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A spherical drop of water has `1mm` radius. If the surface tension of the water is `50 xx 10^(-3) N//m` then find the difference of pressure between inside and outside the spherical drop is : |
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Answer» Correct Answer - `100N/m^(2)` `P_(excess) = (2T)/(R) = (2(50 xx 10^(-3)))/(10^(-3)) = 100 N//m^(2)` |
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