0:00 hello everyone welcome back to dsha Karnataka YouTube channel and today I am going to discuss a very important uh thing that is your model question paper number one so P2 model question paper there are three different model question papers published by your pu board this year and as you are aware that all the three uh uh model paper have slightly different uh blueprint for each and everyone I have discussed all the blueprints in a separate video you can check that out and also the number of question is reduced uh previously I think it was uh 48 or 50 questions now the total number of question is 45 so they have reduced few options that you
0:45 might get in the uh answering section right so uh as they are trying to plug the Gap that was there in the previous year because many student complained that many questions are not uh part of the uh blueprint some questions are maybe slightly difficult out of the syllabus even though none of the questions was out of the syllabus yes there was some uh discrepancy some fluctuations from the uh given blueprint so that's why they have decided okay no more drama I have I am giving you three blueprints okay and any one of the blueprint might come and uh that is why they have reduced all the uh bloop holes them that they might have so uh what we will do we will try to solve the first blue model paper today and we will solve
1:31 the other model papers in our upcoming uh days so all the three blueprints all the three model papers will be solved in our channel so don't go anywhere and if you are first time in this channel don't forget to subscribe to our channel right so let's start with the McQ portions first most students uh struggle in the McQ sections so as usual there will be uh 15 McQ questions let's see one by one the SI unit of electric charge is uh Kum me Kum per meter Kum per Kum okay what is the question SI unit of charge density SI unit of surface charge density oh no it is only electric charge it's not charge density what is the unit of electric charge obviously
2:17 Kum as simple as that so simple people say that uh mcqs are tumba hard tumba hard it is not hard it's very easy right next is the angle between the potential surface and the electric potential so if if this is the uh sorry electric equip potential surface and electric field so this is the equip potential surface electric field will be always 90° so correct answer is 90° now next question statement one the resistivity of metals increases with increase in temperature resistivity means row so row actually increases with increas in temperature yes this is the correct statement increasing the
3:03 temperature of the metal causes more frequent Collision of the electrons so electrons which is present in the metal they help to carry the electric field so that is why current flows from one point to another right so they are reacting to the applied electric field that is why current flows now as as you increase the temperature the electrons are gaining energy and gaining the energy that means they are now colliding with each other more Collision means less electrons are passing so less electron passing means resistivity increases yes statement two is also correct and it is the correct explanation for statement one so this is where student make mistakes uh because they are not able to relate whether the both statements are correct whether the
3:49 statement two is the correct explanation of the two or one or not so this is where student makes uh mistakes so that is why you have to practice such questions statement based questions that is uh very very uh important because at least one uh statement based question will come in your P final okay both one and two are true and two is the correct explanation as one so option A is correct answer next is a moving coil galvanometer can be converted into voltmeter by connecting so you are converting a galvanometer into voltmeter we know that we have to connect a high resistance in series to converted into into a voltmeter this can be converted into a voltmeter by adding a high
4:34 resistance in series so a high resistance in series with the galvanometer is required to convert it into a voltmeter so correct answer is option D you know the answer you have to check or read the options and find the correct option as soon as you can find it next question number five when a bar magnet is suspended freely it points in the direction of east west north south northeast southeast northwest Southwest is it a question this is not a question this is just a very very easy question north south when you suspend a magnet it always points towards the North and South direction right as easy as this the energy stored in an inductor of inductance l in establishing a current i
5:23 in it so now here they have not mentioned AC so be careful if this is in AC the answer would be zero but here it is not AC it is just establishing the current I so answer is half l i s if the same question was asked inductor what is the power uh what is the energy stored in a inductor what is the power dissipated in an inductor the answer would be zero if it is connected in an AC circuit okay be careful here it is energy stored in an inductor it is half Li squ now the direction of current induced in the loop hoop ABC as shown in the figure this is was a very old Cas set question so they have recycled this question for Pub board now so current is
