Predictor JEE Mains 2nd Attempt 🚀 High weightage Sure shot model questions | Physics Chemistry Maths

good afternoon my dear J Warriors this is your captain stest and this is the last session of the J Mains 2024 season and obviously after this you're going to have sessions for J Advanced bitat and you name it VI EAP sat all those things but the major major chunk which most of you are preparing for that's J Mains 2024 this is the last session academic wise right so after this we are going to be on the other side where we are going to discuss the papers already the sessions have been lined up for all of you so if you are on the channel you will be seeing all the J Mains discussion sessions lined up right from Tomorrow shift 1 shift 2 usually they will start at around 1:00 for the first shift and around 730 for the second shift all right so I hope you guys are all ready in full Jo and full Energy new look super sir thank you good afternoon VI Brothers hello roshini hello Amit one last time let's have a look right to all the names because after this you are going to create history after this you are going to mark your name in history you're going to you know make your parents feel proud you're going to make your teachers feel proud you're going to feel good about yourself because you're going to put in your 100% so smash the like button not for me not for the channel but for you if you want to praise yourself if you want to motivate yourself if you want to feel good about yourself if you want to you know feel high about yourself go ahead smash the like button right for you exactly thank you so much good afternoon everyone and this is your captain Shas right and remember as we move on to the second phase of our preparation uh moving on to other examinations also remember there are many activities happening at the vidu offline centers not just that uh there are many batches which are there for the foundation J and need students so if you have some people in your Society or maybe your friends batchmates anybody maybe your friend maybe your enemy and they want to prepare or want to know more about it please let them know about the vedantu Learning Centers there are many many centers right now throughout the country in fact there might be a center right next to and maybe you are not aware so go ahead and take a look at it browse through Google Maps or even you can call on this particular number you can check out the link which is there in the description deson box so you can see so many centers with personalized attention three-way doubt solving and digital classroom smart clickers the best teachers in the country at your doorstep so we have now a complete package of online and offline so don't stay behind and don't let anybody else also stay behind because you should never be selfish if you have faced some difficulties let others not face that difficulties maybe your Juniors or your other friends right so let's begin with the class and here comes the first question on your screen yeah this is the paper prediction for tomorrow and paper prediction means the most likely model questions the most uh uh you know repeated pattern of questions these are the subtopics not chapter subtopics which are frequently found in the last few shifts and even the last few years so that is the meaning of of this particular predictor paper you predict right like having a crystal ball and here comes the first question on units Dimensions the potential energy is given as aunk X by x² + B where A and B are some dimensional constants then what you think is a dimensional formula for a multiplied by B now one thing is clear that we can find a dimension of B very very easily because B and x square are added so that numbers which are added should have the same units same dimensions so one thing is very very clear let me put it up for all of you dimension of B is the dimension of x² so it is L sar's Dimension that's it l Square Dimension now for this a what you need to do is left hand side left hand side is nothing but potential energy so the dimension of potential energy will be the dimension of a into the dimension of root of x and bottom x² + B all of them have the dimension of length Square length Square so that is how you will find the dimension of a my dear Warriors so the dimension of a will be the dimension of U multiplied by the dimension of l s whole thing divided by root X that is L ra to2 but what is the dimension of energy M ra to 1 L ra to 2 tus 2 and then you have l s over here and then here you have L ra to half so if you want take L to half on the top it will become L ra to minus half L ra to- half so what will this become M to 1 L to 2 L2 will become L to 4 so Min - 1 will become 3.5 so which is L to 7 by 2 T to minus 2 that is the dimensional formula for a but the question is not for a or b it is a into B it is basically a into B so you take the dimensional formula for a which is l s and multiply it with this m to 1 L to 7x 2 T to minus 2 what is the option that you're getting many of you started putting b as the answer Ram Ian boy sagui saying a some of you are saying B what is the correct one multiply it and see what do you get see if you getting it as option b my dear Warriors because this is l s this is 3.5 3.5 + 2 is 5.5 5.5 is 11 by two so that should be the answer which is option number B of course for it Bombay if you said a that is slightly wrong it was a into B moving on to the next question I hope you understood which mistake you have made yep so be little bit little bit careful we Brothers okay yes sat option b only yes kameshwar moving on next question capacitor based questions where they arrange the capacitors before and after and then they ask you the change in the energy or they move the switch positions and then they ask you the change in the voltage or the charge or the potential energy so there sorry here the question is find the final charge on basically the capacitor to see all right so what we need to do is in the given circuit the switch was connected to position uh one for a long time then what you do is basically you move it to position number two after some time so the question is the switch is shifted to from position 1 to two find the final charge that means at stady State now the here you know actually the question becomes very simple here you don't even have to find change in anything just imagine switch two is connected and that two for a long time after a long long time what will happen it will be in the study state it will be in the study State and when you come into the study State understand this particular capacitor is of no use because it is not even connected anywhere it is not even connected anywhere in fact only this capacitor and this capacitor are connected across this voltage and both of them are going to be in series both of them are going to be in series what is the equivalent capacitance in series C1 C2 C1 C2 divided by C1 + C2 because you have to use parallel formula for series capacitors so it will become 2 C / 3 is the equivalent capacitance what is the charge then it is C equivalent into the voltage into the voltage C equivalent is nothing but 2 C by3 voltage is nothing but we so that is the charge remember charge is same for all capacitors in series all capacitors in series so hence the answer should be 2 CV by 3 I think the answer marked over here is wrong yes so it should be basically 2 CV by 3 but this is for yeah this is for yeah on the 2C capacitor or C capacitor it doesn't matter yep correct option C is correct very good very good awesomeness everybody has written on option see very good very nice Ram liita so many of you have got the correct answer all right moving on to the next one on your screen now a vessel contains 14 G of nitrogen gas at temperature 27° C how much heat has to be given to Gap the double of the RMS speed of the molecule you have to increase the RMS speed you have to double it so how much do you think heat you will have to give that is the question that is the question think about it right so first of all this 27° C is nothing but 300 Kelvin now in order to double the RMS speed remember RMS velocity is directly proportional to the root of temperature so if RMS velocity has to be doubled root of temperature also has to be doubled root of temperature also has to be doubled that means the temperature has to be made four times do you guys agree with it because root of four is 2 root of 4 is 2 so temperature has to be made four times so hence I would say the new temperature will be 300 multiplied 4 in Kelvin 300 multiplied 4 in kelvin right but that is not the question to find the new temperature the question is how much heat will you give so to understand that the amount of heat that you will give is to change the internal energy of the gas so the formula for that is ncv delt T ncv delta T listen in carefully how many grams of nitrogen is given 14 G and we know molecular weight of nitrogen molecular weight of nitrogen is 28 G so 14 G is nothing but .5 moles of nitrogen .5 moles of nitrogen so n is basically .5 nitrogen is a diatomic gas so what will be the CV for it what will be the CV for it won't it be 5 rx2 won't it be 5 rx2 my dear Warriors obviously 5 rx2 and temperature difference final minus initial so this was 300 into 4 which is nothing but 1,200 Kelvin so just you have to do 1,200 minus 300 understood my dear students just multiply this and tell me what is the answer that you are getting everybody spamming a everybody is spamming a is that so is that so just check it out the answer is 9360 my dear students see if you made a mistake blunder over here 9360 it is not 2229 somebody made a mistake see if you have done something wrong over there be careful be careful let me also calculate it on my calculator you don't use your calculators yes so it is0 5 multiplied by again 0.5 because of this half into 5 into 8.3 multiplied by 900 so it is yes guys 9360 close to that only option C you guys gave a wrong answer I'm not sure what mistake you guys have done see if you have done some big blunder over here I think you have put 3 r by2 or maybe you put the number of moles as one so maybe that is the reason why you might have got a completely different answer be careful okay so these things will suck away all your marks whatever marks you gain previously will take away everything keep this in mind yeah yes yes Sund Rajan both of them will have same same amount of charge when you talk about you know capacitors in series equivalent capacitance C1 capacitance C2 capacitance C3 capacitance all of them have the same charge do not divide it whatever total charge you get don't divide it on the capacitors that will happen in parallel not in series okay so whether it was 2C or whether it was C the charge will be the same okay I hope so ran that you know doubt is very very clear moving on if the focal length of the objective and I lens are this and this respectively and the object is put 1.25 CM away from the objective lens the final image formed at Infinity then the magnifying power of that particular microscope is how much this is the question now read the question carefully the final image is formed at Infinity so the magnifying power magnifying power for normal vision when it is relaxed your eye is relaxed is given by magnification of the objective into magnification of the objective into D divided by focal length of the ipce D divided by focal length of the particular ipce now least distance of distinct Vision I know it is nothing but 25 cm the focus FAL length of the I piece I length is 3 cm so I'll put 3 cm over here the magnification of the object the magnification of the object I do not know remember magnification of the objective lens is nothing but V by U you will be like sir I think I know youu yes I think you know you definitely the object is 1.25 CM away so we are just interested in magnitudes over here please remember that we are not interested in the sign so the object distance is basically 1.25 CM but I do not know V how can I find V well you have to use lens Formula 1 by V right everybody understands you have to use lens formula there is no other option now cool so 1 by vus 1 by U is equal to 1 by F but for the objective so 1 by V I don't know U is nothing but 1.2 5 with negative sign it's a convex lens so positive focal length of 1.2 over here so 1 by V will be minus minus is plus goes on the other side becomes minus so 1.2 minus [Music] 1.25 right you can do this 1.2 is nothing but 6X 5 so this is 6 and this is 5 minus 1.25 you can write it as 5 by 4 5 by 4 so this is 5 and this is 4 solve this and see what is the value of V that you are getting take the LCM it is 30 25 24 come on guys are you getting v as 30 then check you should get v as basically 30 cm everyone very good now substitute it over here substitute it over here and then substitute it finally over here so this is nothing but 30 by 1 25 30 by 1.25 check this out what is the final answer that you are getting check this out what is the final answer that you getting my dear students come on this three will go with 30 10 times 1.25 into 2 is 2.5 uh so basically this will be two times so 1 Zer from there so this goes with this 10 times this will go with this 20 times so this should be 200 so the magnifying power should be 200 yes that is correct oops what is wrong with you all today whatever answer is there you are giving something else only oh my God guys I am scared what happened everything fine or are you just copying whatever somebody's saying one person said c everybody is putting C no no no please don't do that be be careful be careful it is 200 yeah what is d d option is 400 why will it be 400 yeah cool yes yes if you solve these model of questions Beauty definitely you will get 90% L and above cool so remember the magnifying power formulas for all the three things for telescope it is Fe by fo then similarly for microscope scope if it is infinite distance then it is this if it is at least distance then it is this formula into 1 + d by F just like your simple magnifying glass formula magnifying power is just d by f for normal vision for least distance of distinct Vision it is 1+ D byf the problem was this guy magnification which is V by U object distance was given V was not given fine we use lens formula figured this out and put it over here and then finally substitute it to solve and get the answer moving on to the next question guys all right I hope this is clear hey don't focus on derivation man who is that guy don't focus on derivation now you are one day away from the exam and you're asking derivation sir excellent day amazing just go with the formulas and apply it now this is not the time last month is not the time for derivations last month is just the time for applying knowledge applying the formula you should know when to use what that's it okay keep this in mind one month you cannot focus on all these things electric field in the region is radially outward and has the magnitude e is equal to KR the charge in the sphere of the radius a is how much many people try to do this using gos law well I feel that you can avoid gos law completely notice over here electric field is directly proportional to R and there is only one such geometry where electric field is proportional to R and then goes you know as a curved line what is that particular geometry guys if you remember carefully if you remember carefully it is nothing but a solid solid uniform uniformly charged uniformly charged uniformly charged sphere uniformly charg sphere that is how the electric field formula looks like try and recollect it was 1X 4 pi epon Q / capital r CU into small R what is capital r it is the radius of the cube in this case let's say the radius of the cube is given to be I think a so it is basically a cube and this is basically R so comparing both of them substitute electric field as K * of R is equal to 1 by 4i Epsilon Q / a cub into r r r cancels what will Q equal to K into 4 piun into Epsilon into a cub that's it so that is the answer my dear Warriors which is nothing but option number B B for Bombay wow some of you said c some of you said c no it is not C guys it is option b be careful 4 Pi K Epsilon a cub there is no need of 2x3 I don't know why that 2x3 you guys are saying that is wrong it is option number B be very careful yes yes some of you were also worried about Sir how did you put d as 25 that is a standard value least distance at which you can see things clearly before that you will not be able to focus for an average adult beyond that you can focus properly so that is an average least distance of distinct Vision which is a assumed standard value like G is 9.8 or 10 same way okay moving on to the next next one a uniform chain of length L and mass m is lying on a smooth table and 1/3 of its length is hanging down over the edge G is acceleration due to gravity work done to pull the hanging part on the table is how much so let us draw the two scenarios first let us draw the two scenarios over here like this and first of all 1/3 part was hanging 1/3 part was hanging soorry it would be like this then what you have done is brought the entire chain up on the table you brought the entire chain up on the table okay fine great now what I will do is in such problems in chain problems the logic is always choose the top of the table the top of the table as zero potential energy basically this is your zero level this is your zero potential energy level so right now if you see when the entire chain is on the table what do you think is the potential energy of it to find that see where the center of mass is the center of mass of the chain is here it is on that line so clearly the potential energy finally is just going to be zero all right just check the center of Mass to find the potential energy of a complete body see where the center of mass is and then write MGH H H is from the assumed zero line here this is my reference level or the assumed zero level all right so now what is the next thing that you will do my dear students over here if you notice I can break the chain into two parts this part and this part this part has zero potential energy this part has zero potential energy this part has a center of mass here how far below it is check this out this is L by3 guys check this out this much distance is basically L by3 right 1/3 is hanging below so how much distance is the center of mass going to be at how much distance is the center of mass going to be at half of L by3 which is basically L by6 understood till this point how much is the mass which is hanging how much is the mass which is hanging over here how much is the mass which is hanging over here the mass which is hanging the mass which is hanging is the total mass divided 3 the total mass divided 3 because 1/3 of its length is hanging down this much part has zero energy this much part is below so the total energy initially is this Z Plus M which is m by3 g which is g h is L by 6 but keep in mind it will be minus because it is below it is below the zero line it is not above so hence this will become nothing but minus minus mg sorry why did I put 3 this was 6 this was L by 6 yeah minus mg L and 63 is are nothing but 18 63's are nothing but 18 correct so hence the final answer will be the work done work done is nothing but final minus initial energy final is zero initial is minus mg by 18 - mg L / 18 so minus minus becomes plus so it will become Plus mgl divided by 18 which is the for it Delhi finally I'm so proud of all of you that you guys got the correct answer very good proud of all of you finally I was really worried last few questions in Mass you give the wrong answers so now you got it right so sometimes the question can be L by4 is hanging or L by5 is hanging sometimes the question can be this is hanging it is let go it falls off completely what is the new speed or it was hanging completely you brought it on the top so remember all these times you just need to assume some reference level as zero preferably the top of the table what is below is negative what is on the top is positive next thing is if you want to find the energy when you write mg h m is the mass which you're considering G is gravity H is the height of the center of Mass from the reference line if it is on top positive is below negative always choose the center of mass that is why we took this midpoint not lx3 cool let's go ahead to the next question in this experimental setup meter Bridge is a commonly asked question every time you will see almost every shift has either wheat ston Bridge or meter bridge and some question on that or finding the equivalent resistance or capacitance and it ends up being some kind of wheat Stone question is the null point is 40 cm from a and a is a 10 if a sorry if a 10 Ohm resistor is connected in series with R1 null Point shifts by 10 the resistance that should be in parallel okay so let me create a duplicate slide and let me just start with the first scenario null point is at a distance of 40 cm from a so basically This is 40 cm this is basically 60 cm so therefore our r 1 by R2 which is L1 by L2 L1 is 40 L2 is basically 60 so this is 2x3 so R1 by R2 is nothing but 2x3 for the first scenario for the first scenario okay now let's go to the next scenario what happens in the next scenario let me again create another duplicate just in case I need it okay now you connect a 10 ohm resistance in series with R1 so this will become 10 + R1 the null Point shifts by 10 cm so earlier it was 40 it will become more 10 so it will become basically 50 cm here and this will also become 50 cm oh that means 10 + R1 by R2 which is 50 by 50 basically both the resistors are equal so 10 + R1 is equal to R2 10 + R1 is equal to R2 but wait a minute from from here from here do you see R1 is basically 2x3 * of R2 2x3 * of R2 R2 take it on the top right everybody with me on this so substitute R1 as R1 as 2x3 * of R2 let me find the values of R1 R2 now let me find the values of R1 R2 now so let's see what do we get so 10 is equal to R2 minus 2x3 R2 1 - 2x 3 is 1x 3 so r2x 3 therefore R2 will be basically 30 ohms 30 ohms and similarly we can also find R1 R1 was nothing but 2x3 R2 it was nothing but 2x3 * of R2 which is nothing but 30 so 30 by 3 is 10 so it will become basically 20 ohms so we got this we got this we know the individual values okay now the third step of the problem problem R1 is basically 20 R2 is 30 so this is basically 20 ohms this is basically 30 ohms last part what should be the resistance in parallel to R1 + 10 now when it is R1 + 10 it will become basically 20 + 10 which means 30 ohm which means basically 30 ohms okay so you have 30 ohms over here and now you are connecting another resistor let's say small R in parallel with this a parallel with this such that the null Point shifts back to its original state that means this will become 40 cm and this will become 60 cm so do you see the level of the problem wow very good so now if this has to happen so I would say the equivalent resistance in parallel for these two by R2 should be L1 by L2 that means 40 by 60 what is resistance in parallel it is R1 R2 by R1 + R2 so R1 is nothing but 30 R1 is nothing but 30 okay my God what just happened how the Oh okay I don't want this okay hm okay please go inside okay please go inside why is it not going thank God huh uh so where were we RP RP is basically equivalent resistance in parallel it will be R1 R2 so 30 * by R ID 30 + r ID 1 by R2 sorry or divided by R2 R2 is basically 30 over here is equal to 40x 60 which is nothing but 2x3 which is nothing but 2x3 you can see many things are just going to get cancelled like the one over here 30 30 will get cancelled so just multiply now cross multiply guys you will get the answer so it will become 3 * of R will be 2 * of 3 30 + r so basically 2 into 30 is 60 this is nothing but 3 R Take 2 R from here and shift it to this side so small R will be basically 60 ohms which is B for Bombay which is B for Bombay so it's a three-step problem thank you Shas morti means a lot I just did it because the J meanss discussions are here and I won't have time to cut my hair after this Li glad to know