State Board Commerce (XI-XII) - Test Papers
KARNATAKA STATE BOARD XII PCMB Physics Demo Videos
Hello students today we will discuss the concept of magnetism and matter before going to the concept of magnetism and matter I would like to give you certain key points regarding the magnetism that is history of magnetism once magnets was hoarding his sheep in an area northern Greece called magnesia suddenly both the nail of his shoes and metal tip of his staff become firmly stuck to the large black rock on which he was standing . he was very curious to know fact behind it there for to find the source of attraction he dug up the earth and he finds load stones this is how the load stones look like . A load stones which contains magnetite and natural magnetic material Fe3O4 students keep in mind again a load stone which contains magnetite which is nothing but a natural magnetic material fe3O4. Okay fine .So the directional property of magnet was also known since ancient times that is freely suspended magnet and the magnet which placed on piece of cork allowed to float still on water always pointed in the north to south direction this is how can we recognize the direction of property of magnet a freely suspended magnet and the magnet which placed on a piece of cork on floating still water north to south direction ok they have also use this magnet in many application one of the application will see now, you can look at the image the image is chariot with that magnetic idol is suspended the craft man build a chariot on which they place the magnetic figure which swiveled around so that the finger of the statuette on its always pointed south. With this chariot huangti troops were able to attack enemy from the rear in thick fog and defeat them this was one of the application in the earlier days using magnets and the latest science suggest that moving charges or electric current produce magnetic field moving charges or electric current produce magnetic field this discovery was credited to the scientist oersted ampere , Biot and savart . dear students now will discuss the common ideas regarding the magnetism that is the first one is the earth behaves as a magnet with magnetic field pointing approximately from geographical south to geographical north . what is that mean on the whole we can treat as the magnet where the magnetic field pointing approximately from geographic south to the geographic north next, when bar magnet is freely suspended it points the north south direction this property you know when the bar magnet freely suspended it always points in what north and south direction ok that tip which points the geographic north is called north pole and the tip which geographic sounds is called south pole u can observe the tip which point the geographic north is north pole and the tip of geographic south is south pole ok the next point is like an electrostatic here also you can get also properties likes poles ripple each other what does that mean if you bought two bar magnets of same poles , if u bought a bar magnets of same poles each other those two bar magnet tend to ripple one and other why it so because the between those two same poles repulsive force exist in the same way unlike pole attract each other that means if you bought two bar magnets of opposite poles those two bar magnets always attract each other and that is because of the attractive forces exist between the opposite poles so that on whole I can conclude like poles ripples each other and unlike poles ripples attract each other and next point is magnetic monopoles do not exist what does mean magnetic monopoles do not exist which is nothing but if u cut a bar into a N equal pieces each one will be having both south and the north poles that means if u consider a bar magnet and make it into N slices in the number of slices each slices what u have what time will be always facilitated with the two poles that is north pole south pole , north pole south pole there for we can say that magnetic monopole do not exist ok the next point is it is possible to make magnet out of iron and its alloys yes we can manufacture magnet with the help of iron and its alloys for example the manufacture of alnico magnets uses iron and its alloys alnico is nothing but what aluminum nickel and cobalt ok the further discussion will be in the next module . Till then keep learning thank you.
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NEET & AIIMS 11 th PCB Physics -Heat & Thermodynamics Demo Videos
Hello, students, in this module we are going to look at a few important terms related to calorimetry and a 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 numericals. So to begin with the important terms of calorimetry the first term I will be encountering is thermal capacity. Now what exactly is thermal capacity, thermal capacity is defined as the amount of heat required in order to raise the temperature of the entire body by 1 degree centigrade. Whereas if I have now compare this with specific heat capacity, it is the amount of heat required to raise the temperature of unit mass of a substance by 1 degree. 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 is equal to mc delta T, 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 or Joules per Kelvin.
Now, in this list the next important term we have is the water equivalent. Dear 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 its 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 raise 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 mass is m and its specific heat capacity is C and change in temperature is delta T. So how much heat will be related to this particular block. Delta Q is equal to mc plus Delta C. Now let’s say 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 W into 1 into delta T, where the delta T will get cancelled from both sides of the equation. Hence I will get the value of my water equivalent is equal to nothing but the product mass into specific heat capacity. Having done this let’s just go ahead to look at the basic principle of calorimetry. What does it say?
