 All right, so we'll move to the next question try solving this question this one Okay, so this is the viscous fluid We actually ignore viscosity most of the time But then since this is fear in a viscous fluid and if they have velocity Then there will be a viscous force also apart from the buoyant force buoyant force will of course act But apart from that there will be a viscous force also acting on it, which is given by This formula 6 pi eta r to v terminal Anyone got the answer? So few of you might have remembered the formula if it is just one sphere alone, so It is 2 by 3 2 by 9 pi r Row minus sigma times g divided by mu Okay, this This if you remember it will be fine. Otherwise you can always derive it okay, so If let us say I'm talking about P. So I have to take the density of the P and the density of the Fluid signal let us say sigma 2 and we have coefficient of viscosity here Okay So two spheres of equal radii have density row 1 and row 2 The sphere are connected L1 L2 our densities Sigma 1 and sigma 2 so this is sigma 1 and this is sigma 2 Coefficient of viscosity is are also eta 1 and eta 2 So if sphere P alone in L2 if P is alone in L2 Then VP is the velocity. So if P is in liquid 2 then you have to use liquid 2's Density and liquid 2's coefficient of viscosity. Okay, similarly vq is Q alone in L1, which will be what 2 by 9 pi r square row 2 minus sigma 1 Divide by eta 1 okay so this is how you get VP and vq and Now you can compare first of all sigma 2 has to be more than sigma 1 then only You know this liquid will be At the bottom and the upper liquid will be at the top Okay, so this is the first thing you should know and the second thing which There is Density of the second this thing density of q should be More than the density of the first liquid Okay, then only the q is Submuched okay q is not in the first liquid So q is down and also density of the second liquid has to be less than Wait if string is taught okay, and everything is sinking if everything is sinking then density of the second Ball has to be more than the density of the second liquid as well Okay, then only it will keep on going down with the velocity Otherwise, it will just float it will not go down fine. So therefore you can say that row 1 is less than Sigma 1 okay, and you have Sigma 2 is less than row 2 and we anyway have sigma 2 to be Greater than sigma 1 fine So row 1 is less than sigma 1 that is required Otherwise the string will not be taught okay, so Q is essentially pulling everything down Q is making effort so that p comes down So if q doesn't make any effort p will come not come down So basically there has to be a tension in the string Okay, so if there is a tension in the string and if you draw free by a diagram of p Let's say this is tension okay, then you have mg force and you have the buoyant force okay, so mg is what density of the material into volume of the material into g buoyant force is density of the liquid into volume of liquid into g okay, so If suppose velocity is almost zero, so we can ignore the viscous force then Tension is required over and above The mg force so it automatically means buoyant force is more than the mg force Okay, then only it will be pulled down and then it will gain some velocity and then viscous force will also start acting but Primefac the density of liquid one has to be more than the weight Sorry density of liquid one has to be more than the density of p. Okay, so when you analyze all this scenario you'll get option a and d to be correct, okay So we'll move to the next question We have around 15 minutes Okay, let's take a Question on surface tension See these questions Few are beyond mains level, but our focus is learning the concept So that is why I'm making you a little uncomfortable with slightly difficult questions, but You'll learn more Try solving this this one Okay, Simon and Purvik. I I have left that question After analyzing the scenario, so I want you to get option a and d. I have told you the final answer Okay, so after that it is nothing you just spend some time you'll get it I wanted to discuss this question on surface tension. So that's why I'm Not putting this one. So let me know you can WhatsApp me If you're not getting and send me your attempt anyone got the fourth one I think this fourth question is straight forward Kushal, how are you getting see explain quickly? I don't all of you know that surface tension T is force per unit length Okay, and this is also surface energy per unit area Okay So, let us say this is the drop which is formed On a dropper fine Then Let us say that this is the center Okay, I'm connecting the peripheries like this Okay, why I'm doing this because the surface tension will act on this circle Okay, which is from the dropper Okay, and along which direction it'll act it'll act like this Okay, so if this is the surface tension Okay Which is trying to reduce the surface area then Let's say this is theta Okay So if this is surface tension T. So T cos theta Okay T cos theta Into 2 pi r because this surface tension is acting on the entire of this circle Which is the end of the dropper. Okay, so T cos theta into 2 pi r will balance the weight Okay, this is the vertical force. So we need to just find the vertical force only Okay, and also we know that cos of theta is Small r by capital R where this is capital R and if you draw a vertical line like this this This radius of the dropper is small r. Okay, so cos theta where this is theta then this angle This yellow Angle is 90 minus theta then This angle Will become angle theta. So cos of theta is small r by capital R So this will give you vertical force to be equal to 2 pi r square T By small r. Okay So even though I have said that this is very easy one But it was not it was a tricky one which I'm sure many Students who have seen this for the first time in the exam would not have got it Okay, this is again a question which differentiates toppers from others So don't worry if you're not able to get it So just try to learn the concept here What about fifth and sixth? Once you get the fourth one see these passage questions are you know They are linked from each each other if you get one of the question right you will get all of them right because one Answer of the one will be used in the next question. So it will detach from the dropper when The vertical force become equal to mg isn't it And when it detaches, it will be almost spherical drop any answers for the fifth one D Puri is getting D Others so be Be for Bombay Yes No, see the vertical forces what to pi r squared T by Capital R Okay So this Should be equal to the weight of the drop which is 4 by 3 pi R cube Into rho into G Right, so from here you'll get the radius of drop and it will be approximately option a Fine. What about the sixth one? 6th. What is the answer? 6th one is direct Surface tension is surface energy per unit surface area as well So you just multiply surface tension with surface area. You'll get the answer for surface energy Okay, so You have got radius of the drop from here. All right, so R is equal to 1.4 into 10 raise to power minus 3 meters So surface energy will be equal to surface tension, which is given as point one one So point one one into four pi R square 1.4 into 10 is for minus Square this many jewels is the surface energy. Okay, so you'll get option B to be correct over here Fine, so I think today I have made you little uneasy with this chapter by showing you different kind of question that can be there other than your routine ones which I have sort of Intentionally not shown in this particular session Because I wanted you to be aware that there are a few other varieties that exist other than what is there in Etsy Varma Okay, so that's how you have to Take care of the variety rather than the quantity when you practice Questions yourself. All right, that's it for today. So I hope you have learned many things today What I'll do I'll forward you this particular PDF and Few questions. I have just given you hints so that you can solve it yourself. So immediately get the answer for those and what's at me in case you don't get it and Yes, so we will come back to you again in couple of days with another session on revision and this time I'm planning to take Exclusively entire optics. So I think we have already one session on optics now the session Will be slightly faster will be solving other kinds of numericals on Optics in the next class. Okay. Thank you for today. We'll see you. Thank you, sir