Acylindrical vessel of diameter 18 cm contains water. if iron spherical balls of radius 3 cm are immersed in water, then the number of balls required to raise the level of water by 4 cm in the cylinder is?
Answers
Answered by
3
Hi ,
Dimensions of a cylinder :
diameter ( D ) = 18 cm
radius ( R ) = D/2 = 18/2 = 9 cm
level of water raised after iron balls
immersed in water = ( H ) = 4 cm
Dimensions of each iron ball :
radius ( r ) = 3 cm
Let the number balls immeresed = n
Now ,
n × volume of each ball = volume of water raised
n × ( 4/3 ) × π r³ = π R² H
n = ( π R² H )/ [ ( 4/3 ) π r³ ]
n = (3 R² H ) /4r³
= ( 3 × 9 × 9 × 4 )/ ( 4 × 3 × 3 × 3 )
= 9
Therefore ,
Number of iron balls immersed = n = 9
I hope this helps you☺☺✌
#arjun
Dimensions of a cylinder :
diameter ( D ) = 18 cm
radius ( R ) = D/2 = 18/2 = 9 cm
level of water raised after iron balls
immersed in water = ( H ) = 4 cm
Dimensions of each iron ball :
radius ( r ) = 3 cm
Let the number balls immeresed = n
Now ,
n × volume of each ball = volume of water raised
n × ( 4/3 ) × π r³ = π R² H
n = ( π R² H )/ [ ( 4/3 ) π r³ ]
n = (3 R² H ) /4r³
= ( 3 × 9 × 9 × 4 )/ ( 4 × 3 × 3 × 3 )
= 9
Therefore ,
Number of iron balls immersed = n = 9
I hope this helps you☺☺✌
#arjun
Similar questions