 All right, so those who are online, please type in your names. You may like to increase the resolution of your screen. I think these two questions are not very clear, but still I think you can read it if you increase the resolution properly. Just type in your names quickly so that we can start. So I have taken past year comma decay questions and I will be discussing them today. Ritvik is there, Vaishnavi is there, that's it. So Sanjana is also there. Let us start solving these questions. You can see the first two questions in front of your screen. Please attempt these two questions. Are you able to read it or should I dictate, are you able to read it, right? Tell me the answer for these two questions. What is the answer? Should I solve it? Should I solve? Okay, let us see the first question. You have a horizontal metal wire to be prevented from falling under gravity. The wire carries current from north to south. Now external magnetic field should act towards work direction. Whichever direction the magnetic field is, it should create a force that will oppose the gravity. Isn't it? From north to south the current is going. So magnetic force should be upwards. All of you understood this? So which direction magnetic field should be? Inside the screen or outside the screen? Inside is I L cross B, that is the force. Isn't it? This is the direction of length vector L cross B. If it goes inside then only force is upwards. Align your right hand in the direction of current and then go towards the magnetic field. Force will come upward. Okay, now tell me if this is north, what is the inside direction? The inside is east or west. Inner direction is east or west. It cannot be north and south definitely, it will be east. You can think like this, that when you draw it on a plane it is like this, east and west. So when you rotate it like this, where north comes over here, the east goes into the plane and west comes out of the plane. So like this you can analyze the answer is D. What about the second question? Do you get the answer? What's the second one? Second one is D. Magnetic force on a charge is 0 when both charge is at rest or moving parallel. Correct. So this comes directly from the formula. Magnetic force is Q V B sin theta, right? The force will be 0 when either velocity is 0 or angle theta is 0 or 180 degrees. Okay. Fine. So these are the two questions. You can see that CVT questions of the comet K which is of the level of CVT, it is relatively easier. Okay. And that is the reason why you have to solve 60 questions in one hour, 20 minutes. All these two questions, let me know if you are not able to read it, please let me know. I'll dictate the question. Anybody got the answer? Okay. Yes. Answered. Tanjana as well. Yes. The correct answer is C. I'll solve question number three now. Light corresponding to transition n equal to 4 to n equal to 2 in hydrogen atom falls on an alkali metal with work function 1.9. Maximum kind energy of photo electron emitter will be what? Now in order to solve this question, it is, you know, advisable that you should be remembering energy levels, so the energy at different levels, then it will be easier for you. For example, energy in the n equal to 1 level is minus of 13.6 electron volt, right? Energy at n equal to 2 level is minus of 3.4 electron volt and this is minus 1.8 electron volt and energy at the fourth level is minus of 0.84 electron volt, right? So when the transition happens from 4 to 2, okay, from 4 to 2, when transition happens, it will be between these two levels, right? So energy corresponding to the difference of energies between these two will be liberated, okay? So how much energy will be liberated? 3.4 minus 0.84. This much energy will be given away, all right? Now when this much energy is falling on the metal surface, okay, which has a work function of 1.9 electron volt, then the kinetic energy will be what? We have the formula, right, H mu minus work function is the maximum kinetic energy, right? So H mu is equal to this energy and phi is equal to 1.9. So when you subtract 3.4 minus 0.84 and minus 1.9, when you subtract like this, you will get the maximum kinetic energy, all right, which will come out to be 0.65 electron volt approximately, okay? What about the fourth one? So those who are joining a little late, please message your name, okay, Vaishnavi has answered question number four. Others, correct, the answer is B. Intensity is proportional to what is the area of the aperture, okay? And area is proportional to square of the radius or square of the diameter. So if diameter is half, intensity will be 1 by 4, okay? Let us go to the next, these two. Okay, one more thing, guys, if you are serious about writing comet K or CET, please take the mock test which are given in the group every day, okay? Like for example, today also Gaurav sir has circulated one mock test, right? Please take it. Until as you take the mock test, your preparation is, preparation cannot be more than 60 percent. 