Test Papers – National Board Exam (IX-X)
State Board Marathi-Std X Sanskrit Demo Videos
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JEE MAIN & ADVANCE 11th PCM Maths -Sequence & Series-2 Demo Videos
Hello students welcome back in this model will discuss the method of difference now let’s see what does this concept means on method of difference let’s as students I’m given a sequence a1 a2 a3till an and this sequence doesn’t looks like an AP or GP or AGP and we have to find the sum of these sequence now students as we know that if we are able to find the nth term we can use this summation concept understands what you observe where is that the difference between consecutive terms is forming an ap or GP the given sequence is not ap or GP but the difference between the conjugative terms is forming an AP or GP then we can use this concept to find the nth term students what I’m going to do is I written the solve the series as the first equation i’m going to rewrite the given sum and i’m writing that as the second equation now students i’m going to subtract these two series I know subtracting these two series S will cancel out with this will obtain 0 when i’m going to subtract this one see over here s – s will become equal to 0 a1 The first term remains as it is a2 minus a1 have formed in the bracket a3 and a4 from the second bracket and so on till an minus an minus 1 the last an over here which was over here should be minus an has gone over there on the left-hand side Now students so an is over here and now students we know that difference between the conjugative terms is forming an AP or GP there for a 2- a1,a3 minus a2 and so on can be written in form of Ap GP and hands there some could be found out and hence i’ll be able to find an that is the general term and now since i’m able to find a general term everything comes down to this summation series and we know how to do that now students let us take an example and understand what exactly want to do over here now students I’m given the series that is 5 ,7 ,11, 17 and so on and I want to find some of in terms of the series in students note five seven eleven difference between seven and five is two 11 – 7 is 4, 17 – 11 is 6 so differences 2, 4,6 and so on so difference is forming an AP 2,4,6 and so on therefore his students over here to find the nth term I’ll use the method of difference sum s I have already written on again sn unknown but shifting the terms taking the differences students when i subtract i will get tn and you can see over his students on the right hand side I have done this or there on the left-hand side there is 0 tn from here we’ll go on the left hand side so TN will be equal to 5 plus you can see where here students the bracket there are n minus 1 terms 1st term is 2 then 4,6,8 that is an AP that is sum of n minus 1 terms in AP sum of n minus 1 terms an AP write down the formula it’s just simple calculation now students Tn and terms are n square minus n plus 5 so when I were to find the sum of series it’s simple summation of tn and now the sum of terms will be varies from 1 to n summation r square minus summation r varies from 1 to n R plus 5 terms the summation from 1 to n and again one now it’s a simple application of formulas students summation r square into n plus 1 into 2n plus 1/6 summation r is n into n plus 1 by 2 and we know summation on 1 to n and one is simple and so it becomes 5n calculating is simple calculation is students taking out the common and now taking the LCM and then now simplifying it .I obtained the final result n by 6 into 2 n square plus 28 that give us the final answers students let’s take one more example under stranded so let’s different i will find some of the series 1,5 ,11 ,19 ,29 where to find the sum of these nth terms now one in five differences 4, 5 11 differences 6, 11 and 19 differences 8 so as students we can see 4,6,8 the difference again forms an AP so again to find the nth term i use the method of difference rewriting the series can see over us students taking the difference on subtracting i will obtained up in this series this equation is obtained taking Tn on the left hand side Tn will be equal to 1 plus sum of n minus 1 term 1st term is 4 common differences too so n/2 2a plus and minus 1 *T simple calculation now students calculating it expanding the bracket to simplify so when I expand tn comes out so n square plus n minus 1, sum of terms that is SN will be summation are varies from 1 to n TR and that would be equal to some Summation at from 1 to n R squared plus R minus 1 now splitting the summation and we obtained summation r very someone to an R squared plus summation varies from one to n are summation R varies from 1 to N 1 substituting the formulas students that very easy now we have to remember that formula and now its calculation so taking out common and then calculating it to his students we calculate and we obtain the final result and final result comes out as that would be equal to n into n square plus 3 n minus 1 whole divided by 3 students i hope you understand this Example thank you
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JEE MAIN & ADVANCE 11th PCM Physics -Heat & Thermodynamics Demo Videos
Hello students in this module we are going to look at a few important terms related to calorimetry and the very important principle of calorimetry based on which we have a number of questions coming up in the further modules so I want you to pay very, very close attention to this particular module because the concepts that we learn out here is going to be directly applied in the new miracles so to begin with the important terms with calorimetric the first time that I’ll be encountering is thermal capacity now what exactly is thermal capacity now third well capacity is defined the amount of heat required in order to raise the temperature of the entire body by one degree centigrade there is now if i have to compare this with specific heat capacity it is the amount of heat required to raise the temperature off unit mass of a substance by 1D so the difference between thermal capacity and specific heat capacity comes in the fact that it is for the entire body and this is