 those who are online please type in your names who is online please type in okay Saimhir there others only Saimhir is online I will take up only Saimhir's doubts some there is also there fine let's start taking up whatever doubts you guys have sent okay now few of you have taken liberty to send more than 15-20 questions I I want you to try those all questions your own first send me your detailed attempt okay then I will take up individually so if I take all those 20-25 questions from one person only then others doubts will not get clarified okay so I will be selectively taking let's say 3 or 4 questions from each person okay alright so let us take up the first one this is I guess from Sondarya Sondarya question number 11th right 11th is what is your doubt this one right who sent this who has sent this okay so nobody is claiming that who has sent this Snigda has sent this Snigda question number 11th right solve question number 11th all of you try attempting it tell me what is your answer quickly okay do you know that this ray of light will come parallel this line parallel to this line you know this right there will be just a lateral shift right similarly this ray of light will also come parallel to this one there is a small lateral shift okay it will be like this so that is the reason why the angle this angle will be alpha only getting it these two lines are parallel and these two lines are also parallel okay so whatever is the line between the whatever is the angle between these two will be the angle between those two also any two line parallel to this will have same angle okay alpha is the answer got it Snigda understood Snigda you understood I will go to the next one this one all of you this came in 2010 I guess anyone no one the power is equal to v square by r you guys remember this okay and r will be equal to v square by p answer is not a it is not a a is not the right answer yes d is correct how you get d Ramcharan how you get it the resistance is proportional to 1 by power the voltage is same voltage is same for all okay so if power is if p is highest r is lowest in fact 1 by r is proportional to p okay 1 by r100 is proportional to 100 I am assuming they are connected in parallel because in your home all the connection are parallel right so it is a fair assumption you connect all the both in parallel only r40 is proportional to 40 in fact I think option a is correct option d is not correct option a is correct a is correct here if it would have been wait wait wait if we assume that option a is true okay if you assume option a is true the problem is this guys the problem is that resistance will change with temperature change with temperature initially this is correct okay this is correct this seems to be correct but what will happen as temperature increases the resistance will no longer be whatever it was earlier so it not only depends on power it also depends on the temperature also fine so this equation the first one was valid initially and later on it is not valid but option d will be always valid or you can see also that of course 1 by r100 is greater than 1 by r60 is greater than 1 by r40 so both of these options looks to be correct a and d okay but not true if temperature changes okay let's go to the next one all of you understood right okay here diagram is not given wait I think it should be there somewhere this is the diagram this one is the diagram and this is the question okay now let me solve this one rectangular slab abcd n1 is immersed in water of effective index n2 n1 is more than n2 ray of incident ab as shown you need to find the maximum value of alpha so that ray comes out only from the other surface cd I think this we have done in problem practice already so isn't it we have done this already as far as I remember so I will just give you a hint and then you solve your own because this question we have already done or you can watch the video recordings so this is alpha max okay so what will happen is that it will bend towards a normal like this fine so there will be let's say angle i over here okay so if that is i this one will be 90 minus i okay and if alpha m is maximum if alpha m is maximum you know i will also be maximum using snell's law and 90 minus i will be minimum okay so we need to find minimum value of 90 minus i so that the ray comes out from cd so if it has if it should not come from this side then it should at least graze the surface okay so this angle should be equal to critical angle fine so angle of incidence is 90 minus i and angle of refraction is 90 degree and there is a relation between alpha and i using snell's law over here as well okay this is how you solve this question this one all of you specific heat will be same but if we increase the length to 2l the mass will change so just have to use conservation of energy yes who is this okay so alright many of you getting b as answer let me quickly solve this i'm audible right so 3 identical cells of negligible internet systems are connected in series so total potential difference is 3v okay and the wire has a length of l so resistance of the wire is usually what rho l by a right so it is some constant times l fine um so the current i is 3v divided by the constant times l this is voltage divided by resistance this is current and the power that is generated is i square r fine so 9 into v square divided by c square l square into c into l this is i square r okay this into t should be equal to mass into specific heat into delta t okay so this is equation 1 and in equation 2 you have n similar cells connected in series similarly you can find the power as n times v divided by now the length is 2l so c times 2l is the resistance so c times 2l this whole square into c times 2l this is i square r which is power into t this should be equal to now mass of l is m so for 2l it will be 2m 2m into specific heat into delta t so this is equation number 2 fine now just divide it left hand side and right hand side okay m is gone specific heat is gone delta t is gone here t is gone c l c l is gone c square c square l square l square v and v many things are gone this is n divided by n square by 4 this 2 square right so this into 2 this should be equal to 1 by 2 this 2 is also gone so n square should be equal to 36 or n should be equal to 6 fine so