40 percent is just taking a lot of mock tests. And don't worry if you are not 100 percent prepared or you need to study some bit more. You will never be able to prepare 100 percent, that 100 percent will never come. Even if you are like slightly prepared, even then you should take the mock test. Mock test is to mature your strategy of taking exams, right? They are not there to test your knowledge as such. Okay, Atmaish is also there. Please solve this. Today we are doing comet K level questions. Atmaish, I was expecting that you will be taking J advanced mock test today but then you bunged it. Yes. Solve both of the questions and then like, should I solve? So here I will talk about a small trick, question number five. You know that the formula which we will be using is this, h c by lambda. This is the energy corresponding to a wavelength minus work function. This is equal to, you know, maximum kinetic energy which can be written as e times the stopping potential. Okay. Now, you can, you know, right now all of this is in SI system, okay? But you can write V as h c by e lambda minus phi by e, okay? Now this energy is in terms of electron volts, isn't it? And this energy is also in terms of electron volts. So if you subtract the energy in terms of electron volts, you will get the stopping potential, okay? And many a times you will see that you will keep on getting this expression, h c by e, okay? This is very common. So, you know, best thing is to remember the final value of h c by e itself, okay? So otherwise, you know, we will be doing unnecessary same calculation again and again. So if you remember the final value of this, 1.6 into 10 raise to the power minus 19, this will help you, okay? This is around four times. So you can say that this is 12 into, this is 27, 10 raise to the power minus 7, isn't it? 27. Is this correct? Yes. So you can say that this is equal to 1.2 into 10 raise to the power minus 6. So it makes sense to remember the final value of h c by e itself. And then you can directly write down the formula, you know, that stopping potential will be 1.2 into 10 raise to the power minus 6 divided by wavelength, which is 4 into 10 raise to the power minus 7 minus the work function in terms of electron volts, right? And here, the value of v is given as 1.5, okay? Similarly, you can find, you can create an equation for the other wavelength as well, okay? So you will get option b to be correct, all right? But whatever shortcuts we have discussed, don't try to force very hard on yourself to use it. If you are not very comfortable, just fall back to the basics and be using brute force solve it. Okay? 6-1, what is the answer? One more thing, guys, the CET and comment has no negative markings. So please don't leave any question unattempted. But then once in a while, their rules and regulations do get changed. So just read once when you get a CET paper in actual exam, read the instructions and ensure that there is no negative marking and then attempt all the questions, even if like, you know, if they are set. Okay, so first of all, we know that the size of the central maxima for single slit experiment or the diffraction, okay? Size of that, which is, can be presented as radius of that central maxima is proportional to the wavelength of the light used, okay? And we know that the wavelength is equal to, you remember this formula 1.33 divided by root of V and strong, you guys remember this for the electron, V is accelerating voltage, right? So we know that wavelength is proportional to 1 divided by V raise to power half, okay? I can say that this is V raise to power minus half. So lambda 1 by lambda 2 is equal to V1 raise to power minus half divided by V2 raise to power minus half. So this is V2 by V1 power half, okay? And this is also equal to the, you know, ratio of the size of the central maxima, okay? So I can understand there is a little, a different kind of question which, which is as in CET, but then yes, out of 60 questions, there are a couple of questions which are of a different genre which not everybody is expected to solve, okay? So some questions are very easy, some questions are slightly difficult. All of you understood? Should I move forward? Seventh is a fact-based question, either you know it or don't know it. What is the answer for seventh? Yes, B is the correct answer, okay? What is the answer for eighth? See in a humid condition, atmospheric sky would appear, what? If it is a clear sky, then the blue light gets scattered the most, okay? You can see that whenever, I mean the scattering is most when the size of the particle is of the order of the wavelength which it is scattering, okay? So the nitrogen and oxygen molecule have the size of the wavelength of the blue light. So it scatters. But if water is there which has a different size, it may not scatter the blue light as well, as well it was like this thing, okay? So the answer should be A, dull blue. What about ninth? This is also a fact-based question, the answer is B, okay? This is little weird why it is asked here, but then you know, okay? So such kind of fact-based questions do get asked in the physics paper as well, okay? And don't just focus completely on the numericals, your preparation for board exam where you might have you know recollected lot of theoretical knowledge, all that will be useful in CT paper. The nematic liquid is where all the molecules, they are oriented parallel to each other, okay? It is actually not in syllabus but then it is asked over here, okay? So you can say that this question is out of syllabus. All the molecules are, the definition of nematic liquid itself is that all the molecules should be arranged parallel to each other but they don't need to be in a single plane, okay? So that's why option B is correct, orientational order is there because the molecules are arranged parallel to each other, okay? Probably you know this paper is slightly older so that is why you are seeing couple of questions which are out of syllabus, okay? So we will not get into all that, try solving the 12th one, leave 10th and 11th, 12th one. What is R proportional to quantum, with respect to quantum number? R is proportional to quantum number square, right? And