that you are from Viada nice to see all of you out here so see a multi-level approach problem first step for this part second step for the second part over here connected in series third step over here for the last part where you have to find the parallel resistance is that okay so multi-step problem it will take you easily 3 minutes to 4 minutes you have to save time in other questions two long parallel wires at a distance of 2D they carry same current but out of the plane okay towards you the variation of the magnetic field along the line XX Prime is given by which of the best graphs okay magnetic field along the x-axis so here is what we will do just imagine just imagine one of the wire come coming towards you hold the thumb hold the thumb towards you you will see the fingers will curl like this anticlockwise so the fingers will curl anticlockwise like this so this is how the magnetic field lines will basically look like this is how the magnetic field lines will look like right this is how it will be great due to this wire on the left similarly there is another wire on the right side again coming towards you again coming towards you so it will have a current like this again the magnetic field lines will be anticlockwise only so it will be basically like this then you will see like like this maybe somewhere over here oops like this right then maybe like this over here right and this will continue guys this will continue correct this will continue okay now looking at this what is the best graph that you see what is the best graph that you see uh which will be more feasible 100% there is a point somewhere where the magnetic field lines are cancelling out each other so there is a point where the net magnetic field is zero magnetic field is zero in all the graphs you can see 1 0 point 1 0 point 1 0 point 1 0 point is there but interesting thing is the interesting thing is on this side if you notice this wire is Clos this wire is far so the field of green is more and the field of red or that orange is less so one dominates over the other so the net field will be let's say down are you able to understand so the net field is basically downwards over here so in this particular region the net field is downwards in this region orange will dominate over the green so the net field will be on the top net field will be on the top so the sign of the magnetic field has to change the sign of the magnetic field has to change so in this graph it is not changing here also positive here also positive here also negative here also negative so option C and option b gets eliminated option b and option C gets eliminated because when you come close to the wire on this side or this side one wire's effect will be more dominating because of the distance closer wire will have more field so it will give you field in a particular direction is that clear till this point very good now what about what about you know if you uh go on this side of the wire if you go on this side of the wire what is basically happening check this out once you go on this side all the magnetic field lines are up see this it's up green also orange also so the net magnetic field is up so the net magnetic field is up and if you go completely on the left side all the magnetic field lines are down so magnetic field is down so down up down up down up down up up up down down oh up up down down gone up up down down gone do you see down up down up down up down up so hence option A is the correct answer did you get it so it involves a little bit of logical analysis and visualization I hope this is very very clear very nice very nice great sir you know all the concepts you can become a scientist so if everybody becomes a scientist who will make the next generation of scientist somebody has to stay back no right so if everybody joins multinational company everybody goes to us Europe Australia nobody is there in India so what will happen so let's say every everybody in 1990s left that's it nobody left to teach nobody will go to I everybody or even if they go to I then then hardly they will have good knowledge right so we are here to stay back we are sacrificing so many things and not because uh you know of compulsion but out of pure interest right it's not like I cannot go anywhere or anything I left my corporate job after working for 3 years I had a very very nice high paying peaceful job where I used to travel everywhere but I left that job and I'm here to teach you remember that so Guys somebody will have to stay back to Bear the torch in the circuit shown in the figure the value of R is 60 ohms then the current flowing through the condenser meaning capacitor is how much that is the question yep so what do we do what do we do over here LCR circuit question is what is the current flowing through the capacitor in this particular thing so first of all because all of them are in series won't the current in all of them be the same won't the current in all of them be the same because they are in series perfect now if you check the resistor guys the voltage difference is already given and the voltage difference is I into R voltage difference is 15 current I don't know resistance 60 perfect I is 15x 60 which is 1x 4 which is25 a25 ampere perfect it was such a easy peasy lemon squeezy but because it is AC current some students decide to leave it and that is very very surprising okay so it was very L do question Cool shall we go ahead to the next one here it is a string of 5.5 m length has a mass of this much the tension in the string is also given then the speed of the wave is how much so I told you yesterday also and even before the speed of a wave on a string or due to sound or electromagnetic wave or wave equation 100% you will see that kind of a question in almost every shift and you will remember me tomorrow and even day after so guys velocity of the sound is root of tension by mu the value of the tension is 77 and mu that's mass per unit length well mass is 0 or I can just put it as 35 1 2 3 so 10 ^ - 3 divided by the length the length is basically 5.5 m over here okay perfect so I think after this all you have to do is just shift this on the top and and see what is the approx value that you are getting guys just see the approx value that you are getting uh 77 and 55 both have 11 as a factor which is common so 11 into 7 and in fact this also has 11 as a factor in common so in fact you can little bit simplify it there is no problem in that so this will become 7 into 11 and 5.5 I can write it as 5 into 11 into 10 ^ -1 this divided by 35 and into 10 the power basically minus 3 is that okay everybody so 11 goes outside 7 5 35 35 35 cancels 10 the^ uh you know min-2 will remain it will go on the top it will become 10 the^ + 2 so it will become 110 I believe so 110 yes that is the correct answer yes very nice yep is that okay clear perfect oh wow very good C for Captain stus it's not just C it's c for Captain stus it's D for it Del for it Bombay here you have another question coming up on your screen a thin circular uh ring of mass m and radius R is rotating about its axis with a constant angular velocity Omega two objects of mass m are attached gently on the opposite sides of that rotating ring so now the ring will rotate with a new velocity so what is exactly happening over here so this is your ring this is the axis this is basically Omega this is radius R this is mass m after some time after some time you have yeah you have attached two more masses over here so it will have a new velocity Omega 1 here angular momentum is conserved nothing else so conserve conserve angular momentum that's all so angular momentum is nothing but I into Omega I into Omega so basically i1 Omega 1 will be I2 Omega 2 initial moment of inertia is M r² initial Omega is just Omega final moment of inertia will be of the ring and the two masses will be of the ring and the two masses which are attached over here so it is M r² plus small M plus small M r² plus small M r² into the new omega r sare r sare r Square cancels everywhere so M Omega is capital m + 2 small M into the new velocity so new velocity will be capital M Omega / m + 2 * of small m is it there yes why not oh c for captains exactly exactly very good guys very good keep it up you will remember these kind of questions when you see angular momentum conservation based questions in the upcoming shifts moving on to the next one coming up on your screen the diameter of a cylinder is measured using a verer caliper with no zero error verier caliper with no zero error okay it is found that the zero of the verer scale lies between uh uh you know 5.10 and 5.15 of the main scale the verer scale has 50 divisions equal to 2.45 CM and the 24th division of the verer scale coincides with one of the main scale division okay interesting so first of all use this information 50 divisions coincide with or is equivalent to 2.45 CM so one division of verer scale will be how much 2.45 ID 50 cm is that right that is what it will be clear till this point awesome now that is not the end of the story that is not the end of the story you will also have to think what is one main scale division what is one main scale division over here just look at this numbers you will understand one main scale division from here and from here it's between this and between this so one division of main scale is 05 CM because the increment happens with 0.05 there is no 5.11 there is no 5.12 after 5.10 second decimal accuracy the next digit that you get is 5.15 there is nothing in between so one division of main scale is 05 CM which is as good as 5 by basically 100 right of a CM why do I need this for the least count of the verier scale least count of the verier scale which is 5X 100 minus 2.45 by 50 2.45 divided by 50 my dear Warriors just multiply by 2 right so you will get LCM as 100 so this will become 5 so 2.45 into 100 come on my dear students into 100 do it and tell me subtract it and see what is the least count that you are getting subtract it and tell me what is the least count you are getting come on my dear students you should get this as 4.9 right 245 into uh 2 is 490 so this will be 4.9 so won't this be1 / 100 or 10^ -3 of a CM 10 the^ minus 3 of a CM very good awesome now what do you do next once you get this now comes the main step of the problem where you have to find the diameter the diameter which is the final reading of the verar caliper is given by main scale Reading Plus your least count multiplied by least count multiplied by your warer scale reading warer scale Reading Main scale reading is 5.10 verier scale sorry main scale reading is 5.10 CM plus least count is 10 ^ minus 3 great into verar scale reading what is the verer scale reading it is 24th division is coinciding so 24th division is coinciding so what will it be 5124 CM 5124 CM which is option b yes perfect there you go because this is 0 this is 0.024 0.024 + 5.10 is 5124 cool yes yep yeah Kavita what is the question that you have I hope you are understanding everything till this point clear so what and all things you need to do in micrometer or verer problems is you need to know how to find the least count second thing is any kind of reading based question it is main scale Reading Plus least count into the verar or the circular scale reading to find the least count for verer caliper it is main scale division minus verer scale divisions modulus if it is your uh micrometer screw Gage it is your pitch divided by the number of Divisions on the circular scale so accordingly you will find the least count because I need to substitute it over here that is the reason why I need to find this out if it is already given it is a easier problem cool is that fine awesome B for ID Bombay correcto moving on to the next question two steel wires having the same length are suspended from a ceiling under the same load the ratio of the energy stored per unit volume is 1 is 4 the ratio of the diameters is asked load is same tension is same diameters will be different that means stress will be different okay there strengths I don't know I don't know how much is the extension so I should not play with fire that means I should not play with something which I do not know like strain in this particular question so what is the energy density formula energy per unit volume it is half stress into strain but wait a minute strain is nothing nothing but strain is nothing but what since young modulus is stress by strain so it will be nothing but stress by your Young's modulus so stress Square will be there and 2 y will be there below so stress into stress is stress Square stress is force by area and your force is the load in this case is the weight area is PK d s by 4 why because diameter is asked okay so w is same under the same load Young's modulus same material Pi 4 2 everything constants so I can see it is inversely proportional to D s's Square which is as good as saying it is inversely proportional to D 4 that is your energy density so energy density is inversely proportional to D 4 so start with it so row U1 by row U 2 will be D2 by D1 this ratio is given to be 1 is 4 so that is D2 by D1 so D1 by D2 is 4 is 1 so the ratio of uh oh sorry what did I do I did not Square it my bad uh just one second this was D2 by D1 whole ra to 4 so this was given 1 by 4 and this is nothing but D2 by D1 ra to 4 so take the fourth root first take square root square root of 4 is 2 square root of 2 is < tk2 so this will become 1 by < tk2 is D2 by D1 so naturally D1 by D2 take the reciprocal it will be < tk2 is 1 so < tk2 is to 1 is the answer which is option number a is that right everyone very good excellent yes so these kind of questions on ratio of loads ratio of the extension ratio of the young modulus ratio of the stress ratio of diameters ratio of original these questions are very common including thermal stress which is y Alpha Delta t Okay so remember these formulas thermal stress also y Alpha Delta t a tank is filled with water up to certain height hole is drilled toril is law one of the favorite patterns of questions for your NDA examiners so this is our tank a tank is filled with water up to height H okay is filled with water up to height H so let me show the water over here like this okay and a hole is made at a distance a small H below over here so the hole is made here this is small H so what will happen is the water will come out of this velocity of fux will be there what will be the range of the water question is what is the range of the water simple guys the range is range is the velocity of fux into the time of flight the velocity of fux velocity of fux only depends on this height so it is basically root of 2 G small H the time of flight for a normal projectile like thing this is nothing but H minus H so it is root of the height of all divided by basically your G so what will happen my dear students < tk2 < tk2 will come out as two g g will cancel and you will be left with h into H minus H just check this out yes option number a how many of you wrote option a very good har uh what's the name Harish shines wow nice first time I'm seeing this name ual snea very good Dia very good ordinary person game Squad hsh very nice wow we have telu people Manita very good some of the students who are there in the previous shifts are also come over here welcome back and I hope you have smashed the like button if you have not done that already what are you guys doing and yes please subscribe because this is the channel which will also help you for vitle E SRM J EAP sat bits hat and J advance so guys lot of things to come lot of responsibility on our shoulders and maybe there are even 2025 and 2026 students waiting when will these 2024 people pass out so that SL are starts our lectures anybody from 2025 or 2026 silently watching what is happening on this channel what are these dudes up to really over here is this really what J is should I take J or not so any 2026 2025 2027 students watching up there don't worry things will be easy Once We Begin our syllabus right cool what is the uh correct graph which represents displacement time okay is given so what is the correct graph which represents that displacement time in terms of velocity versus time so velocity time is the option displacement time is given wow yes one 2020 78 oh my God I you always you know overshoot my expectations yeah very good Mahindra Singh duni 2031 really shouldn't you guys be watching like Pokemon or something 2050 wow very good so you are in your previous birth if you are going to give J 2050 that means you are in your previous birth and you're are going to get reincarnated very nice so you are preparing from the previous birth itself that is the level of competition for J super super say very good let's see what the next uh answer should be guys come on come on come on my dear students yeah displacement is increasing and decreasing check no the slope increase slope is positive then becomes zero then becomes negative positive 0 negative positive 0 negative so that's all you had to do where did it go yeah here it is positive it is zero then negative so here it is positive here it is zero and here it is negative that's it this this this can't be correct as simple as that you just had to check the slope nothing else because remember velocity is the slope of position versus time so the value of the slope is the value over here is the value over here got it yep Moving On Moving On Moving on oh I showed the answer anyways forget it the ray of light which goes from denser to rarer medium uh the speed of light in the rarer medium is double that in the denser medium okay what is the critical angle so to find the critical angle we need to know the refractive index and to find the refractive index what will I do to find the refractive index of any medium what will I do I will take the speed the speed of light in rare medium and then I will take the speed of light in the dense medium which is usually less so big value by small value will give me more than one as the ratio so mu will be in the rare medium the speed is double so this is directly two times this is directly two times nothing left over here so s of the critical angle is 1 by mu which means it is half so what is the critical angle whose sign is half 30° wow that is the correct answer shall we proceed everybody with me awesome awesome so 90.1 percen enough for qualifying for Advance nobody knows don't play with fire my dear students how do you know 90.1 will be cut off 90 is the approx range it depends on the number of students appearing so when the number of students appearing is more the cut offs will also be more so you know many years back there were like some six lakh s lakh students now this year there are more than you know 12 lakh students appearing so the competition is more seats have not increased so much also they have but not so much so understand it might go up to 91 also so safe score will be 91 to qualify for writing Advanced not for getting I to qualify to write the paper to get into I I hope that makes sense cool okay how to know slope slope is the rate of change of anything slope is basically tan of the tangent drawn tan Theta okay that is how you measure the slope two slits separated by a distance of 1 mm are illuminated with red light of wavelength this much the interference fringes are observed on a screen 1 M away from the slits the distance between the third dark Fringe and fifth bright Fringe is how much okay let's do this when I say third dark Fringe third dark Fringe the y coordinate of that Fringe is given by the formula odd time Lambda sorry or yeah R * Lambda capital D by 2 small D first dark Fringe odd number is one second dark Fringe odd number is three third dark Fringe the odd number is five one is first three is basically second Fringe F five fifth odd number or is basically the third dark Fringe so hence it will be five * Lambda capital D by small D with this number two over here similarly the fifth bright Fringe the formula for bright Fringe is simply n Lambda d by D there is no two and there is no odd over here here the calculation is simple just it is 5 * Capital Lambda d by small D now the question is what is Delta Y what is basically Delta y meaning the distance between third dark Fringe and the fifth bright Fringe so it is basically y5 minus Y3 so Lambda capital d by small D is common in fact even five is also common so 5 Lambda capital D by small D is common and you will be having one from here and half from here 1 Min - half is half so it will become 5 * Lambda capital D by 2D now do you know Lambda yes do you know capital D yes do you know small D yes we know everything what else is left just do the math yeah and see what is the answer that you get you should get it as 1.63 mm approximately right so just substitute you will get it is this logic clear because substitution is not the catch here the catch is whether you substitute the proper numbers or not if you don't substitute it then you're gone is that okay clear shall we move ahead yes we'll be starting just after J Ms okay ABI 1 1212 cool let's let's move on to the next question on your screen now what does it say Wow we have statement based question read everything properly the angle of deviation through the prism is minimum minimum deviation guys what happens incidentally emergent Ray or symmetric refracted Ray inside the prism becomes parallel to the base angle of incidence is equal to the angle of emergence when the angle of emergence is double the angle of incidence okay very simple this is very very simple when the in fact the diagram only shown like that there is nothing to solve here the ray which is inside is going to be parallel to the base yes it becomes parallel to the base definitely B is correct incident Ray and emergent Ray are symmetric in fact that is also true and this is also true because you will see this angle will be also equal to this angle incident angle will be equal to the emergent angle so C and A is correct so naturally it is double is wrong so a b c so option A is correct option A yes option A is the correct answer got it this is directly directly solvable very nice moving on to the next question wow we have come to integer type let's see the angle of polarization for any medium is 60° it's critical angle is this much find that n now the moment you see such question you might get confused things thinking sir what is this polarization angle sir nothing it is nothing but the Brewster's angle Brewster's law do you remember Brewster's law how many of you remember Brewster's law and Brewster's angle that's the angle of incidence when the light which is falling on a denser medium partially gets refracted and partially gets reflected the incident light is unpolarized but the reflected light gets completely polarized at that angle of incidence that is Brewster's angle and that law which gives you is called as Brewster's law so the Brewster's law is tan of IB Brewster's angle is refractive index when the reflected light and the refracted light are perpendicular to each other remember that was your Brewster's law so basically it is given to be 60° So Tan 60 is Mu wow I don't like that mu huh so tan 60 is root3 so root3 is Mu but that is not the question question is the critical angle they have combined critical angle with Brewster's law so s of that critical angle is 1 by mu which is 1 by < tk3 so I is sin inverse of 1 by < tk3 compare and you will see the value of n is three very nice how many of you got three oops some of you said four some of you said two nope the answer is three very good game Squad excellente yes tan of IB is equal to refractive index expect polarization questions guarantee guys Sure Shot malo's law Brewster's law malo's law Brewster's law sure short subtopic I'm telling you tomorrow's paper it's like you know Shar is leaking it types you will feel like sh told so many things and it is just directly okay so malo's law and Brewster's law very very important when the current changes from 2 Amp to minus 2 a in 05 seconds the EMF of 8 volt is induced in the coil the coefficient of self inductance is something into 10^ minus 2 what is that something all right fair enough so