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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KARNATAKA STATE BOARD XII PCMB maths Demo Videos

Hello students welcome to this new chapter probability. Where we are going to speak about condition of probability in this module. Now we should have strong knowledge about probability what you studied in your 11^th std.  You were studied about definition of probability also you have studied about addition theorem rich to tell them. We have proceed further same concept of probability which is chapter for you in the 12^th. Lets consider the definition of conditional probability , if E and F are two  events associated with a same sample space of a random experiment the condition of probability of the event E given that F has occurred is given by P of E given by F is equal to P of E intersection F divided by P of F where P of F is not equal to zero remember this is a formula of conditional probability now I will note taken on that E given F already seen  what about P of F given E what happen what will be happen there P of F given E is equal to P of F intersection E divided by what dived by P of F making sure the denominator is not zero so that is a note you have let me proceed further there are some properties of conditional probability let see one by one and we see proof of it as well the first property says probability of S given F  is equal to P of F given F is equal to one . The second one says probability A union B given F is equal to probability P of A given F + probability B given F – probability A intersection B  given F And third one says probability E compliment given F is equal to 1 – probability E given F this are the three properties let see solution one by one. Let me start with the first probability S given F I understand the condition of probability e given F is equal to E intersection by F what will S given F it will be P of S intersection F divided by P of F remember this what is after that should be in denominator S given F what is second should be in denominator. now what  about the  proceeding part of this S intersection F  what you understand for this  S stands for sample space and I know that F is a subset of sample space F is an event is a subset of sample space and what about the intersection  obviously it is F P of S intersection F,  P of F  divided by P of F it is 1 so I have got first part of first property , what about second part of first property P of F given F it is according the formula of  condition probability I can write it as P of F Intersection F divided by P of F and F intersection F  because F is Same as F it is contained  so I can write F itself  P of F  divided by P of F is equal to 1 so we have got proof of 1^st property . let me go with the proof of second property  . let me start P of A union B given F according to condition probability formula we can write it as P of A union B intersection F divided by P of F now what about the numerator so I can use distributive law A union B intersection C is equal to A union C intersection B union C, same thing I will write here P of A union F intersection  B union c divided by  as it is P of A . now what about numerator use the addition theorem of probability and split the numerator  as it is what I get that I will get it as P of A intersection F + P of B intersection F – P of  A intersection B intersection F divided by P of F . now I will split it into three parts I will get it as P of A intersection F divide by P of A + P of B intersection F divided by P of A – P of A intersection B intersection F divided by P of F now I can recall the condition of probability . earlier we have seen  left hand side right in the right hand side now see the right hand side right the left hand side according to that the first term P of A given F  what is the second term P of B given F  third term will be p of A intersection B giving F and this is the proof of second property  I can take a note regarding the particular property  when A and B are mutually exclusive intersection is zero their for I  can get P of A union B giving P of A giving F + P of B giving F only A and B are mutually exclusive events . let me go the property no. 3 proof , we know that the first property P of S given F is equal to 1 also I know E union E compliment  any event union is compliment same as the sample space let me replace s by E union E’  do you think E union E’ are mutually exclusive yes they are mutually exclusive according to note mentioned earlier  A union B  giving F is equal to P of A giving F + P of B giving F , and I use that particular note P of E Giving F + P of E’ giving  F is equal to 1 there for the taking 1st term to right hand side  I get P of E’ giving F  is equal to 1- P of  E giving F that’s a required proof of the property so these are the 3 properties what we have conditional probability , Thank you . 

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KARNATAKA STATE BOARD XII PCMB Chemistry Demo Videos

