 Those who are online, please type in your names, waiting for everybody to join. Okay, Rithvik is there, Rohan is there. Okay, sure. Fine. So let us start solving the question. So we have solved, I think, 57 questions yesterday. All right. So let us start from question number 58. Okay. Start solving these in front of your screen. These are comment K 2007. What is the answer for 58? Are you guys able to hear me? What is the answer for this? Large selectivity. Large selectivity as in the sharpness of the resonance is not much. It's blunt. Yes. So it can select larger frequency. Okay. Now quality factor of a circuit is omega L by R, right? This represents the sharpness of the resonance. Okay. Now we don't want very sharp resonance to happen. So L should be small. Sorry. Yeah. L should be small and R should be large. Then quality is less and the sharpness is not there. Okay. All these guys quick. Okay. 59 and 50, 59 and 60 also 59 and 60. Force of attraction between two parallel current carrying conductor is F Newton per meter. Current through each of them are doubled and reversed. New force between these conductors is what? Now we know that if current are parallel, they will attract. Isn't it? They will attract or ripple. If current are in the same direction, they will attract or ripple. The magnetic field is into the plane. So L cross V. So that's why there will be attraction if current is in the same direction. Okay. Now you can see that currents are current through each of them are doubled and reversed. Okay. So if current is reversed, we can say that both the currents are reversed. Then again it will be repulsion only or it will be attraction only and the option A should be correct. Okay. But if the current through only one is reversed, then it will be force of repulsion. Option B will be correct if the current only in one of the wires is reversed. Okay. 60. Okay. Others. You know right how to convert emitter into a voltmeter? Alright. So emitter has a resistance of this much and it can read up to 5 ampere. So basically in order to have full deflection through the emitter, the current through it should be 5 ampere. Okay. So this emitter, the current should not go beyond 5 ampere. Alright. So if you have an emitter like this and suppose an unknown resistance is there. Okay. You want to make sure that this complete arrangement emitter plus unknown resistance should read up to what voltage should read up to 100 volt. Okay. Now for 100 volt, there should be full deflection. Right. So for 100 volt, full deflection should be seen. Okay. So how much should be the current over here? This current should be how much? That current should be 5 ampere only. Right. So 5 ampere is needed for emitter to have full deflection. Okay. So 5 into the resistance of emitter that is 0.1 plus unknown resistance, this should be equal to 100. Okay. So r is 20 minus 0.1 or 19.9 ohms in series. Understood all of you? Any doubt? Okay. One thing I have just found out that definition of selectivity is something else. So it is just, I think you should know what the definition of selectivity, there is nothing analytical about it. Okay. So the selectivity, the high selectivity or large selectivity, there is a correction. Okay. Whatever I said in question number 58, I am correcting myself. Okay. So large selectivity means that it should be selecting a particular frequency with high accuracy. Okay. So the peak should be sharp. It should not be blunt. Fine. The peak should be sharp. And in order for the peak to be sharp, quality factor should be higher. Okay. And Q factor will be higher only when L is large and R is small. So that is an option A is correct. Okay. So this definition of large selectivity means that the peak should be sharp. Okay. This is a small correction, question number 58. Please make a note of it. All right. Let us go to the next question. So this is comet K 2007 papers question. All right. So attempt these question number one onwards. I'm assuming you are able to read it. If it is not readable, please let me know. I'll dictate it. First one is two. All of you got first one has two. No. First one is not two. Answer for first one given in the book is three. Okay. Let me solve the first one. Ray of light traveling from glass to air. Refractive index of the glass is 1.5. The angle of incidence is 50 degrees. The deviation of the rays is what? Sorry. It will bend away from the normal, right? It is traveling from glass to air. It will bend away like this. Okay. This is the original path. All right. This is also 50 degrees. And this is the angle of deviation. This angle. This angle is angle of refraction are fine. So Snell's law says refractive index into sign of angle in that medium 1.5 sign 50 degrees. That should be equal to 1 into sign of R. Got it? So R is equal to sign inverse 1.5 sign of 50 degrees. So the angle of deviation is R minus 50 degrees. So sign inverse 1.5 sign 50 minus 50 degrees. This is the angle of deviation. Which is not in the option itself. The option is wrong. It should be 1.5 into sign 50. Then only this is right. All of you understood question number one? Second. Yes. The second one option one is correct. So in the second one, the apparent depth is D divided by root 2 plus D divided by N. This is your apparent depth. Which will come out to be first. Third one. What is the answer? It is a right angle isosceles triangle. So since it is isosceles and right angle, these two angles are 45 degrees. Okay. And