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
Answered by
12
Answer:
Correct option: (C) [{d – d0} / d]
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]
Answered by
8
Answer:
C) is the correct answer
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