The fraction of a floating object of volume V₀ and density
d₀ above the surface of liquid of density d will be
(a)d₀/d - d₀
(b)d - d₀/d
(c)d₀/d
(d)d₀d/d + d₀
Answers
Answer:
Correct option: (B) [{d – d0} / d]
Step-by-step explanation:
Explanation: When object floats, according to Newton’s third law, buoyant force of liquid has an opposite direction and same magnitude as the weight force of the floating object.
From this condition, we obtain equation of force equilibrium.
d ∙ g ∙ Vin = mg ----- where Vin – volume of object which is immersed in the liquid.
Also m = dV = mass of the object
Hence
d ∙ g ∙ Vin = d0 ∙ V ∙ g
∴ d Vin = d0 V
(Vin / V) = (d0 / d)
∴ 1 – (Vin / V) = 1 – (d0 / d)
[{V – Vin} / V] = [{d – d0} / d]
Here V – Vin = volume of floating object
∴ [{Vfloating} / V] = [{d – d0} / d]
The fraction of a floating object of volume V_(0) and density d_( 0) above the surface of a liquid of density d will ...
Step-by-step explanation:
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