Let’s say we have a block which has got a mass m1, specific capacity is C1, it is at a temperature T1 degree centigrade. We have another block whose mass m2, its specific heat capacity is C2, and it is at a temperature T2. Now given is temperature T1 is greater than T2, so heat is going to flow from the body at a higher temperature to the body at a lower temperature and this is the direction of the heat flow. Now after some time there is going to be a state of thermal equilibrium where temperature of both the bodies is going to be equal so the Principle of Calorimetry states that heat lost by a body at higher temperature is equal to the heat gained 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. It is m1C1 into T1 minus T that is the temperature difference for the body at a higher temperature. And what is the heat gained by a body at a lower temperature, it is m2C2 T minus T2. Now equating them this is what is the most important principle that we are learning in this particular module.
Now having done that T is the temperature of equilibrium and if we solve it we get the temperature of equilibrium as m1C1T1 plus m2C2T2 divided by m1C1 plus m2C2.
So having done this, students, we will be using this same concept in the numericals ahead.
Thank you very much.
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NEET & AIIMS 11th PCB 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, that example was that there was this system and we were supposed to find accelerations. Well, we had reached till this state wherein we determined the FBD. We saw what were the direction of friction and also applying constrained equations, constrained methods by using of principle of virtual work, we found that if acceleration of capital M is a right wards, then acceleration of small m will be 2a down, and since capital M and small m are connected therefore acceleration of small m will be also small a right. So we reached till this point, now it is all about writing or using Laws of Motion correctly.
So let’s talk about Laws of Motion, first let’s analyse small m. Now along horizontal direction, students, N1 is responsible for acceleration. So along horizontal what can we say, N1 will be small ma. And in vertical direction we can say if acceleration is down, then we can say mg minus f1 minus t will be m into 2 a but since it is a kinetic friction because there will be relative slipping hence f1 will be Mu1 into N1 you see inter relation.
Now this is small m, let’s talk about or first let’s analyse let’s first write down net equation, if we use N1 from ma and put in the second equation what we will get. We will get mg minus Mu1 ma plus t is equal to 2 into ma. So let’s name this equation to be our equation no.2.
Now we will talk about capital M, students, this was the FBD of capital M. Now along vertical direction first we can see that net force along vertical will be zero because there is no acceleration. Hence what can we say N2, right, this is thus upwards minus capital Mg minus f1 minus T will be zero. That is N2 will be equal to T plus f1 which was Mu1N1 plus Mg. Along horizontal what can we see, we can see 2 T tension right will try to pull the capital M block right. N1 and friction will oppose it. So 2T minus N1 minus f2 that is Mu2N2 will be capital Ma. Now we put N2 into this equation what do we get, we will get this particular expression, so I am solving what we will get. On solving we will get 2T minus whole bracket small ma plus Mu2T plus Mu1Mu2 small ma plus Mu2 capital Mg is equal to capital Ma. On further solving from all the equations we get the result acceleration to be two times small m minus Mu2 into capital M plus small m into g upon capital M plus small m into 5 plus 2 Mu1 minus Mu2.
Students, of course, I was very fast from equation no. 4 onwards and the solving of it, you can always do it. But the main crux or the 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 now in this module, students, please watch previous module and this module, again, again and again because this is a very beautiful case which will help you grow in friction chapter very nicely.
Well, that’s all for now till then thank you, students.
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NEET & AIIMS 11 th PCB Physics -Friction 1 Demo Videos
Welcome back, students.
So, students, we were talking about friction problems let us continue with few more examples.
Now in this given example we have two blocks, A and B. There is a force of 120 Newton which is acting on the left of A, and a force of 100 Newton which is acting on the right of B. The coefficient of friction between A and surface is 0.9 and coefficient of friction between B and surface is 0.3. Mass of A is 10 kg 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 is this situation is given we have to find tension. Now, students, this particular scenario is a very beautiful situation where it will teach us many things. Now what all possibilities can arise?
Now as a result of these two forces, 120 and 100, the possibilities that can arise are (a) A and B move, (b) A and B don’t move. Now for any possibility what we need to know is what exactly will be the limiting value of frictions because in this case MuS and MuK are not mentioned separately, we will take MuS and MuK to be the same and if the friction is static we need MuN, if the friction is kinetic then again we need MuN. So basically first thing is to know what exactly will be the limiting value or Mu times N. Now for A since mass of A is 10 therefore mg will be 100, and Mu between A and surface is 0.9 therefore fA max will be 90, similarly fB max is 60. So the thing is if both A and B move then friction will be kinetic and values of kinetic friction will be 90 and 60, and if both A and B don’t move then whatever the value of friction will be that value of friction should be less than 90 and 60, pretty simple.
