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Deduce expressions for the excess pressure inside a : (i) liquid drop, (ii) soap bubble.

Answer» <html><body><p></p>Solution :(i) Liquid drop:Let us consider a liquid drop of radius <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> and the <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> tension of the liquid is T.<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C07_E03_009_S01.png" width="80%"/> <br/>The various <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a> acting on the liquid drop are,<br/>(a) Force due to surface tension `F_(T)=2piRT` towards right<br/>(b) Force due to outside pressure,`F_(P_1)=P_(1)piR^(2)` towards right<br/>(c) Force due to inside pressure, `F_(P_2)=P_(2)piR^(2)` towards left<br/>As the drop is in equilibrium,<br/>`F_(P_2)=F_(T)+F_(P_1)` <br/>`P_(2)piR^(2)=2piRT+P_(1)piR^(2)` <br/>`rArr(P_(2)-P_(1))piR^(2)=2piRT` <br/>Excess pressure is `DeltaP=P_(2)-P_(1)=(2T)/(R)` <br/>(ii) Liquid bubble:A soap bubble of radius R and the surface tension of the soap bubble be T is as shown in Figure. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Hence, the force on the soap bubble due to surface tension is `2xx2piRT`. The various forces acting on the soap bubble are,<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C07_E01_053_S02.png" width="80%"/> <br/>(a) Force due to surface tension `F_(T)=4piRT` towards right<br/>(b) Force due to outside pressure, `F_(P_1)=P_(1)piR^(2)` towards right<br/>(c) Force due to inside pressure, `F_(P_2)=P_(2)piR^(2)` towards left<br/>As the bubble is in equilibrium,<br/>`F_(P_2)=F_(T)+F_(P_1)` <br/>`P_(2)piR^(2)=4piRT+P_(1)piR^(2)` <br/>`rArr(P_(2)-P_(1))piR^(2)=4piRT` <br/>Excess pressure is `DeltaP=P_(2)-P_(1)=(4T)/(R)` <br/>(iii) Air bubble:Let us consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let `P_(1)andP_(2)` be the pressures outside and inside the air bubble, respectively. Now, the excess pressure inside the air bubble is `DeltaP=P_(1)-P_(2)`.<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C07_E01_053_S03.png" width="80%"/> <br/>In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,<br/>(a) The force due to surface tension acting towards right around the rim of length `2piR` is `F_(T)=2piRT` <br/>(b) The force due to outside pressure `P_(1)` is to the right acting across a cross <a href="https://interviewquestions.tuteehub.com/tag/sectional-2255905" style="font-weight:bold;" target="_blank" title="Click to know more about SECTIONAL">SECTIONAL</a> area of `piR^(2)` is `F_(P_1)=P_(1)piR^(2)` <br/>(c) The force due to pressure `P_(2)` inside the bubble, acting to the left is `F_(P_2)=F_(T)+F_(P_1)` <br/>`P_(2)piR^(2)=2piRT+P_(1)piR^(2)` <br/>`rArr(P_(2)-P_(1))piR^(2)=2piRT` <br/>Excess pressure is `DeltaP=P_(2)-P_(1)=(2T)/(R)`</body></html>


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