Complete Class 11 Physics | 1000+ Practice Questions | JEE Mains 2nd Attempt | Shreyas Sir

good afternoon good afternoon happy holy to all the J students welcome about to the vantu J English Channel and we are keeping up with the one we challenge where we are going to solve lots of questions for you and today's session is the 11th standard physics questions so we'll solve you know maximum questions as much as possible and you have to keep the Jo high and you will learn a lot and lot of things which otherwise you would not have done somebody needs to always push you somebody needs to you know always make you sit down and you know help you solve the problems so here I am just before the exam so making sure that you guys solve the problems uh just a quick reminder or a recap of what this one week challenge is there will be 100 plus problems every class so four classes in 2 days that is today and tomorrow completing the 11th standard syllabus problems so that's 400 problems similar session on Thursday and Friday 400 problems of 12 11th mock test day after tomorrow on Wednesday 12th mock test on Saturday and full syllabus mock test after all of this so totally 1,000 questions we are going to complete in a matter of this particular week and if you want to take more help remember uh you might be probably in a city where there is already a vantu Learning Center present where you can get complete guidance mentorship also your doubts cleared with the amazing teachers at those centers so in fact even I was at one of these centers just recently beat Pune beat Hyderabad and now I'm going to Bangalore and Chennai too so make sure that if you can take help and you need that motivation that guidance or how to crack the exam visit the center and if you have uh your Juniors or you know your batchmates who are looking for or something and looking for an offline batch with the best mentorship with smart classrooms and world class facilities and you will see why vanto offline centers are way different than any other coaching Center in your area so we are already spread out in many cities and classes are running full fledged so I was in Hyderabad and I saw so many batches running you know for the J and then for the next 10th moving 11 C students and so on and so forth so make sure you two are a part of this Revolution and yes there is an amazing amazing event which is happening a grand event in Patna so if you are in Patna or if you know someone in Patna do let them know because there is a big big Grand Event happening with all the best of the best teachers coming in Patna for this mega mega event right so be there all right so the I think the venue is also mentioned March 31st is the date when it's going to happen 11:00 onwards it's in samrat asoke convention center okay right so let's start with the questions get ready my dear students video and audio not sinking why just check if it is sinking right now just s uh see if it is sinking right now hello my dear Warriors can you see me properly hear me just let me know now it is fine yeah now it is fine okay okay great let's go ahead with the questions remember you also need to open your rough sheets and simultaneously you can make running uh notes or running problem solving is what I would say all right okay if you just want to see that is also fine but after the class is over try to solve the same problems on your own and see how much you are able to do here is the first question the length of a sheet a sheet is there one .5 cm and the breadth is 1.23 CM what is the area of the sheet to the correct number of significant figures now the rules when you basically multiply or basically divide is you always choose the minimum you will always choose the minimum of significant figures minimum of the two significant figures right that is how you know the rules go when you are multiplying or basically dividing if you notice over here in 1.5 CM there are two significant figures 1 and five in this one you can see 1 2 0 3 1 2 3 4 there are four significant figures the least significant figures are there in this so the final answer should have two significant figures even without you having to actually multiply 1.23 multiplied by 1.5 which will give you something but you will round it off to the correct significant figures you should see the options this has many significant figures this has many significant figures this has many significant figures this has exactly two significant figures none of these options have two significant figures so even if they are more accurate but using the rules of significant figures they do not make sense so just directly mark this as 1.8 the exact answer will not be 1.8 that is not the point the rounded of value to the current significant digits has to be in two significant digits only did you get it everyone yes Rahul Patel I'll be definitely taking up emat classes as well moving on to the next question on your screen but for now please continue with these classes definitely there is something amazing coming up for 11th M 10 12th 10th M 11th as well as em set and other exams like bitsat so many things are there we'll take care of you but let me tell you these questions will also help you for any other exam like bitsat vle e m set okay so attend these questions you will definitely not feel that you know this is out of context this is properly there in the syllabus the pressure on a plate is measured by measuring the force and the length of the sides if the maximum error of the force and length are 4 and 2% what is the error in the pressure there is some problem with the force and length measurement so there will be obviously some error in the pressure measurement so first of all realize how is pressure related to force and length the length of the plate is given so I can find the area and we know Force per unit area is the pressure so I'll start off with that pressure is nothing but Force divided by area which is force divided length Square which is force ra to 1 length ra to minus 2 why did I write it like this because there is a simple error formula which says the net physical quantity is relative error that means Delta P by P let's say percentage will be take the first physical quantity let's say Delta F by F take the relative error multiply it with the power the power is basically one that's it then add to it the next relative error which is the relative error of the length and multiplied by the mod value always of the power so it is basically 2 into and then percentage error that's all so that will give you the answer my dear Warriors the percentage error in the force is nothing but 4% plus 2 into the percentage error in the length is 2% so 2 2 are 4 4 + 4 is basically 8 so the answer is going to be 8% very good many of you got it correct very good miar very good yashan karik so many of you have got the correct answer proud of all of you moving ahead to the next one the density of a material in CGS System of Units is four in a system of units where uh the unit of length is 10 and the unit of mass is 100 the value of the density of the material in that new unit will be how much so this is a question on conversion of units where you convert from one system to the other system many people try to remember that formula N1 U1 is equal to N2 U2 but there are some confusions which happen when you use the formula so a better way I will tell you what you need to find is density we know density is nothing but always mass per unit volume volume is basically having the dimensions of length Cube so dimensionally density is mass per length Cube that's all you need to convert from any system to any system whatever physical quantity is there at least know the dimensions of it or you can find out the dimensions using some formula which you know now now what is it that is asked you have to find what is the density in that particular unit in CGS you know so think about it the density the density in those units in those units will be one unit of that mass one unit of that mass upon one unit of basically the volume which I can write it as 1 unit of that particular Mass upon volume is nothing but length Cube we know that so it is basically one unit length 1 unit length whole Cube all this I got to know because of the dimensions density was relatively easy let's say it was resistance or thermal conductivity Etc then you need to know the dimensions so that you can write this according to the dimensions Mass was there numerator length was there denominator it was cubed so I'm cubing it if the dimensions are different this would also look slightly different the powers their numerators denominators will change accordingly okay so what is one unit of that mass think about it over here unit mass is 100 g a unit mass is basically 100 G so can I not say this is basically 100 G over here 100 G over here unit length is 10 cm so one unit length can I not just put it as 1 sorry 10 cm and this will be basically Cube so this will be 100 / 10 Cub is basically 1,000 this will be G per CM Cube my dear Warriors z z will cancel so won't it become .1 G per CM Cube .1 G per CM Cub so think about it 4 G CM cube is asked 4 G CM cube is asked just move this 0.1 over here so 0.1 units 0.1 sorry 0.1 comes on on this side it will become 10 10 units 10 units of density will be 1 G per CC 1 G per CC if I take the 0.1 on this side it will become 10 1 by 10 10 goes on the other side so this is 1 G per CC so my dear Warriors what will be 4 G per CC just multiply by 4 won't it be 40 units 40 units of the density won't it be just 40 units of the density which is option C how many of you got it no it is not B many of you made a mistake over there many of you said B option many of you said B option that is wrong so just to give you a quick reminder of how we converted the units whenever any quantity is given let's say for example it was some random physical quantity and the uh dimensional formula is M1 l s tus2 example just Chum taking some example and I tell you one unit of mass is basically 10 G then I tell you one unit of length is basically .1 cm and I tell you one unit of time is let's say 100 seconds example so how do you convert it just if I write down one unit one unit of that particular physical quantity what will you do mass is raised to one mass is raised to one so basically 1 unit Mass this is raised to 1 length is raised to 2 so 1 unit length raised to 2 time is raised to min-2 1 unit time raed to minus 2 what is 1 unit Mass 1 unit mass is 10 G what is 1 unit length1 CM squar what is 1 unit time it is 100 seconds ra to minus 2 that's all now just take the numerical part separately 10 into1 s into 100 S sare into 100 squ into G cm² 2-2 some physical quantity so what will happen over here my dear Warriors this will become 0.1 and this will become 10,000 so it will basically become 1,000 over here 1,000 g cm² 2us 2 so this is how you converted one unit of that quantity into CGS unit it could be done in MKS or any other unit got it my dear Warriors how this has to be done only if you know the dimensional formula you can convert from any system to any other system if you know the dimensions just substitute one unit Mass one unit length one unit time one unit this is how much in MKS or CGS if you substitute take care of the numbers separately numbers separately automatically you will get the conversion formula the moment I got this I know question was 4 G cc 4 G CC multiply 4 multip 4 you got the answer yeah got it time power is min-2 uh oh so sorry time power was min-2 my B I missed that minus two over there I just wrote it as two yeah this would be min-2 so this would technically be 0.1 divided by uh or rather you can put this as 1 14 Z over here so this will just become 10^ -5 G cm² second to-2 yeah correct so this was G cm² 2-2 the power was- 2 there correct all right moving on to the next question two equal vectors have a resultant equal to either of the two then what is the angle between them two vectors you add them the resultant is equal to either this or this so what do you think was the actual angle between the two vectors multiple ways to do the problem one way is formula the other ways just geometrical approach so think about it if the vectors have same magnitude that means it must be an equilateral triangle it must be an equilateral triangle so think about it this Vector plus this Vector might be giving you this resultant Vector this resultant Vector just think about it this is your resultant vector and the other two a bar and B Bar are your vectors being added it should be an equilateral triangle it is an equilateral triangle only then this can happen all the magnitudes are equal remember that this magnitude this magnitude this magnitudes are same because it is given equal to either of the two so question is what is the angle between A and B lot of you will say 60 but be careful always angles have to be measured by keeping their tails together by keeping their tails together or heads together so if this is a vector and this is basically your B Vector you can see the actual angle will be 120° take this Vector slide it take a vector slide it actual angle is 120 it is not 60 many of you wrote option b which is wrong you will get minus one many of you will get minus one be very very careful it is not B it is C option always keep the Tails or the heads to measure the angle have you understood it if you had not attended this class you would have lost many many marks this is what happens in the exam exam and I want you to precisely you know get those marks which you are otherwise going to lose I want you to you know keep a hold of your marks because negative marks is the biggest enemy in the J exam it will just snatch away whatever you know it will just take away whatever you have gone so be very careful is that okay so keep this in mind always angle between the two vectors is measured so if I if I have one vector here if I have one vector here and let's say another Vector here don't measure this angle or if I have one vector here and let's say another Vector here don't measure this angle always measure either when both the tails are together then you can measure it or both their heads are together doesn't matter this angle is also okay this angle is also okay never measure like this never measure the angle like this that is how the angle measurement in vectors has been defined right understood okay another way of doing the same question would have been you know you can say formula for resultant is nothing but is nothing but a s + b² + 2 a COS Theta R A everything is same so it will be x² is = x² + x² + 2 into X into X into cos Theta this will also do I can assume all the quantities are basically X so x x x cancel so 1 will be 1 + 1 + 2 cos Theta 1 1 cancels so it will become -1 is equal 2 cos Theta so basically cos Theta is - 1 by 2 that means Theta must be 120° this way also you will get the same answer this is numerical approach this was geometrical approach okay both the methods have told you is that okay y awesome awesome so I hope you understood this yes this will be available s after the session is over after the session is over you will be able to see the recording also definitely moving on to the next question coordinates of a moving particle is given by x equal to 2tq Y is = 3 T Cub okay 2 T CU 3 T Cub acceleration of the particle is given by what okay let's see if we can solve this first of all from position to find acceleration you will have to differentiate it two times first derivative velocity second derivative acceleration if you look at X over here x is basically 2 T Cub the Velocity in the X Direction by taking just by just taking the derivative of this by just taking the derivative of this will give me will just give me nothing but 2 into 3 into t² which is basically 6 t² so the acceleration in the X direction will be take the derivative of of 6t s take the derivative of basically 6t s it will be nothing but 12T it will be nothing but 12T next Point same