A cylindrical tub of radius 16cm contains water to a depth of 30cm . A spherical iron ball is dropped into the tub and thus level of water is raised by 9cm . What is the radius of the ball ? use pie = 22/7
Answers
Answered by
11
Solution:-
Radius of the cylindrical tub = 16 cm
Depth = 30 cm
Water level rises = 9 cm
Volume of the spherical ball = 4/3*π*r³
Volume of the spherical ball = volume of the water level
4/3*π*r³ = π*16*16*9
r³ = (16*16*9*3)/4
r³ = 1728
r = 12 cm
Radius of the spherical ball = 12 cm
Answer.
Radius of the cylindrical tub = 16 cm
Depth = 30 cm
Water level rises = 9 cm
Volume of the spherical ball = 4/3*π*r³
Volume of the spherical ball = volume of the water level
4/3*π*r³ = π*16*16*9
r³ = (16*16*9*3)/4
r³ = 1728
r = 12 cm
Radius of the spherical ball = 12 cm
Answer.
Similar questions