Hello students welcome to module number 1 let us discuss something about solid state in this module we will discuss some introduction and classification of solids. My dear students we know that anything which occupied space and  has definite  mass as known as matter. A matter exist in a 3 physical state the first one is a solid  second one is liquid and third one is gas and these three physical state are inter convertible by changing the condition of temperature and pressure the solid on heating it is changes to a liquid. On heating the liquid changes into the gas by cooling the gas is converting to the liquid . and again cooling the liquid changes into the solid . There for the 3 states of matter are interconvertible by change in the temperature or pressure now let us scientist has discovered the two states of matter and these are plasma and Bose–Einstein condensate. The discovery of fifth state of matter was done by a great Indian scientist sattindranath Bose and well known scientist Albert Einstein. These 3 physical state of substance are mainly depends on two factor the first one is inter particles attractive forces and second one is thermal energy. The inter particles attractive forces constant particle tends to keep closer but thermal energy tends to keep them apart. Now students what are the solids the solids are the substance which have definite shape and definite volume. For example the TV, door , laptop these have definite shape and definite volume . These are the examples of the solids. Now what are the general  characteristics properties of solids are the first one  the solids have fixed mass, shape, volume  and density for example this box this box has definite mass definite volume definite mass definite shape also. In solids the intermolecular distance is very small take a look here the particles are closely packed there for intermolecular distance are very small. And if intermolecular distance are very small means what is the force of attraction .the force of attraction that is intermolecular force of attraction is very strong and coming to the last point the solids are hard, rigid and incompressible. These are properties of solid. Now the classification of solids based on the nature of orderly arrangement of constitute particles in 3 dimensions. The solids are classified into two types first one is crystalline solid and second one is amorphous solid now the students what are these crystalline solid and amorphous solid .their a long range orderly  arrangement of constituent particles in known as crystalline solid . and there is not long range orderly arrangement of constituent particles are known as amorphous solid .for crystalline solid examples are quartz , sodium chloride etc. for amorphous solid examples are polyvinyl chloride, glass fiber etc. now the properties of crystalline solid  the first property is crystalline solids have Sharp melting point and definite heat of fusion why crystalline solids have sharp melting point because the strength of all the bonds between the particles are equal there for crystalline solids have sharp melting point. Ice is an example of crystalline solid the melting point is zero degree Celsius coming to the second property of crystalline solid that is anisotropy. There is many properties which have different values in different directions for example the thermal conductivity , electrical conductivity , reflective index , mechanical strength are different along different directions in one directions the compositions of solid is  green purple , green purple and other it is purple, purple,  purple, purple since the composition of solid changes with direction has the values of physical  properties also changes therefor it exhibit work and isotropy nature or an isotropy now for this plain solid start and is appropriate in nature now what are the differences between crystalline and amorphous solid coming to the first parameter that is arrangement  there is a long range orderly arrangement of particles in crystalline solid in case amorphous solid that is short range of orderly arrangement i.e. there is no long range orderly arrangement of particles in case of amorphous solid the second property is melting point crystalline solids are have sharp melting point amorphous solid do not have sharp melting point  because the strength of all the bonds between the particles are not equal there for amorphous solid don’t have Sharpe melting point   and third property heat of fusion crystalline solids have definite heat of fusion and amorphous solid  no definite heat of fusion. And coming to the fourth anisotropy and isotropic nature .the crystalline solids are anisotropic in nature   and amorphous solids are isotropic in nature and coming to the fifth point nature. Crystalline solids are also known as true solid and amorphous solids are also known as a pseudo solid and coming to the last point shape crystalline solids have definite geometrical shape but amorphous solid has irregular shape. Thank you students.  

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KARNATAKA STATE BOARD XII PCMB Biology Demo Videos

Welcome students we are here with a beautiful and interesting chapter in biology i.e. health and disease. in this chapter we will talking about the basic different type of disease and then we followed by the very important system of our body i.e. the different mechanism which helps to fight those disease and finally brief talk about the different aspects of addiction. Now lets us talk about very first module in this module we will be dealing with the   erraticallypart what is basically health. Let us talk basically what is health and now it is very simple question for you students and very basic question but we all need to know what is health, yes let us taking an example. In this instance if you see the person here has got some disease yes we say that normally they are not healthy. Then does no disease means a healthy state. Let us talk about that now yes basically health doesnot simply mean the absence of disease or physical fitness then what is health? Yes  Health is basically defined using a 3 important dimension. Can you guess the three important dimension what it could be? Ya first one here is the physical. And second one is very important is the mental and third one is going to be a social. Yes student Let us talk about them in detail now basically health is defined as a state of complete physical well being. Yes student first of all as a state of complete health i.e physical well being. Next important thing and what is this indicate that a person is physically healthy but you know when they are bound with too much of stress or it can be any in balances in the body and it interfused called the depression. Yes student and they play very important role and here if u see the person has to be healthy then it is healthy only if they have what state yes I mean to say a complete state of what complete mental well begin and that mean the person should not have any short of depression also. Next important thing what is this? Yes I am talking about maintaining  good relationship with people in society. And here that is nothing but social well being let us try to summarized them now. Basically health is defined as a state of physical than the mental and social well being and very importantly and not merely absence of disease or infirmity. What is the infirmity? Yes infirmity meaning a state of weakness or debility. Now next important thing how do we remain a healthy? a very basic answer and very important answer here giving  is hygiene. What is the hygiene and write students is the science of health which actually aims at preserving then maintaining and the improving. Yes the health of the it is only individual, ya it is individual and the community as a whole that is how can maintain a hygienic condition. Next important thing here is how to remain a healthy or how to maintain a good  health and then very important thing practiced by us the first one , can you guesses the first one going to be , yes very right that is balanced diet . next important thing that is regular exercise  and yoga meditation etc. and next thing is maintain a personal  hygiene ,next is right attitude of mind and finally a cultivating a good habits  all if this maintain well help us to maintain a  good health now if this all are not maintained  what is going to leading do, yes it might your condition  called disease  what is disease let us now to defined that now disorder of structures  and functions basically other abnormal condition  all of them collectively is called as disease . Now in this module just we have to talk about a brief outline of that disease but later taking about some example also. Now here in this module we have discuss about the health and the basic concept of hygiene and also outline disease. Yes in the next module we will talking about type of disease. Ya  student we will right back in a next module. Bye then thank you.  