thing Vel uh sorry y coordinate y coordinate is 3 T Cub so velocity in the y direction will be just the derivative of this 3 T Cub right in the y axis you are going to get the Y velocity so this will be 3 into 3 into t² which is 9 t Square the acceleration in the y direction will be the derivative of this derivative of velocity gives you acceleration so this will be 9 t² so this will be 9 into 2T which is basically 18t which is basically 18t I got ax I got a y what is the last step going to be so the total acceleration will be nothing but 12 t² + 18 t² 12 t² + 18t Square many of you are saying option b is that right Paka done block yes option b is blocked and it is correct otk 468 time comes outside so this is what you will get this is what you will get understood my dear students this was a two-dimensional problem also of non-uniform motion because the coordinates were dependent on time whenever you go from position to Velocity to acceleration you differentiate if acceleration was given and velocity was asked you integrate from velocity you integrate again you go back to position okay the reverse process will be integration the forward process will be derivatives cool let's move on two projectiles are thrown same speed but different angles to the horizontal which is wrong is the question which of them is wrong read the question carefully two projectiles are thrown at 60° and 30° so realize one interesting thing they are both compliment angles 30° and 60° are complimentary angle so ranges are equal for complimentary angle so first option is definitely correct ranges are same ranges are same ranges are going to be same for complimentary angles for complimentary angles like 20 uh 70 right 45 45 37 53 30 6 10° 80° these are complimentary angles and ranges are same so option A is correct but the question is which is wrong in the exam if you don't read this word you see this as the option First Option you will mark it and you will move ahead without even realizing you have committed a grave mistake so be careful it is wrong which is asked okay height and time if you compare the height formulas height formula is nothing but u² sin sare Theta by 2G basically height is proportional to square of sign so h a by HB will be sin Square 30 by sin sare 60 what is it sin 30 is half half square is 1x 4 sin 60 is < tk3 by2 square is 3x 4 4 4 cancels it will become 1X 3 so 3 goes on the top so 3 h a will be basically HB 3 H is equal to HB yes B option is also correct B option is also correct interesting I think if B option is correct most likely C option will also be correct but we cannot just chuma say it we have to find it out time of FL is 2 U sin Theta by G time is directly proportional to time is directly proportional to sin Theta so time a / time B is sin 30 / sin60 which is 1x 2 by < tk3 by 2 which is 1 by root3 so take < tk3 on the top so < tk3 into time a is equal to TB you take it on the top so root3 ta is TB which is C option that means none of this is wrong none of this is wrong got it what a brilliant question so they can check really your patience they can really check whether you're reading the question properly and it's also going to test whether you remember all the projectile formulas these kind of questions highly probable all the graphs below are intended to show some kind of motion but one of them is not matching with the rest of the three which one is that you decide so pick it up three of them mean exactly the same thing just the y- axis x-axis are different but they are showing the same thing I don't know which one are them but one does not match with the others which one do you think it is I'll give you a hint I'll give you a hint my dear students I think this graphs are most likely suggesting a motion under free fall where you have thrown an object up it goes up stops and comes back so if you realize that it becomes very easy right uh no sumati by seeing the first option can you confirm it is read for example for example just imagine in this previous question just imagine you know it came out to be < tk3 H is equal to HB imagine after solving this is not what comes this comes example but here it says HB is 3 so then B option will be marked then B option will be marked so there is no other way but to actually check the first three options just because a is correct does not mean d is um going to be marked as the option who says option B and C are correct maybe one of them is wrong right so you have to verify it cool clear don't do these mistakes guys over here what you need to do understand is if I throw a ball up it goes up velocity reduces stops and it returns back and hits the ground over here now realize one thing the speed is reducing speed is reducing and then it changes the direction so velocity was earlier positive then it has become negative and here it has become zero at the topmost point so yes B option graph is correct matches it if you look at velocity versus position that is also fine velocity was positive then it was negative this is where you are at the topmost position because this particular length is telling you maximum height this is telling you maximum height it's position a graph on the x-axis be careful position time the position increases increases stops and then comes down so this was the topmost position and this is where you again come back to the same point you started over here and the position came back to the same thing so this also is matching with this description what about this distance distance slowly increases increases and then stops increasing I feel the manner in which this graph is drawn doesn't make sense to me I'll tell you why here the slope is is less here the slope is less then the slope increases and then the slope increases even more this is not acceptable how is the slope increasing actually the speed is decreasing the slope should also have decreased so the slope the slope is increasing which is not correct hence the D option is not matching with the rest of the three the slope should have decreased so actual graph of distance how it should have been means if it was distance if it was distance graph actual graph how it should have been means it should have gone like this like this and then it should have gone like this it should have gone like that okay the distance uh rate will decrease because the velocity is decreasing it becomes zero at the topmost point the slope becomes zero and then again it begins to increase because the speed increases so the graph goes up like that so this would be the correct graph this is is not the right gra got it okay everybody very good Magna very good YT gaming excellente excellente kamini very good Kini awesome awesome keep it up moving to the next question on your screen two balls are there equal masses are there you're throwing it up same line and after 2 seconds Gap same speed then they will collide at which height is the the question they will collide at which height is the question see if one of the ball is thrown up like this with a speed of let's say 40 m/s it goes here and while it is coming maybe it collides so it has traveled for T seconds the other ball the other ball was also thrown at 40 m/s but after 2 seconds so it will collide after T minus 2 seconds why tus 2 seconds because after 2 seconds it has been thrown so take away 2 seconds out of the total time that is the only time for which the second ball is in air or is traveling is in motion is in Flight is that agreeable everybody understands this the first ball was in air for 10 seconds after 2 seconds you throw the next ball so the next ball will have to remain in air only for tus 2 seconds because 2 seconds you have delayed it so this is the first piece of a drawing that you will be doing and then what is same for both of them is their Y coordinates the displacement in the y direction s is u +/ a s u is basically 40 into t+ half into acceleration is basically - g into basically t² same thing you can say displacement in the y direction will be 40 into t - 2 + into - g into uh t - 2 s into t - 2 s these both are going to be equal all you have to do is just solve them just solve them and find the value of time once you find the value of time what you do is find either Sy y or s y from here so s y will be 40 into whatever time you get + half into - 10 into that time squar just see it what do you get the answer as just see what you get the answer as just check it out yep just check if you're getting it as 75 M and the time value as 5 Seconds you should get the value of time as basically 5 seconds so this five you'll put it here also and five you will put it here also so substituting both of them you will get the answer as 75 M you will get the answer as 75 M have you understood the process my dear Warriors yeah Mna we'll be starting very soon very soon just give us some time right now you know J Mains is also round the corner so after close to J mains you know once the papers begin to uh go one by one we'll slowly start preparing full fledge for emat also okay so for now your preparation for J means also helps you for your preparation for Emet so or any other other C or any other exam like we or bit side because the syllabus is the same the same question can be asked in EM set if I just put over here emite session you'll be like oh sir yeah this is emite but remember we'll have separate sessions apart from this as well all right cool moving on the trajectory of a projectile projected from the ground is given by this equation where X and Y are measured in meters the maximum height attained by the projectile will be how much okay that's the question in both are equal because they Collide the question said where they Collide they Collide they Collide means what if they Collide if they Collide means what their displacements are equal their displacements are equal right their displacements are going to be equal right that is the meaning that's why I have equated both of them I hope that was clear if the trajectory equation is given what is the maximum height attained by it now whenever you are given some some trajectory equation meaning there will be a graph associated with it and whenever the height is maximum wherever the height is maximum there is one unique thing which happens and that is the derivative Dy by DX will be zero it is the point of Maxima it is basically the point of Maxima so all you need to do is find the derivative of this Dy by DX will be derivative derivative of x - x² by 20 but at the maximum height this is zero so derivative of x is 1 and this is nothing but 2x shift it on the other side so X by 10 is equal to 1 so X is basically 10 m but that's not the problem the problem is to find the maximum height so maximum height will be that y coordinate which you will get by substituting the value of x as 10 so 10 minus use this equation 10 - 10 s by 20 so I found out the value of x the value of x came out to be 10 m at that x what is the Y that is what the question is now so just substitute 10 square is 100 100 by 20 100 by 20 is nothing but 5 so hence the answer is 5 m hence the answer is 5 m is that okay understood everybody with me on this cool let's move ahead to the next question coming up on your screen if there is an object which is thrown from point A and hit hits the ground after 3 seconds the line from A to B makes 30° what is the initial speed of the object that's the question very interesting so this base is not given the height of the tower is not given speed is also not given in fact that is what you have to find just the angle is given and the time of FL is given try to recollect any formula that you know for projectile from a hill and mainly because time is given that it falls after 3 seconds is there some formula for time which you know already yes the formula for time is root of 2 H by G in fact if you put the value of time as three I think we will be able to find the height also G value is 10 3 square is 9 so this will be 2 H by 10 this will become 5 so H will become basically 45 M now that I know that the height is 45 M I can also find the base if needed because I think for initial speed I will need this base because what else will give me the speed horizontally I need something in the horizontal direction right okay so what do we do we'll use trigonometry over here we can see that tan of 30 which is height by base which is BC so basically BC will 45 tan 30 is 1 byun3 goes on the top we make it < tk3 tan 30 is 1 byun3 so root3 goes on the top we make it 45 < tk3 so hence the base is basically 45 < tk3 do I know some other formula apart from time remember range from the foot of the hill to the place where it lands is speed into the time range we got it as 45 < tk3 speed I don't know time was already given as 3 so V will be 45 by 3 is 15 so 15 < tk3 m/ second that should be the answer 15 < tk3 yes it is the answer got it everyone with me on this right so this is how you apply the formulas this is how you think about a problem whatever is given try to think is there some formula associated with it yes time of light oh I have to find speed but to find speed height is not important the base is important because that will give me the range formula which is U into T I know time I don't know range so to find range I use trigonometry so one thing will lead to other you start from what is given you go reverse see what you need for that you can find that Missing Link The Missing Link was here right cool a stone is thrown at some angle reaches the maximum height what is the time of FL of the stone what is the formula for time of FL if you know that maybe that will help me it is to you sin Theta / G but speed is not given instead maximum height is given maybe we can use something H is nothing but u² sin s Theta by 2G take 2G on the top so 2 g h is basically U sin Theta and this is squared so U sin Theta will be basically root of 2 GH so instead of U sin Theta put over here root of 2 g h rest of the other terms like 2 and G will be common as it is root G root G will cancel root G root G will cancel it will become 2 H by G root of 2 H by G so 2 into root of 2 H by G which is option C which is option C how many of you wrote option C G it should come by this week end I believe yeah it should happen anytime this week don't worry about it you remember right everything comes on time just few days before usually 4 days before 5 days before you know you'll get that admit card also don't worry so for now just concentrate on problem solving City intimation slip usually 2 weeks before it should come that mean sometime this week right moving on to the next question if there is an aeroplane which is going at this altitude with this speed above the ground towards a point directly over the person struggling in flood water at what angle of sight with the vertical should the Pilot release the survival kit if it has to reach the person in water this problem is very similar to projectile from a height or throwing an object you know from some Tower or dropping a bomb from the Aeroplane these are all standard questions if you remember so let's see what do we need to do in this particular situation first of all what you need to do is convert this km/ hour into m/ second so 600 600 kilom per hour into basically 5 by8 right this is what he will do so divide by 6 it will become 3 this will become 500 so 500 by 3 m per second that is in me/ second now let's visualize what exactly is happening this is the ground level this is the person who is struggling in water and there is an aeroplane over here it has to release some food packet this is that Aeroplane okay this is that Aeroplane it is going like this with a very high speed at a very high altitude which is 19 160 m so when it drops a food packet what will happen that food packet will take up this it has to land over here so what and all do we need what and all do we need to find the angle do I know the height yes I think if I just know what this distance is most of my job is done because if I