force is proportional to what? With respect to R, what is force proportional to? It is 1 by R square, yes or no? So from these two you will get force is proportional to 1 by N raise to power 4, that's our option D, okay? So you can see that these CT level questions they are not very, very straight forward although they are simpler than J mains but not so simple that you start comparing with your board level, okay? They are slightly, why isn't force proportional to V square by R? Yes Vaishnavi, it is proportional to V square by R, okay? But when you see and force is also proportional to 1 by R square, okay? What kind of force is there, electrostatic force is there, only that force is there, right? So this is that force is K times q1, q2 divided by R square, okay? This force is equal to MV square by R also, okay? Both are correct but the problem with this expression is that even V is a variable and R is also a variable, okay? But over here only R is a variable so it is easy to deal with, okay? So all three you can solve, all three questions. Okay, so the question is why are you posting such a message on the group? You could have posted that personally to me, it is your choice Atmesh, you can do but then at center now if you come it will not be possible because people have already started the test and they will get disturbed so you can open the website from your home also and you can take a test from your home, okay? The link I have already circulated, all right? All these three questions and let me know what is the answer. As in when you are solving you can answer it, like for example if you solve 13 questions you can just answer the 13th one, answer is there and then 14th one is that, like that. Very good. Energy is proportional to 1 by N square, right? So energy of the first excited state is E2, okay? It is proportional to 1 by 2 square and the second excited state is E3 which is 1 by 3 square. E2 by E3 is 9 by 4, okay, 15th. We know that R is equal to R0, 8 to the power 1 by 3, okay, where A is the mass number, it is not the atomic number, okay? So this is 1.2, 27 power 1 by 3, so this is 3.6, the answer is C, last time how come you got A, what about the 14th, half adder, they are and and zorgates, remember this, it is just a name, okay, you should remember that half adder is a combination of and and zorgates. Any doubts, should I go to the next question, are you able to read these two questions, please confirm 16th, I will read it, activity of radioactive sample reduces to, reduces to that number 16, 1 by 1 4th reduces to 1 4th of its original activity A0 in 12 years, after another 12 years activity will become what? Oh, I see, so during the day time, right, must be any kind of causes from 1250 hours, oh, I see, I see, I see, I see, I see, I see, I see, I see, I see, I see, I see, I see, okay, I am finding the way to the underground, I see, I see. Did you get the answer, what is the half life, how much is the half life, is the answer is C, you can see it has become 1 4th, as in it has become half of half in 12 years, okay, so half life is 6 years, right, 2 half life is needed to become 1 4th, so it is 6 years the half life, now what is asked is after another 12 years what will happen, so after another 12 years how many half lives, 4 half life, so this ratio power 4, so this is 1 by 16, got it, 17th is again a fact based question, either you know it or don't know it, the answer is C, read the question carefully and try to remember this fact, okay, let's go to the next one, this is by the way Comet K 2006 paper, actually when I have created this I thought 2006 is latest but then I realized it is 2018 going on, it is 19 going on, so it is more than 12 years old, time runs fast, this one, okay, so Vaishnavi has answered 18th, 18th is not D Vaishnavi, it is D, D for donkey, let's read the question, intensity of gamma rays from a given source is I0, on passing through X meter of lead, it is reduced to I0 by 8, thickness of lead where will reduce it to I by 2 in meters is what, okay, so intensity you should remember is proportional to 1 by thickness, okay, how much thickness it has traveled, alright, inverse proportion, more thickness it travels, lesser is the intensity, okay, so I1 by I2 is equal to thickness, second thickness by first thickness, okay, and I1 by I2 is I0 by 8, the first intensity and we need to get the second intensity is I0 by 2, okay, so this will be equal to thickness, second thickness divided by the first thickness which is let us say X, okay, so you will get the second thickness as X by 4, this is critical, you should remember this, this D, 19th what is the answer, just above the balance band, okay, good, let's go to the next, solve these, solve this. Okay, 19th, not the 19th, sorry 20th, 20th what is the answer, 20th answer is C, they are out of phase, input and output signal, okay, 21, 21 is also C that is correct, only Vaishnavi got 21, nobody else, how come, so let's say this is mass M1, a constant force acts on it, acceleration A1 is created and if it is acting on M2, acceleration A2 is created, okay, now if the same force is acted on the combined mass M1, M2, then we have to find out what is the acceleration, okay, so we have this equation F is equal to M1 A1 and F is equal to M2 A2, okay and F is equal to M1 plus M2 to capital A, we need to find the value of capital A, right, okay, so solve these questions, solve these things and you will get the answer, all right, for example M1 will be F by A1 and M2 will be F by M2, sorry F by A2, so I'll substitute it over here, M1 is F by A1 and M2 is F by A2 