for a unit mass so if i have the relation here which will give me delta Q = MC Delta C. hence this can also be written as what H into delta T so if I cancel the delta T from both sides of the equation i get my thermal capacity as the product of mass into specific heat capacity so this is a very important relation to be remembered My dear students the unit of thermal capacity is nothing but joules per degree centigrade all joules per Calvin now in this list the next important term that we have is the water equivalent to your students let me tell you water equivalent generally appears in the numerical and I don’t want any one of you to get confused because when we have a mixture present in a calorimeter they don’t mention to us about the mass of calorimeter or it specific heat capacity they give us directly what is the water equivalent of the calorimeter so from there you should not be confused you should be able to use the concept of water equivalent very nicely to find out what will be the Rays in the temperature of the calorimeter so here we try to understand what exactly is meant by calorimeter let’s say we have a block whose masses m and it specific heat capacity is C and the change in temperature is delta T so how much heat will be related to this particular block my delta Q is equals to MC delta T now let’s see the same amount of heat i use to raise the temperature of water by an equal difference so the same amount of heat for raising the temperature of water by equal level then how much water should I be taking that is known as the water equivalent so i can write down this Delta Q is nothing but equals to the w into one into delta T whether the delta t will get cancelled from both sides of the equation hence i get the value of my water equivalent is equal to nothing but a product of mass into specific heat capacity having done this let’s just go ahead to look at the basic principle of calorimetric what does it see let’s we have a block which is got to mass m1 specific heat capacity C1 it is a temperature T 1 degree centigrade we’ve got another block whose masses M 2 its specific heat capacity C 2 and it is a temperature t2 now given is temperature t1 is greater than t2 so he is going to flow from the body at a higher temperature to the body at Lower temperature and this is the direction of the heat flu now after some time there is going to be a state of thermal equilibrium that temperature of both the bodies is going to be equal so the principle of calorimetry states that heat lost by a body at a higher temperature is equal to the heat gain by the body at a lower temperature so having done that what is the amount of heat lost by the body at a higher temperature is m 1 c 1 into t 1 minus t that is a temperature difference for the body at a higher temperature and what is the heat again to buy a body at a lower temperature it is m 2 c 2 t – t2now equating them this is what is the most important principle that we are learning in this particular module now having done that is the temperature of equilibrium and if you solve it we get the temperature of equilibrium as m1 C1 t1 plus m2 c2 t2 divided by m1 c 1 plus m2 C2 so having them the students we will be using the same concept in the numerical ahead thank you very much.
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JEE MAIN & ADVANCE 11th PCM Physics -Friction-2 Demo Videos
Welcome back students so students we were talking about one beautiful example which we left in between let’s continue that example well an example was that there was this system and we were supposed to find acceleration well be received with three still this state wherein we determined the FBD we saw what were the directions of friction and also applying constraint equations constraint method using principle of virtual work we found that if acceleration of capital m is a right words then acceleration of small m will be to 2a down and since captain small I’m are connected therefore the acceleration of small M will be also small a right so we reach still this point now it’s all about writing or using laws of motion correctly so let’s talk about laws of motion first let’s analyze small M. now along horizontal direction students n1 is responsible for acceleration so long horizontal what can we see n1 will be small ma and long vertical direction we can say acceleration is down then we can say mg minus f1 minus T will be M into 2a but since it is a case of kinetic friction because there will be an elective slipping hence f1 will be u1 *n1 you see interrelation now this is small M case let’s talk about or let’s first Analyzed let’s first write down net equation if you use n1 from ma and put in the second equation what we’ll get we’ll get mg minus u1 ma plus T is going to m into 2a so let’s name this equation to be our equation number 2 now if we talk about capital insurance this was the FB of capital now a long Vertical direction Direction first we can see that net force long vertical will zero because there is no acceleration and what can we say n2 write this is the last upwards- mg minus f1 minus T will be 0 that is n 2 will be equal to T plus have f 1hich was u 1n1 plus capital mg along horizontal what can we see we can see 2t tension right we’ll try to pull the M block right n1 friction will oppose it So 2t- n1 then – f 2that is u2 n 2 will be capital m a now we put into into this equation what do we get we’ll get this particular expression so on solving what will gate on solving will get 2t minus whole bracket small M a plus u2t plus mu* mu2small ma plus mutual capital mg is going to M1me on further solving from all the questions we get the result acceleration to be two times smaller2m –u2 M plus small m into G upon capital M plus moment of 5 plus 2 mu 1 minus mu two students of course I i was very fast from Equation number 4 onwards and the solving of it you can always do it but the main character our main heart of this problem is to write laws of motion equations correctly to write constrained relations correctly to mention all the forces correctly and to make sure the direction of friction has been taken correctly so that’s all for on this model certain please what previous module this module again the again, again because this is a very beautiful case which will help you grow in friction chapter very nicely well that’s all till then thank you students
