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Question

A sphere of radius R made up of material of relative density σ has a concentric spherical cavity of radius r. It just floats when placed in a tank full of water. The value of the ratio Rr will be

A
(σσ1)1/3
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B
(σ1σ)1/3
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C
(σ+1σ1)1/3
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D
(σ1σ+1)1/3
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Solution

The correct option is A (σσ1)1/3
Sphere will float if its weight is balanced by buoyant force acting on it.
FB=mg .............(i)


Since, the sphere is just floating i.e. it is floating with fully submerged.
FB=ρwgV
Here, V=43πR3
(volume of liquid displaced)
Mass of sphere,
m=ρsVs=43π(R3r3)ρs
Substituting in eq. (i),
ρwg×43πR3=43π(R3r3)ρsg
ρwρs=1r3R3
1σ=1r3R3
r3R3=11σ
r3R3=σ1σ
R3r3=σσ1
​​​​​​​Rr=(σσ1)1/3
Why this question?
Caution: In this case the volume of liquid displaced by the sphere will be equal to the volume of the entire sphere (of radius R) because the sphere is fully submerged.

Bottom line: Mass of the sphere will correspond to its volume excluding cavity.

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