Now let us assume, in first place let us assume that both bodies move. Now if both bodies move then it is pretty clear that 120 minus 100, that is 120 left and 100 right, so 120 minus 100 is the driving force, which is trying to drive A and B towards left. Now when 120 minus 100 that is 20 is driving A and B towards 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, their values are 90 and 60, therefore the opposing force or the resisting force will be MuA into NA plus MuB into NB that is it will be 150 Newton. 120 Newton and 100 Newton were acting opposite therefore we can say 120 minus 100 is the driving force, friction will be opposing force because friction opposes motion, relative motion, kinetic friction opposes relative motion.
So over here, what can we see, we can see that since resisting force is greater than driving force therefore how can motion happen. Therefore this case is not possible. We can conclude that system will remain stationary it will not move. Our assumption was incorrect, right, students. So now when this assumption is incorrect that both A and B move then what is the next possiblity, A and B don’t move. Now when A and B don’t move, in that as well there can be two possibilities, which one. One possibility will be that friction will be static for both A and B, when I say static friction I mean to say friction will be less than limiting friction for both A and B. This will be possibility 1.
Second possibility will be that friction will be static for one and limiting for another. That can be the second possibility. That is first A will attain its limiting friction and then 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 in 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 Newton friction itself is sufficient, tension will not, even if two bodies are not connected with string, still both A and B will not move, right. So if friction is static then we can say that tension in the string should be zero, right, students. Now if the tension in the string is zero, then to stop A, the friction on A should be 120, and to stop B the friction on B should be 100. But what do we know? We know if friction on A is 120 and friction on B is 100 this value of friction is just not possible because maximum values of friction were 90 and 60 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 limiting value of friction let us see.
Now let us first assume that 10 kg block has limiting value of friction, it reaching its limiting value first, then what can we say, what will be the limiting value – 90. Therefore on A, 120 is the force left, 90 that is the limiting friction becomes right, balance will be taken tension that is tension’s value will be 30. Now if tension is 30, then on B from FBD we can say that T plus f is 100 that is 30 plus f is 100, that is f becomes 70. But again if friction on B is 70, it again is greater than its maximum value which is 60, again this case is not possible. So the only possibility we are left with is B will attain its limiting value first and then balance friction will take care of A. So if B attains limiting value that is on B friction will be limiting and that will be 60. So on B balance will be taken care by tension and it becomes 40 Newton. So if we take the same tension on A then we can say that f will be 80. And if friction is 80 we can easily see that this 80 Newton 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 40 Newton.
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 is not a very simple case always when we are dealing with multiple bodies connected with threads, right, students. Remember, first friction will try to oppose the driving force, then comes tension. Now just make sure friction never exceeds its limiting value.
Well, that’s all for now in this module, we will get back, till then, thank you, students.
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NEET & AIIMS 11th PCB Chemistry structure of Atoms Demo Videos
Hello students, we are studying the chapter structure of atoms. So, in this module I will continue what we were discussing that is the radial probability curves. So, earlier we discussed how to calculate the total probability, right. So, for which we took a shell at a distance r which is thickness dr and we can calculate the volume of it which is 4 pi r square into dr. And probability of finding an electron in that small thickness dr it can be defined as dP which is size square into dv, where size square is the probability density function and which can be written as 4 pi r square size square into dr. So, the total probability we learned that I can integrate that from zero to P dP and it comes to be R1, R2 size square 4 pi r square dr, right, so this is what we learnt. And here that Pr that whatever that has been defined in that integral part is what we call it as a radial probability distribution function. So, therefore to calculate the total probability it is integral R1 to R2 Pr into dr, where Pr is the probability distribution function, right. This is what we learned earlier. Continuing this now we will try to draw the total probability versus the r. So, let us see that first.
So, if at all I draw the total probability versus r here, that P of r is a probability distribution function versus the r. Now, at nucleus, r is equal to zero, so Pr will be zero. So, the graph will always start from the origin. As r square increases size square is actually decreasing. So, there are two functions one is the r square part and the other one is the size square part. Then what would happen, the graph looks something like this for 1s orbital. What is it suggesting that initially size square is decreasing but the r square is increasing but as r is increasing size square is decreasing exponentially. So, initially it will increase because of the r square and then it decreases because of the size square. So, r square is contributing first, in the first half of the graph and the size square is contributing in the second half of the graph and eventually it tends to become zero.
