A cylindrical container filled with water having a cross section 0.04 m^2 is pla

Question

A cylindrical container filled with water having a cross section 0.04 m^2 is placed on a scale. The scale reads 235 N. Now a ball is put into the container. The ball floats on the water and the reading of the scale is 267 N. The density of water is 1000 kg/m^3.

a) What is the rise in water surface after the ball is put in the water compared to before?

b) To immerse the ball totally under the water's surface requires a downward force of 13 N. What is the density of the ball?

c) Now, suppose the ball is hollow inside. The ball is cut into two halves and sinks to the bottom. The height of the water surface compared to when the ball floats would be?

Explanation / Answer

so weight of ball is =267-235= 32 N

so F bouyancy = d V g

V = 0.04*h

32 = 1000*0.04*h*9.81

h=0.0815 m

b)

- Fapplied - mg + F bouyancy = 0

-13 - mg + dwater V g = 0

-13-32 + 1000*V*9.81=0

V=0.00459

so density = m/V = (32/9.81)/0.00459=710.7 kg

c) so when it floated it had to support all the mass now it has to do half that (since all of it is underwater)

h= 0.0815/2

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