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JEE MAIN & ADVANCE 11th PCM Physics -Friction-1 Demo Videos
Welcome back students so students you’re talking about friction problems let’s continue with few more examples now in this give an example we have two blocks A and B there’s a force of 120 Newtons which is acting on the left of a and a force of 100 Newtons ’s which is acting on the right of B the coefficient of friction in the A end surfaces 0.9 and coefficient of friction between B end surfaces 0.3 mass of a is 10kg and mass of B is 20 kg now it is given that we have to find the tension in the string in this particular scenario that does this ,this situation is given we have to find tension now students this particular Scenario that is a very beautiful situation where it will be teach us many Things now what all possibilities can arise now as a result of these two forces when 120 and 100 the possibilities that can arise are, a and B move and a and b don’t move now for any possibility what we need to know is what exactly will be d limiting value of frictions because in this case Newest meu k are not mention separately will take meu s and meu K to be seen and if the friction and static we need meu n and if the friction is kinetic then again we need meu n in so basically first thing is to know what exactly will be the limiting value our meu times n now for a since mass of a is 10 therefore mg will be hundred and meu between a surfaces 0.9 therefore a max will be 90 similarly FB max is 60 so the thing is if both a and B move in friction will be kinetic and the values of kinetic friction will be nineteen sixty and it worked in we don’t move then whatever the value of friction will be that value of friction should be less than nineteen sixty pretty simple now let us assume in first place assume that both bodies move well both bodies move then it is pretty clear that 120 minus hundred that is 120 left and hundred right so 120 minus hundred is the driving force which is trying to drive a and b the words left now when 120 minus 100 that is 20 is driving in with words left friction on both a and B will act right and it will oppose the motion since a and b move friction will be kinetic the values are nineteen sixty therefore the opposing force are the resisting force will be meuA into na plus meuB*nb that is little be 150 newtons 120 newtons and were acting opposite I think are present there for the inside 120 minus hundred is the driving force friction will be opposing force with friction oppose this motion relative motion kinetic friction opposes of relative motion so over here what we can see you can see that since resisting force is greater than driving force their for how can motion happen therefore this case is not possible we can conclude that’s a system remain stationary did it not move our assumption was incorrect right students so now when this assumption is incorrect that both a and be moved what is the next possibility A and B don’t move when the A and B don’t move in bad as well there can be two possibilities which one possibility will be that friction will be static for both a and B when I says that direction I mean to say friction will be less than limiting friction for both a and B this will be possibility one second possibility will be the frictional will be static for one and limiting for another
that can be a second possibility that is first a in its limiting friction and then Oh on b the friction will not be limiting but some lower value or vice Versa so which of the cases are true let’s see let’s assume that both the friction is static that is both a and b the value of friction is less than the limiting value of friction now actually this is a very logical thing if friction in both a and B is static right it means to oppose 120 and 100 Newtons friction itself is sufficient attention will not even if do what is that not connected with string still both a and B will not move right so if friction is that take then we can say tension in the string should be 0 right students now if the tension in the string is 0 then to stop a the friction on a should be 120 and to block b efficient on be should be hundred but what do we know we know if friction on a is 120 and friction on b is a hundred this value of friction is just not possible because maximum values of friction were nineteen sixty respectively so it means this case is also not possible that is friction on both a and B cannot be static one of them has to have a limiting value of friction now which one of them has a limiting value of friction let us see now let us first assume 10kg block as limiting value of Friction it is reaching its limiting value first then what can we say what will be limiting value 90 their for on a 120 is the force left 90 that is limiting friction becomes right balance will be taken care by attention that is tensions value will be 30 now if the tension is 30 then on b from FBD we can say that T plus F is hundred that is 30 + F is hundred that is f become 70 but again in friction on b 70 it again is greater than its maximum value which is 60 again in this case is not possible so the only possibility we are left with is the b eliminating the value for first and then balance friction will take care of a so it’s b tense limiting value that is on be friction will be liberating and that will be 60 so on b balanced Will be taken can get tension and it becomes 40 Newtons so if we take the same tension on a then we can say that F will be 80 and the friction is 80 we can easily see that this 80 newtons is actually less than the limiting friction which was 90 on a hence what do we conclude we conclude that tension between the two blocks is 14 newtons students this was a very beautiful problem you need to watch this module any number of times because this will give you a lot of clarity it’s not a very simple case always when we are dealing with multiple bodies connected with treads right students remember first friction will try to oppose the driving force then comes tension not just make sure friction never exceeds its limiting value well thats all from on this model will get back to till then Thank You students
