a marble is dropped into a cylindrical beaker of water. The beaker is 10 cm tall,though the water only comes up to 6 cm mark,and has a base with radius of 4 cm. the marble is a sphere with a radius of 12 mm. How many centimeters will the water rise in the beaker?

Guest Nov 29, 2014

#1**+5 **

The volume of the water and the marble added together will provide a clue.

V_{water} = pi(r^2) h = pi(4cm)^2(6cm) = 96pi cm^3

V_{marble} = (4/3)pi(1.2cm)^3 = 2.304pi cm^3

So.....the volume of the two things added together is 98.304pi cm^3

So we have the volume of the two things occupying some space in a cylinder =

98.304pi cm^3 = pi(4cm)^2(h) and solving for h, we have

98.304pi cm^3 / (4cm)^2 = h = 6.144 cm

So the water will rise .144 cm

CPhill Nov 29, 2014

#1**+5 **

Best Answer

The volume of the water and the marble added together will provide a clue.

V_{water} = pi(r^2) h = pi(4cm)^2(6cm) = 96pi cm^3

V_{marble} = (4/3)pi(1.2cm)^3 = 2.304pi cm^3

So.....the volume of the two things added together is 98.304pi cm^3

So we have the volume of the two things occupying some space in a cylinder =

98.304pi cm^3 = pi(4cm)^2(h) and solving for h, we have

98.304pi cm^3 / (4cm)^2 = h = 6.144 cm

So the water will rise .144 cm

CPhill Nov 29, 2014