A hollow, plastic sphere is held below the surface of a freshwater lake by a cor

Question

A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.670 m3 and the tension in the cord is 2120 N .

1-

Calculate the buoyant force exerted by the water on the sphere.

Express your answer with the appropriate units.

2-

What is the mass of the sphere?

Express your answer with the appropriate units.

3- The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Explanation / Answer

a) Volume of sphere=v= 0.670 m^3

Density of water= d =1000 kg/m^3

Volume of water displaced by sphere=Volume of sphere=v= 0.670 m^3

Weight of water displaced by sphere=Ww =vdg=0.670*1000*g

Weight of water displaced by sphere=Ww = 670*9.8

Weight of water displaced by sphere=6566 N

The buoyant force exerted by the water on the sphere=Weight of water displaced by sphere =6566 N

The buoyant force exerted by the water on the sphere is 6566 N

b) Mass of sphere= m

Weight of sphere=Ws= mg

The tension in the cord =T= 2120 N.

The buoyant force =6566 N

For equilibrium of sphere

Weight of sphere+ tension in cord = buoyant force

Ws +2120 =6566

Ws =6566-2120 = 4446 N

Mass of sphere = m =Ws/g

Mass of sphere = m =4446/9.8

Mass of sphere = 453.673kg

Mass of sphere is 453.673 kg

c) Let volume inside water =Vi

Volume of sphere= v = 0.670 m^3

wt of displaced water = Vi*density of water*g=Vi*1000*g

Wt of sphere = 453.673*g

wt of displaced water = Wt of sphere

wt of displaced water = Vi*density of water*g

Vi*1000*g=453.673 *g

Vi=0.453673 m^3

volume of sphere =v =0.670 m^3

Vi / v = 0.453673 / 0.670

= 0.6771 or 67.71 %

When the sphere comes to rest, 67.71 % of its volume is submerged.

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