the ray of light incident normally on one face of a right angle triangle. It then grazes the hypotenuse. So since it has to graze the hypotenuse. Second one is direct formula. It is D divided by its refractive index plus D divided by the second refractive index. Okay. So the third one, if the ray comes and hits the surface like this. What will happen to the ray? It will go straight. Undeviated. It will go like this. Because angle of incidence is 0. Okay. Now it has to graze the surface. So this ray will go like this. So angle of refraction is 90 degree. And this is 45 degree. This is also 45. So that is as well 45. Now let's say refractive index is mu. So I can say that mu into sine of 45 degree should be equal to 1 into sine of 90 degree. Okay. So mu is equal to root 2, which is 1.414. Okay. All right. Let us go to the next. Solve these. For 4, option 2 is rewrite 1. So the equivalent focal length formula is 1 by F is equal to 1 by F1 plus 1 by F2 minus D by F1, F2. Remember this formula. This type of question is again and again getting asked. Okay. D is the distance between the two lenses. Fifth one. You need not do the fifth one. Sixth. Let's do the sixth. Correct. It is photoelectric effect because in the photoelectric effect, light is treated like a particle. Right. And electromagnetic theory says that light is a wave. Okay. So that's why for question 6, option 2 is correct. All right. Let us go to the next few questions. These ones. You can solve one by one. First you answer seventh. We'll discuss it and then we'll move one by one like that. Two, three. Nobody is saying four. Okay. So light from two coherent sources of same amplitude illuminates the screen. Intensity of the central maximum is I naught. If the sources were incoherent, the intensity at the same point will be what? Now you need to understand that this I naught is not an individual intensity. This I naught is the maximum intensity. If I is the intensity of one of the lights, then four times I is the I naught. Because four times I is the intensity of the central maxima. Okay. You remember that formula 4I cos square phi by 2. This is the formula for the intensity at any point. Okay. Where phi is the phase difference between two light waves having intensity I. Okay. Now if they are incoherent, they will not superimpose like the way it is in Young's double set experiment. Simply, what will happen is that their intensity will just get added up, you know, algebraically like this. So you'll get on an average two I intensity. So if four I is I naught, so two I is I naught by two. That's our option three. Okay. Now solve question eight onwards. Eight four. Yes. In a Young's double set experiment sodium vapor wavelength. Okay. The half angular width of the central maximum is asked. Okay. So sine of theta is lambda by T. This is angular position. Okay. So from here, you'll get the value of theta time inverse point zero zero one. Yes. Okay. Ninth one. Solve nine. Fran Hauffer diffraction pattern is same as Young's sing. I'm sorry. Singles it experiment. So don't get confused. Ninth is four. That's correct. All of you getting four for the ninth one. This two and minus one. This is the position for the maximum. Isn't it? Right. So from here, you will get singles it when white light is used for what wavelength of light, the third secondary maximum in the diffraction coincides with the second secondary maximum. So this is two and plus one. So third secondary maximum will be two into three plus one. Lambda one D by a should be equal to second secondary maximum two into two plus one lambda two D by a. So from here, you'll get the value of lambda two if lambda one is given. Okay. So remember there, I mean, for diffraction, there will be one question on just direct formula of the diffraction. So don't ignore the single slit experiment. Okay. Because there are 60 questions, so they can ask you from literally each and every important topic in physics. They can give you a question. So 60 questions are, you know, good number of questions. Solve these two. Question number 10 is on the resolution, angle of resolution. You remember the formula for the resolution? Theta is 1.22 lambda divided by the diameter of the telescope. Okay. Theta is what? Theta is the angle two objects are making. And in this case, how will you find the theta? Telescope is your eye only in this case. How will you find theta? Suppose these are the two headlights. Okay. Separation between them is suppose X. Okay. Now here is the person's eye. This will be angle theta. Okay. Distance from the Jeep is unknown. So let's say this is X and this thing is given distance between the two headlights. This thing is given. This is L which is equal to 1.2 meters. Okay. So theta is what? How will you write theta? Theta is 1.2 divided by X. R can divide by the distance. So roughly we have approximated the radius itself as X because X will be very large. All right. So this is 1.22 into wavelength which is 5896 Armstrong. This divided by diameter of the person's eye is 2 into 10 square minus 3. Okay. So from here you will get the value of X. Okay. It will come out to be around 3.3 kilometers. Option 4 is correct. Please solve the next question. 