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KARNATAKA STATE BOARD XI PCMB Physics Demo Videos

Welcome dear students welcome to the chapter thermodynamics in this chapter we start with knowing the basic terminologies  in this module because understanding these basic terminologies  we can understand the chapter better ok let’s see what is the basic terminologies that we are going to studied in this module  very very basic  terms involved in this chapter first of all what do you mean by thermodynamics , thermodynamics  is a branch of physics which deals basically with the conversion of  heat energy into mechanical work or mechanical work into heat energy so it is the branch of physics which deals with a transfer energy from one place to another and conversion of heat energy into mechanical work or vice versa that is thermodynamics for me. Let’s see a simple example I got a cylinder me with me which is fitted with piston you can see on the screen and which has ideal gas in it. I supply heat energy how much I supply heat energy dQ amount of heat energy has been supplied and this energy what it does please observe that then you understand the meaning of thermodynamics this dQ supplied increases the volume of gas that is implies pushing the piston upward so dW amount of work is done    how much work will be done dW amount of work is done so heat energy supply is converted into work and that is what is thermodynamics  all about yeah let’s try to understand different types of system . what do you mean by thermodynamic system surrounding all that thermodynamic system is the quantity of matter certain quantity of matter having fixed identity and mass let’s see what is that A certain quantity of matter part of the matter having fixed identity and mass we call it thermodynamic system example, don’t think big about thermodynamic system is little water you’re got going to boil that and then that water becomes a system for you maybe I have gas in a cylinder that is a system for me look at this example water in a beaker may be 50 cc water Is that it is a system for you , you are expecting that , yes look at that In a container I’ve got gas that is a system for me having fixed identity and mass let’s see what is surrounding you must understand that anything other than the system in this universe become surrounding matter outside the system that becomes surroundings . so you have to understand that thermodynamic system continuously interacts the matter outside the system and that matter outside the system we are calling it as surrounding look at that delicious coffee may be a tea is presented in the cup that is the system for me now this system is a tea or coffee remember it interacts something outside matter outside and that part which is outside the system we are calling it as surrounding we go into different types of system right now what do you mean by open system , open system involves exchange of heat energy and matter between system and surrounding that system I m calling it as open system if there is exchange of heat and matter takes place between system and surrounding call that system as open system look at this piston and cylinder you can see in the picture which has gas in it and you supply the heat energy when you supply heat energy you can see that the gas expands work has been done yes excess heat energy is there what happened to the excess heat energy that will got . so it has got exchange of heat energy as well as matter and that system we are calling it is the open system lets go to the closed system means what a system in which you have got only action exchange of heat but not matter that is absolutely no matter exchange yes, if the exchange is only heat energy not the matter between the system and surrounding we call it as a closed system. cylinder provide with the piston having the ideal gas you supply heat energy remember only the heat energy exchange between the system and surrounding not the matter you can seen that work has been done but there is no matter to exchange let’s see what is that work done dW  amount of work done but no amount of matter exchanged between system and surrounding yes will know it isolated system no exchange of heat and matter between the system and surroundings let’s see that example no exchange of matter and heat between the system and surroundings look at this thermally isolated system nothing is going to inside neither heat not the matter the system is called as isolated system  non conducting wall … thank you.  