know the base if I know the base I think I can find the angle I can find this particular angle whatever is needed so this is the direct line of sight the person who is struggling sees the Aeroplane here that's when the Aeroplane releases the food packet and it continues its Journey Only then the food packet will land here if you leave it before or later on it will go ahead or fall before not near him so that is why drawing this particular triangle becomes very very crucial everybody with me on this very good so with this particular uh uh you know height which is given to be 1960 the first thing you can find is time of FL the time of FL is basically root of 2 into height which is 1960 I think the value of G has to be taken as 9.8 over here right so uh 19.6 is 2 * 9.8 so I can just write it as 2 into 2 into 9.8 and this will make it 19.6 two more decimals so 1 0 0 and divided by 9.8 9.8 9.8 cancels so it will become 2 into 10 which is basically 20 seconds just check if the time value comes out as 20 seconds this will be needed for me to find the base of the triangle which is basically the range range is speed into time speed is how much 500 by 3 500 by 3 into time which is basically 20 seconds into time which is basically 20 seconds is that fine everybody with me very good very good so what will this what will this happen then What will what will you get 500 into 20 by 3 right so that will give me 1,000 into 10 which is 10,000 divided 3 once I get this once I get this the only job that I have is to find this angle or basically to find this angle find the line of sight with the vertical they have asked you so basically they have asked you this angle Theta so you can basically find this angle Theta so hence tan Theta which is opposite by adj will be 10,000 / 3 divided by the height height is basically 1 1960 you can cancel few things you can see where is tan Theta approximately coming out to be is it coming out to be < tk3 or 1 byun3 or 1 byun2 just check it if it is coming out close to 1.7 32 close to 1.7 if it is close to 1.73 that means our Theta should be 6 ° Theta should be basically 60° okay everybody cool to this point awesome great no it is not option A it is not option A just check this out it will be yeah it will be option number c be careful yep is that okay everyone shall we go ahead great great moving on moving on to the next one a ship is going to the east with 12 a woman goes 5 m/s relative to the ship towards the north what is the speed of the woman relative to the ocean relative to the Sea okay okay so what do we do first of all realize this is just like your boat swimmer problem in the river so the ship steaming due e is this much the woman running in the deck at this much so if I talk about the velocity of the woman with respect to the ship it will be the velocity of the woman minus the velocity of the ship so therefore the velocity of the woman will be velocity of the woman with respect to the ship take this on the other side plus velocity of the ship this is vector addition remember that so what is the velocity of the woman with respect to the ship it is given that she is moving towards the north so this is 5 jcap what is the velocity of the ship velocity of the ship is 12 basically IAP towards East Direction so this would be the velocity of the woman but if you ask me what is the speed what is the speed of the woman you just take the mod of this so it will be root of 5 S + 12 squ which will be 13 m/s so actual speed is is 13 m/s but if you look at it Direction wise some components are there 12 IAP is there and five jcap is there five jcap is there and 12 IAP is there so this is the net velocity of the woman net velocity of the woman she's going at an angle yeah slightly towards the North and towards the east yes very good moving ahead to the next question coming up on your screen a river is going from west to east west to east with 5 m per minute speed a man can swim at 10 m per minute what direction should the man swim so so that he takes the shortest path to go south he has to take the shortest path to go basically South think about it a man can swim what direction should he be swimming so just imagine this to be the width of the river like this the river is going this way with a speed of of five with a speed of five if he has to go along the shortest path that means he has to go basically like this he has to swim little bit against the current he has to swim little bit against the current so he will be pointing this way he'll be pointing this way the velocity will be nothing but 10 but the river will pull him over here we'll pull him over here like this and he will end up going resultant like this this will be the resultant this will be the resultant velocity perpendicular to the bank and that's how he goes in the shortest path now if you actually measure this particular angle over here let's say I call it Theta you can see sin Theta is opposite by hypotenuse which is half that means Theta is 30° so basically you should make 30° with the South Direction 30° with the South Direction now if you remember the directions it will be like this this is North this is East this is West this is South so the actual angle that is making is like this 30° 30° with the south towards the West 30° towards the uh West right so basically I think over here man can swim in steel water which person should D to take the shortest path to go to the South I think it should be 30° west of South not east of South I think over here this should have been West they should have been West yeah everything else is fine yep is that clear everybody with me yep very good yep yeah it should have been west of South you can see he has to swim slightly like this it is going from west to east the river so he has to swim slightly against the current that way cool shall we proceed ahead to the next one awesome awesome moving ahead to the next question a man is sitting inside the bus and he's traveling in the direction from west to east speed this much he observes the rain to fall down he observes that means it is an apparent motion relative motion so the person who is standing on the ground how will the actual direction of the rain move how will the actual direction of the rain move is the question so let's use our beautiful equation which says velocity of the rain with respect to the man is nothing but the velocity of the rain minus the velocity of the man this is the actual equation that you need to use the velocity of the rain relative to the man is vertically down so that means vrm Vector has to be like this this is your vrm vector the person is going and going from west to east person is going from west to east that means the velocity of the man is like this but you don't need VM you need minus VM so think of only this Vector minus VM will be exactly opposite of it will be exactly opposite of it so let's say I put minus VM Vector like this minus VM Vector like this you'll be like sir why did you put it here why did you not put it here I'll tell you also why because there is one more Vector which I need to add VR Vector plus the negative of VM Vector so only then I'll get vrm Vector so technically speaking this will be VR Vector so VR plus this negative VM gives you this Vector triangle log a vector plus b Vector is C Vector so this is how the rain should be falling this is how the rain should be falling and remember U this was the west side and this was the east side so how was the rain falling how is the rain falling it's falling down but slightly towards the east side so guys it is falling at an angle going from west to east it is falling at an angle going from west side to the east side yes the correct answer is option b very good if you guys got it till here study up if you're in first year you should be able to still solve all these questions let me tell you if you are done with the syllabus so this will be a good revision for you if you are in the 11th moving 12th standard batch yes Sanjay that is correct cool shall we move ahead why don't you take the velocity of the man in the right direction uh in the previous question you mean this one this one why did I not take I in fact have taken it in the right direction but why did I make left Direction because of this what is Vector subtraction it is nothing but adding a negative Vector so although he was going from west to east when I have to use this equation I have to choose always minus VM see what is the problem with these problems is that many people remember formula but if you seen my actual old classes nurture Pathfinder or any classes I never teach using some formula the only thing you need is this and Vector subtraction you should realize that minus VM means I have to make the velocity opposite direction and then I have to add it so treat it like a new Vector only all together so this is a vector this is only B Vector so a plus b gives me c so you have to align these vectors so that the resultant is down that because that is the net resultant vector of this and this is that okay yeah I hope this is clear cool shall we move ahead to the next question coming up on your screen here it is a fireman wants to slide down a rope the Rope can hold only 34 of the total weight of the person what is the minimum acceleration of the fireman to slide down that is the question so let's draw the diagram only then we'll get an idea this is that rope okay let's say this is that fireman like this is holding on to the Rope like that now what are the different forces acting on him the Rope will be pulling him applying some tension he will be pulling the Rope same tension equal opposite by Third Law but acting on different bodies we are drawing pandu the fireman the weight of the person will be there and obviously the acceleration will be there as well the acceleration will be there as well these are the only forces AC on the person nothing else so from this weight minus tension is basically mass of the person into acceleration it is clearly mentioned that the by the way weight is mg the tension which he can bear is 34th of the weight of the man so this is basically 34 of mg that's it so this is equal to ma m mm cancels so g - 3 G by 4 is equal to a so a will be G by 4 so G by4 where is it option D for it Delhi yes that is the correct answer cool clear shall we move ahead to the next one coming up on your screen ready for it very good yes let's move on guys concentrate on the class don't focus on chatting with each other concentrate exam is just 2 weeks away J means other exams youm that are just one month away so make sure you're utilizing the to the best right so moving on to the next one coming up on your screen the system over here shows that the acceleration of the 1 kg mass and the tension how much it is okay so 3 Kg 1 kg 3 Kg over here question is what is acceleration and what is the tension between block a and Block B simple way I'll tell you simple formula on this side the total mass is 4 kg on this side the total mass is 3 Kg acceleration is is always the bigger Mass minus the smaller Mass minus the smaller Mass upon the total mass the whole thing into G this is the basic formula if you remember this your job will become very easy what is the bigger Mass four smaller Mass 3 total mass 4 + 3 into basically G over here so this will become basically G by 7 G by 7 is only there in option C will you check whether the tension is 6G by 17 I don't think so directly you will Mark option C you will not even check in the exam but let's say this was also G by7 this is also G by7 then you have no other choice but to check it and how do I check that what is the tension between a and what is it A and B right so what is the free body diagram of a going to look like there will be one tension there will be one G and there will be acceleration how much will that acceleration be G by 7 G by write Newton's equation this minus this is total mass into acceleration so 1 G minus tension is basically Mass into acceleration Mass into acceleration this is just net force is ma Newton's law so that's a tension will be equal to G+ uh sorry tension will be equal to g- G by 7 which is 6 G by 7 that's your tension which is there over here it's just double checking it or crosschecking it that is the correct answer moving to the next question moving to the next question here it comes question on constrained motion many people get confused in these questions you are pulling this string it is passing over the pulley it is attached to this Mass if you pull this with v what is the velocity of the block what is the velocity of the block think like this if you pulling this with velocity V even the string over here is coming with v whatever you give only will go there so basically the string is going up with velocity V the string is going up with velocity V but the actual velocity of the block is over here the actual velocity of the block U is actually over here of which V is a component here many people fail to understand who is a component of whom many people think V is the actual velocity and U is the component of it no the block is actually moving here so where is it actually going becomes the resultant where is it actually going is the resultant of the Velocity so this V then becomes this V basically then becomes the component component this velocity becomes a component in fact this has two components this you has two components one here and one here okay if you see it that way it makes sense so if this is Theta obviously this is Theta so therefore can I not say V is nothing but V is nothing but use sin Theta component V is nothing but use sin Theta component that is the answer so the velocity of the block is basically uh V by sin Theta that is what was the question U was nothing but V by sin Theta which is exactly what was predicted cool got it who has to be taken whose component right something went wrong please try again oops okay am I audible guys am I audible am I audible just one second what is my phone saying yeah oh right okay cool let's move ahead to the next one this diagram shows 2 kg vegge on that there is 1 kg Mass there is an angle given also question is if the wed is given some acceleration then will it remain stationary what is the force net force Etc that is the question very nice question let me tell you so in fact let me create a duplicate of this I have to start by drawing the free body diagram and because I'm accelerating the wedge with 2 m/s squar let me choose this as my frame of reference which is going to be now noninertial frame because it is accelerating and it is accelerating with how much 5 m/s squ because of this only the pseudo force will come the pseudo force will be exactly opposite to the acceleration the pseudo force will be Mass into acceleration mass is 1 mass is 1 acceleration is 5 so it is basically 5 Newton this Pudo force will have components one component here another component here by the way this will be 5 this is 37 this is also going to be 37 this is 37 this is 37 so this is 5 cos 37 this will be 5 sin 37 apart from that the actual normal force will be there the gravity will be there whose components will be like this and like this this component will be 1 G sin 37 mg sin Theta this will be mg cos Theta m is basically 1 G is G and this is cos 37 this is what it is so you can see normal force normal force is 1 G cos 37 1 G cos 37 and this is also there check this out 5 + 5 sin 37 just check this out + 5 sin 37 so what is COS of 37 COS of 37 how do I find it very quickly remember this triangle I have told you about this particular triangle 3 4 5 53 will make this 53 53 will make this 53 this angle will be 30 7 this angle will be 37 guys yes so cos 37 will be 4X 5 cos 37 will be nothing but 4X 5 + 5 into sin 37 is nothing but 3 by 5 3x 5 G value if you want you can put it as 10 G value you can put it as 10 5 goes with 10 2 * so it will become 8 + 3 which is basically 11 so the normal force is 11 I can see it is there as B option I won't even check other