11th is, yes, option 1. When the angle of incidence is 60 degrees on the surface of glass lab it is found that reflected rays completely polarized. The velocity of light in the glass we have to find out. We know that velocity of the light in the glass will be velocity of light in air divided by refractive index. Right. Now what is the refractive index? Right. So let's try to find the refractive index. This is the angle of incidence 60 degrees. Okay. And this goes inside the glass from air. Let's say it goes like this. And this is angle of reflection. This is also 60 degrees. Okay. And this is 90 degree. So that is 90 degrees. This automatically becomes 30 degrees. Okay. So now you can use the Snell's law. Reflective index in air multiplied by sine of angle in the air could be equal to mu into sine of 30 degrees. Okay. So mu is equal to root of 3. So if mu is root 3, the velocity will be 3 into 10 to the power 8 divided by root 3. Which is root 3 into 10 to the power 8. Okay. Fine. Let us move to the next question now. Question number 12 seems to be from chemistry. Let's leave it. 13th and 14th. 13th. What is the answer? Okay. The force is equal to 10 mg weight. That is m is for milli. So the force between the two charges is 10 into 10 to the power minus 3 into 10 is 10 to the power minus 1 Newton. This force is equal to the Coulomb force. That is KQ square divided by D square. Okay. So solve this equation. K is 9 into 10 to the power 9. Okay. D is 0.6. You will get the value of Q from here. Okay. What about 14th? Okay. Let's solve 14th one now. We know that potential is 3x square plus 5. Okay. So electric field along x axis is minus of dv by dx. And suppose it was a function of y also. Electric field along y axis will be minus of dv by dy as well. Okay. But then it is not a function of y. So E y will be 0. And similarly E z will be 0. Electric field will be only along the x direction. So this will be given as minus of 6x. Okay. So what is the value of x? X is equal to minus 2. So electric field along x axis is minus 6 into minus 2. Option 4 is correct. Why you guys are saying 3? Understood. Let's move to the next question. Any doubts? Are there any doubts? Solve these two questions. 15th is 2. No. Should I solve? The charge on each of the drop is suppose is q. Okay. So in 8 drops, 8 q charge will be there. Alright. And the radius will also be changing. Why we are 11? Some parents need to add it. Why we are 11? Just take the good messages. Official phone. No. Okay. I will solve it now. Potential of a large drop when 8 drops are combined is 20 volts. So basically k into 8 q divided by r is 20. Okay. We need to find out what is k q divided by small r. Small r is the radius of the smaller drop. Okay. Now 8 small drops are making one bigger drop. So I can equate the total volumes like this. 8 into volume of each small drop is equal to volume of the bigger drop. Okay. So capital R is 2 times the small r. Fine. So basically we need to find out k q divided by small r is capital R by 2. Okay. So this will be k 2 q divided by r which is one fourth of this. The 20 by 4 is 5. Now solve the second question. Next question. Isn't r is equal to how? See 4 by 3 pi is gone. r cube is r with cube root of 8. Cube root of 8 is 2. It's not square root of 2. Square root of 8. Okay. Cube root of 8. Anybody got 16th? 8 joules. No. That's all. Okay. So we have two identical capacitors. 5 micro farad. This is 5 micro farad. And oh they are identical capacitors. 5 micro farad. 5 micro farad. Okay. This one is charged up to 2 kilo volts. And this one is charged up to 1 kilo volt. Okay. Negative ends are connected together and then positive ends are also connected. Fine. So now we will be using charge conservation and the fact that these two capacitors are connected parallel. Okay. Now if these two capacitors are connected parallel you know then let's say charges are q 1 and q 2. Okay. So if they are connected parallel the potential will be same q 1 by the first capacitance 5 micro farad. This should be equal to q 2 divided by the second capacitance that is again 5 micro farad. So from here you will get q 1 is equal to q 2. Okay. Right. And according to charge conservation q 1 plus q 2 which is the net charge now should be equal to the initial net charge which is equal to 5 micro farad q is equal to c v. Right. So that is what I am using. This plus 5 micro farad 1 into 10 is power 3. Alright. So from here you get the value of since q 1 is equal to q 2 this entire thing you can write it as 2 q 1. Right. So you get the value of q 1. Alright. And loss of energy will be equal to initial energy which is half c into v 1 square plus half c into v 2 square this is the initial energy minus final energy. Final energy you can write it like this q square by 2 c 1 plus q 2 square divided by 2 c 2. Okay. So like this you get the answer. And one direct formula is also there for this kind of scenario if you could remember this formula you can directly use it. But you have to be very clear about the fact that this is what the formula you are using as in this is what the scenario is when you are using this formula. This formula you can use as loss in the energy. Okay. So v 1 and v 2 are the initial potentials of the capacitors. Understood. Any doubts? Yeah. Okay. Final answer 16 phase 1.25 joules option 3. So you can see that the power matrix or CET is although it is easier than J mains but definitely not as easy as your board exam. Okay. So don't take it very lightly. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Right. 