is there a formula which relates voltage self inductance time current yes the self induced EMF the self induced EMF is L * change in the current by the time interval magnitude wise of course this is just the magnitude so the self-induced EMF is 8 Vols L I don't know change in the current final minus initial current so 2 - -2 divided by time which is 5 into 10 ^ -2 so what will happen this will become L * by 2 2 uh sorry 2 + 2 is basically 4 and this will be 5 and 10 ^ minus 2 this will be 8 4 goes with 8 2 * 2 5 are 10 so L will become 10 into 10^ - 2 Henry so answer is 10 wow answer is 10 into 10 ^ minus 2 is that clear wow very good very good very good yes there is going to be a question on organ pipes also hopefully I think there must be a question on organ pipes also very soon minimum energy satellite is thrown in the orbit what is the energy change Escape speed ratio gravity at depth and gravity at height standard questions I'm leaking the questions it's like that guys yep abishek sing 2027 student here welcome B welcome I hope you start watching the older classes already don't wait for me to start okay if you are smart if you are preparing for J you should be first of all very smart at least normal smart not about physics chemistry mats you should know ah St sir has made the lectures already St sir has you know kept all these playlists it's there on the channel I'm smart enough to start seeing those playlist start doing that SSR will come for all of you don't worry okay so the question is if you take it to an altitude of two hour what is the energy required for doing so okay see first of all this is radius R you had a satellite of mass m here then you are taking it into an uh orbit of a radius how much radius I don't know altitude is 2 R so basically the radius radius will be basically 3 R radius will be basically 3 R okay so you have to put the satellite over here so first the energy will be initially minus G mm by R only potential energy is there that's it finally the energy of the satellite is given by Min - half G mm by by the radius radius is 3 R this is the total energy of the satellite in that orbit so the question is what is the energy required to go from here to here so that is Delta U which is this minus this which is nothing but final minus initial so final is minus G mm by 6r all right minus this thing minus G mm / R that's it so that should be your answer see what is it going to be g m m / R is common minus minus is plus so it will become one and this will become uh 1X 6 this will be become 1X 6 so answer will be 1 - 1X 6 which is 5X 6 so it will become 5x 6 * g capital M Small m / R so answer has to be given in terms of M Small gr no issues we can do that also what do you do Capital GM Capital GM is gr r² Capital GM Capital GM Capital GM is nothing but small G r² and this is M and this is R so finally what will you get you will get 5x 6 small mg R but you have to write in terms of 12 so no issues write 12 here so multiply denominator by two numerator also multiply by 2 so 10 MGR by 12 so answer is 10 yes that is the correct answer so the final answer is going to be 10 guys everybody clear this 10 and this K that is what is the question K has been asked wow yes hares jalot that is the correct answer shall we go ahead to the next one coming up on your screen ready for it ready for it and here we go fundamental frequency of closed organ pipes somebody was just asking sir take a question on organ pipes and yes I just taken because these are very very important questions very good chance it will get get repeated values will change and you'll remember me the fundamental frequency of a closed pipe is equal to the second over tone of an organ pipe open at both ends the length of the Organ Pipe is how much okay let's go step by step closed Organ Pipe closed Organ Pipe means closed at one end and the formula for such a thing is odd times velocity of sound divided by 4 * length this is equal to the second overtone of an organ pipe open at both ends open at both ends the formula is n v / by 2 L that's it now these both will be equal first of all this velocity this velocity cancelled this two cancelled with this four became two what is the odd number and what is the n value fundamental frequency of closed pipe that means odd number is one because this is fundamental fundamental means the first odd number next second oval tone second overal Tone If you talk for open pipe see the harmonics go like this for open pipe the harmonics go like this for open pipe F 2f 3F 4f all the multiples are there this is your fundamental this is your first overtone this is your second overtone this is your third overtone second oval tone is third harmonic so n is basically Three N is basically three wow very nice and it is also given that the length is 20 cm for the closed pipe so this is 20 so left hand side will become 1 / 2 into 20 right hand side will become 3 V has gone two has gone divided by L2 that's it so L2 will be 3 into 2 into 20 3 2 are 6 6 2 are 12 so 120 CM so that is the answer 120 Yes 120 is the answer guys how many of you wrote 120 wow very good liita I think you only asked it right there you go perfect perfect moving on to the next question how about some Modern physics I can see some Spectrum Photon and excitation so yes modern physics atomic structure 100% you will get questions hydrogen atom is excited from ground state to some State n okay the number of spectral lines in emission spectrum is 10 so if you go from ground state to some nend state in between there might be some levels there will be multiple multiple you know uh what do you say multiple emission lines which will come out multiple emission lines which will will come out so the number of number of lines which are there is given by n into nus 1 / 2 and that lines is already given as 10 that lines is already given as 10 so 10 is n into n -1 by 2 can you guess the value of n you can solve it and get a quadratic equation but I think you can easily get guess it n has to be an integer n has to be integer will it be four 4 into 3 12 12 by 2 is 6 No 5 5 into 4 by 2 5 4 are 20 by 2 is 10 yes n is basically five n is basically five now the question is not to find n the question is the excitation is done by the photon the energy of the photon is so if you go to that state if you go to that state right from the ground state to the N is equal to 5 State how much n energy is required so that energy which is required is energy at fifth State minus energy at Ground energy at fifth state is E1 divided 5 Square E1 divided 5 Square minus minus basically your even very nice so what do you get I will get this as E1 is common 1X 5 squ right is 1x 25 1X 25 so 1 - 1 - 1 by 25 which is which is 24 which is 24 by 25 but with a minus sign but even itself is 13.6 electron volt so into 24 / 25 is that clear my dear students is that clear my dear students perfect so that's it solve the question now there is nothing left over here 13 goes with 26 two times almost so this is 25 almost this will be like uh you know two over here and you should get it close to 12 or 13 around that figure yes the answer is close to 12 or 13 so approximately you will get this as 13 electron volt so this minus will come again and it will become plus okay so there will be minus 2 times actually this will become minus of -3.6 even value is 13.6 I just substituted it over here cool understood yeah it will be around that range yes yes yes game Squad it will be in that range perfect moving on to the next one oh we have come to the uh you know end of the physics part after this you're going to have uh chemistry I believe and then maths and again chemistry okay so that's the flow so this is your first homework question put up the answers in the comments you'll be getting the PDF of the solution with the correct answer key and I think there is one more homework question you can take a screenshot of it quickly on D Broly wavelength highly probable questions right yes so post the answers in the comment section now it's time for chemistry let me just check with the team so uh getting fear of tomorrow's exam motivate us sir okay lucky so first of all let me tell you guys uh watch this particular video which I had made on the channel this is to give you the last minute tips don't do these mistakes in the second attempt please watch this uh literally you might save a lot of negative marks and you might actually improve your score even at the last minute so just check that out number one number two please do not focus on you know what is happening around you and even if something goes wrong like the clock is not working computer is not working remember that you know you might not be alone there might be many students and things might go wrong don't get scared because of all those things anything goes wrong I'm repeating again please ask the invigilator to take a video video of what went wrong because that you can submit as proof last time some students said sir my computer hand I lost 20 minutes like what is the proof nobody will believe you so that is why you should have a proof you can't carry mobile phones so ask the invalidator is this clear absolutely right let me tell you the pattern the models the difficulty will be almost similar to the January attempt yes like nobody can predict something might come really really out of the way but chances are very less remember there will be definitely 50% of the paper which is solvable which you can do and that is what you will Target for first if you are able to get 50/50% at least in whatever your strengths are whatever your strong subjects are you are almost going to get 100 marks for sure in the first you know 45 minutes to 1 hour after that build up slowly slowly go to the next subject go to the medium level questions you don't have to do each and every question don't go onto it and please do not guess all the options guessing should be limited to two or three questions per subject that to maximum don't guess more than two is that clear I'm pretty sure tomorrow and day after tomorrow and even after that you will do extremely well you have been studying there is nothing significantly that you can change in the last one day small things which you can do is watching that video and implementing it so do that save your marks see don't get Negative marks because that will take away all that you have gained so that is my simple advice and just be calm and patient and relax everything is going to be fine and even if things don't be are not fine remember we are there here we will see what next to do whether you have to take a drop or whether you have to re for some other exam and what are the other options which are available everything we will guide you okay so don't worry even if you fail it's okay but go with a mindset of winning go with a mindset of acing the exam go with a mindset thinking that you're going to rock you're going to nail it and you're going to give in your 100% like never before okay so don't worry all the best my dear students I'll be seeing you on the other side but now it's time for chemistry bye-bye see you soon hello students can you hear me and see me clearly hi hi everybody welcome welcome to the organic chemistry session right now we are going to do the most predicted questions from the organic chemistry so here I've curated the best set of questions you can ever get from each and every topic that you can expect in your tomorrow's exam or whatever J uh exam date you have right hello everybody my name is nandini if you do not know me my name is nandini and I teach organic chemistry to J students on this channel hi ch let us begin with guys have a time crunch uh let me find the remote first of all it is here okay CH let us begin are you guys ready are you guys ready please show please show Your Love by liking the session right by sharing with your friends by liking the session so that we can begin first question on your screen from your organic chemistry is here which of these group has the plus I effect another very important set of questions based on your inductive effect you must know which of these group has the plus I effect try to answer this I'm giving you 30 seconds give me the answer in the chat box hi everybody hi Den take yes B yes buta you can expect that actually what happens we should analyze the papers we will analyze the paper which is coming on the fourth of your April right which is tomorrow and we'll be solving those questions over here and what you should be doing you must expect similar questions in the uh incoming shifts right because many of the times the question pattern the type of questions get repeated right so it is very very essential to follow the question solving series which is starting from tomorrow if your exam is either on 5th 6th 9th whatever date right CH I've received so many answers a a a Whoever has answered a is extremely extremely right only ch3 over here will show the plus I effect correct correct right all the other groups this is a alken this is a alkine group these are electron rich but they also have the power to show do not show my plus I effect they do not show plus I effect understood I'm moving on to the next question next question on your screen is this how many geometrical isomer of this compound are possible so you have been given one alken right you have to predict the number of geometrical isomers again a very important kind of question asked in your exam you must expect many questions out of this right right try to answer this try to answer this look at this alen how many double bonds it has one 2 three three double bonds I can see now try to count the number of isomers 30 seconds and then I'll tell you the solution I'll tell you the solution yes try to answer again a very repeating set of question geometrical isomerism a c d okay various answers I have received various answers so look at this alkan over here okay I will tell you a very amazing trick can you see this double bond over here will this double bond the terminal alken will this alken be showing any geometrical isomer no why because this terminal alen is connected to two same groups which is the H group right so this alen over here will not show any geometrical isomer isn't it so the only two alkenes which can show looks like this one as well as this one right only these two alkenes will show your geometrical isomer what are geometrical isomers CIS and trans isomers so only these two will show isn't it so now what is the formula for geometrical isomerism before that you must know is this a symmetrical alen or an asymmetrical alken what do I mean by that I'm writing this alken over here so this is the first alken right the second is c h and then once again here is a c and here once again there is a h over here right here's once again alken here once again A C here's once again a h right and here is CH and here is once again A C double bond C right and here is once again a h h and here is once again a h and here is once again a ch3 right can you see these two alkenes so this alen will not show any geometrical isomer but these two will definitely show they have different groups attached over around each other now can you see B around this alken I'm changing the pen color so that you don't get confused around this alen can you see that there are two different entire groups attached there are two different groups attached similarly around this alen also two different groups are attached here just ch3 is there while here to this carbon there is a whole chain of compound attached so can you see that the uh compounds attached are entirely different to the double bonds and when that happens the formula that you have to use for calculating the geometrical isomerism is 2 to the power n simple easy peasy formula n is the number of double bonds participating in geometrical isomerism what n is equals to number of double double bonds right which are participating in geometrical isomerism so this one is not participating only only two are participating so the answer is 2 to the^ 2 which is equals to 4 so the correct answer here is four whoever answer C is absolutely correct but you must know the reasoning behind it right if the two terminal carbons if the chain around the double bond are different then this formula will be used understood and if and if the two chains are same around the double bond ma'am then what would be the formula then the formula will be this would be this which is 2 The Power n -1 upon 2 The Power P minus one what is n once again number of double bonds participating in geometrical isomerism and the value of P you can calculate using this formula you just have to keep it in mind this is just a easy trick so you have to remember this formula but the good news is that never have ever in your exam this kind of alken has been asked only 2 to the power n type of alkenes have been asked in your J exams yet so that is the trend so you don't have to stress over this take I'm moving on to the next question again a very repeated kind of question which is which of the following orders is correct in everywhere you can see there is nitrogen present right what is basicity able ability to donate electrons right buta basicity is what basicity is ability ability to donate electrons right try to answer this question for me try to answer this question I'm giving you 30 seconds once again this is no more than a 30 second question try to answer this first answer by litha C32 very good litha litha is extremely right game over Tamil Arian s everybody has given me the C answer very very good can you see in the first option there are already so many nitrogens present so many nitrogens with lone pair hence obviously these will obviously donate electrons very very easily right so here the donation capacity is maximum so first will be the most basic compound next in the second option what can you see the nitrogen is connected to the Benzene ring hence here these loone pair are actually participating in resonance hence the availability of the loone pair to donate becomes less because they are participating in resonance within the ring so they cannot donate while your ch3 NH3 is doing what here the alkal group is causing plusi effect isn't it this is donating more electrons to ammonia so this is becoming most basic so first first then your third and then your second this is the order so the correct answer here is C whoever answered C is extremely right understood easy questions these kind of questions you should expect next question on the screen which is the major product of the following reaction once again what are you going to see in this kind of reaction major product you have to predict can you see that here I'm reacting this small molecule with a HCL molecule now let us see the property of each and everything over here HCL is the acid which means what it will easily give away H+ ion isn't it now look at this compound given over here what is the property of this compound there are nitrogen's present hence nitrogen has lone pair of electron isn't it isn't it right now what will happen what is the property of lone pair I just T you to abstract the proton right so which will donate its electron very very easily that is your question question is same that we did just right now of the basicity right here again which will very quickly which of both these two will donate the electrons very quickly that is the question once again H but asked in a very different format so which will donate the electrons very quickly that is your question answer this to me answer answer this to me so which do you think is the answer A B C or D yes liita is telling me ma'am second is the answer liita and Amir is telling me c is the answer so whoever or manvita is also tell me J Jad manvita snea Meha very very good Shiva arind Aran all of those who have given me the answer as B are absolutely right right now let us look at these electrons over here once again what did I tell you the electron pair on here will actually undergo resonance with the Benzene ring isn't it so the availability of these lone pair becomes less available less available to donate to donate isn't it but look at this lone pair on this nitrogen atom what is happening over here this nitrogen has this alkal group over here this plus I group over here this plus I group over here hence this nitrogen very very can easily it donate its electron to H+ isn't it so this will actually donate while that electron lone pair is less readily available so the correct answer is the hydrogen will get attached on this nitrogen hence this positive charge will be formed so the correct answer is B so C the concept is C which will next question on your screen okay we have already done this next question is this which of the following is the most stabilized carbo again this kind of question definitely will come in your exam definitely and there's so much chance that this question can be repeated right whatever questions are here are extremely great questions I'm already telling you these please even if you practice these many questions now your organic paper you can uh expect to score uh like out of 13 questions at least half of them would be correct just from this paper right D D D D D D D very very good whoever is answer D is absolutely right wow wow yes buta yes buta here can you see what the stability of carbocatine depends upon resonance in these cases here this carbocatine do has resonance with this double bond but that is it that is one resonance it's showing now look at this carbocatine over here this has no resonance isn't it now look at this carboca over here again it is showing resonance with this right yes it is showing but look at this carbo over here what is happening in this case in this case something very extremely good is happening can you see these electron pairs can go here through resonance that this electron Bond can go over here right now the next thing is if this Bond comes over here what will happen what will happen 1 2 3 4 5 1 2 here and here was the double bond o m e plus right here was your double bond right isn't it now B here is once again your double bond present isn't it now can you see these two double bonds are once again in conjugation so conjugated double bonds can again undergo resonance so this is showing so much of resonance right so the maximum resonance is shown in option number D so the correct answer is d right even though in a and C there's also resonance but in D there is maximum res resonance present so much of resonance hence D is the correct option understood everybody I'm look to I'm moving on to the next question here what is the decrease in order of strength of the bases very again this is a very important question right what do they ask instead of strength of the bases they can even say strength of the nucleophiles right again a most repeated question so nucleophiles but once again are what something which has a negative charge on it and which can easily give away their negative charge their lone pair of electrons are called nucleophiles and same is a strength of bases what what does the strength of Base determine it is determined by the ability to donate electrons as I'm saying one bar bar H ability ability to donate electrons isn't it that is called that determines the strength of the base and lesser electr negative the atom more strongly it can donate why because more electr negative the atom is for example oxygen Florine those are more electronegative molecules so they hold the electrons closer to themselves and they do not let them go why lesser electr negative atoms like carbon easily gives away electrons understood a a b b so I'm getting various answers the correct answer is to this question what do you think is the correct answer a b c okay I'm I'm also getting d d is not the answer buta D cannot be the answer the correct answer here seems to be a seems to be a right what is happening in a let us look at a here oxygen will be the least strong base least basic why because oxygen here has the most electr negative atom so it will hold its electrons more closely will not donate easily so this is the least basic then comes your SP