Question

a rubber ball floats on water with its 1/3rd volume outside water What is the density of rubber Answer is 667 kg m-3

Answer 1

there!

Let 'g' be the acceleration due to gravity. 'p' be the density of water. So, p = 1 g/cm³ .
'V' be the volume of the Rubber and 'q' be the density of the Rubber.

Volume of Rubber within water = V – (1/3)V = (2/3)V = Volume of water displaced by the Rubber

Weight of the water displaced = (2/3)V × p × g

Weight of the Rubber = V × q × g

Since,
Rubber is floating on the surface,

(2/3)V × p × g = V × q × g

=> q = (2/3)p

That is, the density of the Rubber is (2/3) g/cm³.

2/3=0.66666666667 g/cm³
then change into the kg/m³

2/3 = 666.667 kg/m³

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Answer 2

Answer:

667kgm^3

Explanation:

formula = vpg

Weight of the water displaced = Volume of the body submerged x Density of Water ( 1000 kgm^3) x g

Weight of Rubber should be equal to weight of the water displaced, so

Weight of the water displaced = Volume of the body submerged x Density of Water ( 1000 kgm^3) x g = Volume of the entire rubber (take as V) x   density of rubber (unknown) x g

(2/3)V x 1000 x g = V x p x g

Now V and V get canceled and 2/3 remains

g and g get cancelled

remaining equation =  2/3 x 1000 = p (density of rubber)

2000/3 = 667kgm^3 Rounded of