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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 isA. `0.003 m`B. `0.133m`C. `1.2m`D. `8.9m` |
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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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