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ICSE Class 9 Physics Question 12 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 15

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15
Question

Question 1(xv)

A student is given three liquids A, B and C of densities ρ1, ρ2 and ρ3. He has to arrange these liquids in order of their increasing densities. To do so, he takes an object and dips it in each liquid and observes that the object floats with 1/9, 2/11 and 3/7 parts of its volume outside the surface of liquid A, B and C respectively. The correct order is :

  1. ρ1 > ρ2 > ρ3
  2. ρ2 > ρ3 > ρ1
  3. ρ3 > ρ2 > ρ1
  4. ρ3 > ρ1 > ρ2
Answer

ρ3 > ρ2 > ρ1

Reason

Given,

An object is floating in three liquids A, B, and C and the fraction of volume outside the liquid is :

A : 19\dfrac{1}{9} outside ⇒ 89\dfrac{8}{9} submerged

B : 211\dfrac{2}{11} outside ⇒ 911\dfrac{9}{11} submerged

C : 37\dfrac{3}{7} outside ⇒ 47\dfrac{4}{7} submerged

Making the denominator common for all the submerged fractions:

LCM of 9, 11, 7 = 693

A : 89\dfrac{8}{9} = 8×779×77\dfrac{8 \times 77}{9 \times 77} = 616693\dfrac{616}{693}

B : 911\dfrac{9}{11} = 9×6311×63\dfrac{9 \times 63}{11 \times 63} = 567693\dfrac{567}{693}

C : 47\dfrac{4}{7} = 4×997×99\dfrac{4 \times 99}{7 \times 99} = 396693\dfrac{396}{693}

For a floating object :

Fraction submerged=Volume submergedTotal volume=Density of objectDensity of liquid\text {Fraction submerged} =\dfrac {\text {Volume submerged}}{\text {Total volume}} \\[1em] =\dfrac{\text {Density of object}}{\text {Density of liquid}} ​ Since the object is the same in all three cases then submerged fraction is inversely proportional to density of liquid i.e., less submerged volume → greater liquid density and vice versa.

On comparing submerged fractions:

C : 396693\dfrac{396}{693} < B : 567693\dfrac{567}{693} < A : 616693\dfrac{616}{693}

So, the order of density of liquids is : ρ3 > ρ2 > ρ1