Question

A hollow shell of iron (density of iron 8g/cc) has an outer volume of 100 cc. It just floats in a brine solution of density 1.2g/cc. What is the size of the hollow portion in cc?

Answer 1

Answer:

The density of material is ρ

The hollow sphere will float if its weight is less than the weight of the water displaced by the volume of the sphere This implies mass of the sphere is less than that for the same volume of water. Now, mass of spherical cell

m1 =4πR   ×t×ρ

While the mass of water having same volume

m2 =  4/3  πR^3  

where, ρg = density of water 1 g.cm  −3

 

For the floatation of sphere,  

m1 ≤m2

4πR ×t×ρ≤  3/4 πR^3

 

⇒tρ≤  R/3

​  

⇒t≤ R/3ρ

​ hope that helps

    pls mark as the brainliest