17th answer is 2. So you can see that a parallel plate capacitor with air as dielectric has capacity a slab same thickness as separation is introduced to fill one-fourth of the capacitor. Okay. So it is filled to one-fourth of the capacitor and it has dielectric constant. We need to find equivalent capacitance. Okay. these two capacitors are connected in parallel. We can break the entire capacitance into two. So, let us say this is C1 and this is C2. So, C1 is what? C1 is K times area of cross section of the upper plate will be now one-fourth. So, K a by four separation between the plate is still same whatever was earlier separation let us say D. So, C1 is basically earlier capacitance by Kc this is epsilon naught sorry Kc by four. All of you understood? Now C2 will be equal to epsilon naught 3a by four which is the remaining area of the plate divided by D. So, C2 can be seen as if it is three-fourth of the original capacitance total capacitance will be C1 plus C2. So, that is K plus 3c by four that is an option two is correct fine. So, next one okay 18th how you are getting one all of you getting one how come answer in the book is saying four 10 is power minus three you get right or 10 is so minus two you will get. Just check once again you are all of you getting current is five this is equal to Neavd n is five into 10 is power 26 charge of electron 1.6 10 is so minus 19 and e is four into 10 is power minus six into drift velocity. So, this is 19 and minus six from the five this is 10. So, you will get five is equal to five into 1.6 into four into 10 into Vd five is gone. So, drift velocity is 0.1 divided by four into 1.6 which is 0.1 divided by 4.6. Yeah, it is option one. I don't know why in the book it is saying option four. So, option one is correct okay let us move to next one these two none of you got the answer okay I will solve it two bulb rated 25 watt 220 volt and 100 watt 220 volt you are connected to 440 volt supply then what will happen see the thing is that from the rating you can get the resistance easily all right. So, how will you get the resistance the rating 25 watt is equal to V square by R1 this is resistance R1 and you have 100 is equal to 220 square divided by R2. So, from here you will get the value of R1 and R2 all right. So, R1 will come out to be equal to 1936 ohms and R2 will come out to be 484 ohms all right permissible current through the first bulb you can find out from the power rating as in 25 volt 25 watt will be equal to 220 into I okay. So, from here you get the permissible current I1 to be equal to 0.1136 and you get permissible current through the second bulb by quitting 100 is equal to 220 into I2. So, this will come out to be 0.4545 okay if you connect these two resistance as in these two in series the total resistance will be some of these two which will be 2420 ohms okay. So, the current will be equal to total voltage that is 440 now divided by the total net resistance 2420 okay this will be equal to 0.1818 amperes okay. So, which bulb will fuse this one will fuse right because only this much is permissible. Permissible current you will get from the rating of the bulb itself. So, you can get two things from the rating of the bulb permissible current and the resistance of the bulb 20th these are from current electricity chapter and you know there are just few types of questions that exist from this particular chapter. So, you should make sure that you get full marks in this chapter current passing through the ideal emitter in this circuit is what? The emitter has no resistance okay. So, as if 4 ohm is shunted okay. So, 4 ohm this 4 ohm will not make any difference as if parallel there is 0 ohm connected which is emitter. So, current will go like this will pass through emitter and none of the current will go to the 4 ohm and it will travel like that okay. These two ohms they are connected parallel. So, this is equivalent to 1 and this one is 2. So, 2 plus 1 is 3 and 1 ohm is again the this thing 1 ohm is internal resistance. Total resistance is 2 plus 1 plus 1 which is 4 okay. So, the current is 4 divided by 4 1 ampere how you got 0.5 you have counted 4 ohm as well. This one will not be counted right there is no current through this this is emitter this is not voltmeter emitter is 0 resistance pass okay. 4 ohm is shunted all this 21 is 2 yes also for 21 is option 2. The first of all whenever you see such kind of structure which is which shown bridge you need to check whether it is balanced. So, p is 10 r is 15 q is 20 and s is 30 okay. Is it balanced? You can see 10 by 20 10 by 20 is equal to 15 by 30. So, it is balanced okay. Since it is balanced this entire thing can be treated as if it is like this. Gallonometer will not have any effect because there is no current in the gallonometer. So, this is 6 volt this is 10 this is 15 20 and 30. So, this is 25 connected parallel to 50. So, total resistance is 25 into 50 divided by 50 plus 25 which is 75 50 by 3 okay. So, current is V divided by r that is 18 by 50 okay. That is why option 2 is correct. All right let us go to the next question now. So, all this one 20 second yes 20 second option 1 is correct. Magnetic field is mu naught i by 2 r right. Magnetic field is mu naught i divided by 2 r and for the number of turns n I can write it as mu naught n i divided by 2 r. It may not be safe I don't know who has kept it. The number of turns is 3 to have 3 turns. What will happen to the radius? It will have a new radius right it will have a new radius let us say r n but the circumference will be same. So, 3 into 2 pi new radius will be equal to 2 pi the old radius. Okay. So, new radius will be the old radius divided