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KARNATAKA STATE BOARD XI PCMB Maths Limits Demo Videos

Hi I'm Hari Prasad let us start the chapter limits in this chapter we are going to learn how to find out the limit of the function and what do we mean by limit of a function let us start module number one in this particular module we are going to learn what is the meaning of the word tending to let us start the chapter now meaning of X x tends a is what we have to learn first so immediately the question that comes to my mind is what is the meaning of X tending 2 or X  tending to any particular number a in general so it can be particularly X tending to 2, X tending 3 something like that so what is the meaning of that does it mean x is equal 2  does it mean know that x is equal to something else so answer this particular question will consider the following example consider boy who always covers fifty percent of the total distance that means the boy covers five meter for the first time so where he will be he'll be at five meter mark the boy will cover all fifty percent of the remaining five meter that means he will cover all 2.5 meter for the second time third time he will cover 1.25 meter and so on if this is the case when the boy will be exactly on 10 meter mark can you predict the answer is the boy will be never on 10 meter mark but he's very close to 10 meter mark right similarly when we write X tending 2 we mean that X is very close to two but it is not equal 2 consider the number line we have minus 1 0 1 2 3 and so on when x tends 2 to it can be close to two when I say close to two it can be from the left side or it can be from the right side also write suppose if I say x tends 2 and if X's or close to 2 from the left side then it can take the following values can you get what are the values it can take x is equal to 1 .8 is possible it can take 1.9 yes it is close to 2 ,1.99 yes it can take these values right and one point nine nine nine yeah still very close to 2 and so on there are many infinitely many values that it can take but when X tends to 2 if X is close to two on the right then it can take the following values which are the values can you guess the point 2.8 can take it can take 2.2 it can take 2.1 and 2.01 so on it can take many other values right is so now the question is clear to us let us now talk in case of general elements like X and a, let X be a variable is the meaning of variable quantity which keeps changing and a be a constant with the meaning of constant quantity which has a fixed value right if X takes the value near Nearer  to a then we say that e tendens to a and  we write it as X arrow mark a We write it as does x tends to a note that in this case X cannot be equal to A  and we are not specifying any manner in which X should approach a that means it can approach to a from left as well as it can approach from right now when X tends to a from the left we have the following situation if X approaches a taking all values less than a but Nearer to a then we say that extends to a from the left we write this as extends to a and we put a small minus  over a this minus is not to be used in any of the calculation similarly when X tends to a from the right we have the following situation if X approaches a taking all the values greater than a but nearer to you a then we say that x tends to a from the right and rewrite it as x tends to a plus we write small plus sign over a  like this is the notation that you need to remember thank you.  

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KARNATAKA STATE BOARD XI PCMB Chemistry Demo Videos

Hello students welcome to the new chapter called state of matter in this chapter we are going to learn some interesting facts about the word matter so in this module will learn introduction to state of matter. let me start with the word matter  as we all know matter is anything that occupies the space possesses  mass for example this chair and table or a glass of water or some balloon  containing air so if I concentrate on this I come across with the new classification that matter  is going to get classified into three different types that is called solid other one liquid and one more is called gases so along with these three we having some more state of matter like plasma boseeinstein condensate, fermionic  condensate, quark condensate thought wave along with these three so let me concentrate on these three types called solid liquid  and gases . if u look at these pictures you can come across with the some interesting facts that the faster picture is tightly bond nothing but  solid second one is slightly distributed nothing but liquid last one is completely dispersed nothing but gases so by looking at this picture I think I can make out some interesting points with respect to this point number 1 if I concentrate on shape for solid this is perfect  If I concentrate on liquid there is no perfect shape if I concentrate on gas and similar to the liquid there is no shape for gas also. If I concentrate on volume solid has got volume liquid has got volume but there is no volume for gases. I got there intermolecular force of attraction and  on the word about that nothing but this particular word is very much essential for these two particular parameter for shape and volume so what is intermolecular force of attraction it is the attraction between two atoms if it is very much strong I will be having a solid shape but it is very ,very less then there is no shape and there is no volume so if I concentrate on intermolecular force of attraction it is very much strong in case of solids it is very weak in case of liquids and very week in case of gases so If I want a next point nothing but packing efficiency if I concentrate on packing solid are having closely packed structure if I concentrate on liquid is loosely packed if I concentrate on gases it is very loosely packed if I concentrate on next word called density we know the word density what is density it is nothing but mass by volume . if I concentrate on packing of this particular thing it is closely packed nothing but mass by volume is more in case of solids . so in case of liquid it is slightly less in case of gases it is very very less there for density is very high in case of for solids and it is very least  in case of gases and  is intermediate to solids and gases in case of liquids. So in the next point we are going to learn about the word called flow ability so the word of flow ability means that is the ability to flow which is maximum in case of gases and less in case of liquids least in case of solids. if I concentrate on compressibility it is nothing but ability to get compressed it is very high in case of gases and less in case of liquids  which is very least in case of solids . so there are main two parameters or condition by which I think these particular matters or wearing those two are temperature and pressure . if you concentrate on solid I can interconvert this solid into liquid just by heating and I can interconvert this liquid into gas just by heating so there is interesting phenomena that I can convert ice into water this water into steam just by heating so temperature, so the temperature is the main criteria to convert solid into liquid, liquid into gas similarly I can convert gas into liquid, liquid into solid by applying pressure . so these two are very interesting  condition by which I can convert a solid into liquid , liquid into gas now will concentrate on a new concept where and temperature will play a waiter roll to convert gas into liquid that means at the high temperature and low pressure gases will exist at the low temperature and high pressure liquid will exist there for transformation of liquid to gases can be done by wearing this two condition so we have to concentrate on high temperature low pressure of gases low temperature and high pressure for liquids .  

