The radii of two soap bubbles are `r_(i)` and `r_(2)`. In isothermal conditions, two meet together in vacuum. Then the radius kof the resultant bubble is given by
A. `R=(r_(1)+r_(2))//2`
B. `R=r_(1)(r_(1)r_(2)+r_(2))`
C. `R^(2)=r_(1)^(2)+r_(2)^(2)`
D. `R=r_(1)+r_(2)`

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Correct Answer - C
Under isothermal condition surface energy remain constant
`therefore 8pi_(1)^(2)T+8pir_(2)^(2)T = 8piR^(2)T implies R^(2) = r_(1)^(2)+r_(2)^(2)`

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