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Two soap bubbles, one of radius `50 mm` and the other of radius `80 mm`, are brought in contact so that they have a common interface. The radius of the curvature of the common interface is
A. `0.003 m`
B. `0.133m`
C. `1.2m`
D. `8.9m`

1 Answer

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Best answer
Correct Answer - B
`[P_(0)+(4sigma)/(R_(2))]=[P_(0)+(4sigma)/(R_(1))](4sigma)/R`
or `1/R=1/R-2-1/(R_(2))`
`R=(R_(1)R_(2))/(R_(1)-R_(2))=(50xx80)/30mm=400/3mm`
`=400/(2xx1000)m=4/30m=0.133m`

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