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JEE MAIN & ADVANCE 12th PCM Maths -Relation & Function Module-2 Demo Videos
Hello students, now let’s do few more problems on inverse of a function. The first one is this, we need to find inverse of this piece wise defined function where f is from 0, 4 to 0, 6. Okay, now let me give you the graph of this function so that you get a better idea. The graph looks like this. From 0 to 2, it’s x square, from 2 to 4 it’s x plus 2. Now, how do we backtrack. For 0 to 4 backtracking will be done by red curve and from 4 to 6 back tracking will be done by blue line. Now, say from 0 to 4 we will take 3 there. 3 will backtrack by red curve. While going it was square, so while coming back it should be square root. So, f inverse x here should be root of x. From 4 to 6 let us take 5, while going it was x plus 2, so, while coming back it should be minus 2, so x minus 2. Combining it we get our answer from 0 to 4 f inverse x is root x and from 4 to 6, f inverse x is x plus 2.
Moving ahead we have this example. We are given that fx is equal to its inverse and fx is kx plus 3. We got to find k. Simple, we will first find inverse and then equate it to fx to find k.
Now, for finding inverse, we will express x in terms of y as y is kx plus 3, x will be y minus 3 upon k and hence we have inverse equal to x by k minus 3 by k. Now as given f inverse x is equal to fx, so this should be equal to this. Now, comparing coefficients we have 1 by k equal to k and 3 equal to minus 3 by k. So, we get k equal to plus minus 1, from first equality and k equal to minus 1 from second equality. As there is ‘and’ in between them so taken into section we get k equal to minus 1. So, basically you did nothing you found f inverse x and equate it to fx to get k.
Now, let’s have another example. We got to find inverse of f where fx is x plus 1 by x, okay. I will take it equal to y and try to express x in terms of y. It becomes quadratic and we get 2 values of x, one is this and the other is this. Now, what to do? We know inverse is unique if f is this one way, f inverse is also one way. So, f inverse should be unique. Which of the following is correct? Now, we will use if f is from a to b, f inverse will be b to a. So, f inverse should be from minus infinity minus 2 to minus infinity minus 1. So, taking y equal to minus 3, we find x. We get x equal to minus 0.3815 in this case. Approximately minus 0.38 and minus 2.61 in this case. Now, this belongs to minus infinity minus 1 whereas this does not belong. So, that expression should be wrong and this expression should be right. So, this is wrong and this is my inverse. So, f inverse x is x minus root of x square minus 4 equal to 2. Basically in this case we learnt two things, number one f inverse is unique, number 2 if we need to check which one is correct, use that f inverse will from b to a, okay.
Now, moving ahead we have this example. We got to find m for which this function becomes invertible. Now, don’t get confused, simple condition keep it in mind, very important. A function is invertible if and only if it is bijective. So, we need to know for which values of m this is bijective. Now, cubic polynomial always has range real, I told you. So, it is onto, so you have to only make sure that it is one-one. For one-one we know f dash should be either only positive greater than or equal to zero or only less than or equal to zero. It’s a quadratic with a positive. We can make this quadratic always greater than or equal to zero doing this. We know a condition for quadratic to be always positive. So, a is already positive taking D less than or equal to zero. We get m square is less than equal to 9. That is m lies between minus 3 to 3. So for these values of m it will be one-one. It is already onto be in a cubic polynomial. So, it is bijective and hence it is invertible. This is a good question spend some time as I used to say, spend some time on this question.
So, keep practicing, god bless, take care.
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JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics Module-1 Demo Videos
Hello students, welcome back. So, we were discussing certain properties of charge as well as methods of charging. Now, let’s apply them in some examples. So, let’s start with example number one.
A glass rod is rubbed with a silk cloth. Glass rod acquires a charge of plus 19.2 into 10 raised to minus 19 Coulomb. You have to find the number of electrons lost by glass rod, charge gained by silk and is there a transfer of mass from glass to silk? We are given mass of electrons. Now, students, this is a case of glass rod being rubbed with a silk cloth. Now, in this case we know when rubbing takes place that is charging by friction must be taking place. So, this is a case of charging by friction. Now, when charging by friction takes place there is a transfer of electrons from one body to another, in this case from glass to electrons. Now, how much mass, how many electrons are being transferred can be determined with the help of quantization of charge which says that Q is equal to ne. Now, we know capital Q is 19.2 into 10 raised to minus 19 Coulomb, we know small n is integer and e is 1.6 into 10 raised to minus 19 Coulomb, that is charge on one electron. So, we apply the values we get the value of small n to be 12, that is 12 number of electrons are lost by glass rod. Now, since electrons lost by glass rod will be same as electrons gained by silk as it is the case of charging by friction. Therefore we can say that charge on silk will be minus n into e, where n is 12 and e is 1.6 into 10 raised to minus 19. So, we apply the value we will get Q to be minus 19.2 into 10 raised to minus 19 Coulomb. Now, we know the charge on silk, we know charge on glass. Let’s find will there be a transfer of mass.