by 3. So, here we should write r by 3. So, magnetic field you can see is becoming 9 times the earlier magnetic field that is our option 1 okay. We will move to the next question now. All this 23rd tangent galvanometer is the same device which you have learned as galvanometer. It's full name is tangent galvanometer just like Shreyas full name is Shreyas Bhaktaram. Okay. So, radius is m v by q b fine. So, momentum can be written as in terms of kinetic energy as under root of 2 m into k why I am writing in terms of kinetic energy because kinetic energy is same but the masses will be different. So, momentum of the proton is equal to under root of 2 times mass of proton into kinetic energy and momentum of the deuteron is under root 2 into mass of deuteron into kinetic energy. We know that mass of deuteron is 2 times the mass of proton. Yes or no? Proton has one proton and deuteron has two protons and electrons mass is ignored. Okay. So, now you can compare and find out. See, I will send a write up on questions like 24 how you deal with tangent galvanometer these kind of questions. Okay. So, let's not spend a lot of time in it. It will take 15-20 minutes just to talk about tangent galvanometer. Rather, I will send you a write up on this. Let's go to the next question now. This one 25th is a charge particle is moving in a magnetic field of strength b perpendicular to direction of the field. If q and m denote the charge and mass of the particle then the frequency of rotation of the particle is what? Yeah, that's option 2. Correct. Time period is 2 by m by q b. So, frequency is inverse of that. Correct. This one. Anyone got the answer? Should I wait or solve? Okay. 26th is option 3. It's correct. There will be an attraction between these two and there will be attraction between these two as well. So, total force will be a difference between these two forces. Okay. So, f 1 minus f 2 you have to find out that is equal to mu naught i 1 i 2 divided by in the denominator it was 2 pi d 1 minus mu naught into i 3 i 2 divided by 2 pi d 2. This is i 2 this is i 1 and this is i 3 this is d 2 and this is d 1. Okay. So, when you simplify this you will get force per unit length. Okay. This multiplied by the length of the conductor which is 1 you will get the total force. Remember this is mass see this is force per unit length that you have to multiply with length in this case length was 1. So, the force per length is equal to the force itself but length could be something else also that you must multiply. All right. Let's go to the next question now. These two. 27th none of you it's an alternating current. Okay. Now, I will solve this electric bulb has rated power of 50 watt at 100 volt. If it is used on an AC source of 200 volt 50 hertz a choke has to be used in series with it. This choke should have an inductance of hot. Okay. No, 27 is not 4. So, the situation is like this. You have this is your bulb and the choke is nothing but an inductor. Isn't it? The bulb has a resistance and inductance has inductive sorry yeah in inductive reactance. Okay. So, now electric bulb has this rating 100 volt 50 watt. So, the potential across the resistance should be 100 volts. Okay. This is let us say potential across inductor. Okay. And this is the supply voltage. Let's say V. Okay. So, what is the relationship between V 100 and VL? Anyone? How you relate V 100 volts and VL? V is by the way 200. What is the relationship in 200, 100 and VL? This voltage is 50 degree out of phase of this one. Okay. So, that's why you can't just simply add 200 will be equal to 100 square plus VL square. Okay. So, from here you will get VL to be equal to 100 root 3 volts. Okay. Now, this 100 root 3 should be equal to current into XL. Okay. And XL is what? XL is omega times L. This should be equal to 100 root 3. So, L is equal to 100 root 3 divided by omega which is 2 pi into f that into current. Current you can find out from the rating of the bulk. Current is 50 watt divided by 100 volts because power should be equal to V to I. Okay. So, this is how much? 0.5. So, current is 0.5. F is 50 hertz. Alright. From here you will get the value of inductance L. It's a good question. Understood? Should I move to the next one? This one? 28th. 28th one. Yes, that is correct. So, inductance and the capacitance and the resistance are connected in series with this. The phase angle of this circuit is what? Tan of phi is XL minus XC pi R. Okay. So, from here you can find out. XL is omega into L and XC is 1 by omega C. And omega is 2 pi into frequency. You can see that you get tan phi is equal to 1. So, phi is equal to 45 degrees. You can also write down cos of phi is equal to R by impedance where impedance is XC minus XL whole square plus R square. Then you will get cos of phi is equal to R2 and then phi is equal to 45 degrees. Okay. Alright. So, all these questions. 28th one. Sundarai has already answered. That's, no, sorry, 28th was previous question. 29th is, yes, 29th is 4. I will solve it quickly. Those who got the answer please attempt the next question. Step down transformer reduces the voltage of the transmission from primary voltages 2200 and secondary voltage is 220. The power delivered by it, power delivered as in power in the secondary circuit is 880 volt. The efficiency is 88%. So, power delivered is this. The power in the primary circuit is 1 divided by 0.88 times 880, which is 1000. Okay. So, this 1000 should be equal to voltage in primary circuit into primary current. Okay. So, from here you get 0.465 as the answer. Okay. 30th. Okay. 30th also you have solved. 