Now, here you could see it is reaching at a maximum value and that maximum value is what we call it as radius of maximum probability and that is found to be 0.529 Angstrom which is exactly the same what Bohr’s have calculated. And let me let me assign it as a0. Now, we know that according to the Bohr’s r is equal to 0.529 n square by z, so, if z is constant for every value of n, you should be getting higher value of the radius, right. So, which I can write it as a0 into n square by z. So, from the graph it is evident that the probability of finding the electrons at nucleus and at infinite distance from the nucleus is always zero.
Now, let me draw a similar graph for 2s. For 2s if at all you see I, plotting probability distribution function versus r. Now, this is the graph you will get. Now, here what I would see that, is that we are getting a radial node here where the probability of finding electron is zero. Now, this is the maximum distance or the peak where we have the highest probability and this distance amazingly matches to 4a0 which is again suggesting from the Bohr’s theory that radius will be a0 into n square. So, it is a second orbit, so, n square gives us 4a0. So, radius of maximum probability 1s will always be less than 2s, so that is what we will get. That means electron in 2s is always found at the larger distance from the nucleus compared to that of 1s orbital.
Now, if at all I keep drawing this for different orbitals, what do I get? So, total probability, probability distribution curve versus r, if it is for 2s, this is what we found and it has one radial node, with maximum, radius of maximum probability at a distance. Similarly, if I draw it for 2P, I will get no nodes here; this is how we will get with certain maximum distance. Similarly, for 3P, you would see it is quite similar to that of 2s. Similarly, if at all I keep drawing it for, again with the radial node here and if I draw it for 3d, again you would see it is quite matching to the node of 1s and 2P. Similarly, if at all I draw it for 4d, you would see the graph varies quite similar to 2s and 3P with one radial node. This gives us as a pattern. So, here you would see there is one radial node forming so this gives us a pattern that the total number of radial nodes will be equal to n minus l minus 1. So, therefore to calculate the number of radial nodes for any particular orbital we can use this formula. That is n minus l minus 1. So, from there we can get what is the total number of radial nodes.
As we go for the next module, we will see about the plainer nodes or the angular nodes.
Thank you.
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NEET & AIIMS 11th PCB Chemistry-Redox Reaction Demo Videos
Hello, students, let us continue our discussion on redox reactions. Now, in the last modules we will be discussing about calculation of equivalent weights of oxidising agents and reducing agents and we also defined a very important term z factor. Today we will continue to discuss some examples on equivalent weight calculation and then we will see two particular special cases FeC2O4 and Cu2S. Let us start with some examples.
Find the equivalent weight of KBrO3 and Br2 in the following reaction. Now, you can see that in this reaction 10 electrons are being exchanged. And we remember how we have to calculate z factor. So, let us begin with KBrO3. So, we can say that for 2 moles of KBrO3 number of electrons exchanged or rather gained in this case will be equal to 10. So, therefore we can say that for 1 mole of KBrO3, number of electrons gained. So, we can say that, so, we can say that for 1 mole of KBrO3, number of electrons gained will be equal to 10 by 2 that is 5. Now, the z factor of KBrO3 will be 5 and therefore equivalent weight will be equal to molecular weight of KBrO3 divided by 5. I am not getting the exact weight, I am just representing it symbolically and that is what is important. Now, for Br2 we see that the number of electrons exchanged are 10 and since there is only 1 mole of Br2 we need to see that there is only 10 electrons that we have to use. So, the z factor of bromine becomes 10 and equivalent weight will become molecular weight of bromine divided by 10.
Let us see the next example. When HNO3 is oxidised into ammonia means NH3, the equivalent weight of HNO3 will be how much? Now, HNO3 is being changed into ammonia. First of all we will write their oxidation number, nitrogen in HNO3 will be equal to plus 5, you know the algebraic method to calculate the oxidation numbers. Now, in ammonia the oxidation number of nitrogen is minus 3. So, we can see that, we can calculate the change in oxidation number, since there is only one atom of nitrogen it will be equal to plus 5 minus, minus minus 3 that is equal to plus 8. Keep in mind, change in oxidation number is initial oxidation number minus the final oxidation number. So, this 8 is there, this is what it will become the z factor and therefore equivalent weight is molecular weight divided by 8.
