 do these we already discussed the definition of bulk modulus right I'll write it again this is the bulk modulus so is it option a option a others okay poor week is also getting a yes option a is correct here so volume change by 10% it means that delta v by v is actually 0.1 okay and delta p is also given delta p is p2 minus p1 okay so that is 1.65 64 in fact 1.64 10 to 10 ratio power 4 Pascal's right 1.55 alright so bulk modulus will be delta p divided by delta v by v so it will be 1.55 into 10 ratio power 5 that's our option is correct what about the 11th one so here comes the surface tension concept this is turned this is hemisphere this is sub hemisphere fine so can I say that radius of bubble one is less than radius of bubble two yes or no I can see there is a debate going on so I'll just show you something over here let's say this is the you know diameter of the pipe okay now let me draw a circle over here now here I am drawing a circle whose diameter is equal to the length of the this thing length of the pipe itself a dam the length or the diameter of the pipe itself now let me draw a sub hemispherical shape here so this line no longer be a diameter are you getting it so this is how the scenario will be okay so you can see that I mean I'm not able to perfectly draw this it will be like that okay so you can see that the white circle okay where you can just consider scenario like this now tell me which one has bigger radius the white or the orange one this is let us say hemispherical which is r1 radius this is sub hemispherical which is r2 which has more radius the white one right so that is the reason why sub hemispherical bubble will have larger radius okay now we know that delta p for a bubble is 4s by r okay delta p is what delta p is pressure inside minus pressure outside outside pressure is atmospheric pressure okay this should be equal to 4s by r1 right right now my focus is to find p1 so I'll write it as p1 minus pa is 4s by r1 are you able to see my screen okay so this is let us say my first equation and my second equation is p2 minus pa is 4s by r2 okay now tell me which is more p2 or p1 since r1 is less than r2 okay so p1 minus p2 or p1 minus pa sorry this will be more than p2 minus pa right hence p1 is greater than p2 getting it so pressure here is more than pressure there and that is the reason why the air will flow like this so air will go from 1 to 2 from higher pressure to lower pressure that's its regular tendency right and when it does that then the pressure of 1 reduces pressure of 2 increases okay so air flows from 1 to 2 and the volume of soap bubble at end 1 decreases so that is why option b is correct over here any doubt guys on this question I'll move to next one meanwhile any doubts okay do these two questions okay anyone question 12 Mishra is saying no Mishra answering 13th one okay yes 13th is C so 12th one is clearly a question which will just eat up a lot of your time okay so the smart way is to first answer question number 13 and then go to question number 12 so we'll also be smart and we'll answer question number 13 first one end of the horizontal thick copper wire of length 12 and radius 2r is welded to an end of another horizontal thin copper wire length l and radius r when the arrangement is stretched by applying forces at two ends the ratio of elongation the thin wire to that of thick wire is what is asked so the thick copper wire has length 12 and radius 2r the thin one has length l and radius r now they are pulled by some force let us say force is f okay so if you draw a free body diagram of the thicker wire it will experience same force and the thin wire also experience the same force okay so we have the stress in the thick wire equal to be force divided by pi into 2r square okay so this is f divided by four times pi r square point and sigma of the thin wire is f divided by pi r square all right and we know that stress by strain is constant so let's say thick is sigma one and thin is sigma two so sigma one by epsilon one it's a it's Young's modulus right should be equal to sigma two by epsilon two so from here I will get epsilon one divided by epsilon two to be equal to sigma one by sigma two right sigma one by sigma two is one by four okay epsilon one is delta l one divided by the length of the thicker wire which is 12 divided by delta l two divided by l this is equal to one by four okay so from here I will get delta l one by delta l two to be equal to one by two okay okay so delta elongation the thin wire is two times the elongation the thick wire so don't be in a hurry and you should not mark b okay option c you should mark because to clearly see which one you are finding the ratio all right now let us try to solve question number 12 question number 12 is actually one of those questions which distinguish toppers from others okay so it's not a straightforward one there are a lot of fine things which people ignore okay so anyways a thin uniform cylindrical shell closed at both the ends so you have a cylindrical shell which is closed from both the end is partially filled with water so this cylinder is filled with water okay it is floating vertically in the water half submerged state so it is half submerged fine and if rho c is a relative density of the material of the shell rho c with respect to water then correct statement is that shell is more than half filled okay so basically you know we should first before even thinking first we should do the force balance here okay and then look at the statements because we need to actually evaluate if what happens