Now, students we know when charging takes place it is due to transfer of electrons. It is due to transfer of some finite mass because electrons have some finite mass. Hence we can say yes, there is some mass transfer because an electron has a finite mass. Now, how much mass transfer will take place? That can be given by mass of 1 electron into number of electrons transferred which will come out to be 1.08 into 10 raised to minus 29 kg. Now, students, this was a case of charging by friction and quantization of charge clubbed together. Many problems are formed with the help of these two concepts.
Let’s discuss one more example where charging by friction is clubbed with which body will gain which type of charge. Now, if an object made of substance A is rubbed with an object made of substance B, then A becomes positive charged and B becomes negative charge. If however an object made of substance A is rubbed with an object made of substance C, then A is becoming negatively charged. We have to tell what will happen when B and C are rubbed with each other. Now, students, what are we given, we are given A and B when rubbed with each other, A is acquiring positive charge and B is acquiring negative charge. When A and C are rubbed with each other than A is acquiring negative charge obviously C will acquire positive charge. What will happen, when B and C are rubbed with each other? Now, let’s solve. Now, from information one when A and B are rubbed with each other we see A is acquiring positive charge and B is acquiring negative charge. As a result we can conclude that B has a higher tendency to gain electrons as compared to A. From second information when A and C are rubbed we can see that A becomes negative charge and C becomes positive charge. So, this can help us determine that out of A and C, A has a higher tendency to gain electrons as compared to C. Now, from step 1 and step 2, we can see that out of A, B, C, B has highest tendency to gain electrons as compared to A and C. So, when B and C are rubbed with each other then it is very obvious C will lose electrons and B will gain electrons. As a result we can say that C acquires positive charge and B acquires negative charge. So, students, we can see charging by friction and which body will be charged negative and which body will be charged positive, we have learned one example illustrating this particular concept.
We will come back in next module till then, students, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics module-2 Demo Videos
Welcome back, students.
So, students, till now we have been dealing with examples of field and cases of field due to discrete charges. There was a charge or two point charges or multiple charges placed at different, different points. Now, let us discuss the case of continuous charge systems and to start with we will be finding electric field to a line charge at axial position. Now, students, we will be following a certain strategy to solve these questions or these cases. Number one, we need to choose the element whose field is already known to us. For example if you talk of this line charge. Now, this line charge, we can, we can imagine this line charge is made of small, small point charges and since we know the field to point charge. Therefore we can consider a very small length which will be just like a point charge. Similarly, for this ring, if we closely observe and take a very small part on this particular ring, it will also be like a point charge whose field is known to us. And so if we talk of this disc and this disc, if we closely analyse, it will be made up of small, small rings and if we know the field to a ring then we should be taking our element as the ring. So, first step is to choose an element. Step 2 is to find field to this element and next step is to integrate. So, let’s apply that in this particular scenario wherein we have to find electric field due to a finite line charge, at any point along the line charge. So, what do we have? We have a line charge and we need to find field at point P and what is given to us? It is given all these conditions that is length, distance of this point from one of the edges and we are given that this line charge has a linear charge density Lamda and length L. So, step one will be to choose a very small element whose field is known to us. So, let’s consider a small element whose field is known to us. Now, what will that element be? Remember, for line charges the elements will be a very small length which will be like a point charge. Now, if we closely observe this line charge is made up of small charge plus small charge plus small charge and so on and out of these small charges if we closely observe and then take a very small portion and say it is at a distance of x from O and has a thickness of dx then this element will be just like a point charge and so it qualifies to be an element. So our element is a point charge at distance x from origin having length dx and having a charge dq which will be equal to Lamda dx. Why Lamda dx, because linear charge density Lambda implies charge on 1 unit length is Lambda. So, charge on dx unit length will be Lambda dx. So, we know charge on element is Lambda dx. Now, what is our step 2? Step 2 is to find field due to the small element that is find dE vector. Now, what is dE in magnitude? Remember field to a point charge is k charge upon distance square. So, the charge of this element is dq and what is the distance, distance will be total length, I mean, L plus a minus x. So, hence magnitude of the field will be kdq upon L plus a minus x square. Now, as far as direction is concerned. Since this dq charge is positive. So, field due to dq charge will be in this direction that is this d vector will be kdq in this direction. We can rewrite it as kdq upon L plus a minus x square I caron. So, now we know field due to our element. We have to find field at P, what is our step 3? Step 3 is to integrate. Now, before we integrate remember, students, our all the terms should be in same variables. Now, since dq is Lambda dx, we can rewrite this equation as dE vector is equal to k Lambda dx upon a plus L minus x square I caron. Now, we can integrate now. So, integrating we need to apply limits. Now, what will be the limits be? If we closely observe limits of x will be from 0 to x equal to L, because if we closely observe line charge starts from x is equal to 0 and goes till x is equal to L. So, hence this limit will become 0 to L. Now, what is the integration of this expression? Remember, integration of this expression L plus a minus x raise to power n dx is as shown. Provided n is not equal to 1. That is integral is L plus a minus x raise to power n plus 1 upon n plus 1 and there is a minus sign outside it because in this term there is coefficient of x is minus 1. So, let’s integrate we get E vector as minus k Lambda into a plus L minus x raise to power minus 1 upon minus 1 into minus 1 I caron, and if we apply limits we will get this result. I hope you know how to apply limits. Value of expression at x is equal to upper limit minus value of expression at x is equal to lower limit. So, in this case we will get this expression. On further solving we will get the result to be k lambda L upon a plus L into a I caron.