hybridized carbon once again understand this thing SP hybridized carbons but are very electr negative species right so what do they do they are exactly like oxygen understand Like Oxygen can keep a negative charge on itself oxygen is more electronegative so it is not donating electrons similarly carbon is more electronegative so will not donate electrons when it is SP hybridized then comes a nitrogen here nitrogen here has a negative charge on it hence it can easily donate electrons more than the SP hybridized carbon similarly here this carbon is sp3 hybridized so this can very very easily gives give away electrons because it is least electronegative it is not interested in keeping electrons to itself it will readily give away electrons take away my electrons then nitrogen then P carbon and then oxygen understood this strength bookmark the strength everybody everybody right so the correct answer is a I'm moving on to the next question now the next question is this very very good question once again which of the following is most likely to undergo a favorable hydride shift who can answer this question for me 30 seconds for you to answer hydride shift means the shift shifting of hydrogen can everybody see there is a carbo in each case right and the hydrogen next to the carbocatine can shift in order to for example in order to become from primary to secondary to tertiary carbo why because here as we go the stability increases isn't it the stability increases from primary to secondary to tertiary carbo a a a a a b okay so many answers so many answers the correct answer here is a whoever said a is absolutely right look at the a option here this is a secondary carbo isn't it connected to two carbons but if this carbo if the hydrogen over here shifts over here what can happen the carboca moves over here and becomes a tertiary carbo D which is more stable so hydride shift is very much possible in a now look at B in B there will be a methy shift right and then the carotin can go there so it is not the hydride shift but the methy shift in order to become tertiary similarly let's look at C in C there will no shift because the maximum stable will be just secondary carboca no shift possible now look at D in d what Can happen in D I know there will be a ring expansion happening which means that this Bond understand that this is a box and this is the lid of the box so the lid of the Box will open and it will connect to this carbon so the four carbon ring will convert into a five carbon ring like this and this will come here and the positive charge will come somewhere over here right so here once again a methal shift has happened not the hydride shift a ring expansion has happened so the correct answer is a only no other answer is correct but I explained it to you why others are not correct I hope this thing is clear to everybody yes I'm moving on to the next question again a very important question very very good question which of the following orders is correct for the stability of carbo again the stability stability order will definitely be asked in your exam which of the following orders is correct for the stability of these carbo try to answer this question for me try to answer this first answer received by jadev a and chinai has given D okay chinai has given d and a a liita is liita not right this m sorry sorry so can you see buta in First Option such kind of compounds are called what this compound such kind of compounds are called These are this is called the with three bonds with can you see that three Cycles right it it seems like it seems like this any compound that seems like this this carbon over here where there's the attachment of three bonds is called the bridge carbon what it is called it is called the bridge carbon understand this thing and understand another thing that bridge carbons can never be planer structure understand keep this in your head that bridge carbon bridge bridge bridge carbon can can not be planer planer right and what is the hybridization of a carbo a carbo structure looks like this with a posi charge a carbo is SP2 hybridized which means it is a planer structure isn't it and I just told you that on this bridge carbon the the carbon can never be a planer structure so this positive charge will never be stable on this so this will be the least stable compound so I know first will come at the last this is the least stable compound now we have to compare between third and second easy comparison once again this is a tertiary carbon so this will be the most most stable right tertiary carbons are tertiary carbo cattins are most stable so second is the most stable followed by Third and then first so this is the order second third and first so whoever said D is absolutely right the correct answer is d right please keep it in your mind that this bridge position carbons can never be planer structures and because carbo have the arrangement of a planer molecule because these are SP2 hybridized hence this positive charge is never stable on the bridge carbon even even I'll tell you one more thing on this bridge carbon right even double bonds are not stable double bonds not form they do not even form because they do are not stable on this bridge carbon you will never find a double bond if a double bond is there it means your structure is wrong understood because double bonds are once again what SP2 hybridized isn't it take CH next question on your screen which of the following ring compounds Obe huckles rule who can answer this question for me which of the following ring compounds obeys huckle rule follow form the ring compounds out of this huckle rules is what huckle rule is 4 and + 2 pi electron rule okay try to answer this question for me so let's draw each and every structure what is C4 H4 minus1 C4 means carbon 4 four carbon cyclic ring four means four hydrogen atoms are there one atom two atom three atom right and here is your this here is your uh double bond here and there's a negative charge over here so I will just draw a negative charge over here so this is C4 H4 minus1 structure can everybody see this and so the next hydrogen will come over here right right now this can this structure ever follow the huckle rule this is not even aromatic can everybody see that this is not even aromatic this carbon is sp3 hybridized and this is not even following the huckle rule because how many electrons are present 4 Pi electrons which means this is the anti-aromatic compound hence this will not follow the huckle rule let us form the C4 H4 +1 compound C4 H4 + 1 means what C4 H4 here comes the another H here comes the positive charge isn't it so this is C4 H4 + 1 compound right now here the hydrogen will be there now count the number of electrons in this this is 2 pi electron system isn't it this is 2 pi electron system but this carbon over here is sp3 hybridized which means it is not aromatic so this cannot be the answer now look at C4 H4 minus 2 I'm follow I'm forming C4 H4 min-2 it looks like this C4 H4 min-2 structure but looks like this can everybody see the structure can everybody see the structure C4 H4 -2 now tell me how many atoms does it have it has 1 2 3 three which means this is now a 6 Pi electron system isn't it this is a 6 Pi electron system now take take so what do you think is the answer now similarly you can force C4 H4 structure C4 H4 structure looks like this this is C4 H4 this is once again a four Pi electron system right so even though even though these two are following these two are following your huckle rule take should I move forward D cannot be the answer but D is a 4 Pi electron system which means it is anti-aromatic anti-aromatic and if I put four electrons over here one second let me make it once again let me make make it once again this is a C4 H4 system this is four electrons here four here right now how many electron system it is this is once again a 4 Pi electron system because C4 H4 compound it is right 4 Pi electron system once again not huckles rule so I think the huckles rule is only followed in this system which is C4 H4 + one system isn't it it's followed only in this system right right beta it looks like that c would follow because we have done examples like this but if you form the structure of C4 H4 minus 2 then the structure forms like this and it is actually a 4 Pi electron system even though the carbon over here is sp3 hybridized what is happening it is following the huckles rues even though it might not be following your aromatic rule but it is following the huckle electron Rule and that's the question that that is asked so this might not be aromatic compound but it is following the huckle rule which is 4 n + 2 pi rule right right yes buta B is the answer B is only the answer right we're moving on to the next question which of the following have larger Co Bond length than Co Bond length of ch3 Co ch3 try to answer this question for me so can you see here this Co Bond length is what with this Bond length is a double bond Right double bond because and it is a pure double bond because there is no resonance present in the acetone structure right so this is a pure double bond and you have to tell me which H in on in which the bond is larger the bond is larger okay there can be more than one answer to this question try to answer this try to answer this question for me take C and D I've received answers so let us see let us see first compound number a in the compound number a I taught you that the double bond outside the ring can everybody see this double bond which is outside the ring is called what buta this is called the EXO Bond and when you have to look at the aromaticity of this compound what will you do you will break this Bond can you see oxygen over here is more electronegative so when I break this double bond what will happen oxygen will get the negative charge the carbon over here gets the positive charge isn't it now count the number of electrons present now this becomes a 6 Pi electron system isn't it this becomes 6 Pi electron system okay which means this is a aromatic compound so when it is becoming aromatic when the bond is breaking right so this is definitely a longer Bond when this bond is breaking what is happening I'm drawing it here one once again once again one second yes so what is happening here but what is happening here this bond is breaking into a single Bond hence this Bond will definitely be longer than this B because after breaking this is getting aromatic character so it will definitely do that it will definitely exist as a dipole now look at the structure if I break this Bond like this what will happen the negative charge will come here the positive charge will come here now count the number of electrons this becomes the 4 Pi electron system which means this becomes anti- aromatic which is very unstable so it will never do that it will never break into positive and negative charges it will remain as a double bond because after it is breaking into positive negative it is converting into anti-aromatic so it will never do that so this Bond will not be longer than this Bond now look at the C option over here in C what is present there is so much of resonance present isn't it this Bond can go over here this Bond can go over here then this Bond again can go over here right right so there is so much of resonance present so once again what can happen and the Ulta resonance can also happen because you have to show if this bond is bigger H Ulta resonance can happen how this Bond can go over here and this Bond can go over here right now once again this is also showing your single Bond character after resonance so this will definitely be the answer the answer is two two correct answers are A and C options again resonance is not present in present in D so this will exist as a double bond only understood this will exist as a double bond understood everybody take the correct answer is a and C options both are correct understood everybody I'm moving on to the next thing the next is assertion reasoning on your screen try to answer this for me assertion is Paran Nitro phenol is a stronger acid than orthon nitrophenol and reason is intr molecular hydrogen bonding make the Oro isomer weaker acid than para isomer so first of all tell me is the assertion true or false and is this true or false try to answer both of these questions to me try to answer this question to me yes so but what is Paran Nitro is paranitrophenol stronger acid than orthon nitrophenol what does orthon nitrophenol looks like this I'm drawing this I'm drawing this this this here will be the N here will be the double bond o here will be the O here will be the O and H right this is orthon nitrophenol and between this oxygen and hydrogen there will be a h bonding right so due to this H bonding what is happening due to int molecular intra moleular H Bond can you see in moleular h bond is formed so due to this bonding this hydrogen will not leave right what is stronger acid acids are those which easily give away H+ but because due to this hydrogen bonding it will not give away it H+ and hence orthon nitrophenol is definitely a stronger acid than a Paran nitrol no no no no huh so paranitrophenol is a stronger acid so this becomes a weaker acid now because this hydrogen bonding will be formed and hence it will not break it will not give the H+ easily it will not give the H+ due to your hydrogen bonding hence Paran nitrophenol is definitely stronger acid than orthon nitrophenol due to intramolecular H bonding now let us read the other reason intramolecular hydrogen bonding make Ortho Isom we weaker this is also true statement right both are true and the reason is also the correct explanation of your assertion understood this point understood this point everybody take I'm moving on to the next question here what will be formed in the following coal based electrolysis try to answer this question for me what happens in Co Bas electrolysis so first you're doing electrolysis what will be the answer what will be the answer uh I'm looking at the chat box 30 seconds to answer this question C first answer C to B time very very good freak nation very very good harid also has given me the correct answer pradep it is not D B but it is not D look at this what happens in electrolysis this CO2 will go away this CO2 will go away isn't it isn't it so what will happen what will happen C double bond C here is the ch3 here is the ch3 and after two co2's go away the free radical is formed and after the hydrolysis what will happen it converts into a buttine structure it converts into a buttine structure which is a two buttine structure like this right so the correct answer is C2 buttine whoever answered C is correct right col electrolysis again very very important name reaction please focus on name reactions right right next question on your screen so this is the answer next question on your screen the boiling port point of trans Doro ethene is 48° so trans isomer boiling point is given which number best approximates the boiling point of CIS Doro ethe so trans you have been given the boiling point of trans right trans Doro ethene Doro ethe here's the H here's the H right so this is the trans droro eth and you have to tell me the boiling point of CIS dry Doro ethe what will be the boiling point of this tell me the answer to this I'm looking at the chat box D D D D whoever is answered D very good absolutely correct why buta in trans isomer boiling point depends on two things remember put it in your head boiling point depend on two things number one thing is if there is hydrogen bonding present then boiling point will be high more hydrogen bonding more will be the boiling point number second is it on its dipole moment so can you see the dipole moment in the trans isomer becomes zero so because the dipole moment is zero it will have lower boiling point but in the CIS isomer the dipole moment adds up in this direction so this has some dipole moment so s isomer always have have higher boiling point than trans isomers boil whenever we're talking about boiling point so here 54 is the only higher number given to us so the correct answer is 54 all others are lesser than 48 so they cannot be the answer because for the CIS isomer it will always be higher boiling point will be higher than the trans isomer due to dipole moment so put these two things in your head boiling point depends on two things next question on your screen what is the major product in the reaction what is the major product in the reaction so you have been given nh2 minus 2 equalent of amide minus take try to answer this question again this is the favorite reagent of your examiner every year this reagent is asked nh2 is your favorite reagent try to answer this either a b c d what should be the answer what should be the answer try to answer this question for me everybody try to answer this question for me d d answer I'm getting very very good whoever answered chinai very good madavi see NOA chinai Chinta very good harid and bav banut has answered correctly game over Tamil D very good so what answer is D whenever nh2 minus is used two equivalent means what with one equivalent one hbr will be removed okay so it will get we will get one double bond with the other nh2 minus the other hbr will remove which means we'll get a triple bond so here two equalent are used so we will get a triple bonded this kind of structure right depending upon the equivalence of Amite with one equivalent one hbr removes with the other equivalent another hbr gets removed take moving on to the next question which of the following alkanes is most reactive towards electrophilic addition reaction once again try to think out this through electrophilic addition reactions which will show electrophilic addition reactions a very good electrophile is isn't it isn't it so where will this electrophile attack where will this electrophile want to attack amongst these alkenes that is the question asked try to focus on what is asked so where will this positive charge want to attack amongst your four options C J Jad c chinai c sandep c and madav VRA C Anand no t t s Krishna C Arya C very good Tanya SRI Vidya Rishi everybody has given me this correct answer which is c b c c is the correct answer why can you see here two plus I groups are attached to this uh negatively see here the uh double bond will attack the positive species right because double bond is electron rich now these plus I inductive effect groups these alkal groups are increasing the electron density on the double bond isn't it so this will definitely show the highest electrophilic addition reaction because we want the alken to be very very electron Rich so that it can attack the electropositive species over here isn't it take I'm moving on to the next question over here this is the question can you see here toine when reacts with three chlorine in presence of light what will be happen then Bine in the presence of the Catalyst Fe then zinc HCL what will happen try to answer this question this is very easy question should not take you more than 30 seconds just as you look at the question you should know the answer these easy kind of reactions will also come in your exam okay mark my words mark my words c d try to answer this so sudha has given me D answer and MAV vwar has given me me c Rishi has given me D sudha is saying D option ma'am CH let us try to see what can be the answer so in the first we reacting it with three chlorines in pres of presence of light so but whenever light is given the addition will always happen at the benzilic carbon understand this where the benzilic carbon right so I'm drawing the Benzene ring over here once again right and here was the ch3 now one by one each hydrogen will be replaced by three chlorines because we're using three chlorine so all the three hydrogens gets replaced by three chlorines First Step second step is the addition of br2 in presence of Fe now this addition is what kind this is electrophilic aromatic substitution why because we are using bromine in presence of a catalyst so this is EAS reaction now but where the bromine will get added now can you see the ccl3 makes the ring this ccl3 group is actually meta dirting Group which group is this metad dting why because chlorine here will act as a uh electron withdrawing group so this becomes a metad dirting group so at the meta position your bromine will be added here will be the ccl3 right and here this is the Benzene ring at the Met this is the meta position respected to this here the bromine will be added next is in next you are doing in zinc and HCL zinc and HCL does what but it does nothing it is a reduction reaction it reduces what will it reduce now so here all these chlorines will get reduced by the zinc ET is a very very good reagent to reduce okay this bromine won't be a ected why because this bromine B is in conjugation with this Benzene ring isn't it so this Bond bromine and carbon bond gets a double bond character which is very difficult to break hence this bromine will not leave the ring instead what will happen this structure will be formed here it will come ch3 and here will be here was your Benzene ring and here the bromine will be present so this is the final structure so the correct answer is B the correct answer is B so whoever gave the correct answer with bromine at the metap position with respect to your methy group is correct so the correct answer is B understood every point you must understand here understand what is the function of zinc xcl is a very very good reducing reagent next is coming over here now look at this over here which of the following is the correct structure of the organic product of the above reaction again very simple reaction don't get afraid of it it is again a very simple reaction you have taken asile chloride in presence of alcl3 what kind of reaction is that ay chloride in presence of alcl3 answer me in the chat box what is the name reaction what is the name of this name reaction yes buta try to tell me what what is the name of this which is a very very important name reaction what name reaction is this I want that answer before what do you think is the correct option over here what name reaction is this what name reaction is this I've received so many a b d answers freed craft Bish has given me very good bhavish the only person who has given me the correct answer and then e biology is also given me very good very very good bhavish this is fredal craft as silation reaction isn't it now let us see what will be the product over here so this is the reaction given to you right so what is happening this alcl3 will take away the chlorine from here H and we leave behind a AUM Ion with a positive charge on it now you have two kinds of benzene Rings right two Benzene rings with here the nitrogen is there here the carbonal group is there right what is the function of carbon group this is a electron withdrawing Group which means it is decreasing electron density in the ring while this nitrogen is doing what through resonance it is increasing electron density which means this is a electron releasing group right and what is the attacking species here it is a positively charged ayon so positive charge will be most attracted to the most electron Rich ring so it will attack the this ring B will attack from the par position this positive charge will attack right so on this ring buta this asile atom will be attacked at the par position because par is the least hindered most stable position least thally hindered so the correct answer here is D if you gave d as the correct answer you are absolutely right the correct answer is d everybody understood this yes game over Tamil very very good very very good okay next is again a freal craft reactions such kind of questions you should expect because these have been repeated in like past 5 years at least three times this question has been asked in some of the other format which will undergo Freel craft alkylation reaction which will undergo freedel craft alkylation reaction try to answer this question for me freal craft alation so understand what happens in freal craft you take a Benzene ring and you try to react that Benzene ring with RCL in