like this you get the value of 6 good that many of you have got it nice to see that we will go to the next doubt now whose doubt is this 47 all of you i think this is from akshath anyone wire is massless okay ignore the mass of the wire earlier there was some mass which is not given as a bob now the mass of the bob is increasing okay good let me quickly solve this we know that time period of the pendulum is 2 pi under root l by g okay now this time period does not depend on mass of the bob it is independent of mass of bob okay but what will happen is that the length will change because the wire is elastic it has a young's modulus so it will stretch a bit because the mass is stretching the wire that is why time period will change if mass does not stretch the wire time period will remain unchanged so let's say time period is 2 pi the length will now become l plus delta l this divided by g the power half fine so time period is now tm so time period is changed right earlier time period was this now time period is that if I divide it because if you see the option there is a t divided by tm that is coming in the expression fine so if I divide like this tm divided by t this will be equal to l plus delta l divided by l the power half okay understood as those who are sending me doubts now I may not be able to take up you had like full day to send me doubt on the last moment if you send me doubt it sounds very casual it is extremely casual and I will not entertain the casual attitude okay so this is what you get fine sending me out of doubt okay so delta l by l see the tension the additional tension that will be there because of mass is m into g fine the stress will be this okay so stress by strain will be equal to Young's modulus fine just give me a second I am getting lot of hmm so from here you get delta l by l to be equal to mg by Ay fine so we have tm by t whole square minus 1 equals to mg by Ay okay so that is option A will be correct over here fine I will go to the next question now okay Akshath you are not there is it you asked lot of doubts and then you are not there anyways we will skip other questions from him solve this one this is from Swetha I guess okay Vaishnavi I am very bad at remembering names no one should I solve this okay let me solve this question now so assume there is a tilt of alpha okay and then do the force balance okay so AB this this is AB this is A and that is B AB has a mass m and length l okay so from center of mass there will be centrifugal force there will be centrifugal force also everywhere the centrifugal force will be and that will be all the points are moving horizontally right they are moving horizontally only getting it all the points are moving in a horizontal circle now the centrifugal force will be like this okay at every point it will be horizontal and its value will be equal to let's say suppose this length we are taking r okay so the radius of this element would be what the radius of revolution of that element will be equal to r sin alpha fine so this force over here let's say dm mass is there is equal to dm omega square into r sin alpha fine this is the force now this is horizontal so who is balancing mg force can you tell me mg is vertical but the centrifugal force is horizontal so how mg is getting balanced anyone quickly tell me when you draw the free by a diagram who is balancing mg force yes force from the hinge okay so it is this bar is not only just in a force equilibrium condition but also it is not rotating like this getting it so torque about the hinge should be balanced okay so if you balance the torque about a the hinge forces will not come in play okay so torque because of mg is straight forward torque because of mg about the hinge is what mg force into l by 2 sin of alpha right this distance is l by 2 sin alpha torque because of mg is that okay this should be equal to torque due to the centrifugal force okay now torque will be different at different point because of centrifugal force so we need to integrate to find a total torque first we find the small amount of torque because of centrifugal force which is dm omega square r sin alpha this is centrifugal force right this multiplied by the vertical distance which is this from the hinge which is what r cos alpha right r cos alpha alright and dm can be written as dm will be equal to for total length mass is m so m divided by l into dr you can write it as dm okay so d tau will be equal to m by l omega square sin alpha cos alpha into r square dr everything else is constant only r is a variable and r goes from 0 to l alright you are able to understand right any doubt I am finding total torque because of the pseudo force which is centrifugal force this sin alpha cos alpha into l cube by 3 understood all of you able to get it what I am writing here okay this will be equal to torque because of mg that is mg l sin alpha by 2 fine so m is gone from l square so even that is gone alright one sin alpha is also gone cos alpha is 3g divided by 2 omega square l fine that is how often c is correct clear right shall I go to the next question okay let's go to the next one this one block of mass 3m connected to a mass less rod of length l lying address on a fifth frictionless table a second block of mass m impinges on the system it hits the system okay with velocity b0 strikes the opposite end of the rod at a right angle and sticks to it do you know the answer anybody got the answer it's a straight forward question straight forward yes kundi that is correct by conservation of momentum you have to use momentum okay so momentum of m is m into v0 plus momentum of 3m initially was 0 and when it sticks to the rod total mass is 4m 3m plus m this into final velocity so final velocity is v0 by 4 understood okay let me ask you to find angular velocity of this rod omega is what omega of entire system is what angular velocity oh yeah correct correct correct this is velocity of center of mass this is vcm guys okay so vcm is correct this is not the velocity of the center of the rod you need to find velocity of center of rod you got velocity of center of mass okay okay so you need to find velocity of this center of the rod not center of mass of the rod this velocity is what you need what you got is velocity of center of mass and center of mass is