Calculating Number of Air Bubbles Needed for Sphere to Float in Water

[leena19]

Homework Statement

Water in a tank is uniformly bubbled with small identical air bubbles each having volume V₀.
A sphere of mass M and volume V floats in water due to the attachment of certain number of air bubbles in the surface.If d_w is the density of water,and the minimum number of air bubbles that is needed to be attached to keep the sphere floating in water is n, then

(1) n =( M-Vd_w)/ (V₀d_w )
(2) n > ( M-Vd_w)/ (V₀d_w )
(3) n < ( M-Vd_w)/ (V₀d_w)
(4) n > ( V₀d_w )/( M-Vd_w)
(5) n < ( V₀d_w )/( M-Vd_w)

Homework Equations

The Attempt at a Solution

For the sphere to float,the density of water+ air bubble mixture should be more than the density of the sphere (I think?).
Assuming it is so,
d_w + n/v₀ > M/V
n/v₀ > M/V - d_w

I get
n > [ (M - dw*V)/V ] * V₀
which is not among the choices given

so I hope someone can tell me where I'm going wrong.

Thank you.

 

[kuruman]

Hi leena19,

Never mind the density relation. Go back to basics, namely Archimedes's Principle. For the mass to float, the displaced water must have weight equal to the weight of the object. Here there are two kinds of displaced water: (1) water displaced by the floating mass (we assume that it barely floats, i.e. its entire volume is under water) and (2) water displaced by n bubbles attached underneath the object.

Can you finish it now?

 

[leena19]

OK I think I get it now,

Mg = (V + nV₀)d_w*g
therefore n =( M - Vd_w) / V₀d_w

which would be answer no.(1) ?

 

[kuruman]

I would say the answer is (2). Let's say you are given numbers and you substitute and you get 156.6. Since bubbles come in whole numbers, the correct answer would be 157. With 156 bubbles the object would sink. So the > sign ensures that this does not happen.

 

[leena19]

Alright.
Thank you very much!