30th, the answer is 4. So, let's see how we can do 30th. Current in a coil changes from 4 ampere to 0 in 0.1 second. EMF induces this. Self inductance of coil is what? So, EMF is equal to L di by dt. We know this. Right. So, approximately we can write it as L into delta i by delta t. Okay. So, L is what? L is EMF into delta t divided by delta i. Okay. So, EMF is 100. So, this is 100. So, delta t is 0.1 divided by delta i. Chaining current is 4. So, 25 into 0.1, 2.5. That's our option 4. Oh, 31 also you have answered. Very fast. Okay. First is 1. Okay. All components of electromagnetic spectrum in vacuum have the same velocity. Correct. They all are EM wave. They move with the same velocity that is speed of light. Okay. So, now let us take a small break continuously for 2 hours. We have been solving problems. Okay. So, let us take a break for 15 minutes. We will meet after 15 minutes. Okay. Those who don't want to take a break can attempt this bonus question. So, right now it is 12.30. We will meet at 12.45. Okay. And the class will run till 1.30. Okay. So, just take a break, eat something and come back. All right. All right guys. So, we can start now. Oh, looks like 30 second question. You have already attempted question number 32. It says option 4. Are you guys able to hear me? Am I audible? Okay. Let's quickly solve this one. Which of the following graphs represent the variation of maximum kind energy of emitted electron with the frequency in the photo electric effect correctly? Do we have h mu that is energy of photon minus the work function this to be equal to maximum kind energy? Right. So, the maximum kind energy, so k is equal to h mu minus 5. Okay. So, when mu is 0, k, I mean, you can treat it like a mathematical equation, y is equal to mx plus c, where when x is 0, y is minus c. So, basically k should be minus 5 when mu is 0. Okay. Although it physically doesn't make any sense, but when you look at like just an equation, it is like that and the slope should be positive. Intercept on the yx should be negative and slope should be positive. So, only this one makes sense. Okay. Which is option 2. Option 2 is the fourth graph. Okay. So, straight forward. Go to the next. Solve it. Okay. How discuss another? Okay. 33rd is, say I have got one answer. Others, see these kind of question, you must not get wrong. Okay. Energy of the first frequency, you need to find in terms of electron volt. So, if energy of the frequency is more than the threshold energy, then photoelectrons will be emitted, otherwise not. Okay. So, in terms of electron volt, it will 6.6 into 10 to the power minus 34 hc, sorry, mu is directly given, lambda is not there, h into mu, 1.8 into 10 to the power of 14 divided by, see you can see that if this frequency emits photoelectron, this will definitely emit because 2.2 is more than 1.8. So, I am just checking for 1.8. Okay. So, if I divide this with charge of electron, I will get the energy in terms of electron volts. Roughly, this is around 4. So, this is 7.2 and this one is 10 to the power minus 1. So, this is 0.72 electron volt. Okay. Okay. So, 0.72 electron volt is there. So, A cannot emit the photoelectrons. Okay. Now, the problem is that there is this option, neither A nor B. So, I need to check for 2.2 also. Okay. So, had this been 2.2, then you would have 4 into 2.2 you would have got into 10 to the power minus 1. So, this would have been around 0.88 you might have got. So, B will, this frequency will create photoelectron. So, B alone will create the photoelectron. Understood all of you? Wait, wait, wait. This is not the case. This is the threshold frequencies that are given. So, 0.72 electron volt is the work function of metal A and 0.88 electron volt is the work function of metal B. Okay. These are the threshold frequency as in this much frequency is required by 2 metals to create photoelectron. So, this is the phi A and this is phi B and you have this much energy of photoelectrons impacting the metal surface. So, this is more than 0.72. So, 0.825 is more than the surface A's threshold this thing. So, that is why only A will create the photoelectron. Sorry about that. Option 3 is the right answer. Okay. What about the next one? R is the Rigbert constant. One of you 34 is option 2. Ionization energy means that the electron has jumped from n equal to 1 shell to n equal to infinity. That is what the ionization is. So, 1 by lambda is R into z square 1 by n1 square minus 1 by n2 square. This is the formula. So, 1 by lambda is Rigbert constant into 3 square 1 by 1 square minus 1 by infinity. So, 1 by lambda will come out to be 9R. So, lambda is 1 by lambda is this. So, hc by lambda is the energy required. So, this is 9HCR. This much energy is required. Option 2. 