like this, okay, so F cancels away and from here A will be equal to A1 A2 divided by A1 plus A2, fine, so you get option C to be correct, what about 20 seconds, physical quantity which remains constant throughout the trajectory of a particle, we are talking about the projectile motion over here, correct, horizontal component of the velocity, see what kind of questions your CET paper has, like I think 20 to 30 percent of questions are extremely simple, okay, but then there are one fourth or one fifth number of questions which are, you know, you can say that up to mains level, okay, it's not that entire CET paper is very very simple, also energy remains constant, no energy doesn't remain constant, Vaishnavi energy doesn't remain constant, if you say, if you talk about, you know, entire mechanical energy, let's say we, okay, when you say energy you mean sum of kinetic energy plus potential energy, right, but the option written over here, this, they are assuming that they mean to be kinetic energy, okay, I get, understand Vaishnavi, what you are saying is correct if your definition of energy is kinetic energy plus potential energy, okay, but then here when it is written energy, they are assuming it to be kinetic energy, okay, all right, the question is little vague, okay, let us go to the next one, all of these questions are from mechanics, 23rd is time of flight, direct formula, do you sign theta by g, right, 10 seconds, other two questions, no one, question 24, what I saw, the hint is that when you drop a stone from the balloon, the stone was not at rest, it was moving with the velocity of balloon, it had initial velocity, okay, I solved 24th question, so here is balloon which is accelerating at 5 meter per second square, after 10 seconds a stone is released, right, so after 10 seconds, the velocity will be v is equal to u plus 80, so 5 into 10, so 50 meter per second will be the velocity of stone upwards, it is just released, so because of its inertia, it will try to move with the balloon, okay, and it reaches the ground after traveling a distance x in the air under free fall, the value of x is what, now you can see that its trajectory will be something like this, it goes up, velocity becomes zero and then it will fall down, right, this distance will be the distance balloon has risen up to, that will be equal to half 80 square, right, so half into 5 into 10 square, so this is 100 divided by 250 into 5, so that is 250 meters, this distance is 250 meters, right, so the stone has to travel 250 meters from here and the stone has travelled this distance twice and how much is that distance, initial velocity is this, at the highest point the final velocity should be zero, okay, so we will use v square equal to u square plus 2 As, the velocity in final velocity is zero, initial velocity is this thing 50, 50 square minus 2 into g is 10 into s, this distance is s from here to here, okay, so this is 50 square, 50 into 50 divided by 20, so 25 into 5 that is 125, okay, so 125 up and then 125 down, total becomes 250 and then 250 this also, so 250 plus 250 500, so answer is c, all of you understood, please type in yes or no, try solving these three questions, correct, 26 answer is b, it is theoretical, yes, 27th, nobody got 27th, who has answered c, who has retracted the message, name, okay, 27th the answer is siddhant, siddhant you are attending class, long time, okay, all right, so see the velocity, I like to pull siddhant's leg, don't mind that, siddhant I am not here to chat, we can chat after the class, this is k, isn't it, then this wave equation is in the form of kx minus omega t, this is k and this is 3 pi is omega, right, velocity is omega by k for a wave that is 3 pi divided by 0.01 pi, so that is equal to 300, okay, D is coming out, 27th, yeah, the answer is 300, correct, 28th, 28th, what is the answer, 28th, D, Vaishnavi has said D, okay, so the wavelengths, correct, velocity of sound is let us say v, v divided by lambda 1 is mu 1, okay, mu 2 is v divided by lambda 2, okay, mu 1 minus mu 2 is given as is given as 34 divided by 10, beat frequency you have to equate, 34 beats in 10 second means 3.4 beats in 1 second, so beat frequency is 3.4 not 34, okay, we will go to the next question now, these two, what is the answer for 29th? Okay, so Vaishnavi has answered, Samjana has answered, others what are you doing, so this is open pipe, okay, its fundamental frequency is like this, okay, if it is closed from one end, its fundamental frequency becomes that, okay, so you can see that the frequency of this wave is two times of frequency of that wave, isn't it, the length is same, the length of the pipe is same, but in the same length there are, double the, this thing contains, but here you can say that lambda 1 by 2 is contained and here lambda 2 by 4 is contained in the same length, so when you draw one complete wavelength the size of this one complete wavelength will be two times the size of this complete wavelength, so that is the frequency of this is two times the frequency of the closed pipe, okay, so frequency of closed pipe is open pipe frequency divided by 2, 30th Vaishnavi has already answered, others, okay, two spherical black body of radii R1 and R2 are at a temperature of T1 and T2 respectively, they radiate the same power, then R1 by R2 is what, now the radiation power is equal to sigma EA T ratio power 4, this is radiation power, okay, the two spherical objects are there, radii R1 R2, this will be sigma and they are black, right, so E is equal to 1, so sigma into 4 pi R1 square T1 ratio power 4, okay, so you can see that power