6:08 flowing in this direction okay and you have to find the uh uh mag um induced current due to this coil so first of all when the current is flowing in this uh wire what is the magnetic field produced by this current so the magnetic field produced by this current is into the page of the paper so you know the right hand rule so you have to put your fingers in the direction of paper a direction of current and then you have to point your um fingers towards or someone say that if this is the current then whatever Direction your your palm points at that is the direction of magnetic field so that is also Palm rule so Palm rule means this
6:55 is the current so magnetic field will be into the paper so into the paper means this is the direction of magnetic field this is the magnetic field due to this current right this is the magnetic field due to this current now uh this magnetic field will try to induce a current on this Loop right there is a circular Loop so this there will be an indu induced current now what will be the induced current the induced current will be such that it opposes the magnetic field so the opposes the magnetic field means um it will be the the magnetic field will be in the opposite direction the original magnetic field into the paper original magnetic field is into the paper so the induced magnetic field will be out of the paper so out of the paper
7:41 means it will be flowing the current will be in this direction right so a CB or b a c whatever you consider so the induced current will be in this direction okay so that you get a magnetic field out of the paper okay this is a very uh high level question not super simple as you can see so along ABC if it is increasing along ABC if it is decreasing okay both cases are given so increasing decreasing both are given so it will be along this direction if uh current is increasing okay so current if current is increasing then it will be along this direction okay current increasing means it is increasing so it will try to oppose the current now if it is decreasing so if it is decreasing
8:27 then it will be a different answer if it is decreasing then the the current will be in this direction so blue line signifies the current decreasing so current decreasing means it will be again into the same direction of the magnetic field because decreasing magnetic field gives rise to the uh induced magnetic field will try to increase the field so it will be in the direction of a b c if the current in uh the wire is decreasing if the current I is decreasing then it will be along ABC this is the currect answer okay it will be along AC CB if the I is increasing this is also correct so many student gets confused with this answer let's see once more if this is the original current and due to this there is an
9:13 existing magnetic field now you have placed this coil and now there will be some induced current passing through this coil based on whether the current is increasing or current is decreasing if the current is constant no change will happen so first of all we have found out that if current is increasing so if current is increasing then the induced current induced current will try to oppose this already existing magnetic field keep this in your mind there are a lot of explanations available in Internet I have seen that and many of them are actually uh try to complicate the things first of all you have to find the original magnetic field due to this current so current is passing through this so the original magnetic field in this part of the will be into the page
10:01 of the paper right clear about that so you have to put your fingers like that and put your uh uh Thumb in the direction of current and your other fingers will Point towards the direction of magnetic field I hope you are familiar with this so from this we have found out the original magnetic field is into the paper now if current is increasing then the induced current will oppose this magnetic field so current will be flowing in the direction so that the current opposes this magnetic field so basically induced current will be like this in the red red arrow right the red arrow signifies the uh the red arrow signifies the motion of the uh current right so this is the outward Direction so it will be in the ACB direction if
10:48 the current is increasing okay so it will be along the ACB direction if the current is increasing so option C is correct okay and if the current is decreasing then if the current is uh decreasing that means magnetic field is decreasing then the induced current will be in the same direction of the original field so in that case it will be in the ABC Direction so in this question in the blueprint uh both options are correct okay B and C are correct but there is a mistake from their side because in the they have copied somewhat copied this question from uh CET exam as I said in the old caset exam this was a question now that diagram was slightly different the options was slightly different while typing it from the CET paper whoever
11:35 typed this question they type it incorrect that is why option b and option C both are correct there are many uh uh solution key answer key available in Internet and they are giving you completely incorrect answer okay there are few correct answers as well but many of them are giving you incorrect answer even I have seen one uh paper key they are saying along ABC if I is increasing which is completely wrong okay so be careful of such a uh answer keys if you are following our Channel we will always give you the correct uh answer okay so this is both of the options are correct for this particular question but be sure that in the final exam there will be no such questions which will have both correct answer obviously this is done by only a few uh group of teachers and they
