Question

Pressures inside two soap bubbles are 1.01 atm and 1.03 atm, ratio between their volumes is
@)20:1
(b) 3:1
c)12:01
(d) none of these

(VMMC 13)
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Answer 1

Answer:

Excess pressure as compared to atmosphere inside bubble A is Δp1=1.01−1=0.01 atm inside bubble B is Δp2=1.03−1=0.03 atm Also when radius of a bubble is r, formed from a solution whose surface tension is t, then excess pressure inside the bubble is given by p=4tr Letr1be the radii of bubbles A and B respectively then p1p2=4T/r14T/r2=0.010.03 r2r1=13 Since bubbles are spherical in shape their volumes are in the ratio V1V2=42πr3143πr32 (r1r2)3=(31)3=271 V1:V2=27:1

Explanation:

Answer 2

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Answer:

Excess pressures in the two bubbles will be (1.03-1) or 0.03 atm.

$\frac{p1}{p2} = \frac{r2}{r1}$

$\frac{r1}{r2} = \frac{p2}{p1} = \frac{0.03}{0.01} = 3$

$\frac{v1}{v2} = ( \frac{r1}{r2} ) ^{3} = {3}^{3}$

ans 27:1

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