1.

Derive an expression for excess of pressure (pressure difference) inside the drop and bubble.

Answer» <html><body><p></p>Solution :In figure (a) a gas <a href="https://interviewquestions.tuteehub.com/tag/bubble-904920" style="font-weight:bold;" target="_blank" title="Click to know more about BUBBLE">BUBBLE</a> in <a href="https://interviewquestions.tuteehub.com/tag/liquid-1075124" style="font-weight:bold;" target="_blank" title="Click to know more about LIQUID">LIQUID</a> is shown and in figure (b) this bubble is shown spherical and in figure (c ) a soap bubble in air is shown .<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C10_E01_067_S01.png" width="80%"/><br/>Suppose , bubble formed by this way radius of bubble is r and inside pressure is `P_(i)`and outside pressure is `P_(0)`.<br/>Letus obtain `P_(i)-P_(0)` for bubble in liquid or drop of liquid .<br/>The pressure insidethe drop is `P_(i)` and outsideis `P_(0).P_(i)` is slightly higher than `P_(0)`.<br/>When bubble increase by `Deltar`, new radius becomes `r+Deltar`.<br/>Area of this surface `A_(2)=4pi(r+Deltar)^(2)`<br/>`A_(2)=4pi(r^(2)+2r*Deltar+Deltar^(2))`<br/> `Deltar^(2)`is very small comparedto others terms<br/>`thereforeA_(2)=4pi(r^(2)+2rDeltar)` ....(2)<br/>The change inarea of surfaceof drop<br/>`DeltaA=A_(2)-A_(1)`<br/>`DeltaA=4pi(r^(2)+2rDeltar)-4pir^(2)`<br/> `=4pi(r^(2)+2rDeltar-2r^(2))`<br/> =`4pi(2rDeltar)`<br/> `Deltar=8pirDeltar` ...(3)<br/>The work done for increase area `DeltaA` of free surface of drop.<br/> `W=S_(la)(DeltaA)`<br/>`W=S_(la)(8pirDeltar)` ....(4)<br/>In the process of blowing bubble the force due to pressure <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> `(P_(i)-P_(0)` on the inside surface of area `4pir^(2)` is `(P_(i)-P_(0))4pir^(2)` . Under the <a href="https://interviewquestions.tuteehub.com/tag/effect-966056" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECT">EFFECT</a> of this force , surface moved by `Deltar` displacement , hence work done on the surface, <br/>`W=("force")<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>("Displacement")` <br/> `W=("pressure difference")xx("area")xx("displacement")` <br/>`W=(P_(i)-P_(0))(4pir^(2))(Deltar)`...5<br/>Comparing equation (4) and (5),<br/>`(P_(i)-P_(0))(4pir^(2)Deltar)=S_(la)(8pirDeltar)`<br/>`thereforeP_(i)-P_(0)=(2S_(la))/(r)`.....(6)<br/>Bubble in liquid and drop of liquid has one free surface . For them equation of pressure difference ontain according to equation (6).<br/>Bubble in air have two free surface . For them formula for pressure difference obtained as below.<br/>`P_(i)-P_(0)=(4S_(la))/(r)` ....(7)<br/>From this it can be said that for drop or bubble <br/>`P_(i)-P_(0)prop(1)/(r)`</body></html>


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