12:21 give it to some typewriter said they type it right they have done some mistakes and it was not properly reviewed but when the final paper comes there are there are a lot of teachers there is at least a very uh experienced well educated well experienced teachers are sitting there in the panel they will review your answer uh thoroughly right so don't worry about it now next question an ideal Step up Transformer decreases uh ideal Step up Transformer Step up Transformer means it is uh original purpose of a transformer is to change the voltage right either it will increase the voltage or it will decrease the voltage so if it is decreasing the voltage we say Step Down Transformer if it is increasing the voltage we say Step up Transformer now in this case tape of
13:08 Transformer means it is increasing the voltage now if the voltage increases the other thing which decreases is only the current okay now for an ideal Transformer power remains constant frequency remains constant voltage will increase so current will definitely decrease so the answer for this question is current now question question number nine the displacement current is due to flow of electrons flow of protons changing electric field changing magnetic field so the displacement current is basically due to uh we know that basic the relationship between uh displacement current and uh electric field is that when the electric field is changing as the
13:53 electric field changing the current will Al also change so basically it is due to the changing electric field fi changing magnetic field is the induced current not the displacement current displacement current is between the capacitor plates if you have read the electromagnetic wave chapter there we have seen that when the magnetic electric field changes there is a current induced that is called displacement current so it is that due to the changing electric field now an object of finite height is placed in front of a concave mirror within its focus it forms a real enlarged image so first of all let us understand the question we have a concave metor okay this is the focal length and we have placed an object in front of within the
14:38 focus within the focus means between the pole and the focus okay so definitely it will produce a virtual it will produce a virtual and enlarged image right we know that a concave meter produces virtual image when the distance is less than focal length right convex meter on the other hand always produce virtual images and that is diminished but in case of of convex mirror when the FOC distance is less than focal length it produces a enlarged image virtual and erect so it will be a virtual enlarged image so virtual enlarged and erect image if they had one more option they could have added it but it is not there so virtual and enlarged image is the correct answer for this question next question a beam of unpolarized light of intensity I KN
15:25 it is passed through a pair of Polaroids with their axis inclined at an angle of theta the intensity of the emergent light is equal to very interesting question so first we have a Polaroid right this there is a Polaroid one there is a Polaroid two okay now they are not parallel there is a angle between them so let me correct this one uh just give me a second let me correct this one so they are not uh parallel they have some angle between them so this angle say Theta okay so there is an unpolarized light unpolarized light generally represented
16:12 like this okay this is the representation of nonpolarized light now once it passes through this Polaroid it becomes polarized okay now it becomes polarized now what is the intensity of this this intensity was I not okay so after it passes through the first Polaroid the intensity becomes half so it will be I by 2 okay now when it passes through the second Polaroid the intensity of this light becomes I by 2 cos² Theta this is malo's law okay this is malo's law so you have to remember this it will be I by 2 cos² Theta so answer is D very very careful many students blindly will answer I not cos Square Theta that is
16:58 not correct this is a trick that when unpolarized light passes through a Polaroid first then the intensity drops to half then after that when it passes through any other Polaroid then Malo law applies so if you read the Malo law it says that any polariz polarized light of intensity I not so if the polar light light of Po light is already polarized with an intensity I then it is I cos Square Theta but here the intensity of the Polar ized light is I not by 2 so you have to be careful to answer this type of questions this is a very trappy question so uh even though the answer is easy but it is a very trappy question emission of electrons form a metal
17:43 surface by heating is called so emission of electron by heating a surface is called thermoionic emission okay thermoionic emission that is the name for this I think the spelling is ther IC emission I think spelling is slightly incorrect thermoionic emission is when you are heating a metal and it is giving you electron okay if you are applying uh field electric field and it is emitting then it is called uh field emission and if it is uh due to light then it is called photo emission okay photo electric emission or photo emission okay secondary emission I don't know there is any term called secondary emission this is just a uh bogus option now question number 13 when alpha particles are passed through a thin gold foil most of