is proportional to R1, R2 T ratio power 4, okay, so P1 by P2 is equal to R1 by R2 square multiplied by T1 by T2 ratio power 4 and this will be equal to 1, okay, so R1 by R2 is equal to T2 by T1 whole square, that's our option B, okay, let us move to next question, solve these four questions. Okay, you can answer one by one, 32 expansion of the universe is accounted by saying C, Doppler effect, that's correct, Doppler effect is also red shift, the wavelength is shifting towards the red side of the spectrum as in the wavelength is increased, so the frequency is decreased, frequency decreases when the source goes away, so it is seen that the light received from far away galaxies are, the wavelength of that is towards the red side of the spectrum, so from there you can make out that the universe is expanding or the universe is going away from you, okay, 31 null that is not correct, see gas is compressed adiabatically till its temperature is doubled, so the ratio of initial volume to final volume is what, so basically we need to use this formula T1 V1 gamma minus 1 is equal to T2 V2 gamma minus 1, so V1 by V2 raise to power gamma minus 1 is equal to T2 by T1, okay, ratio of initial volume to final volume, so V1 by V2 is equal to T2 by T1 to the power 1 by gamma minus 1, T2 is equal to 2 times of T1, so 2 raise to power 1 by gamma minus 1, right, now tell me gamma will be, it will be, so sorry V1 by V2 will be more than 1 or not, it will be more than 1, okay, gamma is like approximately let's say 1.4, by the way this option B is more than 1, not more than 2, sorry for that, that is how it is shown, because this value will be more than 2, this thing is more than 2 because 1 divided by gamma minus 1, you can see it is 1 divided by 0.4, 1 divided by 0.4 is greater than 1, so 2 raise to power greater than 1 is greater than 2, so D cannot be correct, it is not less than 1, it can't be 2 weeks, right, so it has to be B, okay, in fact if you write greater than 2 then also B is correct, understood all of you, how 31 question is solved, see gamma tends towards a value of 2, usually gamma is less than 2, okay 33 and 34, Srikant has answered 33rd, others 33rd is C, correct, 34th, total internal reflection happens when the light travels from denser medium to rarer medium, right, so C to A is rare to denser, not possible, C to B rare to denser not possible, B to A rare to denser not possible, so answer is D, A to C, okay, fine, let us go to the next set of questions, these three questions from optics, you can see that there is a heavy weightage of ray optics, two questions we already solved, three more questions are there out of 60. So you can't use here the equivalent focal length concept or the formula, you have to that formula which you remember, okay, so if you know that the equivalent focal length of two lengths are separated by a distance x is this, then only you can use the direct formula, okay, this is the formula, if the distance between them is x, the distance between them is 0, if x is 0, it will be simply 1 by f1 plus 1 by f2 is 1 by f equivalent, okay, by using that and get the answers, answer is A, 36 A, Vaishnavi is getting A 436, 36 answer is 45 degrees, answer given is D, B for Bombay, let's see, we have light incident on an equilibrated prism of refractive index, route 2 suffers minimum angle of deviation, so route 2, mu is equal to sin of A which is 60 degrees plus delta M by 2 divided by sin of A by 2 which is 60 degree by 2, okay, so we will get sin of 60 degree plus delta M by 2, sin of 60 degree by 2 is sin 30 which is half, so half into route 2 is 1 by route 2, okay, then the angle of incidence and minimum angle of deviation is what, the minimum angle of deviation you can find out from here, right, so this thing 60 degree plus delta M by 2, this should be equal to 45 degrees, sin inverse of 1 by route 2 is 45 degrees, okay, so minimum angle of deviation is equal to 30 degrees, all right, delta M is 30 degrees, so it can't be angle of incidence and minimum angle of deviation, yeah, so it can't be B, it can't be C and we know that for minimum angle of deviation, the refracted ray will be horizontal like this, if we draw a normal, this angle is known because this is 60, so refracted angle is 30 degrees, this angle is unknown, let's call it as I, okay, and refractive index is route 2, so 1 into sin of I will be equal to route 2 times sin of 30 degrees, so that I is equal to 45 degrees, so yes, the answer is A, I don't know why it is written B, but answer is A, yeah, correct, what about 37, anyone got 37, should I solve, okay, so I'll solve question number 37, there is a plane of convex lens, the curved surface is silver, tell me, will it behave like a lens or a mirror, the overall thing, it will behave like a lens or a mirror, it will be mirror only because light is reflected back, okay, now if it would have been lens, then if you could have used the lens mega formula, 1 by f is equal to mu 2 minus mu 1 divided by mu 1, 1 by r1 minus 1 by r2, okay, so 1 by f is equal to mu 1 is air, right, and mu 2 is n, so it will be n minus 1, r2 is minus of radius of curvature of this, so it will be this divided by r, getting it fine, this is for the lens, okay, now when light goes inside, okay, when light goes inside, it will experience this lens 2 times, while going inside and while coming out, okay, right, and it will experience the mirror once, so the equivalent focal length is equal to 2 divided by focal length of the lens as in 1 by fL plus 1 by fL, plus 1 by focal length of the mirror, okay, so this is focal length of the lens, so 2 by