where center of mass will be somewhere over here so you got this velocity okay the rod is also rotating let's say it rotates like this so velocity of this point will be equal to vcm plus if this distance is x because omega into x will be the velocity of this point fine so you need to get the omega also now solve this it's a nice question anybody got angular velocity of omega see here you need to conserve angular momentum okay conserve angular momentum about center of mass so initial angular moment first of all you need to find where is the center of mass so let's say m comes and sticks over here center of mass will be where so let's say this is my y axis and this is my x axis so x coordinate for 3m is 0 so 3m into 0 plus m into l divided by 4m okay so l by 4 it comes from 3m okay so this distance center of mass distance is l by 4 from here so that distance will be 3l by 4 okay so angular momentum of this mass capital M mass before collision was m into v0 into 3l by 4 there was no angular momentum of the rod after it sticks it becomes entire single rod this will be moment of inertia i into omega fine so i is odd i is 3m into l by 4 whole square plus m into 3l by 4 whole square right the rods moment of inertia is 0 because it is mass less so moment of inertia is only because of this 3m and m getting it all of you this will come out to be 12 by 16 ml square okay 3 by 4 of 4 ml square is moment of inertia 9 plus 312 understood any doubt quickly tell me 3 by 4 ml square into omega okay 3 by 4 is also gone so omega is simply v0 by l after collision fine now velocity of the center of rod is equal to 3m which is v0 by 4 plus omega into l by 4 sorry omega into this distance this one okay this is also l by 4 because from 3m distance is l by 2 this is l by 4 so this is also l by 4 so omega which is v0 by l into x which is l by 4 fine so you get v0 by 4 plus v0 by 4 so which is v0 by 2 okay what is the correct answer in the book whichever book you have referred is it b who asked this question is it b okay great so nice question after long time I am seeing a good question this one who asked this question all this 16th question okay Ramu asked did not answer did not answer the hint is get the value of get the value of the resistance of gallium meter and the current which should be passed to the gallium meter for it to give the full deflection okay it is very easy to get lost in the words okay so using this information get the maximum voltage and maximum current should I solve moving gallium meter has 150 equal division 10 divisions per milli ampere okay so up to what milli ampere it can measure the current 150 divided by 10 so 15 milli ampere is the maximum current fine all of you getting it right and the voltage is 2 division per milli volt okay so we have 150 division so 150 divided by 2 is 75 so 75 milli volts is the maximum voltage fine so when the when the readings see in a gallium meter it will not be written that the scale is for voltage or current simply the divisions will be made so here you have 150 divisions like this okay so if it swings like this the pointer swings like this this is the maximum scale it swings to this maximum point means 15 milli ampere it could also mean 75 milli volt depending on what you are measuring fine it will not be written that this scale is only for the voltmeter or for the emitter okay it is up to you to assume what is the scale is so you can say that okay 10 division means 1 milli ampere and 2 division means 1 milli volt so like that you know you go about in voltmeter so we get the value of maximum volt and maximum current so the resistance of the gallium meter is 75 divided by 15 which is 5 ohm milli will get cancelled of numerator and denominator we need to reduce sensitivity so go on adding resistance increase it instead in order to each division reads 1 volt resistance see let us solve it using basics let us not remember any formula okay I am going to solve this using basics of what is voltage what is correct okay so there is no question of why this formula is used sensitivity should be increased and like that so I am using very very basics here so now each division should read 1 volt okay so maximum voltage should be what the maximum reading should be 150 volts isn't it so 150 divisions are there and one division should read 1 volt okay by the way we are not reducing the sensitivity we are increasing the sensitivity earlier it was milli volt now it is 1 volt 1 volt is 1000 times milli volt anyways so maximum voltage is 150 now we should measure okay so what is the resistance we should connect in series now it has become just another question we know the resistance of galvanometer we need to find out what resistance should we connect here let's say resistance is s okay so between 1 and 2 this maximum voltage should be 150 okay so if it is 150 volt then the current should be max allowable current it is 150 volt because max current is showing the full deflection the current that goes inside the galvanometer should be max then only for 150 volt full deflection will come okay so the current is 150 milli ampere and voltage is 150 volt right the 150 volt should be equal to the max current which is 15 into 10 is power minus 3 multiplied by the total resistance total resistance is resistance of the galvanometer plus resistance which are connected in series that way okay so this is 1510 so 5 plus s is 10,000 okay so s is 9,995 all of you understood are you understood why are you there are you understood okay oh that one this question this is actually little confusing but since someone has asked this let's solve this question ideally I should have ignored this one there are some questions which are meant to be left when you encounter such thing in your actual j-main paper which is like completely new scenario you have never have thought about it just leave it and move to the familiar question okay try attempting this for 2 minutes hello yes can I call you sometime in a class have you read the question now should I solve okay I am solving it if light goes straight this is geometrical spread to a width is because of geometrical spread but there will be diffraction so light will bend like this okay this this is because of diffraction spread understood this is y and this is y total spread is 2 times