18:31 them go undeviated so when alpha particles are passing through a thin gold foil most of the particles are going undeviated because most of the region in an atom is empty space yes so most of the region in the atom is empty space so we know that atom is basically this is a gold foil atom suppose this is an atom so most space in the atom are empty space we have some electrons orbiting but the entire Mass is located here and most of the space are empty space so when an alpha particle is going it is going like this and it is uh so only few of them are getting deviated which are actually going towards it so they are going getting deviated some are deviated little but most of the particle which is uh passing through this empty
19:18 space so it is not getting deviated so that means if you are passing uh you are throwing I gave you example you are throwing balls towards the wall if the wall is hollow most most of the part is wall is hollow then the most of the balls will pass but if the wall is solid 90% of the ball 99% of the balls will come back right so basically because the most part is hollow that's why you are getting they are getting passed through okay now nuclei with same atomic number is called Isotopes nuclei with same atomic number same atomic number means they are the same nucleus uh same element sorry they are the same element that means they are isotopes okay okay now let's see a matching question the column one of the list of the
20:04 material and the column two is the list of the energy band gaps EG identify the correct match so we know that uh any band gap which is zero or uh uh close to zero they are conductor so conductor is basically when EG is equal to zero right so conductor is basically Gap Band Gap is zero and insulators is higher band Gap do their electron cannot jump from conduction band to bance band so definitely it will be higher so it will be definitely higher than 3 electron volt whereas semiconductors they are basically in between conductor and insulator so they have they are less than 3 electron volt which can come down to zero in certain suitable condition okay and they can behave like a conductor so answer is 1 b 2 a 3
20:54 C uh no sorry 1 uh B b 2 c 3 a okay so 1 B let's see 1 b 1 B both of them as one B so either one of them next is uh two 2 c 2 C is this one okay so this one is the correct answer 1 b 2 c 3 a right so you can either you have to match any two of them to get the correct match right uh all three variations are not generally possible so if you get two matches correctly most it will be correct answer okay that is one of the trick that I am teaching you now let's see the fill in the blanks Photon polar need zero infinite phase phaser a molecule possessing a permanent dipole
21:41 moment is called permanent dipole moment is called polar right it's called a polar molecule the net magnetic flu through any closed surface the net magnetic flu through any closed surface it is not Infinity it is zero okay uh that is gs's law of magnetism net magnetic flux through any closed surface is zero a rotating Vector used to represent the alternating quantities is called the rotating Vector representing alternating quantities is called phaser phaser very careful phaser a wavefront is a surface of constant phase wavefront is a constant surface of phase and Inter and in interaction with matter light behaves as
22:26 the as it as if it is made up of packet of energy called photon is called Photon okay easy question only phaser and phase may be slightly confusing otherwise all the other answers are straightforward now let's move on to the two marker section State and explain the gs's law in electrostatics so gs's law states that the total electric flux passing through a closed surface is equal to 1 by Epsilon * the net charge in enclosed by the closed surface so explanation is fin net equals to uh Q enclosed by Epsilon KN now you have to show some charges Q
23:13 enclosed and this is the surface okay this is the surface the five this is the F net okay so total flux coming out of a closed surface is basically given by total charge enclosed in the by the surface divided by Epsilon this is the explanation simple question very simple question you have to draw this diagram and explain this mention this formula that is sufficient now let's move on to the next question Define drift velocity and mobility of free electrons in conductors so drift velocity the average velocity with which the free electrons drift under the influence of an electric field is called the drift velocity so average velocity with which threee electrons drift drift means if you uh uh
23:59 if you have played that game when I was a kid I used to do that we used to make uh that um uh small uh uh paper boats and then we uh put it into the rain water so the paper boats used to move along the direction of the rain water right so that motion is called drifting it moves due to with the along with the uh Force along with the motion of the water right so since like that electrons move in the in the dire opposite direction of the electric field this is called Drift drifting the mobility of free electron is the unit is meter per second very interesting it is same as the speed okay the mobility of free electron is defined as the drift velocity of electron per unit electric