presence of al3 isn't it this is the fcft alkylation reaction now what happens this al3 being the Lis acid abstracts the CL minus and we are left with the are positive C which means a carbo is getting formed so now we know that our Benzene ring is electron Rich species so it will attack this positive charge and alkylated Benzene is formed as a product okay so what do we look for in this reaction that we want this Benzene to be very very electron Rich species right we want the Benzene ring to be electron Rich only then the freal craft alation will occur understanding this point so can you see this NO2 group over here NO2 is what kind of group this is a electron withdrawing group hence this is making your Benzene ring less electron Rich so this will not show your FAL graft reaction because we want benzing to be electron Rich similarly here this is showing plus I effect which means it is increasing electron density so this will definitely show so second will show now look at Co this is once again a electron withdrawing group right because Co is once again a multiple bonded group right so this is electron withdrawing group hence will not show freal craft this o over here will once again through resonance increase the electron density in the ring so this will definitely show so second and fourth will show so the correct answer is B understood everybody everybody understood B yes B for IIT Bombay this is the best response have received yet by Anand yes B for I Bombay right take understood I'm moving on to the next question over here try to answer this question the question is you have been given a Benzene ring an alken and alkine in the same species and you want to react it with two kind of re agents so what will be the two kinds of products form the first is H2 in presence of Paladium and carbon next is H2 in presence of lindler catalyst so what can be the answer to this question what can be the answer to this question try to form this in your notebooks and then I'll tell you try to perform this in your notebooks okay so I'll tell you how to approach this question so H2 Paladium and carbon is a very very strong reducing agent what does it do it does hydrogenation right it does hydrogenation so it will do hydrogenation irrespective of whatever is happening so what will happen in the first case in presence of H2 Paladium and carbon but this is your Benzene ring given to you right this is the Benzene ring here all the atoms will be converted into alkanes here it will be ch2 ch2 ch2 and ch3 so what has happened all the double bonds and the triple bonds have been converted into single Bond so this has happened with H2 padum carbon right but what happens with lindler Catalyst what happens with lindler this is a selective Catalyst this does Selective hydrogenation I'm writing it here selective Hydro hydrogenation understand these things take very very important so here only the alkine here will get the hydrogenation part so the answer here would look like this is a Benzene ring so this double bond will remain same while your this will again convert into a into a double bond like this so what is happening it has only converted your triple bond into a double bond that is it and your double bond over here remains untouched why because this is a selective hydrogenation Catalyst take this does Selective hydrogenation understood everybody take understood now moving on to the next question now answer this question which of the following are deactivating deactivating but orthop par directing during electrophilic aromatic substitution reaction so which of these are for first of all deactivating but orop parad dting groups try to answer this question can there can be multiple answers there are more than one answer I'm telling you more than one option is correct try to answer this now more than one option is correct a definitely a a is definitely chlorine every hallogen atom every hogen atom is a deactivating group isn't it but it is orop parad directing so a is correct next next but I will also tell you even this your alen group over here your alken group as I told you this is a deactivating group right we just discussed our very first example was this but this is once again orthop parad dting group even though this is deactivating okay this can once again take electrons to itself right but it is deactivating but still it is what or parad dting group so two answers are A and C so keep this in your mind once again you are learning one new thing in each and every question understood now try to answer this question for me in which of the following reactant followings reactant and product are correctly matched reactant and product are correctly matched reactant and product are correctly matched try to answer this question for me everybody everybody try to answer this question in which of the following see first is br2 in presence of ccl4 what will happen will BR get attached over here no in presence of ccl4 the addition happens on the alken right so this is the wrong product in presence of ccl4 the correct product will be this is your Benzene rink and here at the double bond 2 br will be added right so this is is the correct this is not the correct product so a is wrong let us look at B product br2 in presence of febr3 see I am actually explaining to you all the reagents in just one question so again please stay very alert during this question Fe in presence of catalyst so wherever a catalyst is given the bromine will definitely add on the Benzene ring now what kind of group this is this is a electron withdrawing group hence the addition will happen at the meta position so this is absolutely correct reaction right now look at option number c here once again br2 in presence of catalyst so once again it will attack on the Benzene ring only but can you see that there are two Benzene Rings given one is here the other is attached to the nhph group right and we know and we know what that this couble Bond O is a electron withdrawal group while the nitrogen present over here is a electron relasing group right and br2 br2 here is acting as a electrophile isn't it because on your aromatic ring electrophilic aromatic substitution happen so this positive will be attracted more to the more electron Rich rink isn't it so this br will definitely be added at the parap position of this nitrogen added Benzene ring so this is again a correct reaction given next look at this cl2 here chlorine is already there chlorobenzene chlorine in presence of alcl3 again this is what this is a uh in the presence of catalyst your attacking chlorine right so chlorine is orop parad directing so two orop parab results are formed this is also a correct reaction so the only wrong reaction is the first rest all are correct reactions is this this thing clear to everybody is this thing clear take every reaction is correct except the first CH next question choose the major product of the reaction choose the major product of the reaction try to answer this question for me choose the major product of this reaction so here once again what is given to you a bulky base can you see this is a bulky base and what have I told you every time about bulky base whenever a bulky base is there the major product will be the elimination product elimination product will be the major product take the elimination product will be the major product now try to answer this question for me take so whoever is answered B is absolutely right absolutely right the correct answer is B actually the chlorine with go over from here and the hydrogen will leave from here right very very good the correct answer is B take what is happening according to the SF rule the major substituted stable alkan is formed whenever a very very bulky base is used so chlorine and hydrogen will get lost and alken product will be formed which will be the major product take I'm moving on to the next question now the next is this what is true about the following reaction so you've been given one reaction you have to tell me what is true about this reaction yes what is true about this reaction try to answer this this is once again a very very bulky base given isn't it bulky base bulky base is given right what will happen in this case a bulky base is given so I told you the two products are possible here the two products are in bu base there will definitely be a elimination reaction and in elimination the most stable alken is form so this br will be gone and the hydrogen from where will go from here the hydrogen will go so that you get a conjugated alkine as a product conjugated stable stable alen will be formed right so this is the first Aline but but but this is the major product but as a side product you can also get you can also get a substituted product so here was this this alen was there and here this BR can replace by o C2 H5 molecule so this substituted product can also be there substituted product will dominates is the correct answer so the correct option is D everybody understood the correct option is D take next question on your screen try to answer this again we have discussed a similar question cl2 in presence of ccl4 will form what stereochemistry of the product will it form a diomer a miso product a rasmic mixture or pure enantiomers try to answer this question for me what should be the answer what should be the answer B option yes try to solve this question for me guys try to solve this question for me B option miso product no B here what will happen see this cl2 in of CCL for first of all understand this is a anti addition reaction what kind of addition on your double bond anti- addition will happen so this is your cyclopentane ring so here the ch3 is already present for example the ch3 is present on the front side right here the ch3 is present ch3 now what can happen here one chlorine will be added over here one chlorine gets added over here the other chlorine added on the back side right so this is one product the other product is its Mirror Image the mirror image can be like this the ch3 can be at the back the chlorine at the front right and here here the chlorine at the back right so what have we got we have got a rasmic mixture rasmic mixture of products rasmic mixture we are getting we're getting both we're getting both kind of products in equal quantity so we are actually getting a rasmic mixture why because the addition is anti addition is anti addition understood everybody everybody understood yes the correct answer is C absolutely correct try to answer this question for me the major product of the following reaction so you've been this is a very important reaction okay try to tell me what name reaction is this this is a alkal halide given to you in the presence of Na ether right what name reaction is this who can answer this question for me what name reaction is this what name reaction is this who can answer this question question for me who can answer this Wood's very good bhavish bhavish extremely good wow this is Wood's reaction right inwards reaction what happens so this is a very very extremely good example of intr molecular Woods reaction what kind intra molecular mole ular Woods Woods reaction this is a very very good example of intramolecular Woods reaction so what will happen from both the ends this bromine and this chlorine will go away okay now count the number of carbon atoms one 2 3 4 five five carbonate atom and these will join together what will happen join together right that's what happens in woods reaction right two alkal chlorides join together to form a Al here what is happening same in the same alkal chloride you have two uh two halogens attached so you will leave them and instead join it to form a ring so first a five membered ring will be formed like this take and on the second number of carbon there is a methy group now what now you're reacting it with br2 in the presence of H new okay so now where the br2 will add br2 will add over here one bromine will be added over here now is the alcoholic K what does alcoholic Koh do this do elimination reaction isn't it elimination which means it forms very very stable alen so it forms the more stable Aline which will be the B option b option so this br will go away from here the H will go away from here and the alen will be formed here okay so the correct answer is B every everybody understood this was a very fine example of woods reaction isn't it very very fine example of woods reaction correct this we have done now this is the question which is once again your homework question product in the following reaction the reaction is given to you ch3 MGI in the presence of hch take this is once again your homework question so what will happen this is a grignard reagent in the presence of hch what do you think should be the product right so here I say let me know in the comment section below what do you think the answer this is a very simple question and you should expect these questions in your J exam take now it is time to invite Nam ma'am uh for your math session one second okay and before I go away again for your J Mains preparation if you have any doubts anything regarding anything if you have any doubts and you want to send me this question screenshot you're not understanding how the bonds are moving why are the questions so you can reach out to me on my personal telegram Channel which is nandini J you can check it out it's a relatively new telegram Channel and where I answer to all your queries all your doubts except when I'm sleeping otherwise I'm online for most of the time so you can contact me on this till your J 2024 preparation till your j24 exam but if you are someone watching for the J 2025 preparation as well then we can connect on this for the year long of your preparation now let us welcome Nam Ma okay hey Ninja Warriors are we ready for mathematics section thank you uh dges ma uh is your login all right all right mat a yes all right shall we start oh oh oh yes yes yes yes yes all right all right I guess uh ragul is writing tomorrow I have J exam so you have to be super duper chilled out relaxed stressfree and give it your best shot to everyone who are having paper tomorrow and we'll be discussing question paper tomorrow as well all right all right all right so the very first question without wasting any further time yes yes yes all right so the very first question is from vector and 3D geometry right and the question says so so this is a typical question wherein cross product and Dot product is involved so you might see a question from vector and 3D geometry wherein dot product and cross product are involved in single question so this is that type that's why I kept this question over here most predictable question right okay so clearly A and C are having magnitude one B Vector is having magnitude 4 and B Cross C magnitude is otk 15 so I can write B I can write C and I can write sin Alpha is equal tok5 why because B Cross C is what b c sin Alpha mind it understand that this is modulus no need for NCAP Vector okay so B is 4 C is 1 so can you see sin Alpha from here is < TK 15 by 4 right okay now this is given to us b - 2 Cal to Lambda a we are looking for the value of Lambda okay so if I Square this if I Square this I will have b square 4 C square and a DOT product will also be there and in dot product I will want I would want cos Alpha right right b c angle between B and C is Alpha in in in cross product I have sin Alpha but in dot product I would need cos Alpha so from here cos Alpha will be square < T of 1 - 15 by 16 so this will be 1X 4 this this is your cos Alpha right clear now Square both the sides so b² + 4 c² - 4 B do c is = to Lambda Square a² b square is what 4 16 4 c² is 4 - 4 B do c is what b b which which is 4 C which is 1 and cos Alpha so this is b c cos Alpha cos Alpha is what 1X 4 is equal to Lambda squ a square is 1 square now you've got the value of Lambda right so 4 4 cancel 16 + 4 20 - 4 is equal to Lambda Square so Lambda square is equal 16 do I need to tell you that option D is absolutely right okay all right yes option D clearcut thumbs up give me a big thumbs up if you've understood this question the very good question I would say moderate question I would say and easy question obviously because Dot and cross both are involved okay all right and now we are having next question which is from differential equation which is one of the high weightage uh topic right so differential equation may I have uh I have actually heard a lot of problems you guys face in differential equation in identification of the differential equation that ma'am we don't know that how what kind of differential equation is this so what we need to do is we need to go line by line sequence by sequence so first of all by first glance by first look just try to understand that what is it so by first look we are not able to understand in this particular question so right now start counting is this variable separable am I able to uh uh separate the variable y on the left hand side x on the right hand side no no no I can't do that is this homogeneous no actually it is not homogeneous as well Dy by DX I won't be able to represent Dy by DX completely in terms of Y byx no I am not able to do that so this is not homogeneous as well what about linear differential equation so this looks like it is not a differential equation as well this doesn't look like a differential linear differential as well what does that mean what does that mean if everything fails if everything fails we have we still have one type of differential equation left which is again linear differential equation but in X so reverse it out figure out DX by Dy so if I talk about linear differential equation in X right and linear differential equation in y so in y it is we know that Dy by DX plus p y is equal to Q DX by Dy + PX = Q here p and Q are function of X and here p and Q are function of Y so let us try doing this way X x + 2 y Cub upon y right so this will be DX by Dy is equals to can you see so sorry minus sign minus sign we are having - 1 by y * X can you see yes this is your p is equals to can you see 2 y² yes this is lde in X right this is your LD in X how to solve your lde in X so let me just show you how to solve it all right yes so how to solve this linear differential equation in X so this is x * integrating factor is equals to integral of Q * integrating Factor Dy + C this is the solution so x * now what is integrating Factor let's calculate integrating factor which will be e to^ integral of minus 1 by y Dy so this is e^ minus log Y and this is 1 by y 1 by Y is your integrating factor x x * 1 by y right which is equals to Q what is q q q q q 2 y s so 2 Y 2 into 1 by y Dy integral + C okay so now X by y this is going to be 2 y integration so X by Y is = to y² + C is your final answer let's match it out so X = Y Cub + c y option b is absolutely right yes you guys are saying D why D do we uh do we have a negative sign minus c y y minus c y no it's option b right option b why are you guys saying d d is incorrect right because y will go and multiply it with this Cy okay why are you saying B somebody's saying option A somebody's saying option X is oops oops oops sorry it is X = to X = to Y CU + C Y yes sorry sorry sorry a a a a a a my bad my bad my bad sorry my bad yes yes yes option A all right all right so this is xal to yq plus c y all right now we are having another question from differential equation question number three differential equation is again one of the high weightage topics so yes this is your differential equation so I told you by first glance what would you do that okay if you are able to recognize or identify it it's fine but if not then go by line by line step by step number one is it variable separable so try to just shift y uh y things towards the left hand side terms related to Y and try to shift terms related to X so this is 1 by one this is also X yes I am able to separate it out the very first step is very separable and I'm able to do that so this is a variable separable right integration integration this integration is very easy you just have to multiply two and divide by two and this will be T So this whole will be DT so this is 1x 2 integration of DT by T correct left hand side is sort it what about right hand side okay so right hand side looks a little bit shaky let's just first of all 1 + x² let's Club them together so this is 1x x 1 + x² + x upon 1 + x² 1 + x² this is your expression DX can you see this is going to get canceled out right so first 1X X integration is log of mod of X Plus can you see x x is going to get cancel out so only 1 upon 1 so this is tan inverse X plus constant and this is nothing but 1X 2 log of mod of T what is T so this is 1 + T sare right 1 plus sorry y squ okay this is your answer now we just need to write down this particular solution nicely so that we can just identify the correct answer 1 + y sare half will become root so root 1 + y s this will be equal to log of mod x and the C can be written down as log C so log X into c x into C and this tan inverse X can be written down as log of e to^ tan inverse X so this is e to^ tan inverse X log log canceled out and we are left with 1 + y sare root is equal to x c tan inverse X can you can I see yes absolutely right option C is absolutely correct so they can give you variable separable they can give you linear differential equation and all Al they can give you homogeneous equation as well right they can give you homogeneous equation so one more question so three types of question they can give you in your examination right okay so this is homogeneous right uh so I have revealed beforehand only I should have waited for it okay so can you see Dy by DX is equal to Y by X and this log y - log X this type of question was there in January attempt by the way so this is log of Y by x + 1 right so can you see Dy by DX completely has been written in form of Y byx that means this is a homogeneous equation and in homogeneous equation how what do we do we put Y is equal to CX y v y is equal to VX right so Dy by DX is equals to V + x DV by DX right so first of all Dy by DX let's replace V + x DV by DX is equal to Y by X is V this is log of V + 1 now what do we do can you see the moment I multiply this with v this V and this V canceled out and this is again now it is becoming variable separable so this is DV upon V log V is equal to DX upon X integral integral if I put this is equals to T So DV by DT will become DT DV by V will become will become so this is the integration will become 1 BYT integration will become log of T what is T so log of log of V log of log of Y byx directly you can write in one go log of log of YX is equal to log of x c can you see log of Y by X log log cancel is it = x * C this is your final solution log of Y by x = x c log of Y byx is equal to x c which answer is correct so log of Y by X is = to CX so none of these none of this is correct no no no no no no none of these is correct y b log of Y byx is equals to no no no this is incorrect this is also incorrect log of x by this is also incorrect option D what are you guys doing enra enra you have taught me this enra pandra yes log d d d yes log of Y by y x is equal to x c okay yes now we have a question this question we are having okay f of x is equals to integral of this quantity all right f of x is equals to integral of this quantity and we are looking for actually we are looking for a relation between F triple Dash F Double Dash and F Dash Okay triple Dash Double Dash and single Dash somebody is saying why not B why not B Because log Y byx is equal to x c this is Cy c y and CX right hand side is completely different shata I hope that this is clear okay all right FX equal 0 0 to X like this now we we are looking for fle dash so what we will do is first of all we'll figure out FX right we have to be extremely patient in this question first figure out FX then F Dash then F Double Dash then F triple Dash let's do that okay let's do that okay all right so first of all f of x let us write it nicely so can you see this is T sin x and this is DT this is DT so sin x can be taken out this is sin x integration 0 to x t DT correct minus what about this this will be T sin T So integral