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KARNATAKA STATE BOARD XI PCMB Biology Demo Videos

Hello Students, I am Sarita, and, you and me today will study all about the plants. In studying the plants in module 1, will study about the tissues the plant tissues and students before we can go to the study of the tissues we need to study 2 terms that is why we need to study about the tissues and where are we studying about the tissues, look at that histology is the branch of biology where we study all about the tissues and why do we study about the tissues exactly to study about the internal structure and that is anatomy. The branch of biology that deals with the study of the internal structures with that the whole concept of plant tissues originated with work of the pioneers Nehemiah Grew as well as Marcello Malpighi of whom Nehemiah Grew later was called the father of plant anatomy and this rightly a 1mark question awfully asked in the boards students, who is the father of plant anatomy? And that’s Nehemiah Grew. Now what are plant tissues, there are group of cells or aggregation of cells which perform a common function and they have a common origin. Plant tissues basically can be classified into two types based upon the capacity to divide or not and are been classified as Maristem. Maristem which are those cells, which have the capacity to divide and again observe that students it’s a 1 mark question awfully asked in the board. So Maristem are those tissues which are capable of dividing have the capacity to divide they are constantly undergoing division. They had no other work other than dividing. The second group of tissues called them as permanent tissues and these are tissues which have stopped dividing or incapable of division and they are assigned with a particular function to take part in the plant body. Here we go on now. Before we study very clearly about the tissues, let us study 2 more terms one of them is differentiation. Now differentiation is refer students cells that is Meristematic tissues becoming assigned as a function that is less specialize cells become more specialized cells in order to perform a function, observe there Meristerms get themselves differentiated to give rise to Pyrencyma which a permanent tissue, observe that Meristems differentiate to become a permanent tissue. So they are form permanent tissues which are performing a function. I move on to the next term that is Dedefferentiation, now this reverse of differentiation, differentiated cells revert back to the dividing stage that means pyrencyma which is a permanent tissue becomes now a Meristematic in nature, look at that Meristerms permanent tissues divide or Dedefferentiate to become Meristerms that is permanent tissues which are assigned with the function, now start dividing and awfully happens in last trees for increase of growth, 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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NEET & AIIMS 11th PCB-Chemistry-General Organic Chemistry Demo Videos

Welcome back, students. So, since we have learned quite a lot about the phenomena of resonance. Now, this would be the last module where we will learn the application of resonance and proceed to the next electronic effect that is the introduction of a Hyperconjugation, alright.