35th. The Rigbert answer is 35th. That does not seem to be correct. Others? Okay. I will solve this now. Electron at n equal to n level can emit 3 spectral lines. So, if the electron is at n equal to n, then it should emit nc2. So, n into n minus 1 by 2, this should be equal to 3. Okay. So, n equal to 3. Is it? Yeah. So, n equal to 3. So, electron is at n equal to 3 now that we know when they are at another energy level, they can emit 6 spectral lines. So, n into n minus 1 by 2, when you curate it to 6, you will get n equal to 4. This is n1 and this is n2. The orbital speed of electron in 2 orbits are in what ratio? So, we know that orbital speed is proportional to 1 by n. Orbital speed decreases as you move away. So, v1 by v2 is equal to n2 by n1. Option 2 is correct. Is it clear to all of you any doubts? So, you can see that this question is J mains level. So, it is not that Cp level, all questions are straightforward. You can just run them over. So, these ones, a lot of calculation is there, right? Question number 36. You can also see that there is, you know, difference, I mean, there is a good amount of difference between the two options. So, you can roughly also, estimate and get the correct answer. You do not need to be very accurate with the answer. Roughly also, you are getting it, then you will be able to arrive at the correct option. Nobody got it? These type of questions which are very calculation intensive, you can probably mark it and do it towards the end of the paper. But then make sure you are attempting these questions because these type of questions, you know exactly how to go about. Rather than breaking your head on questions which you are not very sure about, so towards the end, you can attempt these ones. Yes, option one is correct. I think you have not used calculator, right? So, lambda is h by p where p is under root 2m into kinetic energy and kinetic energy is q into v. Fine, you just substitute it, you will get option one to be correct. q is the charge of electron and m is mass, sorry, yeah, and m is mass of proton. 37th, what is the answer? Isotope is what? Isotope has same atomic mass or atomic number. It should have same atomic number, right? Now, if these disintegration happens, they will change the atomic number and atomic mass, all right? So, you can evaluate. For example, option one, option one says alpha particle is also emitted and beta particle also emitted. So, what alpha particle will do to the atomic number? It will decrease by 2, okay? And beta particles can increase the atomic number or decrease the atomic number. So, if it is beta minus disintegration, then you can see that it can increase the atomic number. So, 2 beta particles will increase by 2 the atomic number and 1 alpha particle will decrease by 2. So, net-net atomic number remains the same. So, option one is the correct option, okay? No other option balances the atomic number. 38th. Now, you can see that in 38, the options are very close to each other, okay? And hence, you should be very careful with the calculation. You can see that 10,000 nuclei are there and its half-life is 20 days. Number of nuclei at the end of the 10 days is what? So, how many half-life, okay? No. So, here we will be using this formula ln n by n naught is equal to minus of lambda into t, okay? And lambda is equal to 0.693 divided by t half, okay? So, ln of n by n naught is equal to minus of 0.693 divided by t half, that is 20 days and t is 10 days, okay? Now, this is ln n by n naught and if you remember, 0.693 is ln of 2, ln of 2 by 2. So, you can see that this will become ln n by n naught is equal to ln of 2 raised to the power minus half or n naught is equal to root 2 times n, right? Yes. You can equate this with that. So, n is equal to n naught which is 10,000 divided by root 2 which is 1.414. You get option 2 to be correct. It is roughly 7,000. You can multiply the denominator by 7 to check that. It will be 9.8 something. Clear? All of you clear about these questions? Any doubts? Let us go to the next. Question number 40, all options are correct. This is theoretical stuff. So, just remember it, okay? If you know the full form of laser, then also you can answer it. 41, 41, what is the answer? The radius is proportional to a ratio power 1 by 3. Volume is a, okay? So, while, you know, I think we have browsed more than 45 to 46 questions, you might have realized that more than 50% of questions are straightforward, okay? And there are few questions that will just eat up your time. So, the trick is just ignore the questions which are eating up your time and get those questions right which can be solved quickly the first, okay? In the next pass, you can solve the questions which will take more time. And I think you have done enough practice to identify quickly which question will take less time and which question will take more time. Just by looking at it, you can find out, okay? For example, you know, these theoretical questions you can see 42, 43, they will not take a lot of time. Okay, 42, what is the answer? Either you know it or don't know it. So, you can mark, if you don't know it, you can mark anything and move on. It's lepton. How can electron be a nucleon? Inside the nucleus, protons and neutrons are there, there are no electrons. I think 3 is correct. Just another name for the electron. 43rd, 3 electrons. 