options because I have got this as the answer automatically others will have to be wrong it's a single correct one it's a single correct one there is no doubt about it cool make sense right moving ahead over here if you want to find what is the acceleration of the block and all of that here I don't think these two forces will be equal maybe we can find it out 5 cos 30 C cost 37 is 4X 5 so 5 into 4X 5 will make it 4 Newton 1 G sin 37 will make it 10 into 3x 5 so which will make it 6 Newton 6 is acting down four is acting on the top net force will be down because of which because of which you will see that it will have some acceleration down so therefore I can say 6 - 4 is mass into acceleration so 2 is equal to 8 so the acceleration is 2 m/s square this is saying 1 m/s square which is wrong it should actually be 2 m/s square 2 m/s square that is also wrong is that clear everybody with me great moving ahead to the next question coming up on your screen if the coefficient of friction is given between the insect and a bow what is the maximum height the insect can crawl this kind of question has come many times in the actual J exam you would have seen this what is the maximum height sometimes values are given how will the diagram look like this is the bowl this is the bowl this is the ground level let's say this is the ground level what happens is as the insert goes up let's say over here inclination will increase inclination will increase and the tendency to slip will increase so think of think of the tangent to this bowl as an inclined plane and there is a maximum angle till which a block can be kept at rest on an inclined plane supported by friction that angle is called angle of repose if you recollect angle of repose is the maximum inclination before which the block begins to slide so what I will do is I'll just draw a particular tangent over here this is the maximum angle which it can make let's say call it Alpha right I call it Alpha and let me draw the other things which are important over here like the radius like the radius over here and all of that and let me also draw the horizontal let me also draw the horizontal over here and the question that is being asked is only this what is this height what is this height so if this angle is Alpha not doubt about it this angle will also be Alpha you'll be like sir how did you get it so quickly I know that if the tangent rotates by Alpha even this point with respect to this point would have rotated by Alpha think when the tangent was horizontal the radius was like this the tangent rotated by Alpha even the radius would have rotated by Alpha with the vertical or else you can just start marking this point this will be also Alpha this will be 90 minus Alpha and so on and so forth either that or through proper visualization either ways you'll get the same answer now once I know this is Alpha then what can I do I know this particular height I know this particular height because this is radius this will be R COS of alpha this will be R COS of alpha so what will this height be what will this height be because the total height is from here to here is R so it will be r - R cos Alpha r - R cos Alpha r - R cos Alpha R take it as common so 1 - cos Alpha okay so the real question now is what is COS of alpha I know that this Alpha is your angle of reos angle of repose and what is the formula for angle of repose tan Alpha is basically mu tan Alpha is basically mu meaning if you think of a triangle having this angle as Alpha tan which is opposite by adjacent is Mu so this hypotenuse will be 1 + mu^ S this hypotenuse will be 1 + mu^ S so COS of alpha adjacent by hypotenuse adjacent by hypotenuse that is the answer so height will be R common 1 - 1 by 1 + root 1 + mu s under the root yes that is the correct answer got it everybody shall we proceed ahead to the next question yes very good Goro G beautiful question which involves a little bit of trigonometry the concept of angle of repose is also there and yes uh you have to use proper algebra and geometry otherwise you can make some silly mistake over here the next question block on block questions are also very famous where you have been asked what is a minimum force to uh just begin the motion or what's the maximum Force so that they don't slide those kind of questions are also very common now these two blocks are kept over each other the bottom surface is smooth friction only exists between the two surfaces of the block you are pulling it by 15 newton so you don't even know whether the frictional force is static or kinetic because nothing has been mentioned whether the blocks are sliding or moving together or just about to slide with respect to each other nothing has been given but one thing is there if I consider them as a whole system if I consider them as a complete system then they will have some acceleration and I can say net force is equal to the total mass into total acceleration so total force is 15 mass is 6 into acceleration so acceleration will be divide by 3 will become 5x 2 m/s square that's the total acceleration if they are moving as a system together now just consider the 2 kg block just consider the 2 kg block and see what are the different forces acting on it obviously one is 2G Force down another is normal force and it will be having this acceleration and it will also have some friction it will also have some friction so no other force apart from this I'm looking at it from the ground frame so I'm seeing it go with two uh with some acceleration of 5x2 m/ second Square so here I can say f is equal to mass into acceleration that friction I don't know mass is how much 2 acceleration is 5X 2 which makes it 5 Newton this is the friction which is actually there between 2 and 4 kg but you cannot be so confident unless you check whether it is more or less than the static friction so just check what is FS Max which is nothing but mu into n mu is.3 normal force will be 2G so 10 into 3 will make it 3 3 2 6 Newton so 5 Newton is less than 6 Newton that means this friction will be static in nature and they are moving together maximum friction is 6 Newton but that much is not needed now you just need 5 Newton if this value comes out to be 7 Newton that means they are not moving together that means you already overcome the static friction it has gone into the kinetic Zone and kinetic friction will also be mu and you will see 6 Newton of kinetic friction will be there between the blocks they'll have relative motion between them right now there is no relative motion right now there is no relative motion keep that in mind next question two blocks are connected by a metal rod 8 kg the system is pulled up by this much Newton 10 tension at the midpoint is asked tension at the midpoint is asked so you know what you can do to find the tension at the midpoint divide the system into two parts and think they are connected by some Force like this imagine this now the upper Mass M1 is how much 20 kg the whole Rod is of 8 kg so can I say this is 4 kg this is 4 kg and M1 is how much 20 kg and M2 is how much 12 kg so it's as if you have a mass of 20 + 4 which is 24 kg 4 + 12 which is basically 16 kg and they both are connected by a force they both are basically connected by a force so there is tension over here and that is exactly what has been asked what is the tension at the midpoint so what do you do now my dear students think to find the tension at this particular midpoint you have to just draw the free body diagram either of this or of this it's completely up to you if you take the free body diagram of that particular 6 kg Mass the forces acting on it are tension and its own weight which is basically 6G and apart from that there will be some acceleration also can we find the acceleration why not why not I I know if I take the whole thing as a system if I take the whole thing as a system 480 minus the total weight which is 20 plus 20 + 8 + 12 G is the total mass into acceleration is the total mass into acceleration which is 20 + 8 + 12 into acceleration does that make sense to all of you does that make sense to all of you the total force will be 480 minus the weight that force will be the total mass into acceleration that's all I have written from this you can find out the value of acceleration my dear Warriors have you understood it till this point just let me know quickly have you understood it till this point just let me know quickly so what will this be 480 minus uh 12 + 8 20 20 + 20 40 so this this will be 40 and into G that means 400 is equal to 40 into a this will be 80 so a will be basically 2 m/s squ once I get this acceleration use this over here like this tension tension - 6G is equal to mass into acceleration so tension will be equal to 6G means 6 into 10 plus 6 2 are 12 60 + 12 that's 72 Newton so 72 Newton I think something is wrong oh I wrote 6G my bad it was 16g not 6 I wrote it as 6G so sorry this was not 6 it was 16 this was 16 yeah so this would be actually T minus 16 G is equal to 16 a and that a value I know is nothing but 2 so from this you will write 16 G + 16 * 2 16 common G is 10 10 + 2 is 12 16 into 12 which is going to be 192 Newton yeah just check this out 16 into 12 that's 192 m is that okay everyone very good awesome awesome so what is the Crux of this problem is not Newton's Law equation f is ma but the choice of system because the force was asked here if I take the entire thing as a system this tension becomes internal so when I choose the entire system I was able to find the total acceleration which I found over here to find the tension I separated them out into two different bodies I either draw this guy or this guy's fbd and then I write again the free body diagram and the Newton's equation f is equal to ma solving that I get the tension so whenever you have to find the force somewhere in between make sure that the force is external by the right choice of the system by the right choice of the system okay I hope this is clear nitish in the previous question why are we taking height which height this one because that is what was the question the maximum height was asked see this the maximum height to which the insect can crawl so this is the height right maximum height till which the insect can crawl is what is asked that is the reason because it was asked okay moving ahead beautiful question again two people are pulling p and Q points of threads like this this mass will go up it is making Theta Theta equal angles the situation is completely symmetric with what speed does this Mass go up if if you're pulling p and Q points down with a speed of U you're pulling it down like this with a speed of U like this with a speed of you with what speed will this guy go up that is the question that is the question what will you do is either visualize or use my technique of net work done by tension is going to be zero that will be a better technique to do over here see at this particular point if you notice there is for example tension T over here and tension T over here then then downwards at this particular Point downwards at this particular point the tension acting is 2 T cos Theta why because T cos Theta will act up T cos Theta will act up so total tension up will be 2T cos Theta which has to be balanced by downward tension of 2T cos Theta is that making sense very good that is at that particular point so using the concept that for a light inextensible thread the net work done by the tension forces is going to be zero net work done by the tension forces is zero work done by the tension over here because the same string continues here will be let's say t let's say t into basically X same thing over here it will be plus T into X because the same string continues tension into displacement here this string is connected to this guy this string is connected to this guy and the tension acting let's say uh you know uh over here if I call it as 2T cos Theta so let's say this is 2T cos Theta the displacement is y and the total sum should be zero 2T cos Theta is the tension and displacement let's say call it y so Force into displacement is the work done but be careful about the signs at all these places like for example for this particular Mass right for this particular Mass the tension acting on it will be like this the displacement is also like this tension and displacement are in the same direction but for this particular point over here right for this particular Point p over here realize the tension is here the displacement X is down so actually work done will be negative so technically speaking both these are negative works and this is basically positive work this is negative work this is positive work so what will you get as the final answer you will get min - 2 x t is equal to or or rather - 2xt + 2T cos Theta Y is equal to Z TT cancels bring this on the other side even cancel two you will get 2 2 sorry 2 has also gone my B just y cos Theta is equal to X now divide this with time divide this also with time y by T displacement by time is velocity so that will give me velocity of the mass into cos Theta is equal to X by T displacement by time will give me displacement of Point P sorry velocity of Point P which is basically U so therefore V will be nothing but U by cos Theta V will be nothing but U by cos Theta which is option number c so the technique which I have used is the total work done by the tension at all the end points will be zero because the string cannot store energy and work done is force into displacement if this string had tension T this has tension T this string will have some different tension because net force has to be zero so tension cos Theta tension cos Theta 2T cos Theta balanced by 2T cos Theta so this element has 2T cos Theta work done on this mass will be nothing but this 2T cos Theta into y that is positive on this guy tension is up displacement is down so negative work force into displacement Force into displacement so solve it you will get the relationship between their displacements divide with time you will get velocity if you divide with time again you will get acceleration that's how I did it is this concept clear how we did it is this clear everyone cool I only studyed 99 percentile series it is not included in that which 99 percen have you watched my old classes as well I hope you have a Pathfinder nurture series how to solve the questions on you know constrain motion it's all there on this channel right so make sure you're utilizing all these resources which which are freely available the theory which has been taught to you next question particle of certain Mass has been subjected to certain force and it moves as per this particular graph and this is Force versus X graph question is what is the velocity at X is equal to 12 at X is equal to 12 hello JGA good afternoon all right yes yes we will be doing we will be doing yes ukuk they were gems Force displacement not force time what does force displacement graph give you I think the area gives you something not the slope if you calculate the area if you calculate the area that will give you the work done okay the area basically gives you the work done and that work done comes out to be half into of height which is basically 10 into sum of parallel sides this side is 8 - 4 which is 4 + 12 correct 4 + 12 16 16 by 2 8 so this will become basically 80 JW as for the work kinetic energy theorem this work will be the gain in the kinetic energy or the change in the kinetic energy which in this case is half into Mass which is 0.1 into V which I do not know initially it was at rest initially it was at rest it starts from rest it is clearly mentioned over here that's why final minus initial which is z okay so basically a is equal to into1 into v² bring that two on the top bring that point1 here it will become 16 0 0 is v² so V is basically 40 V is basically 40 which is option b yes that is the correct answer cool so this was a question on area under Force displacement gives you work and from that you find the kinetic energy and