old Volume) /27
⇒ old Volume of sphere /Volume of new Sphere = 27
This means 27 new sphere can be casted
Answer:
Explanation:
Diameter of Cone and Hemisphere = 2r
Therefore, radius of Cone and Hemisphere = 2r/2 =r
Height of cone = 2r
Volume of Hemisphere is given by (2/3)πr³
and Volume of cone = (1/3)πr²(2r) = (2/3)πr³
hence, total Volume will be = (2/3)πr³ + (2/3)πr³ = (4/3)πr³
Given the diameter of new Sphere = 2r/3
then the radius of new Sphere = r/3
Hence, the volume of new Sphere = (4/3) π(r/3)³
= (4/3)πr³ / 27
on observing we can say
Volume of new Sphere = (Old Volume) /27
= old Volume of sphere /Volume of new Sphere = 27
This means 27 new sphere can be casted