of T sin T DT 0 to T so we need to integrate this so this is sin x x² by 2 minus now we'll be applying by Parts over here no no no no leits rule no leits rule over here we have to apply Barts why we can't apply leits rule because of this sinx involvement yes because of the sinx I can't apply it okay right uh I mean you can apply lebit rule over here but it will make things complicated we have to apply uh product rule while differentiating right so that's why this is the best method Now by Parts first function constant okay integration of second function fun so that will be Min - cos T minus integration of derivative of first function and integration of so this is plus cos T and this is DT right so this is x² by 2 sin x and this is what plus we are going to have X cos x and we are going to have plus sin x right integrate so we are going to have plus sin x this is your FX so FX is coming out as x² by 2 sin x x cos x + sin x so this is your FX can you see x² by 2 sin X+ X cos x plus sin x right this is your FX 0 to X haha 0 to X now f- X who is going to tell me so in this we have to apply the product rule so first of all let's differentiate this so it is going to become X sin x is constant plus x² by 2 as it is this will be cos x so derivative of this is this now X cos x again product rule we we are going to apply so this is uh okay so cos x is constant derivative of x is 1 minus this time we are going to keep x constant and this will become plus sinx and what about this so derivative of sin x is what cos x so can you see product rule product rule and that's it so if we solve it what do we get x² by 2 cos x x sin x x and 2x sin x and 2 cos x this is your fouble tip right yes we have to cancel this out so this is gone right so this is X sin x this is min - x sin x right because cos x derivative is What minus sin x right that's it so this is x² cos x + 2 cos x f- x is here what about f- X now we have to differentiate it so x² by 2 constant derivative of cos x is - sin X product rule I'm applying plus this time cos x is constant and we have X over here then we have - 2 sin x product rule and this so this is your f- X now what about f- X f-x okay so again we have to apply we have to differentiate this yes we have to differentiate this so Min - x² so can you see fouble Dash what is fouble Dash X cos x - 2 sin x x cos x - 2 sin x yes and minus this yes this is my now fle dash okay fle Dash so x - x² sin x I have to differentiate it so first of all minus of X sin x constant okay I have differentiated this and kept this constant minus this time what I'm going to do is - x² by 2 is constant derivative of sin x is what cos x correct this time correct now cos x derivative of cos x is what Min - sin x and what about this - 2 sin x so derivative is - 2 cos x - 2 cos x this is your f- X now we just have to so this is my third equation this is my second equation and and this is my f-x first equation right now can you see in each and almost each and every option A and B at least A and B looks a little bit similar and C is saying F triple Dash and F Dash so can I check this option C F triple Dash and f f Dash so triple Dash is this and what about F Dash so F Dash is this x² by 2 cos x x² by 2 cos x can you see that if I add these two quantities this term is getting cancel out now - sin x will be there okay - x sin x will be there and also 2 cos x can you see 2 cos x and 2 cos x will also get canceled out so we are going to left with what - x sin x - sin x will be my answer right so if I add fle Dash and F Dash cos x - 2x sin x will be left out so cos x minus cos x - cos x - 2x sin x okay - x sin x all right did I do anything uh wrong over here it's X cos x yes it's X cos x somewhere here fle Dash I am getting Min - x sin sin x and we are having uh - x² cos x right we are having cos x - x sin x oops yeah this is X cos x oh oh oh this is X cos x so X cos x if I have to differentiate I have to apply product rule so if x is constant this is going to get min - x sin x yes now it's fine and if cos X is kept constant so this is going to give me plus cos x now if you add you you going to get option C as the correct answer now it's fine yes can you see if you add these two quantities will get canceled out cos x and cos x is also gone right and uh you are left with what and you are left with what can you see uh yeah now it's absolutely fine right now it's absolute cos x will be left out and - 2x sin x will be left out okay so I missed this X over here I considered this cos x okay all right yes so we have to differentiate this is a little bit lengthier question sometimes it happens that so FX is given first you have to calculate FX properly we have calculated FX and and it took us some time in that then we have differentiated then twice differentiation then Thrice differentiation so we will leave this type of question because it will irritate us it will irritate us it will actually give us negative marking sometimes okay and it will consume our time as well so we are going to leave such questions okay because can you see first we have to figure out FX F dash fle dash Double Dash okay all right so we are not going to do that but we have done it anyway all right now we are having next question okay next question is from definite integration which is one of the high weightage topics so by the moment I look at it now the moment I look at it I feel that yes this is a problem of King's property I have to apply King's property in that but something is actually uh not clicking in this question because of the x square so what I'll be doing is I'll be converting this I'll be putting x² is equal to T let's do that yes so x² is = T 2x DX will be equals to DT correct so this is X DX so can you see X DX will be DT by 2 s of what is this x² so sin T whole upon this is also sin t + S of log of 6 - t right now wait a minute wait a minute wait a minute wait a minute wait a minute wait a minute over here you have to change the limit as well so x² is = dut X = root log 2 so this is log of 2 this is log of 3 now can you see this question is clearly from that shortcut the moment you apply King's property and add numerator denominator will get cancelled out right so the answer will be what half will remain as it is the answer of this integration is upper limit minus lower limit by 2 that's it you are done so this is what 1X 2 1X 2 will be 1X 4 log of 3x2 where is the correct answer option A all right all right all right yes so I love this question even though it was not the direct shortcut application but a little bit tweaking just a normal substitution converted the integration and then directly we can apply the shortcut over here I have told you this shortcut okay all right all right yes B minus a by 2 yes Hass absolutely right okay now we have another question because King's property is n's favorite property you all know that okay all right so now can you see this integration this integration is I = to a b can you see a plus B minus the moment we see that it is also we we can understand this this is from King's property so let us write it down f of x DX F of a + f of a + b - x now the moment you replace X by a + b - x what will happen what will happen over here so this is yes this is your I so this is a b and this will become F of wait a minute is this yes sorry this is your X this is not a this is X this is a misprint yes now apply the king's property so this x will be replaced by a + B- X this will be F of a + B- X and what about this this will become f of x integral DX right I + I if you add these two quantities together 2 I will be equal to the moment can you see the numerator Den denominator denominator same so you can add if you add you are going to get A2 b 1 DX basically this is a shortcut question that's what I'm saying so this the answer of this integral is b - A by 2 which is equals to what 10 so can you see value of B minus a is nothing but 20 direct direct direct direct direct direct question okay direct question with the shortcut all right yes cool Perfect all right now we have another question again from uh integration okay uh indefinite integration this is okay so F of something is given as x + 2 right and we are looking for integration of FX so how to approach such questions think about it because this thinking is not that difficult thinking of this question is not that difficult right so first of all we have to put this 3x - 4 upon 3x + 4 is = T okay all right so if this is T so this will become something else so we need to uh figure out the value of X in terms of T by using component and dividendo we can do that or by using cross multiplication also we can do that okay so this is 3xt + 4T x x terms together so this will become 3x t - 1 and this is - 4 - 4T right so now can you see the value of x is what 4 by 3 and this is 1 + T upon 1 - t is = to X x value of x in terms of T so now can I replace this so this is f of T because this is T right so this is f of T is equals to now x x this time will get replaced by what x will be replaced by uh uh 4X 3 1 + T upon 1 - t then we have + 2 right so that's how we figure out F of t or f of x 4x3 now one 1- T can be taken LCM and it will become 1 + T and we are going to have 1 + T um sorry this is going to be this is going to become 4 + 4 T 3 - 3 t + 2 now take LCM 3 - 3T 4 + 4 t + 3 2 6 - 6t so this is going to become 10 - uh 10 - 2 T and 3 * 1 - t or rather if I take two common 2x 3 1 5 - t 5 - t upon 1 - t this is your F of T finally you've got your F of T finally you've got your F of T is equal to 2x3 5 - t upon 1us T okay now let us just write down f of x clearly is what 2x 3 and we are going to have 5 - x upon 1 - x how to do that so can I 4 can I write 4 + 1 - x 1 - x upon 1 - x so this is going to be 2x3 and we are having for now integrate integration of FX integrate so this is for log of 1 - x upon we are having -1 and then we are having 2x 3 X Plus constant okay all right and uh this is going to be okay now let's match the answer 2x 3x can you see this is 2x 3 x is matching 4 2 - 8x3 log of 1 - x option b is absolutely matching yes all right clear I hope that this is fine right yes okay so this question is done and dusted nicely another oh this is my favorite question this is my favorite question this is actually my favorite question I call it coutur putur coutur putur kurur putur means manipulation okay kurur putor is manipulation C putor is manipulation okay all right so whenever we see such questions now we have that mental blockage that we are not able to do even First Step ma'am because all we know is substitution or applying some formula or understanding some type but this is not making any sense this type of question the very first step the very first step we are not able to figure out okay yes hassini so cter putter means manipulation so we have to manipulate this expression in such a way that we have to take something common from the denominator and plug in in the numerator so that substitution method works so substitution method we have to make it work so what we will do is we will take maximum power common from the denominator and push it in numerator understand take maximum power power from the denominator common and push it in numerator how will we do that let me just show you so maximum power I can see X ra to^ 7 can be taken common so x to^ 7 but there is a square so x to^ 7 Square can be taken out from this x^ 7 if I take common so it will become x^ - 5 then x^ - 7 then this is + 2 right and this is 5 x^ 8 + 7 x^ 6 okay it might create a some confusion because the powers are becoming negative over here now x to^ 14 so this will I will delete this from here this is x^ -4 now push it inside the numerator 5x -4 + 8 8 will be uh what - 6 and 6 - 14 is what so 7 x^ - 8 DX upon we are having x^ - 5 + x^ - 7 + 2 and we are having a square over now put this denominator is equals to T let's put the denominator x^ - 5 + x^ - 7 + 2 is = t - 5 x^ - 6 + - 7 x^ - 8 right DX is equal to DT can you see if I take minus sign common this will become minus DT so this is becoming minus DT upon this is your t² integral so this is going to become t^ minus 2 so 1 by t + C is your final answer what is T this so 1 upon x^ - 5 + x^ - 7 + 2 + constant is your final answer so they are saying that FX is this that means this is your FX right f of0 is z they are saying so why don't we multiply X to^ 7 over here in numerator denominator what do I get so my FX will become FX will become If I multiply X to^ 7 over here and here so this is x to^ 7 upon x to^ 2 + 1 + 2 x^ 7 plus constant F of 0 is given as has Zer so the moment I put zero over here so this is zero that means constant is zero okay now they are asking F of 1 so for f of 1 1 upon 1 1X 4 is the correct answer right so 1X 4 is given in the option b this is your uh F of 1 minus missing no no no minus sign we have removed why 1 by T sare integration is - 1 by T 1 minus was already there that that's why it is a plus is a plus sign right okay all right good question C put understood sometimes you see that this type of question all right where we have to extract something out of denominator and push it in the numerator okay all right now we are having a question from application of derivative two types of question are extremely extremely highly frequent I would say uh in previous year jman we have seen from application of derivatives especially first increasing decreasing second maximum Minima okay so FX is equals to this is given as a decreasing function so this cubic equation is a decreasing function in this particular interval so uh that means f-x is less than 0 for X belongs to -2 to min-1 okay and one more thing one more information they are giving Which is this large possible interval that means Beyond this so basically F Dash is less than zero for this only for this okay only in this interval it is not negative Beyond this interval so we have to be very strict about it okay now f-x 3 2 are 6 x² + 9 18 9 2 are 18 x + Lambda should be less than zero yes so uh now if I take 6 common what do I get x² + 3x + Lambda by 6 less than 0o so this this equ this quadratic expression should be less than zero for this interval what does that mean that simply means that uh this particular interval right so basically the product of roots product of roots so uh basically what they are saying is you should get this as your final answer x + 2 x - uh x + 1 less than 0 as per this this should be your quadratic inequality so can you see that product of root is what two so Lambda by 6 Lambda by 6 should be equal to 2 so the value of Lambda is coming out as 12 easy peasy easy peasy value of Lambda is coming out as 12 all right yes okay because the quadratic equation they are suggesting is this because x + 2 x - 1 less than 0 will give me this interval now method of interval say so option A is absolutely right I hope that this is clear all right we have another question from application of derivatives by the way this function has so we need to figure out this function as what uh so this is a cubic equation which is having either uh repeat last step okay so basically it's not the last step in the beginning only this is a cubic equation right now it is a decreasing function in this interval that means f-x if you solve f-x less than zero it should give you this as your answer so f-x is coming as quadratic okay now think about it think about it no think about it that how do you get this interval tell me - 2 -1 + - plus decreasing how do you get this interval who is going to tell me obviously by solving this is reverse method of interval can't you just uh can't you just make quadratic equation if I give you this can't you do that if I give you this is reverse wav cve or reverse method of interval so this should be your expression so you just need to compare that's it okay all right now we are having this it's a cubic equation we are looking whether it is having two Maxima two Maxima which cubic equation is having two Maxima no two Minima no either it has one Maxima one Minima like this H and also the leading coefficient is positive or it is having no Maxima and no Minima that's what we need to figure out so for that f- x 3 2 6 x² - 6 x - 12 that's it 6 if I take common x² - x - 2 x - 2 2 x + 1 * 6 okay so one Maxima and one Minima clearly clearly one Maxima and one Minima easy peasy easy peasy question right okay so one Maxima and one Minima and they are not asking us at what point we are having Maxima at what point Minima so we don't have to solve it okay all right FX now continuity and differentiability we have another question right right right continuity and differenti ility questions are pretty much doable usually so let's start doing it first of all let's talk about the continuity so clearly I have to check the continuity of function at F of 1 at X = to 1 what is f of 1 who is going to tell me x = 1 the moment I put one over here 1 - 1 is zero so the value is 1 what about 1 plus that means when my when my X is greater than one so the moment I put one over here 1 + log so this is 1 + log 1 log 1 is zero so left hand uh limit what about 1us 1 minus will give me same so that means yes function is continuous right at least my function is continuous so the C option is rejected okay now we just need to check about the differentiability okay so for differentiability what do I need to do is uh I need to do this f-x equal to differentiate this one derivative is 0 - 1 upon Square < TK of 1 - x² and we are having 2x correct correct the moment I put x equal to 1 what do I get so the moment I put X oh the moment I put xal to 1 what do I get the moment I put x equal to 1 non differentiable it is clearly visible okay or we can do what we can do is f d 1 + H is equal to F of 1 + H minus F of 1 upon H this is right hand derivative Now 1 + H for 1 + H we have to go over here so 1 + log 1 upon 1 + H okay X will be replaced minus F of 1 we can know one this is H so can you see so 1 upon 1 + H H is zero so log 1 is 0 by H so this is what yes so what we can do is this is minus of log 1 + H upon H so this is minus one so right hand derivative we can calculate this is minus1 what about left hand derivative f- 1 - H = to F of 1 - H - F of 1 will be this what about 1 - H for 1 - H we have to plug in the value over here let me Calculate 1 - H from here so 1 - H if I want to replace it will become what 1 - H let me just replace this from here so 1 - H so this is 1 - s < TK 1 - 1 - h² 1 and 1 are canceled out minus minus canceled out the moment you put put H is equals to 0 now you are going to get a zero o o oh so that means 0 by H so non- differentiable non- differentiable option b yes continuous but nondifferentiable right okay all right so continuous but non now we have a question from determinants determinants you all know matrices and determinant is one of the high weightage topic this is a very good question from the determinants because because it is having some glimpse of permutation and combination as well okay all right so in such questions what we will be doing is in such questions first of all in row number one I will be multiplying 1 * X and uh X Cub y so x x Cube Y and X Cube Z so R1 multiplied by X and R2 with y so R2 multiplied with Y will look like this and R3 with zed so x z Cube y z Cub Z CU + 1 Z this is my first step what did I do I multiplied Row one with X row2 with Y Row three with zed so it will become what I have to divide it as well right in the whole determinant what I have done is I have multiplied with X Y and Z so I have to divide it as well so in row one can you see one extra x one extra y over here one extra Zed over here why did I do that because now I'm going to take from column number one this is a magic from column number one I can take X common so this is X from column number two can you see y can be taken common and from column number three Zed can be taken common so basically I have actually done nothing to this determinant okay but now this determinant looks a little bit symmetric let me show you how so this is X Cub + 1 x cub and X Cub y Cub y Cub + 1 and y Cub Z Cub Z Cub Z Cub + 1 now solving this is very easy now what we can do is we can just uh okay we can actually uh replace Row one go to Row 1 + row2 + Row 3 why because can you see sum of these three values is equals to sum of these three values is equals to some of these three values right yes so now is the time Row one has to change and this Row one row one will become X CU + y CU + z Cub + 1 this is first element the second element will also be the same and third element will also be the same same to same and rest this this this this this this will become same and this is nothing but 11 so this can be taken common right so now this can be can you see this can be taken common and we are left with what if this is taken common so we are left with 1 1 one and row two and Row three are exactly same now we need to open this determinant let's open it if you open this determinant the value will be one let's do that so 1 y Cub + 1 Z Cub + 1 minus times we are having y Cube into this and plus times we are having y CU Z CU minus this minus y so plus y CU Z CU minus uh Z Cub y CU + 1 can you see y Cub Z Cub is getting canceled out from here so here y Cube will be left out okay which will get canceled from here Z Cube will also be left out which will get canc from here so overall one will be left out if you solve it now very easily you can see one so what does that mean that simply means that okay simply what this simply means that my answer my final answer is what x Cub + y CU + z Cub + 1 into determinant value which is 1 is equals to 11 so this is nothing but 10 x CU + y CU + z CU value is nothing but 10 okay X Cub + y CU + z Cub is nothing but 10 and now we are looking for what we are looking for positive integral Solutions so we have to think about it positive integral Solutions right so basically positive integral solution 1 has to be 2 because 2 cube is what 8 8 + 1 + 1 10 or 8 + 1 + 1 10 so 1 has to be two and rest remaining two should be two so either X X is 2 either Y is 2 or either Zed is 2 is the final answer okay all right yes clear have a look understand so what we did what was the catch in row one we have multiplied by X row one row two Row three is multiplied by XY Z respectively and from column 1 column 2 column 3 we have extract the same XY Z so that's why XY Z XY Z getting cancel out and now R1 goes to R1 + R2 + R3 why because there is a symmetry and take this first first thing common and solve the determinant you will get the equation equation solving is extremely intuitive and easy all right another question ready ready for another question are you solving with me do you solve mathematics with me do you uh have pen paper or you just watch it like a web series it's not a web series right because mathematics is all about pen paper practice and passion right four piece of mathematics what are they pen paper practice mathematics yes okay and Mr W is saying ma'am determinant is deleted properties of determinant are deleted but you know in January attempt they have given such questions where properties of determinant a little bit property is involved so we will do it okay all right now we have a question hent and and we are a question from uh system of linear equation where they are saying that not all zero XY Z not all are zero what does that mean not all are zero that means we are having nonzero solution or we are having non-trivial solution for non-trivial solution Delta is zero non-trivial your nonzero solution Delta is zero so it's very easy so 1 we have Min - c minus B then we have minus c 1 and Min - A then we have - A Min - B and 1 determinant is equals to Z so you have to arrange it you have to arrange these three equations in normal right so x - c y - b z so 1 - c - and then we have - CX - plus Yus a z and like that similarly now you just need to just solve it out so this is going to be 1 and minus AB plus we have c c * we have minus C and we are having minus a and we are having minus B so B C + a is equals to 0 correct so the sorry sorry sorry sorry minus a minus B this is minus B and this is minus a now so this is it this is it so 1 - a square is this