A vessel in the shape of a cuboid contains some water. If three identical spheres immersed in the water, the level of water is increased by 2 cm. If the area of the base of the cuboid is 160 cm ² and its height 12 cm, determine the radius of any of the spheres.
Answers
Answer:
The radius of the sphere is 2.94 cm.
Step-by-step explanation:
Given :
The area of the base of cuboids = lb = 160 cm²
Level of water in the vessel increased , h = 2 cm
Volume of the increased water in the vessel (cuboid ) = (lb)h
= 160 × 2 cm²
Volume of the increased water in the vessel (cuboid ) = 320 cm²…………….. (1)
Let the radius of each sphere be 'r’ cm.
Volume of each sphere = 4/3 πr³
Total volume of 3 spheres = 3 × 4/3 πr³ = 4πr³ cm³ ………..(2)
Since, the volume increase water in the vessel (cuboid ) is equal to the total volume of 3 spheres.
Volume of the increased water in the vessel (cuboid ) = Volume of 3 spheres
320 = 4πr³
320 = 4 × 22/7 × r³
r³ = (320 × 7 )/ (4 × 22)
r³ = ( 80×7)/22 = (40 × 7)/11 = 280/11 = 25.45
r³ = 25.45 cm
r = ³√25.45
r = 2.94 cm.
Hence , The radius of the sphere is 2.94 cm.
HOPE THIS ANSWER WILL HELP YOU….
Answer:
Step-by-step explanation:
Area of the base of cuboid = 160 sq cm
Let the radius of sphere be 'r'
Volume of one sphere = 4/3πr³
Volume of three spheres = 4πr³
Increased height of the cuboidal vessel = 2 cm
Increased volume of the water = (160 ×2) = 320 cu cm
Since the volume is increased after immersing the 3 spheres in the vessel, so increased volume of water will be equal to the volume of three spheres.
⇒ 4πr³ = 320
⇒ 4*22/7*r³ = 320
⇒ r³ = (320*7)88
⇒ r³ = 2240/88
⇒ r³ = 25.45
⇒ r = 2.94 cm
So, the radius of each sphere is 2.94 cm
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