Question

Question #025b7

Answer 1

I found `24.8 cm`
I called the outer radius `r_1=50/2=25cm` and the inner radius `r_2`.
The sphere displaces a volume of water whose weight is equal to the upward buoyant force that maintains the sphere floating (all submerged).

The volume of water displaced by the sphere is: `V_(water)=4/3pir_1^3` and the corresponding mass is `m_(water)=eta_(water)*V_(water)` where `eta_(water)=`density of water`=1 g/(cm^3)`.
The weight of water displaced is: `W_(water)=m_(water)*g=eta_(water)*V_(water)*g=eta_(water)*4/3pir_1^3*g`
The upward force is balanced by the weight of the sphere which is:
`W_(sphere)=eta_(Fe)*4/3pi(r_1-r_2)^3*g`
For floating (all submerged) must be:
`W_(sphere)=W_(water)` displaced
`1*r_1^3=7.9(r_1-r_2)^3`
`r_2=12.4cm` and `d=2*12.4=24.8 cm`

Please check my maths.