is 1 - a square let us write it one more time yes now 1 - A S + C and we are having c - c minus a correct and we are having minus B and we are having a + b okay all right so 1 - A S - c² - AB b c minus AB C again - b² equal to 0 so shift everything everything towards the right hand side 1 will be equal to a sare + b² + c² + 2 A B C all right all right option A without any doubt we are having option A as the right answer okay some people are saying C some people are saying a please do not like make that mistake 1 is equal to a square + b square + C square + 2 a c all right yes got it getting it okay all right we have another question from limits chapter yes we are having a question from limits and if you are aware about if you have practiced enough questions on limits now then you'll be able to solve it using the shortcut not exactly the shortcut because this question is clearly saying so this is 0 by0 form right so what we need to do is in such question is rationalize but you know what the moment you rationalize it what do you get so you are going to get rationalized both ways numerator rationalization and denominator rationalization okay so okay let me just write it down so over here you will be having first of all let's talk about numerator this is x² + 8 + 10 - x² upon same thing over here and the denominator rationalizing so uh we are having < TK x² + 3 + < TK 5 - x² and same thing over here right okay can you see that if I talk about a minus B A + B so it will become a square - B sare so this will become x² + 8 - 10 - x² upon again can you see a minus B and A + B so x² + 3 - 5 - x² now two terms two uh two things will be left out which one so these two club together this is left out and these two club together this is also left out so for that we have to put one over here so we need to put one over here let's put one over here 8 + 1 so this is 3 this is also three if you put one over here what will you get < TK 4 is 2 + 2 and can you see now can you see x² + x² so this is 2 x² 8 - 10 is 2 x² + x² 2 x² - 2 so these two are getting canceled out we are going to be left with 4X 6 which is 2x3 yes 2x3 is absolutely right okay all right multiplication yeah lital also you can apply if you want to it's fine okay 2x3 is the correct answer okay all right now we are having a question from binomial theorem binomial theorem we are having a question okay all right binomial theorem so sixth term is 5600 let's calculate sixth term sixth term will be what 8 C5 1 upon x to^ 8 by 3 ra to^ 8 - 5 3 x² log X to the base 10 ra to power we are having five this will be 5600 8 C5 who is going to tell me the value of 8 C5 so 876 upon 3 2 1 multiply can you see this and this are getting canceled out x to^ minus 8 this is x to^ 10 and log of x to the base 10 ra to^ 5 this is nothing but 5600 okay 3 2 are 6 8 7 are 56 and this is so this will be 100 right okay which color should I use 5600 okay correct so this is X ra to^ 2 and we are going to have log of x X the base 10 is = 10 Square all right by looking at the options x = 10 if I put x = 10 if I put so this is 10 Square 100 it will match and X = to 10 so this is RA to^ 5 log 10 to the base 10 will be 1 so x = 10 will satisfy definitely this option C is absolutely right okay all right yes it is very clear very easy question I to love this question because 5600 and 100 this gives us High okay yes 56 5600 100 100 this is a like very God level moment okay all right so we are having a question from quadratic equation right quadratic inequality which says a quadratic inequality is less than zero then a belongs to which interval right so if this is a quadratic inequality and this is less than zero that means they are hinting about downward facing Parabola whose discriminant is less than 0 a less than 0 D less than 0 for a x² + b x + C so what is the leading coefficient so this a + 2 a - 1 less than 0 so - 2 and + one so this is my interval right this is my interval and now discriminant less than 0 to b² - 4 a c less than zero a + 2 if we take out a + 2 + 4 a - 4 less than 0 okay so a + 2 and we are going to have 5 a - 2 less than 0 so - 2 and 2x 5 Okay so 2 and 2x 5 so this is my intergral right so what we need to do is we need to take intersection so final intersection will be what - 2 to 1 - 2 to 2x 5 So my answer would be - 2 to 2x 5 but wait so this is a belongs to but wait wait wait wait wait wait now if leading coefficient is what if leading coefficient is zero and what if this is not a quadratic equation right so so that means if a is equal to -2 then what will happen tell me if a is equal to -2 this will become zero okay if a is equal to-2 this will also become zero that means min-1 is less than 0 that will be a true statement so Min -2 I have to include please keep that in mind they have not given that this should be a quadratic equation so when leading coefficient is variable or leading coefficient is not fixed you have to check that leading coefficient can be zero as well okay I hope that this is clear so that's why minus 2 for a equal to minus 2 you have to check separately okay and now we are having a last question yes we are having a last question let me just ask nabamita ma'am whether she is ready okay all right so this is your complex number equation and and and X is equal to KY you are looking for K so this type of questions actually are very very important very very actually not important important because we have seen such type of questions in previous years right so let's start solving it so first of all 3 to^ 49 x + Iota Y is equals to now can I take root3 common if I take root3 common it will be root3 to^ 100 and if I take root3 common this will become < tk3 by2 plus Iota and this will become 1x2 because root3 is G this is power 100 okay so can you see this is 3^ 49 this is 3 to^ 50 so x + Iota y will be equals to 3 and we are having < tk3 by 2 + Iota by 2 ra to^ 100 okay all right yes now whenever you see some complex number raised to some power which is power is bigger always think about polar and iser's form right so this is nothing but < tk3 by2 can be written down as cos so cos 30° that means < by 6 + Iota sin < by 6 right because sin 30° is half ra to^ 100 three is already there so we can write it this way and since Theta is nothing but e to^ Iota Theta right cos theta plus Iota sin Theta is what e to power Iota sin e to Iota Theta so we can write WR this as 3 e to power Iota and we are having < by 6 ra to^ 100 so this is 3 and this is e to power Iota 50 piun by 3 or we can directly now we can convert so this is cos 50i < by 3 plus Iota sin 50 piun by3 and this is nothing but x + Iota y now you can compare X is what 3 * this Y is what 3 * this this so you are looking for what x by y that means X by y will be what 3 cos 50i by 3 upon 3 sin 50i by 3 3 3 cancel so we are looking for cot my answer will be C 50 < by 3 now C 50 < by 3 C 50i by 3 50 piun by 3 can be written down as 48 + 2 so 48 by 3 will be what tell me 48 by 3 will be 16 so yes C of 16 piun + 2 piun by 3 correct C 16 piun + Theta so basically c 2 piun + Theta will be C Theta so we are looking for c 2 piun by 3 so c 2 piun by 3 will be nothing but minus C < by3 so cot 60° C 60° is 1 byun3 so 1 byun3 with a negative sign is matching with option D all right clear yes okay so we are done and I'm am uh going to hand over to nabam M okay all right chemistry ready we all are chemistry ready yes yes yes done done done okay all right all right all right all right we are having last question sorry sorry sorry we are having last question yes it SK slipped from my mind so this is actually a good question wherein you have to substitute so I'll quickly give you a hint and then we'll jump to the chemistry part okay so how to tackle such questions if it is was X now then we have to just substitute but it is not X unfortunately it is x square it is also x square so how are we going to deal with it how are we going to deal with it so we have to substitute it xal to sin Theta is going to give me the substitution if x is sin Theta DX = cos Theta D Theta right so yes now please convert it out so 1 - x² so 1 - sin square is what cos² Theta ra to^ 3x 2 x² is what sin sare Theta and DX is what who is going to tell me cos Theta D Theta okay limits we'll talk about it later on so square square cos Cube so this is sin Square upon cos Square Theta Theta D Theta now is the time we can convert the limits are we ready for the limits yes so this is 0 to half the moment I put xal to 0 I get Theta 0 the moment I put xal to half I am going to get 30° that is < by 6 0 to < by 6 So Tan sare Theta is nothing but SEC Square thet Theta - 1 and we are going to just plug in the limits SEC squ < by 6 so that would be 2x < tk3 Square isn't it so 4x3 minus this is oh oops oops oops oops oops oops So Tan sare Theta integration we need to figure out right tan Square Theta is integration of SEC Square Theta - 1 D Theta Now integrate so this is tan Theta - Theta and now you have to plug in the limits So Tan < by 6 is what 1 by < tk3 minus you are having > by 6 yes this is your final answer and uh yes so this is equals to K by 6 1X < tk3 - < by 6 is = K by 6 and 6 by < tk3 - < is = to K give me a big thumbs up if you've OD this okay all right cos Cube yes cos Cube but cos Theta will get canceled out now right cos Theta and cos Cube so that's why cos Square I hope that this is clear okay all right CH see you guys all the best for tomorrow's papers e e e e e e e for e e finally yes I'm live hi everybody as I always say namaskar everyone and a very warm V to English Channel quickly go ahead and let me know if I'm Audible and visible I guess I'm Audible and visible now yes I'm hoping that I'm Audible and visible so shall we start with our inorganic and physical chemistry session yes we are just about to do uh you know 12 to 15 questions not more than that because you had already done with the organic part I am just going to go ahead and take you through okay I am just going to go ahead and take you through your uh what do you say it uh physical and inorganic question yes guys am I Audible and visible now has the stream stopped yes we are here right okay all right everybody so I'm going to take about you know 40 minutes or so not more than that so let us start with the session everybody and the first question is right on your screen everyone the first question is one mole of an ideal monoatomic gas is subjected to changes as shown in the graph this is the graph everybody yes the magnitude of the work done by the system or on the system is Dash ju and nearest integer you have to tell given log of two value is given to you and Ln of 10 value is 2.3 this is also this is also written here take all right everybody to tell let us start with the session what do we have to do the first thing first everybody let us read the let us try and understand that what is there in the graph everybody so I'm going to move a little more this side so that you can check the graph all right so look at the graph from 1 to two what is happening pressure is a constant right from one to two when it is going pressure is constant that means what can we say we can say that from 1 to two this is isobaric isn't it this is isobaric process now when the graph is moving from 2 to 3 what do you see the pressure is decreasing if it is pressure decreasing then what can we say can we say that this is isocoric this is going to be isocoric and from 3 to 1 everybody when the graph is moving from 3 to 1 what is happening here here this is going to be isothermal as you can see the temperature here is going to be yes temperature is not changing so much so we're just saying that this is isothermal am I clear everybody does this does this does this does this you know this question very good um maheswari very good very good B yes so now what do we have to do we don't have to do much guys we all have to do only one thing that is work done is going to be W from 1 to 2 plus Yes W from 2 to 3 plus W from 3 to 1 isn't it this is going to be your work done so now everybody for isobaric process what can we say what is going to be the work done everybody yes what is going to be the work done here what is going to be the work done come on everybody yes it is going to be minus P yes V2 - V1 + 0 and for the last one that is 3 2 1 it is going to be Min - P1 V1 Ln of V2 ided by V1 can we write it like this yes now the now the moment we have written this all we have to do is just substitute the values as per the graph whatever is given in the graph we are just going to substitute the value so what will we see we will see here that Min -1 multiplied with see what is written here see pressure is one here isn't it one is the pressure so I'm writing here minus one okay now if you see 40 to 20 yes the volume is changing from 40 to 20 so 40 - 20 if I do yes then what will happen plus here it will be zero yes as it is we are writing it down once again - P1 is going to be min-1 multiplied with 20 natural log of 20 / 40 if we write then this is all that we have to calculate correct everybody this is all that we have to calculate absolutely right everyone yes or no guys yes or no yes why is isotomic because you can see you can see look at the graph and understand what is happening here guys look at the graph and understand what is happening am I Audible and visible guys or have I gone away because I am not able to understand something is going on with the live streaming today something is going on with the live streaming guys am I live am I live here everybody I am yes okay all right sorry I was not able to see myself and I thought that the stream ended here CH no problem okay so 40 - 20 and here is minus that means this is going to be minus 20 yes plus here what is happening here 20 Ln of two can I write yes because 20 - 1 minus so 20 right 20 Ln of 2 Ln 20 by 40 right so here if we solve it it will be 20 ln2 now as for the question we are given some data isn't it what is the data that we are given we given minus 20 plus plus 20 now from Ln can we change it can we convert it to log yes we can what do we have to do we just have to do multiply here 2.3 and what else is given to you see 2.3 Ln of 10 and here log of 2 is 0.3 so 2.3 multiplied with 0.3 if we do what is the answer that you guys are getting you will be getting you will be getting - 6.2 which is equivalent to 620 Jew and who was who said that yes this is the answer yes somebody had said right this is the answer no namash sh the answer is a little bit different because you are asked that the unit has to be Jewels okay the unit has to be jewels and yes that is why 620 Jew will be the correct answer see option oh there is no option this is an integer based question now why are we doing this question this is a pyq here everybody and let me tell you that from the solution chapter a graphical question coming into the into your paper there is a high chance so there are possibility either from Solutions chapter or they will ask you a graph from thermodynamics chapter okay so as for the analysis as you can see that in the thumbnail we have written that as for the fiveyear analysis yes so past fiveyear analysis shows that Solutions or chapter or thermodynamics there is a higher chance for a graphical question to appear in your exam so today I have not brought you thermodynamics graphical question but I got you a Solutions graphical question is that okay everybody yes to CH moving on to the next question here everybody now look at this this is also a pyq and this question has repeated so many times okay this was repeated in 2019 it came in 2016 it came in 2008 and it also came in 2006 so this type of question may appear okay this type of question may appear so CH let us read the question what is the question about the question says that at room temperature a dilute solution of Ura is prepared by dissolving 0.6 gam of Ura in 360 G of water if the vapor pressure of pure water at this temperature is 35 mm AGG lowering of vapor pressure will be lowering of vapor pressure they are asking everybody yes lowering of vapor pressure they are asking everybody okay guys all the students who are asking that ma' we are not able to see the other subjects that's because see we all are taking from different different place shre s is taking it from his home durgesh mam dures nandini M and am M took it from office I'm taking it from my home so that is why when we using the same stream key for the same stream what happens is it takes a little bit time to process after the class is over wait for half an hour after that if you watch it the whole video will be processed you will be able to see sha dures ma Nam M as well as me all right you will be able to see everybody take studies I hope that you understood so here what do we have to do we have to do Delta p is equal to P multiplied with we know that it is n ided by capital n + n okay now let us do one thing so guys Delta P see what is it asking lowering of vapor pressure so Delta P yes now what is given to you see you are given you are given pure water the vapor pressure of pure water at this temperature pure water is given to you so that is why we can write here that it will be 35 isn't it yes now 355 multiplied with C given mass of UA is 0.6 yes 0.6 and what is marol mass of Ura everybody it is 60 if you do not know know this okay then what do we have to do what is going to be what is going to be 360 G of water yes 360 G of water that is the given mass and what is the mass of water everybody 18 18 18 18 isn't it plus 0.6 / 60 now all we have to do is just solve this my people yes so we are going to do 35 yes multiplied with can I write it as 0.6 instead of 0.6 can I just write it as 6 here if I write it as 6 here then what I have to do is 60 another 10 we have to multiply in the denominator so 600 it will be yes divided by 18 multiplied with 20 will be 360 is that right absolutely and this 0.6 divided by 60 I can write it as 6 ided by 600 am I write 6 divided by 600 I can write if I write this as 6 divided by 600 that means I can also write it as 1 ided 100 am I right yes so 20 / 18 + 1 / 100 if we do and then if we try to solve it my people yes then if we try to solve it so what what will happen what will happen everybody yes if we try to solve it now what will happen yes what will happen everybody we will get 35 multiplied with 100 yes this 100 will go to the numerator part correct divided by 2001 because 20 + 1 here if we if we solve it then we will get 2001 multiplied with 100 so this 100 this 100 will get canceled if we do 35 divided by 21 what will you get you will be getting 0 point 0 point how much guys yes yes move move move come on come on check it out let me know check it out calculate this and let me know what is the answer you are getting 0.0 I have written because all the options have 0.0 but you calculate and tell me what is the correct answer guys come on yeah I I I moved I moved I moved sorry sorry guys I moved sorry sorry moved I moved come on go ahead let me know come on go ahead let me know C are you guys getting C prad Sunny is getting 0.028 are you getting 0.028 or are you getting 0.017 majority of the students are saying 0.017 so I'm clicking on option C but you guys calculate and tell me if this this is the correct answer or not yes is this the correct answer is this the correct answer well done well done well done everybody amazing amazing all right so now let us go to the next question shall we let us go to the next question everybody yes this is done the next question is the five successive ionization enthalpies of an element are given okay what are the what are the ionization and P here you can see that it is 8800 then it is 242 2427 then it is 3658 after that the fourth ionization energy is 2,24 guys do you see 2,24 and the fifth ionization energy is 32824 now they are asking you the number of valence electron in the element is how many do you think are the valence electron guys number of valence electron here it is going to be three you know why because you see that the first ionization energy the second and the third it does not have much difference as much as the third and the fourth ionization energy is there so three valence cell for that yet three valence electron for that yes the ionization enthalpy will keep increasing definitely but the jump is not so much as much as you can see in the fourth ionization enthalpy and the fifth ization enthalpy that is why option C three is our correct answer yes everybody yes guys cool CH moving on to the next question moving on to the next question everybody here once again another solubility product question here is okay all right yes so the solubility product of pbi2 is 8. 