fL will be equal to 2 divided by 2 into n minus 1 by r plus focal length of the mirror is r by 2, right, so it will be 2 by r, so 1 by f will be equal to 2n by r or r by 2n will be the equivalent focal length of the, it will behave like a concave mirror, okay, any doubts? Whatever we did, see, Rithvik, it is not that you are targeting 100% marks, understand that, we are not here to score 100% marks, this is not your board exam, okay, this is a CT exam where, you know, you just target the easier questions initially and try to maximize your marks with whatever given amount of time you have, don't target to score full, okay, otherwise whatever you know, even that you will not be able to answer and it is very rare that somebody gets full, okay, so decent marks like, you know, 70% or, you know, near about 70% is a very good mark in a CT exam, okay, don't worry too much about it, just target the easier questions and the concepts which are, which you feel that you are strong at, just target questions on that, I think, you know, mechanics can be very difficult if it is a J advance paper, but in CT, I think mechanics will be easy to solve because you guys have already practiced a lot of mechanics questions which are up to mains or advance level, so CT level mechanics question you will find very easy, so my suggestion, although it may, it may not work for everyone, my suggestion is to target the mechanics question first in a CT paper, not in J mains or J advance, but in CT you can target J, you can target the mechanics question first and then you can target the electrostatics, you know, magnetism, you don't need to go sequentially also and it is very easy to talk about strategy and you'll see that when you try to apply the strategy in the heat of the moment, all the strategy will fall apart if you have not practiced enough mock test, okay, so you should take at least 5 to 10 mock tests, then only sit for any exam, okay, for example, your CT exam is coming in, so you must, you know, you must be, you must be taking at least 10 mock tests of CT before writing the actual exam and similarly, if you are writing J mains in April again, then there also you must, you know, take at least 10 mock tests, okay, revising syllabus is of no use just before the exam, it is the strategy which you have to work on, all right, these are on the wave optics I think, try to solve this, okay, anyone? We'll take a break in 10 minutes, guys, attempt these three questions, should I solve or should I wait? Tell me quickly, okay, 38th, salt, salt, okay, 38th, when a thin transparent plate of refractive index 1.5 is introduced in the path of one of the interfering beam, 20 fringes shift, if the plate of is replaced by another plate of refractive index 1.6 and half the thickness, the number of fringes that will displace is what? See the fringe width will be lambda d by d, okay, so fringe width will be same in both the conditions, all right, yes or no, but the center maximum will shift, if you have slab over here, central maximum will shift somewhere over here, this is the point where the net path difference is zero, now the location of this point is let us say y, the number of fringes up to the distance of y will be y divided by beta, okay, so how to find y, you can see that at a distance y, the actual path difference is this, which is d tan theta, tan theta is y by d, so you can see that this is y d by capital D, this is your, this thing, this is your actual path difference and there will be a path difference because of this slit as well, sorry, because of this glass slab as well, which is mu minus 1 into the thickness, right, so when the path difference because of this and that, they are equal, the path difference at this point will be zero, okay, so y d by d, when this is equal to mu minus 1 into t, you will get location of the central maximum to be mu minus 1 t d by small d, okay, so number of fringes in this distance y is this divided by the fringe width, okay and when you remove it, sorry earlier the central maximum was here, so when you place it central maximum shifts here, so this many fringes which are over here will shift, right, that is what the number of fringes shift is, the number of that has got shifted is y divided by beta, so that is lambda d by d, so you will get number of fringes that has shifted is mu minus 1 t by lambda, okay, so this is 20 initially and this is lambda 1, okay and this is mu 1, okay now plate is replaced by another plate of refractive index 1.6, let us say 1.6 is mu 2, so mu 2 minus 1, this is t 1, t 2, now the wavelength of the light is same in both the cases, so we will keep it as lambda 1, this is let us say n 2, okay, t 2 is t 1 by 2, so I can write it as t 1 by 2, all right, so you can divide this and that, we will get the answer, okay, can you solve it now, 38 the answer is b, all of you understood, right pain, great, solve these two questions then we will take a break, these two, okay leave the 39 Newton's ring, okay, just solve 40th, you know the formula for the dark band, where it is, this is a location of dark fringe, okay, in Young's double sheet experiment, this is a location for bright fringe, but in a single sheet experiment, same similar kind of expression you get for the dark fringe, but here you are getting c, yes, c is the right answer, okay, so third dark band, so n equal to 3, lambda is 5000 Armstrong, capital D is 1 meter and a, the width of this latest 0.1mm, okay, right, so like this you have to solve these questions, we will take a small break, right now it is 1pm, we meet at 1.15, those who don't want to take a