KVPY SA Exam 2014 Question Paper

Question : 67 of 80

Marks: +1, -0

A machine is blowing spherical soap bubbles of different radii filled with helium gas. It is found that if the bubbles have a radius smaller than 1 cm, then they sink to the floor in still air. Larger bubbles float in the air. Assume that the thickness of the soap film in all bubbles is uniform and equal. Assume that the density of soap solution is same as that of water (=1000 kg m

^{-3}

). The density o f helium inside the bubbles and air are 0.18 kg

^{-3}

and 1.23 kg m

^{-3}

, respectively. Then the thickness of the soap film of the bubbles is (note 1 µm = 10

^{-6}

m )

0.50 µm
1.50 µm
7.00 µm
3.50 µm

Validate

Solution:

Let X is thickness of soap film for equilibrium.
Gravity force = buoyancy force

4/3π(10^{-2})^3× 0.18 + 4π(10^{-2})^2 (x) (1000) = 4/3π (10^{-2})^3 (1.23)

⇒

4π (10^{–2}) (x) (1000) =4/3π (10^{–6}) (1.08)

(10^5) x = 0.36

x = 0.36 × 10^{–5}

x = 3.6 × 10^{–6}

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