So, when you talk about the application of resonance is basically the phenomena which helps us to understand the stability in the compound. So, here we are going to learn the acidic strength of carboxylic acid. As we know the fact any compound is acid if its conjugate base is stable. So, let’s look at the structure of carboxylic acid and if it donates a proton it forms a conjugate base which is called as a carboxylate ion, as we have seen earlier. Whereas now we are having two structures one is carboxylic acid and other is carboxylate ion. We will try and draw resonating structure of both of these because both of them are having lone alternate pie condition present in them. Which is one of the condition, which make the compound eligible to show phenomena of resonance. So, look at the carboxylic acid structure, one which is to the left. If we show delocalisation here, we will get a resonating structure which can be simply drawn like this. So, since the lone pair from the oxy atom has delocalised it is experiencing loss of lone pair, so it will carry a positive charge and the double bonded action gaining one addition electron pair so carry a negative charge. Which means now the newly obtained resonating structure carries a positive and negative charge, which means this structure is a charged separated structure. Now, try to work on the phenomena of resonance in this structure right here, which is nothing but the carboxylate ion. So, we again have a lone alternate pie condition present over here. So, if we show delocalisation will get one more resonating structure, which is equivalent of this carboxylate ion and you can see that there is only negative charge which is getting dispersed, delocalised. So, I will say that in this structure, one and two, which are the equivalent structures, you will find that these equivalent structures are charged delocalised structure. And whenever you get a charged separated structure, it is relatively less stable structure, on the other hand charged delocalised which means if only charge is getting dispersed, delocalised over several atoms then that delocalised structure is always a more stable structure than the other. And now you think about why are we calling this as an application? We say that any compound is acid if its conjugate base is stable. So, in case of carboxylic acid we saw that the conjugate base was a carboxylate ion which is giving us charged delocalised structure which is more stable. The more stable the conjugate base is, more stable is the acid from which that conjugate base is obtained. So, likewise, students, if you try and work on the sulphonic acid or if you try and work on the phenol, you will find that the charged delocalised structures are going to be more stable due to phenomena of a resonance. So, phenomena of a resonance can simply justify the acidic strength of the compounds. Well, having known this, now we are going to learn or see a simple sequence of acidic strength but the most important one. So, remember if we talk about this several acids the sulphonic acid is the strongest acid among all because its resonating structures are more stable. And after sulphonic acid the next acidic is carboxylic acid followed by we have a carbonic acid and then we have phenol, after phenol the structure or the compound which comes most acidic is methanol. Check out here we have methanol then relatively weaker acid is water and then we again have any other alcohol. Students, please look at this, methanol is only exceptional alcohol which is more acidic than water, rest all other alcohols are weaker acid than water and after this we have alkyne then we have ammonia and then we get alkene followed by alkane. So, this is a simple order of acidic strength of several compounds and you got to remember this, alright, students.

Now, having said that we are going to learn application of resonance and this is what, was the application of resonance. Now, we will see introduction to phenomena of hyperconjugation and there are several more names coming up right up there. We will justify each one of thes names but for the time being let’s look at this reaction.

So, we have tertiary butyl halide, so when this tertiary butyl halide is brought in presence of a polar solvent like water, again we will learn what is meant by a polar, then the C-X bond heterolytically breaks and which means the bond pair between C and X have been completely taken by X, so, the carbon has lost bond pair and therefore produces a positively charged species like this. This positively charged species where positive charged is carried by carbon is called carbocation. Now, let’s take that carbocation over here and remove all these methyl group. We will get the structure like this. So, this is a general representation of carbocation. Let’s take a replicate of that, so we are taking four such general representations of a carbocation. And now, in structure number one, right there we are fulfilling all the valences let’s say by methyl groups. So, now that carbocationic carbon which is carrying a positive charge is attached to three more carbons. So, the nature of the carbon carrying positive charge is tertiary. So, we call it as tertiary carbocation. Same way if we take two methyl groups attached to such carbon you will call it as the secondary carbocation. Since it is attached to two other carbons. So, if we move further, if I am fulfilling two valences H’s and one with methyl the nature of carbocationic carbon is primary and I fulfil all the valences with H’s, I will call it as the methyl carbocation, since it looks like methyl. So, the nature of carbon carrying positive charge decides the nature of carbocation and it is said that among all this the order of stability is found to be tertiary the most, then secondary, then primary and methyl is least stable carbocation and why is the order so, can be proven by the phenomena of hyperconjugation.

So, in the very next module, we will try and prove this order and with respect to this we will learn the phenomena of hyperconjugation but we will stop here in this module, thank you. 

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NEET & AIIMS 11th PCB- 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, be it a uniform circular motion or non-uniform circular motion. The value of centripetal force is equal to MV square by R or M omega square R. We also discussed that the particles would be acted upon by a tangential force which is M alpha R for non-uniform circular motion. In this particular module, we will discuss about the examples of the same.

The statement of the first example says that a particle is projected with a speed u at the 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 projectile. 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 a 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 with the horizontal with speed u. If we try to resolve this speed we get the value of horizontal speed to be u cos theta and the vertical speed here is u sin 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 since the horizontal speed is u cos theta it does not change if air resistance is neglected, at the highest point the particle would have a horizontal speed of u cos theta.  Now, we know that for a particle moving in circle the velocity is always tangential. So, therefore at the highest point what we can say is, the horizontal direction is our tangential direction that is e theta cap and the vertical direction is our radial direction that is er cap. Let’s say that the centre of the circle here is C and the radius is R. We are required to find out this value of R. Now, if we recall this 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 projectile and therefore it is going under the influence of gravity. Thus the only force that acts on the particle is mg and that is also vertically down. That means it is towards the centre and the net force which is required for a particle in circular motion towards the centre 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 value of R to be equal to v square by g and we are putting down the value of instantaneous speed at the highest point which is u cos theta.  So, we get finally that the radius of curvature at the highest point as required in the question is u square cos square theta divided by g. This is the answer to this particular example.