break, I will move the slide to the next one, you can attempt those questions, so yes, okay, so we will have a break from 1 to 1.15, so see you at 1.15pm, okay, okay guys, are you there, who are there, please type in your names, are you going to hear me, okay, one of the three is there, okay, try solving these questions, you don't need to solve 41 out of syllabus, just solve 42, you remember that polarizing angle, the reflected and refracted light should be 90 degree to each other, okay, fine, so this is the interface, light comes like this, it will get reflected like that and refraction will happen, okay, so this is the angle of incidence, even that is the angle of incidence and this is the angle of refraction, okay, now the division angle is 22 degrees, how will you do this now, this should be 90 degree, right, so i plus r should also be 90 degrees, okay and this is the deviation angle, okay, the deviation angle is what, this angle is i minus r, so i minus r should be equal to 22 degrees, okay, so 2i is 112 degrees, so i is 56 degrees, understood, Ritvik you understood, you go to the next question then, solve this one, I am assuming you are able to read, if you are not able to read anything, any particular thing please tell me, I will read it out to you, 45 is B, 44 is the idea of continental radiation is used to explain photoelectric emission, correct, you can see that in CT many questions are theoretical in nature, so you can target to first identify those theoretical question and finish it off quickly, 45th, electric intensity due to electric dipole varies with the distance as r square n where n is equal to minus 3, okay, right, 46th, what is the answer, 46th is a, we will confirm that 46th, yes 46th is a, conducting sphere of radius r has surface charge density sigma, the electric potential on the surface is what, we know that on the surface the sphere will behave like a point charge, okay, so potential due to a point charge is what, potential due to a point charge is 1 by 4 pi epsilon naught q by r, okay, and q is what, q is sigma into 4 pi r square, okay, this divided by r, so you will get here sigma r by epsilon naught, sigma r by epsilon naught, right, so this is what you get, option 8, okay, any doubts, should I move to next question, okay, these three, they are from electrostatics, so electrostatic also has a good weightage, minus q, 47th question it is minus q, anyone got the answer, okay, should I solve, so conducting sphere of radius r carrying charge minus q is connected to an uncharged conducting sphere of radius 2 r, the charge that flows between them is what, see when you connect them, when you connect them just the fact that they are connected their potentials will be same, okay, and I am assuming that the wire that connects them is of infinite length, as in the two spheres are not influencing each other's potential, okay, so let's say after some time this reaches charge q1 and that reaches charge q2, then q1 plus q2 should be equal to minus q, this is charge conservation, okay, and also potential on this surface should be equal to potential on that surface, okay, so k times q1 by r should be equal to k times q2 divided by 2 r, right, so you will get q2 to be equal to 2 times of q1, alright, and you can substitute it over here, you will get q1 to be equal to minus of q by 3, right, so earlier the charge was minus q and now the charge is minus q by 3, so how much charge has flown, final charge minus initial charge, so this will be 2 q by 3 charges flown, so option D is correct, understood all of you, any doubts? Only Rithvik is there, is it? Nobody else, okay, this question, okay, to the 48th question, should I solve or should I wait? Yes, the answer is C, okay, so I don't need to solve it, right, so when N analytical capacitor are grouped in series, the equivalent capacitor becomes C by N and when they are in parallel it will be N times C, these two groups are connected in series, so equivalent capacitance will be N by C plus 1 divided by N C, this is equal to 1 by C equivalent, so C equivalent will be equal to N C square, sorry, N into C divided by N square plus 1, solve this 49th, okay, should I solve? Two air capacitors, 5 microfarad and 10 microfarad, they are charged to 10 volts each connected in parallel, the space between first capacitor is filled with material of dielectric constant 3, potential difference across the capacitor becomes what? Okay, so you can see that, alright 49th, I will tell you the final answer, answer is D, D for Delhi, let's see, so they are charged to 10 volts each, so charge on the first capacitor is C1 into V, so that is V is 10, so Q1 is 5 into 10 is minus 5 coulomb and Q2 is 10 into 10 is power minus 5 coulomb, okay, now they are connected in parallel like this, so charge will be conserved, yes or no, so let's say final charge over is Q1 dash and over there is Q2 dash, so Q1 dash plus Q2 dash should be equal to sum of these two charges, they can exchange the charge but charge is not going anywhere other than these two capacitors, okay, so Q1 dash plus Q2 dash will be sum of these two which is 15 into 10 is power minus 5 and they are connected in parallel, so Q1 dash divided by, now dielectric constant is 3, right, so the capacitance becomes K times the earlier capacitance, so Q1 dash divided by K which is 3 into the capacitance, earlier capacitance, there is a potential difference, this should be equal to Q2 dash divided by 10 into 10 is power minus 6, 10 microfarad, right, so solving these two, you will get the value of Q1 dash and Q2 dash, the potential difference across the