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23/12/2021 12:58 pm
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There is a soap bubble of radius 6 cm and it has another soap bubble of radius 3 cm inside it. Find the equivalent radius of the soap bubble which has same excess pressure inside it as 3 cm bubble.
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 9 cm
1 Answer
1
23/12/2021 1:11 pm
Correct answer: (b) 2 cm
Explanation:
P1 - P0 = \(\frac{4T}{r_1}\)
P2 - P1 = \(\frac{4T}{r_2}\)
P2 = \(\frac{4T}{r_2}\) + \(\frac{4T}{r_1}\) + P0
Now new soap bubble of radius x having same excess pressure as 3 cm bubble
P' - P0 = \(\frac{4T}{x}\)
P' = P2
\(\frac{4T}{r_2}\) + \(\frac{4T}{r_2}\) + P0 - P0 = \(\frac{4T}{x}\)
\(\frac{1}{r_1}\) + \(\frac{1}{r_2}\) = \(\frac{1}{x}\)
x = \(\frac{r_1r_2}{r_1 + r_2}\) = \(\frac{6 \times 3}{9}\)
= 2 cm
This post was modified 4 years ago 2 times by Reyana09