of a plus y and this spot is central maxima find the location of central maxima you know that it is d lambda by a okay and here d is l this is l lambda by a this will be equal to 2 times of a plus 2 times of lambda l by a this is b okay now I need to find the minimum b in order to find minimum b let me differentiate with respect to a and equate that to 0 so when you differentiate with respect to a you get 2 minus 2 lambda l by a square to be equal to 0 from here you get a is equal to root over lambda l getting it just substitute the value of a over here and over there okay you will get you will get 2 root which is under root 4 lambda l okay so it is between this and that and a should be equal to lambda l we just derived it this is option 3 is correct understood although the solution is straight forward but I can understand it is confusing to confusing to solve because the way the problem is put forward so the best would be that you don't waste any of your time on such kind of question and move ahead okay understood all of you which particular thing you did not understand location of first minima both sides should be minima l lambda i by a l lambda by a okay plus 2 a plus l lambda by a is a total spread fine then you take the derivative to get the minimum value of the bit okay we will go to the next question now who has asked this question who has asked solve the third one third one anyone it is from ncrt only something very similar to what you have in ncrt this is your earth this is the tower and this is the range tangentially it will go okay so beyond this point it is not able to detect anything right the center is here and this is radius of earth this is radius of earth this is what we need to find d is what we need to find this height of the tower is given which is 500 meters okay and this is 90 degree alright just you have to use the Pythagoras theorem okay so this is let us say h so h plus r the whole square is equal to d square plus r square fine now you need to understand the approximation over here h is very less compared to r r is like 6400 kilometers and this is just half a kilometer so h square you can ignore alright plus 2rh is equal to d square plus r square so d is under root of 2rh okay so now you will get the answer quickly 2 times h is 1 1 kilometers and r is 6400 kilometers okay so that is first fine now this is something which is not in the syllabus actually but then still they have asked something beyond which again is a direct formula substitution only yeah Vaishnavi it is independent of that it is independent it is a line of sight communication now in the next question number 4 you need to know this frequency sorry you need to know this formula that maximum frequency that is detected will be equal to 1 divided by 2pi modulation ratio into r into c direct formula here the modulation is 60% so ma is 0.6 r is known c is known 2pi is a constant when you substitute all that you will get option number 4 to be correct what is this height of the antenna you deleted the message okay we will go to the next question alright guys these are solved questions I guess this is from centre module only right why you want me to solve the solved question something unsolved you are not able to get then it will be better anyways attempt this one question number 71 this is from I didn't know that solved this question others central maxima of single slit diffraction pattern how much width it has 2 lambda d by a this is a width of maxima central maxima and the fringe width the fringe width in case of young's opposite experiment is lambda d by d this is a fringe width for the young's opposite experiment so number of number of bright fringes that are inside this how will you find if I divide the fringe width from this what should I get tell me if I divide this with that what I will get this is what this fringe width of young's opposite and this is the central maxima of the single slit so if I divide it what should I get number of fringes I will get okay now number of fringes will have you know similar amount of dark and bright fringes fine let's first simplify this lambda d is gone so you will have 2 times of d by a so this is around 12 okay d is 6.1 times a so this is around 12 12 fringes will be there 6 will be dark and 6 will be bright that's option b okay now I will go to the next question we needed to find intensity maxima so maxima how many maximas are they going to find not how many fringes this one yes swetha it is the distance between the 2 bright fringes that's how fringe width is defined no no no swetha you are not getting it this is suppose this is bright the central one is bright okay so fringe width is this this is the fringe width fine distance from here to there 71 it says c okay I will come back to this 71 fine so this is on me I will come back to this 71 question fine actually the definition of fringe width we are taking something but yeah I can understand it could be from the center of this the center of that if this is the definition of fringe width then you have in one fringe width you have one complete width of dark and one complete width of bright right so I will come back to that immediately after class I will message salt 70 second why you are solving meanwhile also I can see this is the fringe width this is the pattern of maxima and minima this is the pattern of maxima and minima so this is the maxima maximum intensity and this is also the maximum intensity so the fringe width here it is not clear this is clear thing I forgot actually so this is the minima this is the minima and central one is maxima so fringe width covers half the width half the width of minima this side so total one fringe width consist of one full minima width and one full maxima width this is the maxima this portion is maxima so you can see that this fringe width has half the minima from top and half the minima from the bottom it consist of one full minima width and one full maxima width so yes one fringe width has one fringe width has one maxima and one minima so we should not have divided by 2 so this 12 means we have 12 dark and 12 bright okay so yes see is the correct one okay shweta you are also correct answer is 12 so one fringe width has one full width of bright and one full width of dark the weight is defined nothing