24:46 field applied so drift velocity of electron per unit electric field applied and its unit is M Square per second per Vol inverse or m/ second Newton inverse Kum okay so Newton per Kum is the electric field and me/ second is the D velocity so Mobility is basically drift velocity divided by electric field okay so this is another definition okay now let's move on a long air core solenoid of 1,000 tons per unit length carries a current of 2 aamp calculate the magnetic field at the midpoint on its axis so uh there is a
25:32 long solenoid carrying some current uh number of turns per unit length is 1,000 so what is the magnetic field at the center okay so magnetic field for a solenoid we know that is given by mu n i okay so mu is 4 piun into 10 ^ of - 7 n is given 1,000 and current is given 2 Amp so this is the magnetic field so answer will be Tesla in Tesla okay now multiply 2 into 4 is 8 piun into 10 power of minus uh 1 2 3 3 uh - 7 + 3 is
26:18 -4 so 8 piun into 10 ^ of -4 Tesla as easy as that you can use the calculator to find the value otherwise if you leave it here you will get full mark only thing you have to mention that the magnetic field inside a solenoid is given by okay B is equal to Mu not n as simple as that if you do it you'll get full full marks now question number 24 give with the principle of AC generator why is a current induced in an AC generator
27:03 is called alternating current let's see the answer quickly an AC generator converts mechanical energy into electrical energy by changing the flux linked with the rotating coil the direction of the current produced in an AC generator changes periodically and therefore the current is called alternating current AC so this simple so AC generator basically works in the principle mechanical energy converted into electrical energy so that is why it is called generator generator means it's generate and uh the uh current that is generated in generator they are basically changing their Direction with every passing second that's why this is called alternating current as simple as a very simple answer now next move on write any two uses of ultraviolet radiations another simple question UV
27:49 radiation can be focused into very narrow beams for high Precision application such as lassic lassic is nothing but uh laser eye operation so if you suffering from some eye defect and you want to do some operation lassic is one of the option next is UV lamps are used to kill jums in water purifiers if you have a water purifier in your home you know that there is a UV lamp people call it UV lamp Okay uh UV lamp is basically a device which can kill germs uh present in the water okay UV filter if you have heard the name UV filter UV filter means just a UV lamp which is killing the the germs present in the water next question number 26 name the objective used in the refracting type of
28:35 telescopes and reflecting type of telescope so in the refracting type of telescope convex lens is used in reflecting type of telescope concave miror is what is the objective objective means from where the light falls the first time okay so there may be some other Arrangement as well so when the light whoever is gathering the light for the first time is called the objective in case of uh concave mirror is the uh objective for reflecting telescope and for refracting telescope obviously we use uh convex meter generally all the meters in IPS also we use convex meter for refracting telescope but in case of uh concave miror there are different Arrangement okay now let's see the next part write
29:22 two conditions for the total internal reflection to occur okay I think this is a two mark question somehow it is mistaken it will be a two mark question okay what are the conditions light must travel from denser to rarer medium the angle of incident must be uh greater than that of the greater than the critical angle between the medium right so this is the answer now next question name the majority and minority
30:08 charge carriers in an N type semiconductor N means negative type right so negative type means majority will be electrons minority will be holes as simple as that you have to identify n type means uh in negative type again this is a two mark question somehow some of the two Mark questions are uh unfortunate marked as three marks okay so this is also two mark question so majority is electrons minority are holes now write any three properties of electric field lines this is a three mark question let's see uh another easy question common question so uh I think there is a mistake here so this is not electric field lines this is the properties of charges uh this answer is incorrect okay uh electric field line properties we know that two electric
30:54 field lines never intersect uh then we also know that electric field line starts at positive charge ends at negative charge or at Infinity okay uh similarly uh higher number of electric field lines represents higher electric field higher streng stronger magne electric field whereas less number of field lines represents uh weaker electric field okay so these are the very three basic properties of electric field lines you can find the answer I have discussed this multiple times and very easy question so even though by mistake there are properties of charges are written but but not is electric field lines are also very very easy right now obtain an expression for the effective capacitance of two capacitors connected in parallel uh so I think this