44, there are 2 terms in transistor, input resistance and output resistance, okay? So, it is input voltage delta V by delta I, this will give you the resistance, right? So, delta V is the 1 and change in current is 0.5 into 10 is power minus 6, okay? So, you will get the option number 1 to be correct. Okay. So, all this question, what do you think this table is for which 45th is 3, correct? You can see that just 3, 4 questions, how fast you have solved, right? So, identify these questions quickly. You don't need to go sequentially the way you do the board exam. So, this is the nand grade, option 3. Finally, class 11 syllabus questions are coming in. Mechanics in comet care or CT will be very simple. So, make a point to attempt mechanics a little earlier. So, impulses what? Force into time or impulses simply of dimension of momentum because that is going to change in momentum and momentum dimension is mass into velocity that is mLt minus 1. So, option 2 is correct. What about 47? No one got 47? Okay. Maximum height attained by a projectile when it is thrown at an angle theta with the horizontal is found to be half the horizontal range. Maximum height is what? u square sine square theta by 2g. This is equal to half of the range which is u square sine 2 theta by g. So, when you equate it you will get tan theta equal to 2 or theta is tan inverse 2. So, that is formula application 48. Okay. So, what is the answer to 48? Should I solve or should I wait? Okay. Okay. A shell of mass 20 kg at rest explodes into two fragments of masses are in ratio 2 to 3. A smaller fragment has mass has velocity. This kind of larger fragment is what? It is a direct application of conservation of momentum. Initial momentum is 0. This is equal to 2 to 3. So, 2 by 5 into 20 and 3 by 5 into 20. These are the masses. So, 8 and 12. So, 0 is equal to 8 into 6 plus 12 into v. So, v will come out to be minus 4 meter per second. So, kinetic energy of the larger fragment is half mass is 12 into v square. It is 16. So, this is 96 joules. Fine. Let us go to the next. Solve all these three versions quickly. Then we will go to the next one. Okay. Okay. 49th is 2. That is correct. Capillary effect is because of the surface tension. All right. Next one. 50th. So, accession due to gravity will become g by 2 at what height? Variation of gravity with height. Yes or no? Can I use the direct formula? Can I use the direct formula that was there? You remember the direct formula? G0 1 minus 2h by r. Can I use this formula? I cannot. This formula is an approximation. Okay. Variation of gravity with depth is not an approximation, but with height it is an approximation. I cannot use it because the height is much larger than the radius of the earth. Okay. So, how will you do this now? You have to find out the gravitational force g m into mass of earth divided by radius, not radius, distance. We need to find the height, right? So, radius of earth plus height whole square. Okay. Now, the accession due to gravity will be this force divided by mass. So, that will be g m e divided by r e square 1 plus h by r e the whole square in the denominator. Okay. This thing is accession due to gravity. Fine. So, g will become g0 by 2. This is g0 divided by 1 plus h by r e whole square. G0 and g0 gone. We have 1 plus h by r e the whole square is equal to 2. So, I am getting 1 plus h by r e as root 2. So, h is equal to root 2 minus 1 times r e. So, root 2 minus 1 is 0.4 times r e. Okay. Actually, in this book, they have used the direct formula which is actually not correct. Okay. We have solved this without any approximation. So, suppose you have solved it like this and then you are getting 0.4 r e which is not in the option, then you can check with the direct formula because somehow they want you to use this formula even though this is an approximation. If you use this formula, you will get h is equal to r by 4. All right. But of course, r by 4 is comparable to the radius of the earth. So, this formula itself is not valid but then they expect you to use this formula which is incorrect. Anyways, let us take one more question before we end the session. So, I will pick that. Okay. Thermodynamics is there. Solve this one. 53. 53, you are getting 2. 53, yes. Correct. Let us say this temperature is T. Of course, T will be between 100 degree Celsius and 25 degree Celsius. Okay. So, heat will flow from higher temperature to lower temperature like this. So, if I treat it as two different rods, dq by dt over here will be equal to dq by dt over there. Okay. So, dq by dt which is the rate of flow of heat between A and C can be written as thermal conductivity of A to C area of cross section into delta T that is 100 minus T divided by the distance A to C which is midpoint, is it? C is a midpoint. Two identical rods. So, identical rods. So, length are same. So, L. This should be equal to thermal conductivity of C to B area of cross section, then T minus 25 degree Celsius by L. L and L gone, A and A gone. Then thermal conductivity is R in ratio of 2 is to 3. So, if this is 2, then this is 3. So, I can say that 200 minus 2T is equal to 3T minus 75. Okay. So, 5T is equal to 275. Okay. So, T is 55 degree Celsius. Option 2 is correct. Okay. There are I think six or seven more questions that are remaining. I will share this file with you. You can attempt the last six questions that are remaining. Okay. And I think now you have good understanding of what kind of questions you can expect in comet payor's utility. All right. Also, please take the mock test which are shared by Gaurav every day. All right. We, I mean, taking mock test is very, very important. Don't just ignore it. Okay. Thinking that you'll be taking it later, later and later. Okay. Your syllabus will never get over. Don't think that first you should read everything, all the formulas and then you take the mock test. Just take the mock test. Nobody need to know how much marks you scored and all because test is more than knowledge. They will test you on how statistically you attempt a question paper, what you will do if there is a time pressure and all that. So, it's very, very important. Okay. So, we'll end the session now and we'll meet again sometime in the next week. Bye-bye.