Students, if you feel that I have been fast here please pause re-watch this particular thing because you need repetitions to understand it nicely. So, we will get back in next module till then students, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics Module-3 Demo Videos
Welcome back, students. So, students in previous module we had just mentioned potential due to various continuous charges. We learnt potential due to ring of charge, due to an arc of charge. More over we also talked about potential due to a line of charge. Now, let us extend the discussion and study potential due to spherical charge configurations.
So, let us start with hollow spherical charge. Let us say we have a hollow sphere whose charge is capital Q and radius is capital R. Now, what will be potential at any point inside, say at a distance r from the centre? Students, potential at any point inside as a matter of fact is k capital Q by capital R. It is constant, it is independent of small r, right, surprised. So remember students, potential inside a hollow spherical charge distribution is constant, it is same as that of potential at surface. So, if we draw the graph of potential versus R for inside the region it will be a constant. So, till r small r does not become capital R, potential will be a constant, as shown. Now, what will be potential at any point on the outside? So, at any point on outside potential will be equal to k capital Q by small r. So, its graph will be a hyperbolic graph as shown which will be inversely proportional to small r. Students, I hope you remember what was electric field due to hollow spherical charge distributions. Electric field outside was kQ by small r square and inside it was zero. So, we had learned that for outside points the sphere of charge behaves like a point charge placed at centre. The story remains the same even for potentials. But for inside electric field inside was zero and here electric inside it is k capital Q by capital R that is constant.
Now, let’s talk about solid spherical charge distributions. Again let us consider we have solid sphere having charge capital Q and radius r. Now, what will be potential at a small r distance inside the solid sphere, will it be constant like the case of hollow spheres? No, students, it will not be constant. It will be equal to some (2:39) some tough complicated formula but quite easy to remember. k capital Q by 2 capital R into 3 minus small r square by capital R square. It is very big complicated formula but still not that tough to remember. Remember, repeat it 3 times with me, potential inside a solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Two more times, potential inside due to solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Third time, potential inside due to solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Students, you please repeat it, I will have to move on but you can repeat it as many times as you want. Now, if you have to draw the graph of this potential. You can see in this case, potential is a quadratic function, it is proportional to minus r square. So, its graph will be something like this. Potential at o will be basically potential at centre having small r is equal to zero. So, it will be 3kQ by 2R and potential at surface will be at small r is equal to capital R it will be equal to k capital Q by R. So, potential inside is, we are aware of, what will be potential outside. Now, at outside point potential just like hollow spherical charges, potential is constant, is like potential at centre. Its capital KQ by small r. So, at outside points potential due to the solid sphere is same as that of potential due to a point charged placed at centre. It will be equal to k capital Q by small r. So, if you have to draw the graph it will again be a hyperbolic function. So, students, I hope you remember potential due to hollow sphere inside is constant, due to solid sphere inside is this, tough, this big formula. Potential outside due to hollow sphere and solid sphere is same.
We will get back in next module. You please repeat this module as these particular formulas are important. Till then thank you, students.