31:39 is a three mark question again uh some um mismatch happened this is a three mark question we also know this is a very easy part so uh two capacitors connected in parallel so what is the meaning of parallel means okay so here we have two capacitors say uh C1 C2 connected in parallel with the battery say V okay what is the thing that is constant in parallel and if you draw an equivalent Circuit of this then there will be only one capacitor and the same battery okay same battery C equivalent
32:26 and the same battery V okay now uh when I say equivalent capacitance what is the meaning of that equivalent capacitance means uh I can replace these two capacitors with a single capacitor but the everything in the circuit will remain the same so uh suppose initially charge total charge flowing in the circuit was Q the same charge will flow in the circuit again okay nothing will change only the two capacitors will be replaced by one capacitor that is called equivalent capacitor basically uh now uh cies and parallel they behaves differently in case of uh series capacitor uh charge through all the capacitors are same in case of parallel capacitors voltage across each capacitors are same because they are
33:13 connected in parallel so you can consider uh this is a path independent path this path is independent from the below path right both are separate paths separate path means individually both the cap capacitors are getting the same voltage right now if the charge through this one say q1 and if the charge through this one is Q2 okay then from the conservation of charge we can say Q is equal to q1 + Q2 right charges cannot be created or destroyed so uh definitely whatever the total charge now remember that in case of resistance you have to use current but in case of capacitor you have to use charge now uh since voltage is same for both capacitors what we can write is
33:58 that uh q1 is basically uh C1 V okay and Q2 is equal to C2 V similarly from the uh series capacitor the equivalent capacitors we can write Q is equal to C equivalent into V very simple okay this is for capacitor 1 capacitor 1 capacitor 2 this is for C equivalent so total we can write we can put these values here C equivalent into V = to C1 into V + C2 into V now you can take V common so C equivalent is equal to C1 + C2 right this is the uh derivation simple easy derivation that's why three
34:44 marks now let's move on to the uh next part what is lowen force another three marks question okay what is lowen force okay L force is the force between um uh magnetic field uh when a charge when a charge is moving through an electric and magnetic field
35:34 the total or the net force experienced by the charge is is called lowen Force right what is the expression low range force is given by F = to Q common electric field e+ V cross B right Q is the
36:21 charge e is electric field B is is magnetic field V is velocity of the charge right so this is uh you have to write it in full I just write it in shortcut so that I don't waste too much time on this this is easy question so L force is very very very important from this uh chapter uh moving charges and magnetism um McQ question two mark question three mark question all sorts of question will come from this concept only okay so now write three differences between diamagnetic and paramagnetic uh substance again another three mark
37:08 question okay let's see the answer uh so diamagnetic and paramagnetic magnetic behavior these are weakly repelled by magnetic field paramagnet they are weakly attracted by magnetic field magnetic susceptibility have small negative magnetic susceptibility K Less Than Zero have small small positive magnetic susceptibility Kai Greater Than Zero Effect in a magnetic field induces a magnetic moment in the opposite direction of the applied field induces the Magnetic Moment in the same direction as the applied field now uh now this is corrected describe an experiment to demonstrate the phenomenon of electromagnetic induction using a bar magnet and a coil let's see the experiment so when a bar magnet approaches the coil the
37:54 galvanometer needle deflect indicating a current is is induced when the magnet is pulled away the needle deflects the opposite direction showing that the current reverses no current is induced when the magnet is stationary related to the coil this much answer along with this should be sufficient but if you want you can add this conclusion as well uh maybe uh sometimes a teacher wants something more so this is the part where you have to mention that okay there they have only mentioned that demonstrate explain the experiment they have not mentioned any conclusion but I I have given this volunteer this experiment demonstrate electromagnetic induction a current is induced in the coil when the magnetic flux through the
38:41 coil changes this confirms farad's law of induction and the direction of the current follows lenses law opposing the charge in the magnetic flux right give any three results of the experimental study of the photoelectric effect so let's see see this is the same question as H Leonard experiment observation by H Leonard uh interpretation of H lard same uh question is uh is this for a given metal no electrons are emitted if the frequency of the incident light is below a certain minimum value called The thresold Frequency new not regardless of the intensity of the light photo electron emission occurs without any noticeable delay after the light strikes