then the speed if by chance this was time then understand the area under this will not give you work it will give you the change of momentum Force time area gives you change of momentum or impulse okay so that's how you would have solved the question so we have come already to work power energy and all that if there is a mass there is this Force how much work is done to move the object from 0 to five a coordinate whenever force is given as a function of x to find the work you just integrate it using this formula work is basically integration of force into displacement if this is force due to only X component this will be also X component or X displacement only so what is that force it is basically 3 + 2 x and this is into DX so 3 integral DX is basically x 2 x DX's integration is basically x² by 2 but you have to put the limits also do not forget that you have to put the limits also 0 5 0 5 so this will become 3 into 5 - 0 + 2 2 cancels it will become 5 2 - 0 square so this will be 15 and this will be 25 which will make it 40 so answer is option b yes 40 Jew 40 Jew is the answer how many of you got it very good good gorov excellent day excellent day awesomeness let's continue very good and I hope you guys have liked the video already and if you have not or if you have forgotten just a friendly reminder please do it right now show your support you're watching all these classes for free you want to gain knowledge but you don't want to smash the like button come on a block of mass 10 kg moves up with this speed it reaches the top and comes back to the same position and stops what is the work done by the friction in the whole process this is a brilliant question if somebody can solve this within 5 to 10 seconds I would be really really proud I'm waiting to see all your answers come on how many of you can solve it within 10 seconds let me just get my water for Naga saying option D May saying option b work by friction guys lot of you might be thinking it is zero but zero is not there as the option it is going up going up going up and it is also again going to come down in this entire process what is the change of the gravitational potential energy guys it went and came back so zero gravity wise there was no change but kind itic wise there was a change kinetic energy there was a change half into Mass which is 10 into V which is 20 and the zero actually final is zero initial was this much so it is 0 - half into 10 into 20 s 20 squ 400 400 by 2 200 200 into 10 2,000 so the change is - 2,000 now think about it potential energy did not change kinetic energy changed in fact it decreased so where did the decrease of kinetic energy go so kinetic energy has gone down where did it go obviously the work done by friction has sucked it away so the work done by the friction has come by the loss of the kinetic energy potential energy did not change so it should be- 2,000 J directly I can say option D many of you wrote B but some of you also said c also but no it is D no work done by friction is not zero friction is a non-conservative force please remember that friction where is it friction is nonconservative and for non-conservative forces the work done in a closed path is not zero so the total work done in a closed path is not zero so even if you go and come back even if the displacement is zero work done is not zero if it was gravity spring electric elastic which are all conservative forces for them the answer would have been zero got it understood the catch in this problem let's move ahead a block is released from some height the maximum compression in the spring is how much that is the question if you release the block from this height right and it is going like this it's a smooth surface the entire energy of potential will convert into kinetic and that kinetic will convert into spring so the transition happens like this there is gravitational potential energy that completely converts into kinetic energy that again completely converts into spring potential energy so why not equate these two rather than these two first and then these two later on gravitational potential energy is MGH H spring is half K x² so x² be 2 mg H by K so X will be root of 2 m g h / K root of 2 mg h / K which is option number a got it my dear students exactly so you don't have to equate these two and these two oh sir what is the speed here then from speed I'll find compression no I know this goes into this this goes into this directly equate these two MGH is half KX squ you can solve for x very beautiful question there is a uh you know a track which has some part smooth some part you know which is uh having some friction the flat part is 3 m the curved portions are smooth only the flat part has a friction of2 coefficient if you release this Mass from this particular location a 1.5 M where will it stop okay you know what I do to solve this problem I'll imagine it like this when I release it it comes here all the energy of potential gets converted into kinetic now that kinetic energy will be slowly sucked away by the friction because even if it goes here it will go up stop and come back but you know it will come back with the same speed here is where it loses that kinetic energy so how much precise kinetic energy it has how much kinetic energy does it have the same amount of potential energy does it have so which is MGH h mass is how much I think it is not even given or something it's not needed I guess okay M into G which is 10 into basically H which is 1.5 so it is basically 15 M now this kinetic energy slowly goes away against the friction this kinetic energy slowly will go against the work done against the friction and the work done by the friction is the frictional force into the displacement frictional force is Mu mg and displacement is s mu is basically 2 Mass I don't know G 10 displacement also I don't know 10 into 2 is 2 so 2 Ms is the work done by the friction or the which is going to be sucked out by the friction and this will be equal to also 15 M mm cancels mm cancels so from this I'll get S is 15x 2 which is 7.5 the total distance traveled on the flat track is 7.5 now you know what to do next the flat track which is having the friction is nothing but how much in length it is given 3 m so first time when it comes it will go 3 m like this and it will still not lose all the energy because totally it has to travel 7.5 M next time it will travel like this nothing will happen remember it has come from here gone like this it will go up and it will again come down by the same amount it will again come down by the same amount then again over here it will go for some amount up and again it will come down by the same amount and now again it has covered 3 m total 6 M only 1.5 m is left worth of travel so it will travel only for 1.5 M and finally it will stop so it will stop exactly in the middle the distance be will be 1.5 so 1.5 into 2 will make it three so the answer is three answer is three because the question is two times the answer is how much got it everyone very good yeah moving ahead to the next question if there is a uniform chain of 3 m in length mass something it is hanging with 2 m lying on the table the kinetic energy of the chain when it completely slips off the table is how much let's try to visualize this first two situations this is the table before this is the table afterwards first 2 m was lying on the table that means most of it was on the table only a part of it is hanging below like this later on the entire thing is like this the entire thing is like this so if I assume this to be mass m this will be mass of 2 m this will be mass of 2 m this total chain will have a mass of 3 m correct understood everybody oh actually the mass is given also and the mass is given to be how much much 3 Kg so this is basically 3 Kg so only 1 m is hanging means 1 kg was hanging here and the remaining 2 kg was over here perfect now that I know it was at rest over here but at this situation it has some speed where did it get that kinetic energy from how did it get ktic energy H think about it because it has lost potential energy how sir okay let's do one thing where is the center of mass for this particular Rod or chain every time this kind of question comes this has to be solved by a center of mass the center of mass is here in fact the length of the chain is how much 3 m the length of the chain is 3 m so technically the center of mass is 1.5 M below the table Point 1.5 M below the table point for here most of the rod or the chain is on the table only this part is hanging so that gu's Center of mass only you consider this one kg Center of mass this is 1 M this is 1 M so technically the center of mass the center of mass is only5 M below the table is Just5 M below the table that to only for the 1 kg Mass so if I have to write an equation saying like this initial potential plus initial kinetic is final potential energy plus final kinetic if I take this level as zero level this level as zero level of potential energy this part of the mass has no energy because it is on the reference line you can choose anything as the reference this part is below the reference so instead of plus MGH you'll use minus MGH so the initial energy is only of this part that to negative because it is below so it will be minus Mass which is 1 G which is 10 into H which is basically .5 kinetic energy is basically zero because it was at rest finally the entire mass is 3 G is 10 height is 1.5 but negative why negative because it is below the zero reference line plus half into Mass which is 3 into v² that's it solving this you will get the answer everybody with me understood Clearo what we had to do the main concept is about Center of mass only the mass which is hanging consider that part as the mass and find the center of mass of only that part reference level take it as a table accordingly all the masses which are below will have negative potential energy now you have to just solve this so this will be nothing but 5 with negative sign this will be 3 into 1.5 or 1.5 decimal will go 15 3 are nothing but 45 and this will be nothing but 3x 2 v² so take it on the other side this will become 40 is = 3x 2 v² this will become 80 by 3 is = to v² so V will be basically otk of 80 by 3 V equal to root of 80 by3 right that is what you'll get but I think here they had asked what is the kinetic energy not the speed so if you look at this quantity over here if you just look at this particular quantity over here this is basically your kinetic energy this is basically your kinetic energy so the value of the kinetic energy comes out to be 40 JS the value of the kinetic energy if K is the value of the kinetic energy that just comes out be 40 J Speed if it was asked it would be root of 80 by3 Right Moving ahead shall we go ahead to the next question guys here it comes a body of certain Mass accelerates uniformly from rest to V1 in time T1 the instantaneous power to the body as a function of time is how much it's going from rest to certain speed in time T1 let's visualize it's starting from rest goes to this velocity V1 in time T1 definitely there was some acceleration in fact that V1 I can say is U plus acceleration into time therefore acceleration was V1 by T1 now the question is what is the power as a function of time so at any random instant you take any random instant when it has certain velocity that velocity will be U + a into that time acceleration is V T1 by T1 and time is basically T this is velocity at any instant I will need it because I want to find the power I want to find the power and what is the formula for power power is force into velocity force is mass into acceleration mass is M only acceleration is V1 by T1 velocity is V1 by T1 into T so this will become M into V1 by T1 s and this into t so V1 by T1 whole square into T that is option number B yes that is the correct answer that is the correct answer shall we go ahead no it is not D my dear students I think you missed that t you missed that one T1 squ so be careful T1 is there here also T1 is there here also it'll make it T1 Square below denominator beautiful question on power power is force into velocity that is the formula which we need to use next question two blocks are there connected by a spring if both are given some speeds what is the maximum elongation of the spring that is the question what's the maximum elongation of the spring okay so if you give both these blocks certain velocity what will happen the spring will start elongating and finally when they stop finally when they stop all the kinetic energy because both of them had kinetic Energies will get converted into the spring potential energy why two because there are two blocks what is the kinetic energy half M v² and spring energy is half K x² question is find the elongation okay 2 two cancels mass is given as 250 g 250 g means 25 kg into speed is it given no it's just V is equal to half spring constant it is given as 2 into x² 2 2 cancels so basically 25 is 1x 4 V is nothing but x² so X will be V uh sorry this was v² V by 2 x is nothing but V by 2 which is yeah B option b for it bombing just check this out no it is not D B option yeah cool so this two is very important many people miss this two they just write kinetic energy got converted into potential but both the blocks had some speed both the blocks had some speed so 2 into kinetic energy is converting into spring because once they get fully extended once they get fully extended they would have stopped they would have stopped keep that in mind next one conical vendum based question length is L it is fixed at one end it is carrying some Mass the string makes 2x Pi revolutions per second then question is find find the tension in the string this is a direct formula based question tension formula is M Omega S L for a conical pendulum the only question is what is Omega remember that your Omega is nothing but 2 pi f 2 pi F right it is nothing but 2 pi F and it is given that it is making 2 by pi revolutions per second so frequency itself is 2 by Pi Pi Pi cancels so Omega will become four Omega will become four so you just put the number four over here have to square it and L so this will become 16 ml 16 ml should be the answer which is D option yes it is correct Omega is 2 pif it's already given it makes 2 by pi revolutions per second you'll get Omega is 4 substitute here you'll get the answer right shall we move ahead to the next question guys ready for it very good uh Karthik you don't know this formula I'll tell you where to find it how many of you have seen my formula revision class how many of you have seen entire formula revision class I had taken this I had taken lot of effort just do one thing just search for physics formula on the channel just search over here you will see you see this lecture complete physics formula revision J Mains 2024 click on this new syllabus based in that I have given a cheat sheet of all the formulas all the formulas okay and I don't know how many of you know this special gift which I had given in this session wait a minute yeah this is a big session almost 7h hour class the biggest advantage of this class is not just the video but also this PDF which are given it's a cheat sheet click on this link click on this link enter your details you'll be able to download it for free of cost okay so use this complete PDF of the formula sheet got it understood clear loved the surprise loved the gift I hope you now will download it yeah yeah I'll be taking it as a YT class correct yes s CA definitely cool moving ahead to the next question a frictionless banked track has a curve of radius this much the speed of the car is this much what should be the banking angle what should be the banking angle the formula that you need to use is Optimum speed formula is root of RG tan Theta Optimum speed formula is root RG tan Theta the optimum speed is 90 < tk3 and radius is uh uh what is it oh sorry the speed is my bad speed is not 90 < tk3 the speed is 30 speed is 30 radius is radius is 90 < tk3 radius is 90 < tk3 G is basically 10 and uh what else is there uh angle angle is what I do not know so Square both both sides this will become basically 900 is 900 < tk3 tan Theta because 10 into 9 is 900 900 900 will cancel root3 you bring it down over here that will be tan Theta naturally Theta will be how much 30° Theta will be basically 30° uh what is the bank angle why have they put 60° oh the there is there's a misprint over here if you look at the solution they have put 90 byun3 so over here I think one slash was missing maybe that's the reason why the answer given over there is 60° but if it is 90 < tk3 answer is 30 there is a slash missing then what will happen that root3 will go on the top that means Theta will become 60° cool is that fine right let's move ahead to the next one a block of mass at the end of the string is World in a oh my God already the answer is seen a block of mass is World in a circular path vertical circle of radius are the square of the critical speed at the top so that the string would slacken before the block reaches the bottom okay think about it the critical speed when you are in a vertical Circle and you are at the topmost position there is a critical speed for which the tension roughly becomes zero it almost becomes you know slack almost just is on the verge of becoming slack that critical speed is root RG if you remember so if you square that critical speed it will become just RG in fact at the bottom the speed is root 5 RG root of 5 RG right and when it goes on the stop top slowly the speed reduces the speed reduces and finally it becomes root R right at the bottom it is rooot 5 RG at the top it is root RG many students end up writing the answer as option A which is wrong which is wrong okay Clearo shall we go ahead here it comes next question if there is a body at rest and it starts sliding on a frictionless track from some height if it is able to complete the vertical circle of diameter D the height H is how many times the diameter it's again based on the previous concept that you need to give it certain speed so that it is able to complete the vertical Circle in fact it so happens that the speed at the bottom it so happens that the speed at the bottom should be root of 5 RG for it to complete the track now remember radius is nothing but diameter by 2 radius is nothing but diameter by 2 so that is the minimum speed required at the bottom where will the speed come from it comes from the potential energy so I can say the potential energy gets converted into kinetic so potential is MGH H kinetic is half M v² so it will be 5 D G by 2's root square the root vanishes m m cancels m m cancels even GG cancel so H will be M into 5 sorry M has canceled out 5 into uh two and this D is also there and this 2 is also there so this will become 5X 4 d so height is 5x 4 d which is the fourth option D option right over there yes very good Niti Sanjay yes it is not B CNU be little careful ni it is not B it is option D how to derive Optimum speed formula you have to use you have to use the free body diagram only then you get it basically over here the tension is zero there is mg and circular motion means circular Accel cental acceleration so mg will be equal to mass into cental acceleration tension is zero so no tension calculation mm cancels so G will be AC centripetal acceleration is v² by R so therefore v² is RG this is how you get it using Newton's Laws diagram circular motion concept cool let's move to the next one two small particles equal Mass start moving in the opposite directions their speeds are V and 2v as shown between the collisions the particles move with constant speed after making how many elastic collisions at that other than a will these people come back to point a I'll explain this question these two start from here with different speeds they will collide somewhere here maybe then they will exchange their velocities maybe they'll Collide somewhere here maybe then again they'll Collide here maybe again they'll Collide here maybe Collide here and then finally again Collide here eventually come back to point a so the best way to solve the problem I'll tell you what we can do just imagine first situation second situation third situation fourth situation I'll just show some circles just to make you understand it just started this is going with v this is going with 2v where will they Collide think about it where will they Collide after how much angle will they Collide think about it like that okay if you're able to guess it very good if you're not able to guess it no problem this is moving with v this is moving with double the speed so this will cover double the angle as this one this will cover double the angle as this one before they eventually meet so imagine this moves X this will move 2X but the total but the total angle by the time they meet somewhere should be how much 360° should be basically 360° so if a goes if this ball goes x amount this would go this would go 2x amount before they meet this would go 2x amount before they meet and X De + 2x de is 360 60° so 3x is 360° so X is 120° so if they start their journey together in opposite directions after how much degree will they meet 120° they will meet and one thing is standard guys if there are two masses going in the opposite directions with speed V and 2v after the Collision this guy's speed will be 2v and this guy speed will be V they will exchange their velocities right they will just what exchange the velocities in elastic Collision in case of elastic Collision this is the property for equal masses use this concept over here these masses will finally come over here you will see one is coming like this and the other one is coming like this with 2v and V speed where at 120° this is just before the Collision after the Collision you will see they would have exchanged their velocities so this will get 2v speed and this will get V speed this will get V speed correct now this is just like this diagram v2v so what will happen this will move more angle this will move less angle and after 120° they will meet again 120° relative to this that means they should meet over here that means they should meet over here so think about it think about it they have covered 120° and they will meet over here this 2v will come like this this V will come over here like this just before the Collision just before the Collision they will again exchange their velocities this guy will have V velocity and 2v velocity will be there over here perfect now this diagram is just like this diagram so this will move more this will move less this will move more this will move less and you have to cover 120° so that they meet again you will see that this 2v would come here approach like this this V would approach like this so that they would have moved 120° again that's it this is the final time where they come back so guys they have collided once here and they have collided second time over there so there are basically two collisions before they meet at Point a before they meet at Point a this was the first Collision right this was the first Collision which happened this was before this was after this is again before and this is after this is the second Collision after this would be the third Collision after this would be the third Collision So Gone from here to here to here to here to here and so on and so forth beautiful question isn't it the concept is only very nice let's let's go ahead to the next type of questions on gravity two particles distance D apart third particle in between no resultant gravitational force question is what is the distance of the particle from M1 let's draw the diagram to understand what is happening there is a mass M1 here and the distance D apart you have placed Mass M2 somewhere and you have kept a mass over here you have kept a mass over here there is no net gravitational force on that Mass which is kept on the mass which is kept right so you have kept a third particle on the line joining them and there is no resultant Force meaning what would be happening is this M1 would be attracting it with Force One this M2 would be attracting it with Force 2 these both are equal that is why the net force is zero that is why the net force is zero what is F1 what is FS2 well for that let's assume this as X distance this will be basically D- X this is D this is X this is X D minus X fub1 can I say it is G M1 m / x² G M1 M2 by distance Square what is Force 2 G M2 m / d - x² G and M G and M got cancelled so M2 I can bring it down so M1 by M2 will be uh or else we'll do one thing let's take D- X squ on the top so D- X whole Square ID x² this bring it on the top M1 you shift it down M2 you keep it as it is GM GM has got cancelled take roots on both sides so d - x by X is equal to M2 by M1 whole thing under the root so take uh separate out the denominator ators and numerators like this split the terms so d by X - 1 is root of M2 by M1 shift one on the other side so d by sorry d by X is equal to 1 +un of M2 by M1 take LCM so this will become root of M1 + root of M2 by root of M1 so what will X be equal to shift X on the top take these two people below so X will be equal to dunk M1 byun M1 +un M2 that should be the answer that should be the answer yes D root M1 upon root M1 + root M2 this is the correct answer cool Splendid question wow ah wow looks like we are going to have a matchmaking process very soon guys very good let me tell you the kind of relationships that you might get over here is nowhere comparable to the relationships that you might get when you get out of I or nit or any other go Government College so please stop wasting time be a sigma don't be a gamma or something really whatever whatever nonsense you are doing here is of no use nor are you going to get the inst ID nor are you going to get any pleasure nor are you going to you know do anything great by doing this nonsense over here so might as well study go to a good college and definitely you'll get much better options than whatever you are getting over here yeah cool let's move on to the next one coming up on your screen and here it is find the depth and the height from the surface of Earth where acceleration due to gravity is half the value of the surface question on the formula application let's do separately first let's go down then let's go up when you are at depth D the gravity formula is g 1 - d by R because this is given to be half so I'll put G by 2 1 - d by R this is G on the surface D is the depth R is the radius of the planet GG cancels half is equal to 1 - d by R so d by R 1 - half is a half so D will be basically R by 2 r i take it on the top so R by2 I can see either in b or in a so it should be one of them either A or B for the next part for the one on the top when you go on the top the gravity at height H is given by GM by GM by r + h² GM you can also write it as GR squ that's a standard formula GM is g r² r + h² but this gravity is given to be again half the value on the surface so GH is G by2 = G r² by r + h squ g g cancels take roots on both sides so 1 by < tk2 is R upon r + H shift it so you'll get r + H is R < tk2 cross multiply okay and that's what I've have done so therefore H will be R Comm < tk2 -1 R Comm < tk2 - 1 that is option number a cool everybody yes very good Sanjay Mina yes danh lakmi very good so after 10th I would want to take sir please tell me what do you want to take bunny with bunny Allin one shorts what do you want to take please tell me PCM pcmb engineering what do you want to take coaching the change in the value of g at a height H is the same as depth D below both are smaller than the radius which is correct which of the following is correct Okay cool so think about it my dear Warriors what will the formula be when you are going down the gravity at depth D is g into 1 - d by R take G separately over here so or else um you can do one thing take GD on that side so G minus GD this GD by R take it on the other side g d by R take it on the other side this is change in the gravity when you go down by some depth so that comes out to be g into D / R this is the formula but when you go on the top but for a small height the formula is not the same the formula is not the same don't use this big nonsensical formula use a smaller formula which comes after some derivation using binomial approximation I have given that to you in the formula session also that comes out to be as g 1 minus 2 H by R so what you can do is just keep G as it is over here bring that GH over here bring G2 H by R on the left hand side so G2 H by R over here G minus GH is Delta G is nothing but 2 GH by R as for the question is the uh the change in the gravity is the same whether you go below or whether you go on the top as per the question is the same so that means this is the same as this so basically these two should be equal so what will you get g d by R from here is equal to 2 g h by r g r gr R cancels so what will you get D is equal to 2 h d is equal to 2 H where is it option number D got it D is equal to 2 H very good uh no today is only 11th right I had mentioned this clearly in the launch video Al and then we had uh planned to do 12th problems cool let's keep it fast guys coming up on your screen here is the next question the radius of a planet is n * the radius of the Earth a satellite revolves in a circle of radius 4 NR with angular velocity Omega the acceleration due to gravity on the planet's surface will be how much that's the question think which is a Formula which we'll need clearly it is mentioned that you know the radius of the planet is n times the radius of the Earth so new planet has more radius end times satellite revolves around it in a circle of radius 4nr with some angular speed and the acceleration due to gravity on the planet's surface G on that planet will be how much okay so if angular velocity changes because the radius has changed can I relate it to G I think so see whenever a satellite is going around a planet there is a centripetal force which is acting with because of gravity and it causes centripetal acceleration as it revolves with some angular velocity Omega so can I say the gravitational force is M into Omega squ into radius R so this radius happens to be 4 NR 4 * NR and the gravitational force will be G mass of the planet mass of the s upon upon basically 4nr that's the distance of the planet till the satellite two planets distance it's nothing but yeah this one right so what will I do I'll do one thing small M Small M cancels g m will be Omega Square take this over here 4 NR whole Square over here it will become 4 n r whole Cub now this is the interesting part because what is capital GM I use that substitution even over here if you realize GM is gr Square GM is gr Square it's a standard formula which you should know so I can just write it as g into planet's radius square is equal to Omega Square oh but wait a minute the planet's radius is also n * of Earth so what do we do should I put g into n n r whole Square should I put g into NR Square no this GM stands for understand the uh you know acceleration due to gravity on Earth into basically the radius square of the Earth this is a standard formula Capital GM is small gr R square is a standard formula understand irrespective of what planet you are taking you always use it as a Conant and write it as small g into radius of Earth whole Square always so this will be equal to your Omega squ and 4 NR whole Cub is that clear my dear Warriors is that clear my dear Warriors very good so now you just have to substitute it over here and you will get the value of uh G on that particular Planet so just solve this and you are going to get it what is the answer that you get my dear Warriors you can see 4 Cube automatically you can see the number over here 4 cube is nothing but 64 yes you should get the answer as 64 over there 64 over there perfect Clearo shall we move ahead to the next question coming up on your screen right what is the minimum energy required to