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Pressure inside two soap bubbles are 1.0...

Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes is

A

`102:101`

B

`(102)^(3):(101)^(3)`

C

`8:1`

D

`2:1`

Text Solution

Verified by Experts

The correct Answer is:
C
Excess pressure `DeltaP = P_("in") - P_("out") = 1.01atm - 1atm " "=0.01atm " and similarly " DeltaP_(2)=0.02atm`
and volume of air bubble `V=(4)/(3)pir^(3) " "therefore V prop r^(3) prop (1)/((DeltaP)^(3))`
`["as" DeltaP prop (1)/(r ) "or " r prop (1)/(DeltaP)]`
`V_(1) (DeltaP_(2))^(3) (0.02)^(3) (2)^(3) 8`
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