39:28 the surface even for very low light intensities right uh maximum kinetic energy of the emitted electron increases linearly with the frequency of the incident light above the threshold frequency easy question let's move on to the last question I think of the three Mark series write the three postulates of B's atomic model another very very common question I have seen this question multiple times in your preo Series right so most expected question V model is applicable to hydrogenic atoms it cannot be extended even me to uh two electron atoms such as helium while bore atomic model correctly predicts the frequencies of the light emitted by hydrogenic atoms the model is unable to explain I think this is the drawback of
40:14 the B atomic model not the uh uh postulates postulates are like uh electrons are orbiting in the fixed I will I will give you the PDF while I give the PDF I will correct this questions as well don't worry there are two questions by mistake there was answers were wrong but I will give you the PDF with the correct solution don't worry about it uh we will provide the PDF for all the blueprints uh once we solve one by one I'll give you the PDF uh so what is the first uh Point P postate that electrons are orbiting in the stable orbit okay uh secondly uh the orbital angular momentum is integral multiple of uh NH H by 2 pi uh right uh that is the integral multiple of H by 2 pi right and electrons when they jump from one orbit
41:01 to another they emit energy and if they gain energy they go to the higher energy State and if they uh go from higher energy state to lower energy State then they emit energy uh they emit the excess energy right these are the main three postulates you can write it down okay otherwise this is not the correct answer okay this is not the correct answer this is the um drawback of Bo's atomic model okay now let's move on find the energy equivalent of one atomic mass unit in Jewel MB this is also a thre mark question so uh we know that uh energy is basically given by energy is given by mc² e = to mc² what is M here m is 1 U now in this
41:47 question also in your uh blueprint the 1 U value was given 10 the^ of minus 2 such a blind mistake they have made okay one U is is not 10 the^ of -2 it is 10 ^ of - 27 okay uh so uh 1 u means one Atomic unit right so one atomic mass unit AMU or U right so what is m m is 1. 6605 into 10 ^ of - 27 uh into c² C we know that c is 3 into 10^ 8 or whatever value they have given if you use the exact value you'll get more correct answer anyway does not matter we uh know the answer so we don't bother too much about these values okay whole
42:36 Square uh so this is the energy right this is the energy in Jewels you can uh find whatever is the value of this injs applying the multiplier okay you have to square this term and then multiply 1.66 to calculate it in Jewels because in Jewels also you will get one Mark now to calculate in uh uh EV e in EV what we have to do we have to uh divide this entire term 1.66 5 into 10 power of minus 27 into uh it will be approximately if I remember correctly it will be uh
43:22 1 4 9 into 10 the power of - 10 I think - 10 this will be the answer okay in jewles okay so let me write that only let's not waste too much time so uh 1.49 into 10^ of - 10 divided by uh electron charge of an electron so that is uh 1.6 into 10^ of - 19 which will come out to be 0.93 uh 15 9315 into 10 the^ of - 9 10
44:10 the power of + 9 sorry sorry sorry sorry sorry extremely sorry + 9 right um Min - 10 this will go up and will become + 10 so what we will take we will take three of the zeros uh for my own use so 93 1.5 into 10^ of 6 electron volt electron volt now 10^ of 6 can be written as 9 31.5 Mega electron volt okay this is the answer this is a very common useful constant uh if you are preparing for any competitive exam you have to remember this value right I have told you multiple times so this is the thing that you have to solve very easy question now next question derive the expression for
44:55 the electric field at a point point on the axis of an electric dipole let me quickly do it I have done it multiple times uh so this is a dipole uh so which one I'll take positive let's say this is negative this is positive okay we're trying to find the electric field at some point say p okay the distance between the p and the midpoint of the dipole is R okay the length of the dipole is a this is a this is a total length is 2 a right now electric field due to plus is in this direction so this is e+ and electric field to negative charge is in this direction this is e minus right this is the diagram now uh
45:43 dipole moment is defined it is from the negative to positive charge so the dipole moment uh is given by 2 q a okay or Q into 2 a this is dipole moment and direction is from negative to positive this is the thing we know that now let's find the electric field 1 by one e+ is basically 1 by 4 Pi Epsilon charge divided by the distance distance in this case is this distance what is this distance this distance is total and the midpoint from the midpoint it is R from the midpoint the distance is R so from here it will be r r -
46:29 A so r - A squ okay and E minus is 1 by 4i Epsilon Q by r + a whole squ so these are the two electric Fields so let's add them okay now uh what is the direction of the electric field electric field if plus is in the direction or right side along the positive x-axis so and negative electric field is the negative x-axis is opposite direction so even though we are adding them the net electric field while calculating it it will be subtracting right and um if you notice that the p is from negative to
