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A drop of water detaches itself from the exit of a tap when [sigma= surface tension of water, rho= density of water, R = radius of the tap exit, r = radius of the drop]

Answer» <html><body><p>` <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> gt (2/3(Rsigma)/(rho g))^((1)/(3))`<br/>` r gt (2/3(Rsigma)/(rho g))`<br/>`(2sigma)/(r ) gt "atmospheric pressure"`<br/>`r gt ((2)/(3)(R sigma )/(rho g)^(2/3))`<br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/upward-721781" style="font-weight:bold;" target="_blank" title="Click to know more about UPWARD">UPWARD</a> force due to surface tension <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P2_U07_C03_E15_032_S01.png" width="80%"/> <br/> `=Fsintheta=sigmaxx2piR sin theta` <br/> `=sigmaxx2piRxxR/r=(2pisigmaR^2)/(r )` <br/> Weight of the drop of water = <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> `=4/3 pi r^3 rho g` <br/> The drop will detach if, <br/> `4/3pi r^3 rho g gt (2pi sigma R^2)/(r ) or, r^4 gt (3 sigma R^2)/( 2 rho g)` <br/> `thereforer gt ((3)/(2) (sigma R^2)/(rho g))^(1/4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> of the options are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a>.</body></html>


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