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JEE MAIN & ADVANCE 11th PCM Physics -Circular Motion Demo Videos
Hello students welcome back to the chapter circular motion in the previous module we discussed about centripetal force which is necessary for a particle to move in a circular motion we need a uniform circular motion or non-uniform circular motion the value of the centripetal force is equal to mV square by r or M Omega square R we also discussed that the particle would be acted upon by a tangential force which is M alpha r non-uniform circular motion in this particular module will discuss about the examples of the same the statement of the first example says that the particle is projected to the speed u at an angle theta with the horizontal and we are required to find out the radius of curvature at the highest point of the trajectory for the project right now we know that in this particular case the particle is going in a projectile motion it is not a circular motion but every point in the path can be treated as a point of circle so we can find out the radius of curvature in this particular case as well so what we know is that the particle is initially thrown at an angle theta that will be horizontal with a speed u if we try to resolve the speed will get the value of horizontal speed to be ucos theta and the vertical speed here is usin theta the trajectory that the particle would follow would look something like this and at the highest point we know that the particle would have a horizontal speed the vertical speed vanishes and sends the horizontal speed is ucos theta it does not change if air resistance is neglected at the highest Point the particle would have a horizontal speed of ucos theta think that now we know that for particle moving in circle the velocity is always tangential so therefore at the highest point what we can say is the horizontal direction is a Tangential direction that is e theta a gap and the vertical direction is a radial direction that is er cap let’s say that the center of the circle here is C and the radius is R .we are required to find out this value of R now i will recall these steps that we have to follow to solve this kind of question is first we have to identify the forces and we do that by drawing the FBD in this particular case the particle is thrown as a projectile and therefore it is going under the influence of gravity that’s the only force that acts on the particle is mg and that is also what Vertically down that means it is towards the center and the net force which is required for a particle in circular motion towards the center is equal to mV square by r thus this mg is the only force and we can conclude that mg here is providing the necessary centripetal force so equating this we get mg is equal to mV square by r which on solving gives us the value of R to be equal to v square by G and we’re putting down the value of instantaneous speed at the highest point which is ucos theta so we get finally that the radius of curvature at the highest point as required in the question is u square cos square to theta divided by G this is the answer to this particular example let us move forward with the another example which says that there’s a bead which is placed the over a rod which is hinge at one end now this bead is at a distance of R from the hinge and the Rod is rotating with a constant angular acceleration of alpha we are also given the this coefficient of friction between the Rod and the bead and the values nil we have to find out the value of time and we angle after which the bead starts slipping and we have to neglect the gravity in this particular question let us try to visualize what is happening in this particular case the rod actually rotating horizontally in this fashion and we can conclude that The bead would be moving in a circle of radius R before it starts slipping so if we try to view this particular motion from the top end we’ll get the motion of the Rod looks like something like so we can conclude that the bead is moving in a circle of radius R and for a particle to move in a circle what is required is a radial force and this radial force should be equal to M Omega square R that means if it has a angular acceleration then we can conclude that the bead will also experienced and then tangential force which is equal to m into alpha are now what provides the bead this radial force there is no gravity and the normal reaction would provide the tangential force so the friction in this particular case provides the necessary centripetal force the radial force therefore we can write down and m omega square is equal to F which is equal to meu n even let us mark this as equation 1 also we discuss the tangential forces provided by the normal reaction and thus we can say that m alpha r is equal to n, this is a second equation now if we divided the equation one and two what we get is the value of meu the coefficient of friction is equal to omega square r / alpha r. now what we can conclude from here is that the value of omega is equal to under root mu into alpha so this is the value of Omega where the bead would actually start slipping why because after this Omega the friction would not be able to provide the necessary centripetal force the value of friction would be less than the required value so what we’ll do is now we have to find out the value of time and angle will use the fact that we can use the three equation of motion for the constant angular acceleration where the first equation is omega is equal to Omega naught plus alpha t which gives us the value of T to be equal to under root of meu by alpha now use the third equation we get omega square is equal to omega knot square plus 2 alpha theta which gives us the value of angle to be equal to under root of meu by 2 these two values are the final answer to this particular example i hope u understood the module will continue R discussion on circular motion thank u
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