launch a satellite of mass m from the surface of the planet of mass uh capital M radius R in a circular orbit at an altitude of 2 R that is the question yes so let's see if we can do this let's see if we can do this first of all you are having a mass on Earth on the surface and then finally what you want to do is place it in the orbit place it in the orbit over here when you are on the surface the potential energy only is there there is no kind itic energy so the total energy which is there initially without kinetic is nothing but minus G mm by R nothing else when you bring it on the orbit the total energy of a satellite is minus half G mm divided by the radius of the orbit it's clearly mentioned that the altitude is 2 R the altitude is 2 R that means the radius that means the radius is basically 3 r so this will be 3 R over here standard formula for the energy of a satellite it is - half GMM by 2 into radius yep is that clear very good so now what will you do now what will you do just take the difference of these two just take the difference of these two because that difference is what is the energy what you need so the Delta e value will be nothing but uh - g m m by 2 into 3 r- - g m m basically by R so G mm by R is common minus minus becomes plus and this minus is there as 1X 6 1 - 1X 6 is 5X 6 so 5 G mm / r that is what you will need yes correcto yes 5x 6 gmf M by R 5x 6 sorry the 6 was there 5x 6 GMM by R is that clear everyone with me cool shall we move ahead to the next one here it is the work done to raise the Mass from the surface of the Earth to a height which is equal to the radius of the Earth is how much okay very similar question just that the difference is here the mass was in orbit the mass was in orbit that's why you use this formula for the energy of a satellite right but here it's not in the orbit you just taken it till height H so the work which is done is the change in the gravitational potential energy the gravitational potential energy at height H is minus G mm by the distance which is r + H and initially it is just Min - G mm by R so when you are at when you are at a height which is equal to the radius of the Earth r + r will make it 2 R minus minus will make it plus so G mm by capital r so this will become 1 -/ which is half so it will become half g m m / R but the questions options are not having capital G but wait a minute GM is gr squ remember that constant which I told you is nothing but gr r² into M divided by basically 2 r r r cancels I'll have mg R / 2 MGR divided 2 which is option number B of course option number B is that clear shall we move ahead no it is not D it is option number B be little bit careful be little bit careful yes let's move ahead to the next question coming up on your screen the Escape speed of a body which is thrown up is 11 if the body is thrown at an angle of 45° then what do you think is the escape speed then what do you think think is the escape speed that's the question come on think about it my dear students think about it my dear students what do you think is the answer going to be it doesn't matter exactly very good some of you got it no these are not pyqs this could be inspired from pyqs but some of them by coincidence might be pyq because last 3 4 years you see there are like you know every year we have close to you know 10 to 15 shifts so so many questions right so some some things might get repeated here and there yeah so it does not depend on the angle it does not depend on the mass it just depends on the mass of the planet right and the radius of the planet that's all that matters yeah correct okay right now the next question says the two particles of equal Mass go around in a circle under the action of mutual Gravity the speed of each particle is how much how to do this particular question uh this is an exclusive problem solving session if you want to watch any of the concepts you can watch my any of the older classes number one number two there is a link in the description box at 9.99 rupees you can take the crash course you can clear all your Concepts over there also both the options are there with you okay so in those crash course classes concept and problems concept and problems happen but if you think that no I want to see only from YouTube that option is also there with you where all my older classes I have explained all the theory part everything this is exclusive problem solving just 2 weeks away before the exam now is not the right time to teach all the concepts from all the chapters if you have a need for a particular chapter watch that video or use the crash course video both options I have given you so over here when these two particles are separated by some uh distance and they are going in a circle of radius R they're going in a circle of radi radius R okay so let's say this is the center this particle goes like this with some speed this particle goes like this with some speed they are going like this this is going like this okay they are going in a circle like this okay I just shown their motion why are they able to go in a circular motion because they attract each other number one with some force that Force acts like the centripedal force you have thrown it with some speed the force is there so the speeds are always perpendicular to the force and that force is always attractive because of gravitational pull so these two masses always keep pulling each other by gravitational pull and that acts like the centripetal force which changes their direction does not change their speed so I can simply say my f is equal to M into centripetal acceleration that force between them is g m m both the masses are equal distance is is please mind it it is not R but it is actually 2 R it is actually 2 R right it is basically 2 R be little bit careful over there the two masses are separated by 2 R diameter as the distance Mass into cental acceleration centripetal acceleration I can just write it as v² by the radius of the orbit Okay small Mass small Mass one of the r just gets cancelled out so what am I left with GM is there four is also there 1 R is there v² is there this R has canceled out so therefore V is equal to root of GM by 4 R root of GM by 4 R so that 4 can be taken outside as 2 4 can be taken outside as two no it is not B option Mina yes game over Tam it is option number c correct moving ahead to the next question coming up on your screen okay we are going to center of mass and rigid body mechanics beautiful question two masses are going at an angle over here okay on inclined planes the question is the question is if both of them are left what is the combined Center of Mass's acceleration because these two masses go on the inclined planes they will have their own accelerations which is just given by G sin 45 and this is also going to have basically G sin 45 correct each of these G sin 45 will have components one vertically one vertically and one horizontally one horizontally like this but the horizontal accelerations will not cause any motion of the center of mass this Mass goes like this this Mass goes like this components are there the horizontal motions are opposite and equal they will not cause the center of Mass to accelerate anywhere but the ones in the vertical direction that will cause the center of Mass to go down vertically only so it will definitely go vertically down upwards does not even make sense eliminate it so what is this acceleration it will be G sin 45 45 component or cos 45 component which is G 1X < tk2 into 1 by < tk2 which is basically G by 2 so this is G by2 this is G by2 even this is going to be G by2 so what will the center of mass come down by G by2 which is 10x2 which is 5 so the acceleration will be 5 m/s squar uh 1 second H now this is how where I wanted to tell you that there will be a catch over here and lot of people make a small mistake saying that sir this is G by2 this is G by2 so the answer should be you you know 5 m/s square but you need to be little careful over here you need to be little careful over here that when you are taking components when you are basically taking components right this y component this y component that you are going to get uh I don't know why they have not given over here that's five it should have been five M I think they have made a mistake over here guys ignore this I don't know why they made a mistake over here this is my G sign 45 component smooth double incline blocks and this was the component of that yeah so guys the answer is five only it is not 10 ignore that this is five now the thing which I wanted to tell you is Sir can I use this formula Acom is nothing but M1 A1 + M2 A2 by M1 + M2 yes you can use this no problem but it will unnecessarily be lengthy reason being this mass is equal this mass is equal this mass is equal both the masses are same this is also M this is also M so M M M M cancels so you'll have A1 + A2 by 2 now A1 and A2 themselves A1 and A2 themselves are equal to each other A1 and a two themselves are equal to each other so it will become 2 A1 by 2 basically it will become A1 basically it will become A1 why are the accelerations equal because masses are equal angles are equal it's a symmetrical scenario so you will get the center of Mass's acceleration as even so there was no need to do this lengthy method I hope this is clear is that understood there was no need to do this lengthy method so both these masses were going down like this it will have one horizontal and vertical components this horizontal components no use if they go like this what is the point they might just keep on going like that the center of mass will neither go there nor go here because both the masses are going down the center of mass will also come down along with it it will also come down along with it keep this in mind so this component will be G sin 45's cos 45 component that's why G sin 45 into cos45 five understand this so we have the acceleration we are taking a component of that acceleration how much is that y component so that is why it is this guy's cos 45 component this is your A1 which is also equal to A2 this is your A1 which is also equal to your A2 is that clear everybody understood very good very good thank you prad SAS great let's move ahead to the next question here it comes a semicircular ring of mass m and a disk of mass 2m are joined together as shown find the center of Mass from the geometrical Center this is a very unique kind of question because many of you would have seen circle is there square is cut out and all those things this is a semicircle and then there is a uh semicircle but it is solid so like a disc Ender ring where would the geometrical Center be the ring has a mass of M so let me put M over here the dis has a mass of 2 m be careful the disc has a mass of 2 m now where does the center of mass of a dis or a ring lie remember there was a formula if you have an arc if you have an ark and in this case the ark subtains Pi radians the center of mass for a ring this distance is given by R sin Theta by 2 by Theta by 2 sin 90 is 1 so 2 goes on the top so it will become 2 R byun but if it is a solid dis if it is a solid disc this angle is pi radians the center of mass will be here it is at a distance of 2x 3 R sin Theta by 2 by Theta by 2 2 goes on the top becomes 4 by 3 R sin 90 is 1 4 R by 3 < that is the answer 2 r R by pi and 4 R by 3 Pi for a ring and this is for a dis semicircular disc use these in the previous problem so basically you will see this distance is 4 R by 3 Pi by 3 pi and this distance this distance is just 2 R by Pi this distance is 2 R by pi so you have now converted it into a two point Mass system separated by certain distance this is the y-coordinate up and this is the y-coordinate down masses are also given where will the where will the center of mass B so just use our normal regular formula from the geometrical Center you have to find the distance so the y coordinate of the center of mass will be M1 y1 so 2 m into y1 which is 4 R by 3 piun into M2 Y2 so + m into Y is negative so- - 2 R by piun upon M1 + M2 that's it so 2 m + m m m m cancels everywhere so you will have 8 R by 3 piun minus here you will have 2 R by piun whole thing divided 1 + 2 is is basically 3 so you can see the LCM over here is 3 Pi so it will become 8 R by 3 piun - 6 R by 3 piun the whole thing divided by 3 so you will definitely get 9 piun in the down denominator 3 piun into 3 will make it 9 piun 8 r - 6r is basically 2 R so basically 2 R by 9 pi should be the answer 2 R by 9 pi should be the answer yes it is there you go perfect got it everyone yes we have finished more than 50 plus questions if I'm not wrong yeah 50 plus questions 100% we would have finished a similar kind of question can also come where you have taken out something or you have added something and then you have to find where the center of mass lies these kind of questions are also common I'll just give you the hint for these kind of question what you need to do is you need to use the formula X Center of mass is M1 X1 - M2 X2 upon M1 minus M2 where you have taken out something now you are worried what is X1 what is X2 what is M1 what is M2 M1 is the total mass M2 is the removed mass the mass which has been taken out because it's a circular dis the masses are proportional to the area so Mass one which is of the entire circle will be mass per unit area I can call it Sigma into the total area which I can say p piun r² Mass 2 is the mass which has been removed over here see think about it this is radius R this is side this is also side this is also side this is also side the radius is the diagonal the radius of the circle is the diagonal of the square so remember the diagonal was always < tk2 * the side so therefore what will the side be diagonal by < tk2 what is the diagonal R so the side length is basically R by < tk2 so what is the area of it Side Square so R by < tk2 squ into Sigma will give me mass of the square which comes out to be Sigma r² by 2 I got to know M1 I got to know M2 is this Point Clear how we found it mass will be proportional to the area more the area more the mass always write in terms of density if you are worried sir what is this Sigma it's mass per unit area will it bother me no because when you finally substitute it here everywhere it will get canceled from numerator and denominator so don't bother about it lastly what is X1 and what is X2 X1 is the coordinate of the center of mass of the entire circle which is the origin so it is zero what is X2 2 is the center of mass coordinate of the square it is here if this is R this is R by2 so I'll just put R by2 over here X1 X2 M1 M2 everything you know substitute you will get the answer after substituting everything you'll get the answer as R by 4i - 2 you'll get R by 4i - 2 is that clear everyone yes very good awesome let's move on to the next question coming up on your screen a ball of mass 50 g is dropped from certain height it rebounds losing 75% of its kinetic energy if it remains in contact with the ground for this much time what is the impulse of the impact Force this ball was dropped from certain height it rebounds to 75 uh no it rebounds losing 75% of its kinetic energy that is very important so if the kinetic energy was K when it rebounds the kinetic energy was K by 4 because it has lost 75% 1/4 is only remaining 1/4 is only remaining correct so if the speed over here was V kinetic energy becomes 1/4 kinetic energy is half M v² Speed will become V by 2 how did I realize this think about it kinetic energy is half M v² so 1/4 kinetic energy will be 1 by 2 m new speed square because the kinetic energy has been 1/4 speed will have to become half if you don't agree just substitute it over here just substitute K as half M v² which is half M new V1 s so half M half M cancels so v² by 4 is V1 s so V1 is V by 2 so if it hits the ground with speed V it will rebound back with speed of V by2 so to find the impulse G which is change in momentum it's Mass into final velocity which is V by2 minus Mass into