capacitors could be any of these two, we have equated potential difference only, so potential difference across both the capacitors will be same, once you get Q1 dash or Q2 dash, you can use any of these two to get the potential difference, okay, you will get option B as the correct one, understood, I will move to the next question now, solve this one, no Ashnavi, no Sanjana, that is not correct, let us see how, now you have two capacitance in series, okay, so you need to find the breakdown voltage of the combination, so up to what voltage you should apply here, so that what is the maximum voltage that can be applied, so that none of these two capacitors will break down, now breakdown of this capacitance is 100 volt, for that it is 100 volt, okay, capacitance of this is 6 microfarad, for that it is 4 microfarad, now any of the capacitor, if it reaches 100 volt, it will break down, okay, now if they are connected in series, the charges will be same, right, so let us say Q and Q are the charges, okay, so Q divided by 6 into 10 to the power minus 6 plus Q divided by 4 into 10 to the power minus 6, this is the breakdown voltage V, okay, now which will reach the 100 volts first, this one will reach 100 volts first, right, so Q divided by 4 into 10 to the power minus 6, if this becomes 100, this will be less than 100, so this capacitor will break down first, so for this to become 100, Q should be equal to what, 4 into 10 to the power minus 4, okay, so the value of breakdown voltage, overall voltage will be 4 into 10 to the power minus 4 divided by 6 into 10 to the power minus 6 plus 100, okay, so we will get 2 by 3, so we are getting 200 by 3 plus 100, so around 166 volts we are getting, okay, the answer is wrong actually, none of the options are correct, the correct answer is this, 166 volts if you apply this capacitor will break down, if these 2 capacitors are of same capacitance, then option C is correct for the breakdown, all of you understood 50, 50 first, what is the answer, the resistance of the conducting wire is R, resistance of another similar wire of twice the length and twice the diameter, the diameter is twice, the age of procession will be 4 times, okay, so 4A divided by 2 times L, so resistance is 2L, 2 times the earlier resistance, sorry rho L by A is the resistance, resistance is rho L by A, so area becomes 4 times the earlier area okay, and length becomes 2 times the earlier length, so R is becoming R by 2, 52, just give me one second, okay, so 52nd IV graph for a metal at the temperature T1, T2, T3 are given, so which will be higher, we know that the resistance should increase with the temperature for a metal, right, okay, and here you can see that voltage is on the x axis and current is on the y axis, the slope is not the resistance, if voltage is on y axis then the slope is the resistance, okay, so T1 is higher, try option A, okay, we will go to the next question now, don't need to solve this, 54 and 55, please solve quickly, what is the answer 54 and 55, secondary circuit is passing through the jockey, primary circuit is with the scale meter, we know that the current through the secondary circuit will be 0, okay, when the Wheatstone bridge is balanced because the jockey will not have any current when it is balanced, so that's our option A's current is correct for 54, 55 what is the answer, wire of resistance 5 ohm is stretched such that the longitudinal strain is 200%, so if longitudinal strain is 200%, the length has become 3 times the earlier length, okay, and we know that the length into area is equal to the length L2 is 3 times L1 and L2 into A2 should be equal to L1 into A1, right, so A2 will come out to be L1 by L2 times A1 and this will be A1 by 3 if you increase the length, okay, so resistance is rho L by A, so rho L2 by A2 is the new resistance, L2 is 3 times the L1 and this is A1 by 3, so this is 9 times the earlier resistance, earlier resistance was 5 to 9 times of 45 ohm, okay, this is how you solve this question, we'll move to the next, solve this one 56, what is the answer 56, 56 is the last question that we'll be doing today, try to get it correct, I'll be sending you this file as well, this file has 60 more questions, you can attempt it at home, see we know that current is equal to NEA into drift velocity, right, current is 0.2 ampere, this is number of free electrons per unit volume which is 8.4 into 10 square 28, this is charge of electron 1.6 into 10 square minus 19, that is area of cross section which is 10 to the minus 6 meter square, so substituting all those values in this equation you will get the value of drift velocity, okay, the answer for this question is 1.5 into 10 to the minus 5, answer is B, okay, so I hope now you have a fair idea of how your Comet K exams will be, what type of questions will be asked, you can see that the entire paper has a good number of questions which are theoretical in nature and also good number of questions which are direct, formula based, okay, but then there are 10 to 15 questions that are up to J mains level also, so the trick is to get the easier questions and the theoretical questions right first and then look at the difficult questions, okay, now that kind of trick you will only understand when you take a lot of mock test, so try to do one thing, like I think Gaurav has already shared a mock test with you, right, so attempt that mock test today and share the solution with the Gaurav sir and he will be sending you the solution, then you can understand exactly where you stand, okay, all right, fine, so that's it for today, we'll meet again soon, bye-bye.