else I mean there is no as such logic towards it just the definition is such are you able to understand what I am trying to say here fringe width is distance between these two this distance which is double of this distance are you able to get it okay so remember this the naming fringe width definition itself was little confusing so anyways sol 72 yes lord that if there are two only two waves are there they have to have a destructive interference then that is correct there are multiple waves but you can take two at a time tell me what is the path difference between one and two between one and two what is the path difference suppose only these three waves exist so one and two will undergo destructive interference so it has to be at least lambda by two right one and two will have a destructive interference if part difference between them is two and three will undergo destructive interference part difference between two and three is lambda by two okay so two is lambda by two ahead of one three is lambda by two ahead of two the path difference between one and three becomes lambda p2 minus p1 is lambda by two and p3 minus p2 should be lambda by two this is how we analyze a single set experiment right you guys remember you cannot just have a destructive interference between one and three all the waves should have a destructive interference so you take two at a time and make sure all the waves are covered see it is like this you take these two points these two points first you take care of these two interference then you move this point down and this also down after some point and this point will reach here that point will reach there so you need to cover all the waves from here to there no no no that is not correct it can't be that path difference between one and two will be lambda by two see it is a single set experiment so let's say this is wave one this is wave two okay they will have a destructive interference okay and then if I move this point down it reaches here this point comes down like that and I keep on moving this down so I am covering all the points right I am covering all the point as this going down and this goes down I am covering all the points taking two at a time getting it and all the whatever pair I take if I take one and two then if I take this point and that point all those pairs should have a path difference of lambda by two for destructive interference getting my point what I am trying to say here all of you able to understand I am taking two waves at a time for destructive interference how many waves are there there are infinite waves okay there are infinite waves that I need to be taking care of okay I can't just have interference between this point and that point fine so I need to make sure see if I say that path difference between one and two is lambda by two okay let's say this and that these two waves have a destructive interference at some point in time then I cannot say that there will be a destructive interference between this point and that point okay at a particular location where one and two meet okay these two waves this point and this point these two waves will not be having destructive interference fine so the minima will not occur there alright but if I maintain the distance between the two points let's say here is where they are meeting these two points they meet the path difference between these two waves is what if this is a by two and this distance is y so it will be a by two sin theta which is y by d right this is the path difference between point one and let's say this point two okay this path difference only depends on y alright so at a particular location if this has to be destructive interference okay then if the source if two sources are at a distance of a by two the path difference is fixed if a by two is a distance between the two sources okay so what I will do now I will yeah I am coming to that so what I will do is I will be moving this down and moving that also down fine so it will be after some point in time you take these two points so this distance is also a by two the path difference over here will be again this only so if this was equal to lambda by two then you take any two points separated by a by two distance will have the same path difference lambda by two okay and loiter it is not that point two will have just one wave in fact there are infinite waves you can see only three waves over here but between one and two infinite waves are there between two and three infinite waves are there okay so just about two there will be one wave just below two there will be another wave fine just a representation it is all in fact there are infinite waves so if you have paid attention to when we have been discussing single-seat experiment the derivation of single-seat experiment you would have understood this is how we derive single-seat experiment where the constructive and destructive difference happening others are you able to understand nobody else is there is it only loiter is there there are infinite waves it is not just about one two and three infinite waves are there just about two there will be one wave just below two there will be another wave so you can take like that loiter you got it okay so this is what I have then any other doubts let's see is not correct okay the answer is lambda anybody else sent me doubt I will take this one this is during the class so this is done Rohan has asked me about hysteresis curve so I will do that hysteresis curve Saimi here Saimi you have sent me entire book is it so I will take few questions whatever I feel like I will take those other questions I will send you over whatsapp later on fine solve this one this is not j mains level this is advance level so let's not take this right now unnecessarily will be complicating things so let's not take it right now guys remember there is no pride of solving difficult questions if you are not able to get the simple ones it is actually a shame so make sure you get the easy ones correct in j mains we will take care of j advance later on try solving this one sign which book you have taken this one these questions which book you have taken okay this is from your okay again this is advance level only this type of question you will not see in mains how you get it lohitya how you get that answer