initial velocity which is V but there is a catch put a minus sign here also because impulse is a vector even momentum is a vector signs have to be considered many people miss the minus sign MV by2 final momentum MV initial momentum but opposite in direction so minus of that so minus minus becomes plus MV by 2 + MV is 3 MV by 2 Mass you know um velocity you know I think you can find the answer velocity is < tk2 G mass is 50 g so velocity is root of 2 GH height you know 10 m right Mass you know is 50 g substitute everything you will get the answer as 1.06 as the answer cool will you be able to substitute and get it understood the concept behind it main problem happens here second problem happens here that many people read it as the final energy is 75% rather than reading it loses 75% so only 1/4 of the energy is left 1/4 of the energy is Left Right very good let's move on to the next question coming up on your screen a wooden plank is basically resting on the floor a man of mass 60 K starts moving from one side to the other side find the magnitude of the displacement over the floor when the man reaches the other end of the plank this plank man questions are very very common let me tell you that so let's visualize this quickly so there is this plank whose length is basically 10 m and there is this person over here okay and the mass of the person is 60 kg and the mass of the plank is 20 kg so what happens as soon as the person moves the plank moves behind the plank moves behind right the center of mass has to be conserved it will stay there so when the person reaches here the plank has moved here the person moved here the plank has mov here that's all question is find the magnitude of the Plank's displacement the person goes to the other side first of all realize this if you sit on the plank and observe the entire motion from plank frame the relative displacement of the person displacement of person with respect to the plank is displacement of the plank sorry person minus displacement of the plank V12 is V1 - V2 a12 is A1 - A2 S12 is S1 - S2 that related displacement imagine you are the plank you look at the person the person has gone 10 m ahead yes you have gone little bit behind but for you on the plank on the plank the person has moved how much meters 10 m for the ground it is completely different story from here to here the person has moved 10 m relative to the plank so this 10 is nothing but S1 - S2 another formula which you need to use is for plank person problem or balloon monkey problem is M1 S1 + M2 S2 is equal to 0 Mass 1 mass of the person is 60 displacement I don't know Mass 2 20 displacement of the person uh sorry plank I don't know this is what the equation will look like perfect so 60 S1 so cancel 20 so this will become 3 S1 + S2 is equal to 0 rather S1 S1 will be become - S2 by 3 S1 will become - S2 by 3 substitute it over here instead of S1 put it as - S2 by 3 that's all what will you get - S2 is equal to 10 1X 3 + 1 is 4x3 so - 4x3 S2 is equal to basically your 10 out here perfect right so this will basically cancel here this will become two this will become 5 so S2 will become S2 will become minus 15 by 2 which is - 7.5 M so the plank will go 7.5 M behind because the person has moved ahead because the person has moved ahead here if you ask me what is S1 S2 these are the actual displacements of the person and the plank as observed from the ground frame if you sit on the plank you will only see the relative motion so the person has gone from this side to this side the separation between them is just 10 m that is why S12 I put it as 10 m nothing more nothing less okay I hope this is clear shall we move ahead to the next question here it comes then 7.5 M yes the moment of inertia is given as this the body is at rest in order to produce a rotational energy of this much the angular acceleration must be applied about the axis for a duration of T seconds find the value of that time that is the question okay so first of all inertia is given rotational energy of 1,500 uh is also given angular acceleration is also given question is how much time do you need to apply it so kinetic energy formula is half I Omega squ we know this kinetic energy is 1,500 moment of of inertia well that is 1.2 Omega I don't know bring it on the top so this will become 3,000 is equal to 1.2 Omega Square so Omega Square will be bring it over here 30,000 / 12 I can write it if you want this will become 4 so this will become 10^ 4 by 4 or Omega will be 10 s by 2 which is 50 Omega is 50 once I get Omega I can find the acceleration and time relation because by kinematic equation final angular velocity is initial angular velocity plus acceleration into time final is 50 initial is 0 Alpha is 25 time I don't know so T will be 50 by 25 which is basically 2 so the answer will be 2 seconds yes it is 2 seconds perfect got it my dear students amazing right these questions are very good let's move ahead let's move ahead we're trying to practice as many questions as possible a certain block some Mass hits another block unknown Mass if it is elastic and the first block continues to move with three what is the mass of the other block just draw the diagram and use conservation of momentum I feel so 172 G moving with 5 m/s it's another unknown Mass which is at rest after the Collision this 172 G moves with 3 m/s and with and the second block what does it do that is not even given to me okay question is find this Mass first things first use this concept that relative speed of approach is relative speed of separation since e is 1 separation speed is equal to approach speed they are separating with a speed of vus 3 both are in the same direction so vus 3 they're approaching each other with a speed of 5 - 0 so basically V will be equal to 8 m/s that's the first thing that you get next thing use conservation of momentum Mass which is 172 into velocity which is 5 plus no momentum Mass 172 velocity which is 3 plus Mass unknown velocity basically 8 there is nothing over here left which is just Mass which is unknown 17 into 5 172 into 3 bring it on this side it will become 172 and then divided by 8 right you will get the answer as 43 G just check this out if you're getting the answer as 43 G basically you will get 172 into 5 - 172 into 3 the whole thing divided by 8 which is 172 into 5 - 3 is 2 divided 8 this will go four times just check it out if it is 43 yes it will be 43 into 4 is 172 yes 43 G is the answer cool clear shall we go ahead right moving on to the next question coming up on your screen there is a dis which is free to rotate and uh what is happening is uh the axis is perpendicular to the plane passing through the center torque is being applied acceleration is given the net acceleration at the end of 2 seconds on the rim point that means on the tangential point on the circumference what is the acceleration all right so let's try to do this problem guys if you are going to do some time pass you will be blocked guys be careful about it yeah remember that No Nonsense will be tolerated here yeah so what do we do for this problem acceleration is given uh torque okay is applied so from this the first thing we can find is tangential acceleration which is Alpha into R is the radius given yes Alpha is 2 radius is5 M so it is 1 m/s square okay next I need to also find centripetal because there will be both kinds of acceleration it is rotating so it has tangential as well as centripetal so what do we do how do we find cental for that we need Omega for Omega what we can do is Omega is equal to initial angular velocity plus Alpha into time Alpha into time Alpha is basically 2 time is 2 so Omega is 4 Okay so Ral will be Omega squ R Omega is 4 so 4 Square R is 50.2 4 S 16 16 uh right yeah sorry. five did I write 2.5 yeah 16 so this will be 8 m/ second Square now comes the final part it is the total acceleration which will be cental square plus tangential Square whole thing under the root so 8 square + 1 s which is < TK 65 which is approximately 8 which is approximately going to be 8 so answer will be because question said approximately so I can put it as option number c make sense yes very good yes very good Clearo clear Clearo Clearo very good very good keep it up moving ahead to the next question coming up on your screen in this diagram there is some ring which is rolling without sliding with some speed on the surface all surfaces are smooth B has no speed what will be the maximum height reached by a on B wow very nice question this is based on conservation of momentum and stuff like that conservation of momentum and stuff like that so you have to think over here that what and all principles what and all formulas you might need and what might exactly happen when it reaches on the top right now only this Mass right now it is only rolling with some speed this ring with on the horizontal surface everything else is smooth all surfaces are smooth B has no initial velocity also so this kinetic energy will slowly be converted into potential is what you might think but be little careful because there is no friction involved no external Force so whatever is the momentum should also be conserved so right now this mass is moving this is at rest so there is a momentum towards the right as a system so when it goes on the top also there must be some momentum towards the right there must be some momentum towards the right so even when even when this guy reaches over here if this guy reaches over here this entire system will still be having some velocity so conserve momentum conserve the momentum in X Direction initial momentum is just MB this B Block is at rest final momentum uh ring has a mass of M and I think both both of them have the same mass if I'm not wrong uh yeah same mass as a correct so m + m into u m + m is 2 m it cancels it cancels over here so you'll get U as V by 2 so even when it has reached on the top both of them are continuously moving with the same speed you can think of it like this you can think of it like this this ball has reached here it still has this velocity this also has that velocity so that is V by2 so this U comes out as V by2 this U comes out as vx2 that means there is still some kinetic energy left that means there is still some kinetic energy left everybody with me very good so now now what about the rolling motion because it is rolling there will be also some kind of rotational kinetic energy involved and the best part is this rot ational energy will not go only anywhere because all the surfaces are smooth no way you can change the rotation whatever rotational energy was there here the same rotational energy is there here also so kinetic energy of rotation kinetic energy of rotation is there here as well as here there is no change because all surfaces are smooth so now let's write down the energy equation now let's write down the energy equation so initially half m v² translation is there plus your rotational kinetic energy is there which will not matter only because anyways it is not changing then when you go on the top then when you go on the top still that rotational kinetic energy is still there but translation wise translation wise only this component horizontal is there there is no vertical motion because it has reached the maximum height so plus half m u square plus this block also has gained some energy this block has gained half m u Square amount of kinetic energy and this ball has gone up by height H so MGH H MGH H now U is basically vx2 instead of U you can put V by 2 instead of U you can put V by 2 V by 2 that's it so you can see KR and KR just got cancelled over here this and this just got cancelled over here shift all these people over here and cancel out M cancel out M right later on so M will be half M v² minus half M v² by 4 - M v² by 4 mmm you can cancel from this you will get the height solving it as v² by 4G v² by 4G okay just check this out and cancel out this m this m this m M this m just do the match you will get this V uh or sorry this height as v² by 4 G that's all main thing is you should realize rotational energy did not change although rolling was mentioned no concept of rolling was actually inv worlded over here second thing momentum will be conserved in the X Direction so when it reaches on the top still there is some momentum in the xer direction so all these things were there in this problem another beautiful question this is the last question which you'll be doing because after this you will have naita m's class yeah yeah just check this out uh answer is see vishu just check this out answer comes out as a answer comes out as a this will be v² by 8 v s by 8 so that will make it b² by 4 so 1X 2 - 1X 4 will be 1X 4 no it is option A only it is not option C I think you would have made some mistake just check it out okay uh mass of the rod is given it is hinged at Point O it can freely rotate question is the particle strikes sticks to this Rod the angular speed just after the Collision whenever rotational things are there particles are sticking hitting Etc and it begins to rotate instead of linear momentum conservation it will be angular momentum conservation that's all I'll give you the idea about this problem so initial angular momentum will be equal to the final angular momentum will be equal to the final angular momentum now what is the initial angular momentum this mass is going along this line this is your R perpendicular so remember the formula mass into perpendicular distance which will be now d by 2 into the velocity which is V this is initial what about the final guys this whole thing will behave like one single system it will have some moment of inertia so into Omega because all these things will rotate together with some angular speed what is that moment of inertia going to be well it will have two things moment of inertia of the rod plus the moment of inertia of that small M Mass because it is getting stuck over here is that right what is the moment of inertia of that small M Mass it is just m into distance Square which is d by 2 whole squ d by 2 whole Square what is the moment of inertia of the rod it is hinged at the center so it will be Mass length Square by 12 Mass length Square by 12 what is the mass of the entire Rod it is given as 6 M so it is 6 m l s by 12 that's it this equation will directly tell you what is the value of Omega so that value of Omega comes out as 2v by 3D exactly the same thing has been done exactly the same thing has been done cool so the Omega is 2v by 3D amazing yeah so now my dear students shall I share the PDF of the remaining problems as well yeah do you want the PDF one thing I need comment and like comment and like and join the telegram channel so if you are not yet aware where the telegram channel is just as you watching this particular video just as you are watching this video go in the description box of this particular uh video right where is it yeah this one what just happened going into the description box yeah of this particular video okay and uh you can see join our telegram Channel just click on this just click on this link with Anu J English over there I'll be sharing the PDF of all the problems for all of you you have discussed many problems and like you saw solving each and every problem takes a lot of time and we have Justified that time also it's not like just because for the sake of doing it I have done it whatever time was needed for every problem I have given for that okay so remaining problems please take it up as homework I'll be posting the solutions also of that okay I'll be posting the solutions also of that in the uh same PDF okay so make sure you download the same bye-bye thank you so much all the best now attend naita M's class okay and tomorrow you will be having one chemistry and one maths class and day after tomorrow we'll be having a mock test okay so take care bye by by Captain sh signing off as

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Length: 169min 21sec (10161 seconds)

Published: Mon Mar 25 2024