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 0.650 m3 and the tension in the cord is 830 N.

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

B) What is the mass of the sphere?

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

Answer 1

Solution:-

Given –

Volume – 0.650 m³

Tension in cord – 830 N

a)      Calculate buoyant force,

Take density of water d – 1000 kg/m³

Volume of water displaced by sphere = volume of sphere v = 0.650 m³

Weight of water displaced by sphere Ww = vdg = 0.650*1000*9.8

Weight of water displaced by sphere = Ww = 650*9.8

Ww = 6370 N

The buoyant force exerted by the water on the sphere = weight of water displaced by sphere = 6370 N

The buoyant force exerted by the water on the sphere = 6370 N

b)     Calculate mass of sphere ,

Mass of sphere = m

Tension in the cord T = 830 N

The buoyant force exerted by the water on the sphere = 6370 N

For equilibrium of sphere

Weight of sphere + tension in cord = buoyant force

Ws+830 = 6370

Ws = 6370 – 830 = 5540 N

Mass of sphere = m = 5540 / 9.8

Mass of sphere = 565.3061 kg

c)      Calculate volume

Let volume inside water = Vi

Volume of sphere v = 0.650 m³

Weight of displaced water = vi*density of water * g

Vi*1000*g

Weight of sphere = 565.3061*g

Weight of displaced water = weight of sphere

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

Vi*1000*g = 565.3061*g

Vi = 0.5653061 m³

Volume of sphere v = 0.650 m³

Vi/v = 0.5653061 / 0.650

= 0.8697     or 86.97 %

When the sphere comes to rest, 0.8697 or 86.97 % of its volume is submerged.