okay this is the area of the loop don't do that fine anyways so this is the area of the loop right equilateral triangle so if switch is open nothing will happen when switch is closed there tends to be a current that flows like this right and charging of capacitor goes like this right i is equal to i i is equal to what e by r 1 minus e to the power minus t by rc okay and r is 0 r is 0 fine so current will r is 0 so capacitor will get immediately charged capacitor will get immediately charged and this e to the power minus t by rc if r is 0 this will become r is 0 so this will become e to the power minus infinity 0 so current tends to infinity immediately which formula this one this is charging of capacitor this is q sorry this is q q is equal to this remember charging of capacitor this formula q is equal to q0 1 minus e to the power minus t by rc okay if you charge a capacitor against a resistance we did this right you remember we did this charging of capacitor where q is equal to q0 1 minus e to the power minus t by rc and q0 is c v0 you don't remember so current which is dq by dt will be equal to c v0 into derivative of this derivative of that will be equal to 1 by rc e to the power minus t by rc okay so this is v0 by r e to the power minus t by rc okay so this is how current will grow and current will grow instantaneously okay so r is 0 so current this is 0 by 0 format i is equal to v0 e to the power minus t by rc divided by r so r is 0 and numerator is also 0 so can you find this when limit r tends to 0 what is the value of this v0 power minus t by rc divided by r this is the current this is a stupid question unnecessary making complications fine so you get the current that comes immediately when you put limit r tends to 0 fine are you getting it torque will be equal to m cross b m is current into area so i into area cross b now b is making 90 degree with the area vector so it is just i a into b okay i think they mean angular acceleration not angular velocity because this torque should be equated to i into alpha and you get the value of alpha i think there is a correction it means angular acceleration probably okay suddenly angular velocity will not appear out of nowhere until there is an impulse there is an angular acceleration but all sudden done this is a stupid question you will not see these kind of questions in j main don't worry okay and do not put your head on even this kind of question will not even be there in j advance as well they are unnecessarily complicating the stuff unnecessary i hope this one is not like that solve question number 26 again a word of caution do not practice these kind of questions which you will never seen j mains unnecessary okay these kind of question you should not practice just before j mains the one which you have solved just now 26th okay let me solve this fine so we have should i solve or should i wait quickly tell me should i solve or should i wait i will first give you a hint okay the force is applied not all the questions for fitji are bad this is probably a good question okay so this is a force so the basically heat is produced is asked right heat produced is asked so heat is produced due to friction now you need to find out the rate of work done by the friction okay now friction will do the work over here definitely there will be a work done in the bottom surface so let's say the friction force over here is fr1 so fr1 into velocity is work done by the friction from the bottom one and if there is a slipping happening over here between 2 kg and 4 kg if there is a slip that happens then the work done by the friction over the power delivered by the friction from the top surfaces fr2 into v okay so fr1 into v is the power by the friction lower surface fr2 into v is the power at the upper surface the power this power will appear only when there is a slipping happening between 4 and 2 okay so you need to check that if there is a slipping that is happening this plus that is the answer so do you want to solve now try your own at t equal to 40 seconds the force that is applied to the bottom surface is 0.6 into 14 newtons that is 24 newtons if you apply 24 newtons over here can you check whether there is a slipping between 2 and 4 the maximum possible acceleration 2 kg will have is 0.2 into g g is 10 so this is 2 meter per second square this much acceleration 2 kg can support now let us assume that there is no slipping between 2 and 4 kg and external force of 24 newton is applied 24 minus 0.1 into 4 plus 2 into g this is equal to 6 times a this is 24 minus 6 is equal to 6 times a this is j advance level this is clearly not main level unnecessary 24 minus 6 is 18 divided by 6 so 3 meter per second square if they have to move together acceleration should be this much but the maximum possible acceleration of 2 kg block is 2 meter per second square so there is a slipping that is happening fine getting it so because there is a slipping so both the both the values will come over now another problem over here is the velocity of 2 kg is not same as velocity of 4 kg fine so now you need to get the velocity so I will just give you a hint because anyway this type of thing will not appear in mains this acceleration will not directly go to 3 meter per second square so there will be time up to which acceleration will go to 2 meter per second square fine so till acceleration goes from 0 to 2 meter per second square both the blocks will move together fine then from 2 meter per second square till it goes to 3 meter per second square block moves independently of the lower one fine so when they are moving together from 0 to 2 meter per second square you can find the velocity of 2 kg and 4 kg together they will have same velocity so get the velocity and after 2 meter per second square you need to treat them differently to get the velocity of 2 kg and then you need to find out friction into velocity of the first block then second friction divided by the velocity of the second block fine this is the work done by the rate of work done by the friction that will be seen as a heat produced so now Saimir please attempt this and send me where you are stuck then I will send you how you can go about but my advice is just before 2-3 days of J mains do not do all these kind of things but if you really insist now