8.9 I guess 8.9 into 10^ minus 9 the solubility of lead iodide in 0.1 mol solution of lead nitrate is Dash Dash or X I think it is X x into 10^ - 6 mole per liter what is the value of x here what is the value of x here to let us figure it out okay let us figure it out here come on everybody what is given to you you are given KSP the solubility product and the KSP everybody is 8 9 did I write no I think it is 8.0 only it's 8.0 only it's 8.0 into 10^ minus 9 okay this this is the KSP that that we are given all right yes now according to the question PB I2 PB I2 the solubility of lead iodide okay so for pbi2 everybody and lead iodide and 0.1 mol solution of lead nitrate take so there are two things basically now first first of all let us do it for Pb I2 everybody for Pb I2 you know how it will dissociate it will dissociate at SP PB + 2 right and there will be 2 I minus okay so let us consider that for this it will be S and for this it will be 2 s raised to the power two am I right everybody yes correct yes correct everybody all right or we can just write here 2 s right now as in when we are calculating for KSP at that time we will do it as 2 s² okay all right right now check it out everybody the question says that 0.1 M solution of lead nitrate there is lead nitrate as well so now we can write it as pb3 whole tce forms yes how will this dissociate this will dissociate as PB + 2+ 2 N3 minus here okay 23 minus here what is given here 0.1 mol solution so here that means this is 0.1 initially there will be 0 0 right after some time what will happen this will get over and we will find here 0.1 mol for this and we will get 0.2 molar for this am I right absolutely okay moving on from here guys now we all know everybody we all know that our KSP is given to us our KSP everybody is PB + 2 multiplied with I minus and that's will be squared because I has a stochiometric coefficient of 2 now guys for Pb + to everyone do you see that 0.1 was written see do you see that 0.1 was also there everybody yes see 0.1 is here 0.1 is here so that means that we can consider this to be 0.1 plus s am I right plus s everybody yes but s has a very small value s is is so small for pb2 the concentration is going to be 0.1 + S but s is such a small value that we will consider this to be 0.1 okay so here now continuing this process we can write it as 0.1 multiplied with 2 s² okay that means how much will it be it will be 0.1 multiplied with 4 s² and here we have the value of KSP as well so now let us calculate s Square s² will be 8 into 10 ^ - 9 divided by divided by 0.1 can be written as 10^ minus 1 multiplied with 4 so 4 1 are 4 4 2 are 8 10 ^- 1 canel here this will become 10^ minus 8 that means we have 2 into 10 the^ - 8 everybody now we can write this as < tk2 into 10 ^ - 4 for S correct yes everybody great but in the question in the question people can you see everybody yes can you guys see here what is written they have written that X is X multiplied 10 the^ - 6 what do we need we need the power as 10 the power - 6 everybody not 10 the^ minus 4 so instead of writing it as < tk2 can we write it as see root2 value is also given to you as 1.4 y so let us write it as 1.4 1 into 10 the power if it is - 4 that means 10^ - 2 we will have to do right 10^ -4 or we can write it as 141 into 10^ -6 am I right everybody yes yes yes guys here here is your answer so the answer is everybody the answer is 141 into 10^ minus 6 the value of x is this okay the value of x is 14 1 clear everyone yes any doubt here any doubt here I hope so not I hope so not yes yes ver we will do that yes we will do that we will definitely be doing that okay CH guys moving on to the next question the next question is the N3 weighted out to make 50 mL of an Aqua Solution containing 70.0 mg na+ per Mill yes Dash gr you have to calculate it in nearest integer and you are also given the atomic weights of sodium atomic weight of N and there is oxygen as well all of these are given to you CH let us understand this in this question they saying that 50 ml the N3 weighted out to make 50 mL of an Aqua Solution containing this much is per what okay all right so let us suppose let us suppose that 1 ml contains okay 70 mg of Na plus ions right can we consider this yes absolutely we can do this after that what will happen guys after that we can also see what are they saying they're saying that 50 ml if we have 50 ml that means it will be 70 multiplied by 50 and this will become 3,500 mg everybody yes this will become 3,500 mg very good now from here 3,500 mg yes so this will be 3.5 G of na+ 3,500 mg can be written as 3.5 G of na+ in 50 ml am I right everybody yes yes everybody very good okay now guys let us calculate the mar mass of N3 let us consider mol mass of N3 which we can easily calculate because all of it is all all the values are given so na a is 23 plus here it will be 14 yes plus there will be 16 but 16 there are three so 16 mtip with 3 it is going to be all right if you calculate this how much will you be getting guys if you calculate this how much will you be getting come on calculate everyone calculate everybody how much will you be getting yes how much will you be getting 23 + 14 + 16 into three how much will you be getting guys I'm I'm guessing 16 into 3 16 into 3 3 6 18 3 1 3 19 23 + 14 I think it will be 85 am I right guys yes will this be 85 G per mole yes right everybody 85 thank you so much pradep yes thank you so much 85 now guys what is the molar mass of sodium marolar mass of sodium is given to you it is 23 g per mole am I right absolutely now let us calculate the percentage composition of sodium okay let us calculate the percentage composition of sodium in N3 if we do that we will be getting 23 23 ided 85 into 100 okay so if we do 23 instead of 23 if we do 8 uh 25 25 by 85 multiplied with 100 how much will you be how much will you be getting 25 Divided 85 how much will you be getting guys yes 55 25 and uh 5 1's 5 35 5 7 are 13 so it will be 5 divided by 5 ided by 17 5 divided by 17 so it will be 30% not almost 30% little less than 30% I think it is 27 28% approximately is it 27 28% approximately everybody yes I think it is 27 28% so 27% I'm writing here 27% if it is 28% let me know I will change the calculation okay for now let me write here 27% okay if that is the case now tell me what can we do we can write here for 100 G of for 100 G of N3 27 G of na+ ion is what we have right absolutely okay now what do we have we have 3.5 G of Na A+ and that will be my dear students 3.5 ided 27 multipli with 100 if we do how much will you be getting 3.5 divided 27 so it will be it will be 27 3.5 uh who has given me the answer Naruto has given me the answer is it 12.94 guys is it 12.94 and the nearest integer is going to be 13 so your answer is 13 G okay your answer is 13 14 G all right everybody yes or no guys are we understanding are we understanding everybody yes is this clear is this clear shall we go to the next question everyone shall we go to the next question CH let us go to the next question okay the next question my dear butus is the number of species having a square planer shape from the following is X ef4 sf4 SI F4 BF4 minus brf4 minus then you have some coordinate complexes as well so CH let us calculate everybody yes what do we know guys we have xf4 isn't it xf4 we know the structure already so we can just draw it as F here there are four FS yes there are going to be four FS here here you have a lone pair here you have a lone pair what is the shape this is definitely a square planer okay this is definitely a square now this is something that we know okay now let us come to sf4 everybody so CH let us calculate how many electrons how many valence electrons do we have in Sulfur six everybody yes so it will be six plus surrounding there are four atoms so 6 + 4 ided 2 if we do we will be getting five if it is five then what is the hybridization the hybridization is going to be S P3 so four and there will be one D if sp3d can it be can it be everybody 5 sp3d plus there is also one lone pair right there is also one loone pair so that means our structure is a slightly like this okay the our structure is like this because there is a lone pair here so this is your See Saw shape all right this is not your Square planer okay this is not Square planer all right everybody yes guys now next what do we have next next we have sf4 so CH let us calculate it for S4 S4 is going to be four silicon silicon will have four yes because it belongs to the carbon family so 4 + 4 ided by 2 yes that means how much will we get four if it is four than sp3 sp3 hybridization is always going to be tetrahedral am I right yes this will be tetrahedral absolutely this will be tetrahedral right next what do we have next we have BF4 minus okay for BF4 minus boron has valency three + 4 there is also another extra electron so let us add that also here and divide it by two this will give give us 8X 2 8 by 2 means 4 that means sp3 if it is sp3 guys if it is sp3 once again this is going to be tetrahedral correct this is going to be tetrahedral right next moving on to brf4 minus okay we we have brf4 for brf4 we have seven electrons in the outermost shell of bromine plus there are four Florine plus there is also one charge on the on the F minus so 7 + 1 is 8 8 + 4 is 12 / 2 is equal to six if it is six then it is going to be sp3 D2 and there will be two lone pair okay there will be two lone pair so bromine will look like this everybody yes bromine will look like this so yes this is definitely your Square Plan R so till now we have only figured out that there are two Square Plan R guys after vf3 what do we have we have a coordinate complex okay we have a coordinate complex which is cu NH3 4 here and 2 plus everybody let us try to figure out the shape here everyone okay first of all let us calculate the oxidation state it is going to be X MTI x + 4 * with 0 is equal to plus 2 that means X is equal to +2 so copper is in plus2 oxidation state okay if copper is in plus2 oxidation state from 29 atomic number that is 3d10 4s1 copper in plus two oxidation state is going to have 3d9 now what happens is NH3 is a strong field liand guys okay NH3 is a strong field ligand so we will have 1 2 3 4 5 6 7 8 and nine right then you will have your s you will have your P right and here you have the D you know what in fact we do not have to draw the D here now this this this P will be pushed here okay when it is pushed here you will be getting a hybridization which is dp2 okay and for dp2 what is the shape that we have we have square planer once again okay what do we have we have square planer once again again all right yes okay guys moving on from here now what do we have we have fecl4 this is going to be 2 minus guys for fecl4 if you calculate you will find that this will be sp3 hybridization that means it will be tetrahedral and for pt42 minus if you calculate you will find that this is also going to be square planer so we have 1 2 3 and four the answer is four the number of species which has Square planer is going to be four everyone yes it is going to be four am I right everyone yes yes or no yes guys number of yes Naruto absolutely right absolutely right CH guys next question is this is from the reduxx reactions chapter guys yes and from reduxx reaction you will definitely be asked about one of the equations that either deals with iodine or it deals with potassium dichromate or it deals with potassium per manganate so make sure that you again do this okay all right yes CH okay now moving on everybody no B that Toki that particular coordinate complex that we had that is going to be square planer in this case NH3 does act like a strong fiig and it pushes the last electron to the p orbital and then you have Caron hybridization of dsp2 if you want you can read about it once again but that is the answer okay that is the answer there is no mistake I'm absolutely right here so CH here guys what do you have you are given an equation 2 I3 minus + x IUS + 12 H+ uh which forms 6 I2 + 6 H2O okay this is the equation that is given to you you have to balance it and you have to tell us what is the value of x okay to CH let us understand see here what is happening to I3 minus is forming io2 isn't it io3 minus is forming I2 correct as well as we see that I minus is also forming I2 let us balance it here it is I2 so here let's write 2 I here it is I2 so let us write here 2 I now we will also have to balance the oxygen so 2 3 are 6 there are six oxygen that means we will write 6 H2O the moment we write 6 H2O here that means 6 2 are 12 so here we will have to write 12 H+ am I right everybody yes now here you see uh 12 H+ okay 12 H+ plus if you calculate 3 into -2 it will be 3 2 are 8 here is 1 electron and two so I think it is going to be 10 electrons here right we will have 10 electrons here correct and there is also 12 H+ there is also 12 H+ so subtracting and adding everything we will be having 10 electron here now here we have we have only two electron here from 2 I minus when it is getting converted to I2 we will have we will have two electron isn't it so let us multiply this with five if we multiply this with five our overall reaction is going to be see it is going to be 2 io3 minus plus 5 2 are 10 so it is going to be 10 I minus yes then what do we have we have here 12 H+ plus 10 electron and 10 electron this will get cancel this side we will have this side we will have I2 and uh I2 multiplied with 5 I2 so it will be 6 I2 yes and plus 6 H2O everything else is balanced that means your correct answer is option D 10 yes that means the correct answer is option D 10 okay yes all right everybody all right guys all right all right all right yes or no yes or no guys yes this is the correct answer CH all right ch ch ch all right guys so the correct answer is option D now next question everybody is low oxidation state of metals in their complexes are common when legans have good P accepting character do they have good Sigma donor character are having good P donating ability or is it are having poor Sigma donating ability okay first of all let us understand everybody are the lians yes are the lians by accepting Sigma donating what are the lians guys what are the lians everybody yes if you remember the metal and the carbonal yes the metal carbonal Bond everyone do you remember what happens carbonal is the LI here this initially gives some electrons to the metal the metal this is when they form a sigma Bond right this is when they form a sigma Bond after that the metal donates the electron to the antibonding empty antibonding orbital and that's how they form Pi bonds am I right yes that's how they form Pi bond that means that the lians are supposed to have good Pi excepting character now if you look at the metal carbonal such as f E4 and ni3 such kinds you will find that they have zero oxidation state that means that whenever a ligant has better P accepting character the metal will exist in its lowest oxidation state and sometimes it can also be zero synergic bonding ja very good yes Sygic bonding is absolutely the correct answer clear here why did I take synergic bonding I know that there are plenty of questions that are related to Magnetic Moment which we have done plenty in abundance we have done there are questions that are related to nomenclature we all know the rules we all know how to do this yes hence I thought Sygic bonding we have not done many questions that's why I have got you this so there is one question that is either going to be Sygic bonding or it is going to be bor Magneton or it is going to be isomerism oneus Theory or nen clature these are the only questions that can be here okay all right yes okay CH guys moving on to the next question everyone the next question is the current order of basicity of oxides of vadium guys I told you this I told you a super trick and what was the trick it was not a trick actually we all know that whenever the oxidation state increases as the oxidation state increases acidity increases so out of v23 v24 v205 whichever has the lowest oxidation state that is going to be the most acidic that means that v23 is going to be the most basic yes then you will have v24 v24 is going to be second basic and then you have v25 v25 is going to be the least basic that means most acidic as you increase the oxidation state remember that basicity decreases acidity increases am I clear here everybody yes am I clear guys am I clear everybody did we understand this that when basicity increases when acidity increases yes so v24 right this v24 you know what this will also have slight bit of aoic nature this will also have slight bit of aoic nature this is going to be highly acidic here and this is going to be absolutely basic am I clear everybody yes ch let us go to the next question and we are almost about to be done the next question is okay this is from our this is from our rate constant guys chemical kinetics okay so here we have the rate of a certain reaction increased by 2.3 times when the temperature is raised from 300 Kelvin to 310 Kelvin if K is the rate constant at 300 Kelvin then the rate constant at 310 Kelvin will be what okay so this is actually a very easy peasy question you know this is actually a very easy easy question you know when you read it you probably think that oh my God this is going to be so tough but it's actually not going to be so tough because we all know a simple formula and the simple formula is temperature coefficient okay temperature coefficient is equal to see what do we have we have rate constant here okay rate constant at t+ 10° C divided by rate constant at temperature T okay yes and this you know is equal to what this is equal to rate of reaction this is equal to rate of reaction rate of reaction when t + 10 is the temperature divided by rate of reaction okay rate of reaction when the temperature is capital T now what is given to you the temperature coefficient is given to you as 2.3 isn't it so it will be we will write here 2.3 2.3 is equ Al to 310 Kelvin or it can be 300 + 10 Kelvin yes yes yes everybody great okay divided by we can write it as 300 Kelvin yes that means that this is equal to rate of reaction at 300 divided by rate of reaction at 300 now rate constant at 300 is given to you what is it it is K they saying K is the rate constant for 300 so that means now if we increase it by 2.3 times it will be 2.3k for 310 that means that option D is your correct answer yes option D is your correct answer all right guys yeah CH guys one last question I guess I guess one last question take one more question oh no no we have we have few more questions we have few more questions guys don't worry yes we have one or two more questions after that we will be done after that you must go ahead revise a little bit and you should calm yourself down and be prepared for tomorrow's exam so in a second order reaction 20% of a substance is dissociated in 14 minutes the time taken by the 80% of its dissociation is okay this time I'm taking a second order reaction because I believe that you guys already know the first order reaction you also know zero order reaction and how to calculate all the different scenarios right you already know but second order reaction is one such reaction where we do not practice so either it can be second order either it can be first order either it can be zero order maximum chances are that it will be first order but first order formulas I have done it I have sent it everything is done so let us calculate this in a second order reaction 20% of a substance is dissociated in 40 minutes okay so let us consider X is equal to 20% if x is equal to 20% that means a - x is equal to 80% and a is equal to 100% am I right so 100 - 20 is equal 80% % correct now we know that K is equal to 1 by T yes x / a - x multiped with a am I right yes all we have to do is just put the values 1 divided by 40 what is X x is 20 divided by a minus X is 80 80 multiplied with 100 here so 20 and 40 cancel here we will have 20 that means our okay let's write this as a K let us write this as Ka okay not ka CH let us write this as KY okay just just for the sake of it just for the sake of it writing it as KY so our KY is is is 1 divided by 20 into 80 into 100 correct yes absolutely now guys moving this side now let us calculate it for 80% okay so now when X is equal to when X is is equal to 80% when X is equal to 80% we have a - x is equal to 20% right yes now if we calculate KX is equal to 1 by T yes 8 x is 80 80 / A- X is 20 20 multiplied with 100 am I right yes absolutely so now what do we have what do we have guys yes so if we calculate if we calculate for this KY yes if we calculate this for KY what will you get guys what will you get everybody can you calculate and tell me can you all calculate and tell me yes can you all calculate this and tell me what is it that you will be getting for T yes come on guys everybody from here can you calculate how much will you be getting how much will you be getting guys 80 divided 20 into 100 so this cancel this cancel yes that means 2 four are 8 and then you have here so it will be 25 here that means I can write it as 1 / 25t is equal to 1 / 20 into 80 into 100 okay all right so from here 25 and 4 I can write yes now 4 into 20 4 2 are 8 8 into 8 is 640 so 640 minutes is the correct answer that means option b is your correct answer yes option b is your correct answer am I right everybody yes option b is your correct answer very good very good very good next question everyone is your next question is the dissociation constant of NYC acid is 1.6 into 10^ minus 5 and the mol conductivity at infinite dition is 380 into 10 ^- 4 S m² mole to the^ minus 1 the specific conductance of the 0.01 M acid solution is specific conductance is something that we have to find here okay what do we have to calculate we have to calculate specific conductance so what is given to us dissociation constant Ka is given to us which is 1.6 into 10^ minus 5 right okay Lambda M at Infinity is given to us which is your 380 into 10 the power minus 4 yes with its unit that is s m² mole to the power minus1 okay and C concentration is also given to us which is 0.01 everyone 0.01 mol am I right very very good now we have to find we have to find Alpha is equal to Ka / C and root over that means 1.6 into 10 ^ - 5 / 0.01 can you calculate this everybody and tell me what is it that you going to get see it is 0.01 here that means that 10 ^ minus 5 this will go up it will be 10 the power - 10 the power whatever 1.6 here that means that 1.6 if we do root over 1.6 if we do root over I think it is going to be 0.04 0.04 is is that the correct answer again 0.04 into 0.04 if we do we should be able to get 1.6 correct yes now we also know that Alpha is equal to Lambda M ided by Lambda M at Infinity correct now from here guys from here guys we do not know what is Lambda M we do not know what is Lambda M but we do know what is f 0.04 multiplied with Lambda M at Infinity we know yes that means multiplied with 380 into 10 power minus 4 calculate this how much are you getting guys calculate this how much are you getting yes how much are you guys getting everyone how much how much are you guys getting come on come on come on calculate quickly quickly quickly tell me tell me everybody how much are you guys getting 0.04 multiplied with 380 yes you are probably getting 15.2 into 10^-4 that means 0.152 S m² no no no no no is it going to be huh okay s m² per mole correct yes if that is the case we can also write it as 15.2 s CM squ per mole am I right yes if we do this if we do this everybody okay once we do this now what do we know we know that Lambda m is also equal to 1,000 multiplied with Kaa divided by C yes now we know that Lambda m is equal to 15.2 which is equal to 1,00 Kaa divided by 0.01 from here we can calculate Kaa is equal to 15.2 0.0 1,000 that means it will be 1 52 into 10^ -4 s cm ^ -1 if we want to change it to meter then we will have to write 1.52 into 10^ -2 s m the ^ -1 that means option b yes to all the students who are saying that m'am it's option b it's option b yes you guys are absolutely correct okay yes 1.52 option b is your correct answer and wow we are done everybody so with this note guys guys tomor for tomorrow all the very best all the very very very very very best to you guys do well everyone you couldn't find the formula sheet which I have sent it today I've sent it today in the telegram Channel yes I have sent it today in the telegram channel so do not forget to click on the Bell icon yes because from tomorrow we will be doing paper solving so be there once after you have finish the session after you are done with the exam be there in the paper solving so that you understand that which is the which are the correct answer and how much marks you are getting what is the percentile you are getting with this note lots of love all the very very very best to you guys may you all rock your exam may you all get the dream college and may you get all the you know the highest percentiles that you can achieve the best of the ranks and please come back tell us when you are getting your result come to the vantu office we will all celebrate together with dle cake Samosa anything and everything in fact Biryani as well from my side okay last minute tips nothing for today those of you who have a session tomorrow if your exam is tomorrow I need you all to not study anything right now don't study anything have an early dinner have your dinner by 8:30 9:00 sleep calm down and if you want you can just go through the formula sheet for the last time that's it okay yes nothing more than that okay nothing more than that don't do anything more do not Sol try and solve paper now do not try and you know uh pick up something else like pick up something new some New Concept if you're trying to learn it's not the right time everybody okay so all I need you to do is just calm yourself down think that it is going to be a great exam it's nothing it's just going to be a white paper on that you know black ink say the question paper the questions will be written yes and remember if you are panicking just remember that there are many many more exams yes and there are many many many more options it's okay just one paper just one paper where some marks is written it cannot decide your whole future your future is in your hands you will be able to do good but if you are panicking remember that there are many other exams there are many other options there are many other countries apart from India from from where we can go and study so it's okay take a chill pin study well don't study today you know take a good dinner have a light dinner also don't eat like chicken Tika Biryani and everything today only then tomorrow you are sitting in the exam Hall and your stomach is going don't do all of that eat a light dinner have some Cur R have some you know uh what do you what do you say you have some Papu have some uh maybe uh you know just Dal Rasam something like that have a light eat dinner and sleep and tomorrow also you have you know maybe two chapatis in the morning for your breakfast keep it light so that you are more active and you don't feel sleepy take all right so with this note guys T bye-bye I'll see you very soon I wish you all the very best from all of my heart guys not just bottom of my heart from the whole of my heart all the very best I really hope that you guys do well and I will also be very very very proud and I'm waiting to meet you you all in Bangalore in the vantu office when we have the felicitation program yes so CH guys

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Length: 272min 18sec (16338 seconds)

Published: Wed Apr 03 2024