Let us see the next example. The equivalent weight of H2SO4 in the following reaction is. They have given us a redox reaction and in that they are asking for weight of H2SO4. Now, can I say that the z factor for H2SO4 will be equal to 2? Actually it would be wrong, why, because it is not an acid base reaction. It should be actually analysed like a redox reaction, which means we have to focus on the loss and gain of the electron. So, what should be your strategy? First we will find out the overall loss or gain of electrons in the redox reaction. It is like balancing, and with that process we have to get the loss or gain of electrons. To do this, we can either focus on SO2 or sodium dichromate that is Na2Cr2O7. Now, we will use this information to calculate the z factor of H2SO4. Just remember that there is only 1 mole of H2SO4 involved in the reaction. So, let us focus on the solution. The oxidation number of chromium in K2Cr2O7 is plus 6 and that in Cr2SO4 thrice is plus 3. Since there are two atoms of chromium we can see that the number of electrons exchanged overall will be equal to 6 and therefore we can say that for 1 mole of H2SO4 number of electrons exchanged will be also equal to 6. So, z factor of H2SO4 is 6 and the equivalent weight will be equal to molecular weight divided by 6. So, remember that in redox reaction acid and base equivalent weight will be got from loss or gain of electrons. Now, this calculation would also have been done, if you have just analysed SO2 to SO4 2 negative.
So, let us take a look at the next example. The equivalent weight of HNO3, molecular weight is 63, in the following reaction is. So, we have given a reaction between copper and HNO3 to give you copper nitrate and NO and H2. And we are having four options. Now, to analyse this reaction understand that HNO3 here is not just acting as an acid but it is also acting as an oxidising agent. So, HNO3 has dual roles, it is going to act as an oxidising agent and it will also provide the acidic medium. Now, we will find out the overall loss or gain of electrons. And to do this better would be copper because only oxidisation of copper is done. HNO3 is reducing, so, in HNO3 we have to observe NO very carefully. So, using this information we will calculate z factor and then we will get to the equivalent weight part. Let us see the solution. So, we can see that the oxidation number of Cu in copper is zero and that in copper nitrate is plus 2. Since there are 3 copper involved, we can see that the change in oxidation number is minus 6. Now, if we had to do NO, remember we have got only 2 NO corresponding to the change in oxidation number. So, we will find that nitrogen in HNO3 is plus 5 and that in NO is plus 2. So, the change in oxidation number is plus 6. Even if I do through copper or from HNO3 that means NO, from both the loss and gain of oxidation number will be 6 and it should be like that. So, number of electrons exchanged overall is 6 and therefore we can say for 8 moles of HNO3 number of electrons is equal to 6. For 1 mole of HNO3 it will be 6 by 8 that is 3 by 4. So, equivalent weight will be equal to molecular weight divided 3 by 4 or we can say it will be 63 times 4 by 3.
Now, let us focus on some special cases. We will focus on ferrous oxalate, FeC2O4. FeC2O4 is a good reducing agent. Now Fe2 positive is oxidised to Fe3 positive. Now, in this reaction we can see on the reactant side the charge is 2 positive and on the product side the charge is 3 positive. Now, to balance the charge since atoms are already balanced, to balance the charge we simply need to add an electron to the product side. Now, C2O4 negative 2 is oxidised to CO2 and we can see that here we will have to balance the atoms first. So, we just multiple the product side by 2. Now, to balance the charge, reactant side has got 2 negative charge and product side has 0. So, to balance the charge we simply add 2 electrons to the product side. Now, I am seeing both the reactions here that overall FeC2O4 both the components are oxidised. So, FeC2O4 overall loses a total of 3 electrons and hence equivalent weight of FeC2O4 will be equal to molecular weight divided by 3.
Now, let us see another special case, Cu2S, cupra sulphide. It is also a good reducing agent. Now, in this case Cu positive is oxidised to Cu 2positive. Again we will balance the charge. On the reactant side charge is 1 positive and on the product side charge is 2 positive. So, to balance the charge, we simply add an electron to the product side. Now, since there are 2 copper, we can say overall 2 electrons are lost for copper. Now, sulphide iron is oxidised to SO2. Now, first we will balance the atoms. So to do that, we have to add 2 H2O on the reactant side to balance the oxygen and 4H positive on the product side. Here we have assumed that medium is acidic in nature. So, we add these and then to balance the charge we see that the reactant side has a total charge of 2 negative and product side has 4 positive. To do this, we will now add 6 electrons to the product side and this balances the charge. So, 2 electrons of copper and 6 sulphide iron. So, Cu2S loses the total of 8 electrons and hence its equivalent weight will be equal to molecular weight divided by 8.
Thank you.
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