please try your own and send me across and I will help you fine so there was another request from Rohan about hysteresis loop one more advice is that do not attempt any new questions now just leave the new questions you have to revise what you have done in 2 years okay Shweta has sent some doubts Shweta these ones see when I ask you to send me your doubts I did not expect that you will be sending me questions okay solve this solve this that's not how you should ask doubt you should ask doubts the way Shweta has asked see she has tried this and she is asking like how they get this equation in 30th how did they assume this and that so this is the way you ask the doubts okay the solution is there in your solution book also so the proper way is like this Kushal has also Kushal where is Kushal where it is I did not get any which number you have sent from okay this one yeah I told you I will take this one fine so let me first talk about hysteresis loop fine then I will get into other doubts this is better yeah let me go one by one what happened to him so we will scribble on the website only this is the hysteresis loop okay give me some time right now focus on the hysteresis loop this is also important so there is an external field which is H and this M is the field due to the matter so what you are doing is the H is because of the solenoid okay and M is because of the let's say bar magnet or let's say iron and iron bar inside the solenoid okay so H is the strength of magnetizing signal now strength is calculated in terms of magnetic dipole moment per unit volume so for a solenoid M by V is H and magnetic dipole moment per unit volume inside the iron bar is capital M what is happening is that you are creating H and you are getting M fine so this is how permanent magnet gets created okay the M is generated from H okay so you initially take this matter you have H is equal to 0 and M is equal to 0 then you pass the current inside this solenoid so H increases so what will happen is that slowly the material also get magnetized so even M increases right but then there is a limit up to which it will get magnetized it will get magnetized till all the dipole moment inside the material are perfectly aligned fine so once they get perfectly aligned the material cannot have any more dipole moments per unit volume okay earlier what was happening is that dipole moments were randomly aligned so net dipole moment was 0 slowly and slowly they were getting aligned so once perfect alignment is set then you can't get more M even if you increase H M will get saturate so you get a horizontal line so M will not increase now if you decrease H surprisingly it will not come back from the same path it will go sideways like this and hit here so this line will represent H is equal to 0 so even if you are not creating any H you have 0 current in the solenoid then also you have the magnetic field inside the bar so this is how the permanent magnet is made you create H, saturate it and then decrease H to 0 but magnet will not have or matter will not have M equal to 0 magnet will still have some dipole moment left so that is why it is called remnants okay now if you want to decrease M to 0 you need to create H in opposite direction so the current which magnetize it suppose it was clockwise you need to generate anti-clockwise current now so this much extra H you should produce in opposite direction then only M of the matter goes to 0 and if you keep on increasing H in opposite direction now the saturation will happen in opposite direction so earlier it was happening in this direction the saturation now saturation will happen in opposite direction fine and then you decrease H and this is how the loop will get completed and once the loop gets completed same thing will repeat again and again fine this is what this hysteresis loop is about getting it all of you all of you understood okay so we will take a small break now right now it is 12 55 okay we will meet at 1.10 pm fine then remaining 20 minutes we will take up other doubts as well Arun, Rohan there is formula okay H is H you remember was N i okay and M is equal to X times H okay you remember this and X was equal to C by T and mu R is 1 plus X there are lot of formulae Arun it comes in magnetism and matter chapter so that we have discussed while if you want I can send you the lecture link of magnetism matter so you will get the detail theory related to M and H there you want that I will send you after the class just remind me okay welcome all of you I hope you are there okay so let's take up the other questions this I have downloaded right who all are online 7, 19, 33 you have sent the screen of your laptop okay looks to be a nice question all of you please attempt this I hope it is visible here the hint is you need to use ampere circuit a lot okay and because of symmetry the magnetic field will be circular and you need to use super position principle of super position can you first tell me which direction the magnetic field will be positive y or negative y that you can get easily read the question carefully read the question carefully okay now suppose there is current that is coming out current is coming out so magnetic field lines will be how magnetic field lines will be circular isn't it it will be circular like this and if you align your thumb in the direction of current the magnetic field line you will see that it will be like this this is how magnetic field lines will be okay now suppose there is another current density adjacent to it like this okay now because this one is current that is coming out and this one current is going in fine so because of the current that is coming out the magnetic field is like this at point A is where what is point A point A is a midpoint by there is nothing given about A so we will assume A is a midpoint okay A is between these two okay so because of this this current can you tell me which direction the magnetic field will be which direction the magnetic field will be because of this again the magnetic field will be circular like this only and it will go like that if you align your thumb in the direction of current which is going in the magnetic field will be upwards so magnetic field will get added up due to both fine so magnetic field will be positive by direction