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Question

The pressure inside the two soap bubbles is $$1.01$$ and $$1.02$$ atmosphere. When they are in air , the ratio of their volumes is (both have same surface tension):


A
8
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B
4
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C
16
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D
64
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Solution

The correct option is A $$8$$
Pressure difference between outside and inside is
$$P_{diff} = \dfrac{4T}{R}$$
Given,
$$P_{1diff} =1.01- 1=0.01= \dfrac{4T}{R_1}$$
$$P_{2diff} =1.02-1=0.02= \dfrac{4T}{R_2}$$
$$\therefore \dfrac{R_1}{R_2}= \dfrac{0.02}{0.01}$$
$$\therefore \dfrac{V_1}{V_2}= ( \dfrac{R_1}{R_2})^3 =(\dfrac{0.02}{0.01})^3=8$$

Physics

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