Let us move ahead with another example which says that there is a bead which is placed over a rod which is hinged at one end. Now, this bead is at a distance of r from the hinge and the rod is rotating with the constant angular acceleration of alpha. We are also given that there is a coefficient of friction between the rod and the bead and the value is Mu. We have to find out the value of time and the angle after which the bead starts slipping and we have to neglect the gravity in this particular question.

Let us try to visualise what is happening in this particular case. The rod is 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 will get the motion of the rod looks something like, so we can conclude that the bead is moving in the 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 would also experience tangential force which is equal to m into alpha r. Now, what provides this bead the 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 m omega square r is equal to f which is equal to Mu N, let us mark this equation as one. Also we will discuss that the tangential force is provided by the normal reaction. And thus we can say that m alpha r is equal to N, this is our second equation. Now, if we divide the equation one and two, what we will get is the value of Mu the coefficient of friction is equal to omega square r divided by alpha r. Now, what we can conclude from here is 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 will do is, now we have to find out the value of time and angle. We will use the fact that we can use the three equations of motion for the constant angular acceleration, where the first equation is omega is equal to omega 0 plus alpha t which gives us the value of t to be equal to under root of Mu by alpha. Now, using the third equation we get omega square is equal omega 0 square plus 2 alpha theta which gives us the value of angle to be equal to under root of Mu by 2. These two values are the final answer to this particular example.

I hope you have understood the module. We will continue our discussion on circular motion, thank you. 

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NEET & AIIMS 11th PCB Biology-Control and Co-ordination-Nervous Demo Videos

Hello students, so in this module we will be dealing with Hindbrain or Rhombencephalon. And in Hindbrain we will be taking care of Cerebellum in this module.

Now, first let’s see where is your Hindbrain? That’s your Hindbrain. It comprises of three parts. The first part is called as Cerebellum. Right ahead is Pons, the second part of Hindbrain. Below Pons is the last part called as Medulla Bblongata.

In this module, students, we will be talking only about Cerebellum. Now Cerebellum is this structure, students. So where is it present in your brain? This is the side view of the brain. So, Cerebellum is present right below Occipital lobe or you can also say it is present behind the Medulla. Now, let’s see what is the structure of Cerebellum, how does it look like externally as well as internally? Externally it has got two lateral lobes which are known as cerebellar hemispheres and right in between is a worm like lobe which is known the Vermis, the median lobe is called as Vermis. Now, Cerebellum, students, it comprises of 11% of the total brain weight. This is the second largest part of the brain. And which is the largest part of the brain, that being Cerebrum. Cerebellum, students, outside has got folds which are known as Folia.

Cerebellum, let’s see how it looks inside when we cut this section of Cerebellum. It looks like this. These are two halves of the cerebellum. Now, what are these two halves? The outer peripheral region is known as Cerebellar Cortex and it is made up of grey matter, whereas inner called as Cerebellar Medulla, is made up of white matter. But white matter in this structure as you can see, students, is in the form of a branching tree like structure and that branching tree like structure is called as Arbor Vitae known as tree of life. Arbor Vitae an important structure. Let’s see, how it actually in real brain looks like? It looks like this, this is Arbor Vitae the internal structure, the white matter of Cerebellum.

Students, what is the function of Cerebellum? Important functions which are co-ordination of voluntary muscles, also maintaining your equilibrium.  That means what? That means it helps develop motor skills, the movement like walking or even finer skills like knitting. Now, there is a very well trained Cerebellum in this picture. Let’s observe, now you can see the co-ordination of voluntary muscles and maintaining equilibrium through beautifully trained Cerebellum in here.

Cerebellum, students, is highly sensitive to alcohol. Alcohol makes a person to lose the two important functions of cerebellum, co-ordination of muscular movements and loss of equilibrium. So, when Cerebellum is affected by alcohol. This is the impact of alcohol on Cerebellum, loss of equilibrium and co-ordination of muscles. In law it’s a criminal offence to be driving under the influence of alcohol.

So, in this module, students, we took care of Cerebellum. Next module we will be studying Pons Varolii as well as Medulla Oblongata. Till then take care.  