then what will happen so if then else statements are this the the options are in terms of if then else statements right so we need to first get some equations then only we can assess the statements so if you do force balance then then there is downward force okay so there is this downward force let us say mg okay and then there will be a buoyant force okay so simply buoyant force has to be equal to mg buoyant force should be equal to m into g now mg can be divided into two parts mg of the shell this is shell okay plus mg of the water this should be equal to the buoyant force okay now buoyant force is what buoyant force is the weight of the liquid that is displaced okay so if let us say if this this is if the total height is h then about h by two height we have you know the cylinder occupied because it is written that in half summer state it is floating okay so if area of cross section is a okay then I can say that buoyant force is a into h by two into rho of water into g okay so this is for the cylinder and also there will be some amount of liquid that is displaced because of the because of the thickness of the cylinder you know so if let us say the entire volume of the shell okay is v if you ignore the thickness of the shell then because of the thickness of the shell you can ignore the volume displaced okay but if you say here if you look from the top if the cylinder is like this it has some thickness okay if it has some thickness then area of cross section is this inner area okay why I am distinguish the distinguishing two areas because the relative densities are different okay so this plus the area sorry the volume which is displaced let us say v by two where v is the volume the shell okay this volume of the shell divided by two into density of water into g okay this is the total buoyant force this is because of the cylindrical hole and this is because of the thickness of the cylinder this is the weight of the liquid that is displaced now this will be equal to the weight of the shell which is volume the shell divided by now it you have to take total volume the shell volume the shell into density of the shell into g okay plus you have mg of the water inside so let us say up to x level the water is inside okay so I can say that area into x area into x is the volume the water inside into density of water into g okay so let me tell you that this question may not have been solved by the toppers also they might have also left it okay so this is meant to learn the concept but not to actually you know solve so I mean this is a good opportunity for you to understand how the concepts are applied okay okay Purvik is asking me to explain again see when the cylinder is submerged inside the water okay it will displace the volume of the water okay now what I am doing is I am treating the cylinder okay I am treating the cylinder as if it is hollow but it has some finite thickness okay so area of cross section is the area of the whole of the cylinder okay so I am treating that the hollow cylindrical part separately here because I have to use a density of the material over here okay so the volume of the water that is displaced because the cylinder is going down is equal to is equal to the volume of the hollow portion of the cylinder plus the volume of the material which is inside the water waiting it so that is the reason why we have let us say volume the shell to be equal to vs which is the actual volume of the material then we shall divide by 2 into rho w into g this is the volume or weight of the water that is displaced which is the volume which corresponds to the volume of the material only okay but then not only that much volume is displaced but also the volume corresponding to that hollow portion is also displaced I hope I am making sense here let me pull it are you able to understand this or should I explain again see this is the scenario what I am saying is that this cylindrical thing has some thickness okay so let us say this is the thickness of the cylinder so if I just consider this thickness itself it will have some volume yes or no just this thickness or the volume of the material that is used to build the cylinder if I just consider that it has some volume so that volume of the material used in the cylinder is vs fine but that volume will just give you the volume of the material used but when this cylinder is inside the water it will not only displace the volume corresponding to the material but it will displace the entire volume from here till there getting it so the volume the water displaced how I am finding I am first finding the volume which is corresponding to this and then I am finding volume which is corresponding to the material which is there I am separating them apart because the density of the material is given separately and the conditions or the options are given with respect to that so that is the reason why we are doing it and it is okay that you are not able to do it right in the first go so it is expected so but then you should understand how it is doing it is done right now that is more important okay so you will get an equation you get an equation corresponding to x now okay so once you get an equation corresponding to x all you have to do now is to evaluate those options okay so now here onwards try to now analyze the situation further and get back to me in case you have any further doubts because again it will take more than 15-20 minutes of discussion just on this question itself so try this out at your end and let me know if you get stuck okay so I will move to the next one