so C is not possible D is not possible E is not possible so it is between A and B getting it now let us try to find out what it is now I just need to know the magnitude of one of this magnetic field if I know this B then the double of this B is the answer because this magnetic field because of this one magnitude wise same as due to that one because of symmetry right so I will just find the magnitude of the magnetic field because of this white this thing current density so let us say that I have an ampere loop which is at a distance of D by 2 from the center fine so I am going to use ampere circuit law inside this circuit of radius D by 2 and ampere circuit law is what integral of B dot D L is equal to mu naught into I B dot D L becomes B into 2 pi into radius which is D by 2 this is equal to mu naught into I now current density is J current per unit area so J into pi R square is pi D square by 4 ok so you get magnetic field as this one of the D's will get cancelled will get mu naught J D divided by 8 1 2 will get cancelled so there will be 4 over here so total magnetic field is 2 times of B which is equal to mu naught J D divided by 2 this is option A mu naught this is pi in the numerator and denominator also fine understood all of you nice question understood so I can take only few more so these are sent I think 50 these are sent during the class so this I will not entertain before the class only percent then who else said that there is a question Sondarya Sondarya has sent saved as Rangarajan that's one try attempting this one ok so I will take one from Sondarya and one from Shweta for other questions I will send the answer after the class today only ok and I have sent you few questions on EM wave right we have discussed the theory here as in yesterday right so I have sent you few questions for practice from EM wave on the group only please attempt those and in case of any doubts you can talk to me this one all of you attempt this one so when you close S1 it will be like a charging of capacitor only so Q is equal to C times V 1 minus e to the power minus T by RC this is the formula we just you know understood that right so tau is equal to RC ok you are saying 3 tau yes when T is equal to 2 times of RC you will get Cv times 1 minus e to the power minus 2 as the charge so option 3 is correct ok so this is charging and discharging of capacitor ok alright all of you understood right fine let us see what else we have we can probably take up this also quickly and this question Shweta for your doubts I will create a video separately and send it across to you all your doubts I will create a video and then post it on youtube and send you the link is it fine Shweta all of you please try 14th quickly ok let me solve this now so initially suppose this is heater ok this heater was not there initially bulb was there ok bulb is 60 watt rating right now this rating happens at this voltage ok so resistance is V square by R so that is 120 into 120 which is V square divided by 60 so you get 240 ohms as the resistor ratio of the bulb ok now resistance of the heater this bulb resistance of the heater also again it will be 120 into 120 power is V square by R so R is V square by P into 240 so this will be 60 ohms ok yes yes there is 6 ohm this is 6 ohm fine but when ratings is assigned it is assumed that potential across the bulb is the supply potential alright similar for heater when rating is assigned we are assuming that supply voltage now bulb is separate bulb is a different entity altogether see you are confusing 60 watt is written on the bulb have you seen the bulbs CFL or any bulb you see there will be some rating given lets say 15 watt or 100 watt it will be written on it fine these ratings assume that the voltage across the bulb will be supply voltage in this case the supply voltage is 120 ok so it will take an ideal scenario when there is no resistance in series and you are applying 120 volt and this much power will be given fine now you can connect this in series with this ohm 100 ohm 200 ohm whatever you may like but the rating was given assuming that there was no resistance understood so this is rating this has nothing to do with this circuit circuit can change I can put you here 12 ohm but if I put 12 ohm this 60 watt will not become 200 watt it will not change fine so once you get the resistance now you just treat bulb as a resistance and heater as a resistance ok always remember that the power is given with respect to the supply voltage and whatever power you have should be equal to V supply square divided by R so this will give you the resistance of the bulb or the heater ok now this power can change power can change depending on what voltage you are supplying fine but the rating is done for the supply voltage ok but whatever is the case resistance will not change whether you change the voltage or not if you change the voltage power can change but resistance won't change so that is the reason why I am finding the resistance first because resistance is something which remains constant ok now ohm first assuming heater is not there the current is supply voltage 120 divided by 6 plus resistance of the bulb which is 240 fine this is I1 so potential difference across the bulb is 120 divided by 246 into 240 this is the initial voltage now if you connect the heater also which is 60 ohm then you can find the equivalent resistance 60 into 240 divided by 240 plus 60 just 300 so this is 48 ohm this is equivalent resistance so V2 is 1 to 3 first I find current now current is 1 divide 48 plus 6 ok are you able to hear me now you can see the screen ok hmm so this is I2 now potential drop across the bulb is same as potential drop across the combination of bulb and heater so V2 is equal to 120 divided by 48 plus 6 divided by 4 multiplied by equivalent resistance which is 48 so V2 you have got and V1 you have got so V1 minus V2 is the answer understood now clear all of you understood right alright guys so I think there are few doubts and I think it is from Shweta only Shweta I will send you the recorded solution of your doubts ok so today I could not take Shweta's doubts today I will be sending the recording of that soon today only ok thanks for coming in I hope you have learnt many things today and do well ok prepare well for J mains bye