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A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `sigma` is the surface tension the diameter of the drop of the liquid isA. `sqrt((3sigma)/(t(2rho-d)))`B. `sqrt((6sigma)/(g(2rho-d)))`C. `sqrt((4sigma)/(g(2rho-d)))`D. `sqrt((12sigma)/(g(2rho-d)))` |
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Answer» Correct Answer - D In equilibrium force due to surface tension `+` force of buoyancy `=` Weight of the spherical liquid drop `2pirT+(2)/(3)+(2)/(3)pir^(3)d_(2)g=(4)/(3)pir^(3)d_(1)g` `T=sigma,d_(1)=rho,d_(2)=d` `2pirsigma+(2)/(3)pir^(3)dg=(4)/(3)pir^(3)rhog` `impliesr^(2)=(3sigma)/(g(2rho-d))implies[(D)/(2)]^(2)=(3sigma)/(g(2rho-d))` |
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