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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-3 Demo Videos
Hello, students, now let’s try to find the induced EMF in a rod which is placed in a time bearing magnetic field. So see, here’s the situation, there is a rod of length L which is kept at a distance a from the centre. And this is a cylindrical magnetic field region. Now what will happen, since the magnetic field is changing and electric field we induced in the region and because of this electric field basically an EMF will be developed across the ends of the rod. We know the formula for induced electric field it is half rdB by dt. This is the induced electric field inside the cylindrical region. Now what will we do, we will analyze the situation, see this is a rod, electric field is at an angle theta with the rod. So how will we calculate the potential difference? So dv is equal to E.dl, or magnitude of dv is magnitude of E.dl. Now what we can do E dl, that means E.dl is E cos theta into dl because l is along the length, dl is along the length and E is at an angle theta. So d is equal to E cos theta into dl. And the value of E electric field is half rdB by dt so we replace here half r dB by dt cos theta dl.
Now in this formula what is r, r is the distance of the point from the centre. So in this diagram r will be basically a divided by cos theta. So we can replace r which is a variable actually by a cos theta. Now when we do that, r is a by cos theta and we had cos theta dl. So cos theta gets cancelled and we have a very simple expression a by 2 dB by dt integral dl. Now integral dl will be l so basically we have induced EMF across the ends as aL by 2 dB by dt. Now also note we have taken dB by dt outside the integration. That means we have assumed dB by dt is constant that means rate at which magnetic field is changing is constant, then only the formula is aL by 2 dB by dt. What is a, a is the distance of the rod from the centre. L is the length of the rod. So we have to remember this formula aL by 2 dB by dt is the induced EMF across the ends of the rod placed in a cylindrical time bearing magnetic field.
Now what about the direction, the direction basically can be given by the direction of induced electric field. So suppose here induced electric field is clockwise that means it will try to make the current flow in clockwise direction. So the direction of induced EMF will be such that A is at high potential, B is at low potential.
Now let’s take a few basic examples. See in this situation, a rod of length L is at a distance a. Now what about the direction, B will be at high potential or Q will be at high potential. See B is decreasing in this region, so the cause of induced electric field is decreasing B inside the plane. That means the induced electric field should be clockwise since that the B due to is also inside the plane, so the induced electric field is clockwise and induced EMF will be such that it will try to make the current flow in clockwise direction. So we know the formula is aL by 2 dB by dt, and t will be at high potential that means VP minus VQ is half aL dB by dt. a is the distance of the rod from the centre and L is the length of the rod.
Let’s take another example, very similar situation, P and Q length a, distance is a. Now P is decreasing, again if B is decreasing the direction of induced electric field will be clockwise. So that means the formula is aL by 2 dB by dt so VP minus VQ is half aL dB by dt, again the same result because the distance of the rod is the same and the length is L, so again half aL dB by dt and P will be at high potential such that it will also try to make the current flow in clockwise direction.
Now let’s take a different situation. Here one end of the rod is at the centre, other end is P, we have to find the induced EMF across the ends. So let’s first take the direction of the induced electric field, Now B is increasing in this situation so direction of induced electric field will be anti-clockwise, such that B due to is outside so it is opposing the Len’s Law. So basically induced electric field is in anti clockwise direction. Now the formula was aL by 2 dB by dt. Now what was a? a was the distance of the rod from the centre. Now this rod itself is passing through the centre. So basically distance of rod from centre is zero that means induced EMF in this case will be 0. So if a rod is passing through the centre, induced EMF in that rod will be 0.
Now what we can do, we can analyse it in a different way. We have shown the induced electric field, now if the rod is passing through the centre, the induced electric field will be perpendicular to length, now if induced electric field is perpendicular to length, we know dv is E.dL, dot product that means cos theta and cos 90 is 0 that means E.dl is 0. So we can look at it in this way because electric field is perpendicular to the length, induced EMF will be 0.
Now let’s take another example, here we have two rods of length a and it is symmetrically placed above the centre. And here dB by dt is k, k is greater than 0. That means magnetic field inside the plane is increasing. So we have the formula aL by 2 dB by dt we can analyze the situation, a is the distance of rod from the centre. So here how do we find the distance of rod from centre? See because it is symmetric angles will be 60 degree, half angle will be 30 degree. The length is a by 2 half the length of the rod so basically the perpendicular distance here will be a by 2 tan 30. Using trigonometry we can see a by 2 tan 30 is the perpendicular distance. So a by 2 root 3 is the perpendicular distance now just use the formula aL by 2 dB by dt so half as it is, a is the perpendicular distance which will be a by 2 root 3, l is the length of the rod, here it is a, so a and into dB by dt which is equal to k in this case. So we have induced EMF is ka square divide by 4 root 3. Now the direction, see, because field is increasing, direction of induced electric field will be anti clockwise so induced EMF will be such that it will try to make the current flow in anti clockwise direction. So here we can draw the equivalent batteries like this in the left side and in the right side, such that if it is connected it will try to make the current flow in anti clockwise direction. Now VP minus VQ in the diagram we can see VP minus VQ will be 2 times E, that means VP minus VQ is ka square divide by 2 root 3.
Thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-1 Demo Video
Hello, students, now let’s study magnetic flux before we move onto electromagnetic induction. Now same as electric flux, magnetic flux is also defined as the surface integral of the normal component of the magnetic field passing through that surface. And here the SI unit of magnetic flux is weber. So the mathematical formula is flux is integral B.ds. Like in electric field we had seen flux was E.ds. Here it is integral B.ds. So if this is a surface, the normal component of B.product takes care of that. So flux the mathematical formula is integral B.ds. Now B is the magnetic field vector and ds is basically the area vector. Now if we have an open surface the area vector can be taken either in the upward direction or in the downward direction, so there are two possibilities. But if we have a closed surface we always take the area vector as the outward normal. So if we have an open surface you have two choices, if we have a closed surface only take the outward normal as the area. Now see, if the magnitude of magnetic field that is B is constant and also the angle between B and area is constant, then we can simplify our equation. See flux is integral, B can be taken out of the integral. So B.integral ds. Now if we integrate ds, ds is the area vector. So if you integrate ds we will get the area vector so flux is B.A. And since the angle is constant we can also write flux is BA cos theta. So this is also the formula for flux but this is valid only when B is constant and the angle between B and area is constant over the entire surface. Then we can use this simple formula otherwise we have to use flux is integral B.ds.
Now let’s calculate magnetic flux in some simple situations. So in this figure you see there is an area and magnetic field lines are passing through that area. What we can do is first of all draw the area vector so area vector we had drawn perpendicular to the surface. We can also take the area vector towards the left, it is upto you. Now sometimes the diagram can be drawn like this, this is the side view. If you see this diagram from the front we can draw like this. If you see the same diagram from the left hand side or basically call it the front view, then see the magnetic field lines are crossing the area perpendicularly. So this is also a way of drawing. Now flux has two formulas, integral B.ds or BA cos theta. Now how do we decide what formula to use and how to calculate flux? So basically follow three simple steps. First check the magnetic field is constant over the entire surface. So in this question, see B is constant over the entire surface, okay, fine. Now let’s move on to checking the angle between B and area vector. So if you see the angle between B vector and area vector is 0 degree. So angle is 0 which is also constant, then find the flux. So flux is BA cos 0 that means flux is equal to B into A.
Let’s take another situation, here the area is tilted a bit, B is horizontal and which is constant. So again if you draw the area vector we can draw perpendicular to the surface. The other possibility is also outward but any one you can choose. Now if we mark the angle suppose the angle of the area vector was theta from the vertical. So just calculate the angle between area vector and B vector. So we can see the angle between area and B is again theta. So B is constant over the entire surface, first step is valid. Angle between B and A is also constant theta. So we can write flux is equal to BA cos theta.
So basically this thing we have studied in electrostatics, so in magnetism also or EMI also we need the concept of flux. We can also see, if we divide the area or basically take the component one will be A cos theta and other will be A sin theta. So the component of area along the field will not contribute in flux, because the normal will be perpendicular to B. Only the component of area which is perpendicular to B contributes in the flux, so we can also write flux is equal to B into A perpendicular. In this case A perpendicular is A cos theta. So flux will be BA cos theta.
Now let’s take another situation, let’s take it one by one. See this is an area vector and D is as shown in the figure. Now directly you can see, D is constant over the surface, now angle between D and A, see the area vector is inside the plane or outside the plane. So you can take any of the two options, let’s take the area vector out of the plane. So angle between B and A is 90 degrees, you can see in the diagram the angle between B and A is 90 degrees. So we can write flux is BA cos theta that means BA cos 90 which is equal to 0.
Let’s take this situation, now see the magnetic field lines are inside the plane. Area vector can be taken outside or inside. See B is constant over the entire surface, the angle between B and area vector is either 180 or 0 degree depends on how you take the area vector. So we have taken the area vector like this, the angle is 180 degrees, so flux is BA cos 180 that means minus BA.
Let’s take this situation, see this is a tricky situation, this 30 degrees is useless because if you see angle between B and area vector is basically 90 degree. B is constant over the entire surface that is very obvious. Now angle between B and A is again 90 degrees because if you draw any line the plane is perpendicular, the area vector and B vector plane is perpendicular. So if you draw any line at 30 degrees, 0 degrees, 90 degrees, it will all be perpendicular to the area vector. So again flux is equal to BA cos 90 that means zero.
Thank you.
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