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NEET & AIIMS 11th PCB-Biology-Body Fluids & Circulation Demo Videos

Hello, students, in this module we are going to be drawing a so called very difficult diagram of biology. Which is the diagram I am talking about? I am talking about human heart and that too the internal structure of human heart. But you are going to observe within minutes this diagram is not going to be staying difficult for you. In fact you will enjoy drawing this diagram. But before we start with the diagram, students, you must know which side is which one. Now, in biology, if this is a sheet of paper in front you. Now, this is going to be the right side and this is going to be the left side of the diagram. So, let’s start drawing our diagram with first drawing the upper chambers and which are the chambers I am talking about the right atrium, students. So, first draw the right atrium. Similarly, a beautiful curve for the left atrium also. Now, this completes the upper two chambers of the heart. Now, we need to draw the lower chambers of the heart. So similarly, a nice curve for the right ventricle in here and then the left ventricle, students, on the other side. Now, this completes the outline of our heart.

Now observe, students, this is the base of the heart and now here this is the apex of the heart. Apex should be slightly tilted towards the left side. That’s a structurally correct diagram. Now further, students, draw the septum in between the ventricles. This is the inter-ventricular septum, students. After this a beautiful blood vessel arising from right ventricle, which is the pulmonary trunk, students, we are needing to draw. Now, pulmonary trunk, this trunk is dividing into right and the left branch. At this stage you need to draw only the left side of the pulmonary trunk, like this branching it beautifully in here, this is the left pulmonary artery. Now how about the right side, not drawing right now, students, because we need to draw another vessel right now which is the vessel arising from the left ventricle. I am talking about your systemic aorta arching beautifully in here. Three branches you draw in here and then curve it beautifully down here like this. This is your systemic aorta. After this, students, draw the superior vena cava in here and once we have drawn superior vena cava, draw down in here the inferior vena cava here. Now after drawing superior and inferior vena cava, students, now you need to complete the pulmonary trunks right side and observe in here, students, now herein right side of the pulmonary trunk. This is the right pulmonary artery branching beautifully just like the left side. So, this completed the pulmonary trunks both the sides. Now, observe the walls of all the four chambers. You need to draw the walls and how to draw the walls. Make thinner walls of atria as you can see in here. Here the atria have got thin walls, students, compared to the ventricle. Observe the walls of ventricle in here, students, the wall is uneven because of muscular ridges present and there are three papillary muscles also present in here and observe right ventricle wall is thicker than the wall of the right atrium. Now, left ventricles, students, when we need to draw, observe the wall of the left ventricle. It is uneven having two papillary muscles in here. Now, also note, students, when you are drawing the four walls of these four chambers, the thickest wall you will be making of the left ventricle. And I talked about the uneven line, don’t make a smooth line here. Why this uneven line because there are muscular ridges called columnae carnae present in here, also the muscles down here these are the papillary muscles in here, students.

Now, we need to see, we need to draw in each chamber what structure has to be present. Right now starting with the right atrium here, you need to draw the opening of the superior vena cava. Similarly you will be drawing the opening of the inferior vena cava also in here and also, students, in the right atrium you observe a depression down here. Now this depression is called as Fossa Ovalis. Now this was the right atrium structure.

How about the left atrium, in the left atrium, students, you need to draw four openings in here, two on one side and two on the other side. Now, these are the opening of the four pulmonary veins. Now, students, observe another important structure you need to draw. At this stage you will be drawing the valves which are the semi lunar valves. Here you can see this is the semi lunar valve of the pulmonary trunk and another semi lunar valve here of aorta, students. Now, these are the two pocket like valves we call as the semi lunar valves, the half moon shaped valves you need to draw.

After this an important structure we are drawing which is the valve present in between the right atrium and the right ventricle. Now these valves, students, this is known as the tricuspid valve because it is got three flaps present in it. On the other side that means in the left ventricle you can see, say in between the left atrium and the left ventricle, the valve which is present, this valve is the bicuspid valve because there are two flaps present in here. Observe in here, students, this is the bicuspid valve in here. Now, bicuspid valve is attached to the papillary muscles. These are the papillary muscles. By some fibres, now these are the fibres which are present, these fibres are known as chordae tendinae. So, three structures in here, this is the bicuspid valve also called as mitral valve then chordae tendinae, these fibrous structures, and down here you can see the papillary muscles in here.

Now, students, here in we are going to be colouring our heart. You don’t need to colour the heart. You need to just draw the pencil sketch. But let’s see how beautiful our heart looks in here. Now, look at that, pulmonary trunks dividing into pulmonary arteries, that systemic aorta then superior vena cava and the inferior vena cava also observe, students, the walls of all the four chambers, the thickest being the left ventricle wall, the openings of the atria, the bicuspid and the tricuspid valve and the pulmonary veins.

Now, this is the diagram of internal structure of heart, students, and the labelling which are required by you are these labelling only.

So, enjoy drawing the structure of human heart, take care.  

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