The air pressure inside a soap bubbles of radius R exceeds the out side air pressure by 10 Pa. By how much will the pressure inside a bubble of radius 2R exceed the out side air pressure.
A. 20 Pa
B. 40 Pa
C. 2.4 Pa
D. 5 Pa

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1 Answers

Correct Answer - A
Excess of pressure inside the soap bubble,
`p = p_(1) - p_(o) = (4S)/(R) rarr p prop (1)/(R)`
`:. (p_(2))/(p_(1)) = (R_(1))/(R_(2)) = (R)/(2R) = (1)/(2)`
`p_(2) = (1)/(2)p_(1) = (1)/(2) xx l_(0)`
`rArr p_(2) = 5Pa`

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