47:15 positive and the positive electric field is also uh in the direction of P now positive electric field not only in the direction of P this is also bigger than negative electric field so we will take this one first 1 by 4 Pi Epsilon this is common what we will do we will write Q by r - a² minus Q by r + a s considering the direction so we are adding here vectorially uh so here it will be along the direction of X cap you can say that so now 1 by 4 Pi Epsilon if if you don't want to avoid Vector what you can do you can simply
48:02 instead of doing that you can add here negative sign that's that's it okay while calculating the electric field never use uh sign of the charge this is very very important while calculating the electric field never use sign of the electric charge only thing that you have to use is the direction if they're in the same direction they will be added if they are in the opposite direction they will be subtracted that's it while calculating the magnetic uh field sorry electric uh potential on the other hand you have to use the symbol of the charge because that is a scalar quantity this is a vector quantity so Q is also common see Q is also common now let's see so what is the uh this r + a whole s - r - A S / r - A in whole square into r + a
48:51 whole squ right this is the simple thing right the way we do it now what we will do we will simply Square them right now so Q by 4 Pi Epsilon so a a + b square - a - b square that is 4 a b so 4 r a divided by so it here inside if you see a + b into a minus B so that is a squ - b square whole square right now let me write it down here so you have to
49:36 practice this two three times to uh solve it quickly so net electric field is basically given by uh 4 R A into Q ided by uh 4 Pi Epsilon uh r² - a sare whole Square now don't cut this four it will be helpful uh if you don't cut this Force okay so R let keep this R what we will do we will uh make this 2 into 2 A Q into R ided 4 piun Epson r - a r s - A S square right now
50:22 we know that uh R is very very greater than a why so because the definition of electric dipole says that two equal and opposite charges separated by a very small distance so in that case a is very very small compared to R so we can approximate this value to be let me write it here so e net is basically uh now 2 AQ 2 AQ is basically the uh dipole moment so 2 p r divided by 4 Pi [Music] Epsilon into r^ 4 why r ^ 4 if R is very
51:09 very greater than uh a then r² - a square nearly equal to only R square right this is a very very small number so you can ignore that term in terms of uh importance now one of this R will cancel so you will get e net = to 2 p by 4 piun Epsilon R CU now if you find the direction so electric field and the dipole moment will be in the same direction we have already seen that electric field is also in this direction dipole moment is also in this direction so this is the common derivation now you have to practice this derivation two three times to master it now next question two cells of different emfs uh and different internal
51:55 resistances are connected in series derive the expression for Effective EMF and effective internal resistance of the combination we do not have too much time so I have to move little faster right you don't want to sit here entire day to watch me solving the model question paper right so we have to solve it little quicker this is Epsilon 1 this is Epsilon 2 this is R1 this is R2 okay so this is the uh say uh sorry uh this is not my mistake this will be the resistance
52:42 right this will be the resistance R uh what is the cells in series series CS in series let me do redraw this once more so we have two cells in series one uh cell is this then another cell is this and then they are connected with the resistance okay so this is Epson 1 R1 Epsilon 2 R2 now if we find a equivalent circuit equivalent means we replace everything with only one okay valent means we replace
53:28 everything with only one so this is Epsilon equivalent this is our small R equivalent and we are trying to find this and all the other things in the circuit will be constant so if the current was I same current I will pass again right now now question number 38 two cells of different emfs and different internal resistances are connected in series so here we have uh two cells we know that cells in cies basically they are asking to okay find the equivalent EMF and equivalent resistance for the combination right so how to do that we have to first connect them so this is Epson 1 R1 epon 2 R2
54:15 generally EMF is represented using uh not exactly V1 V2 that represents potential so for uh emfs we uh do it using Epsilon right and the equival circuit what is an equivalent circuit equivalent circuit as I say is basically everything will remain same only instead of having multiple components we will have only one component for example here we had uh two uh emfs so total current say I the current in the circuit will remain unchanged now instead of Epsilon 1 and Epsilon 2 we will have only one epon equivalent and R equivalent this external resistance will remain as it is so what we have to do we have to find uh this uh this is an equivalent circuit so
55:01 we have to find Epsilon equivalent and R equivalent from this given value right so first what we have to do we have to apply cop's law and find whatever it is there so they are connected in series so we can write current through the resistance is IR so I uh plus because all the resistances can be considered to be connected in series we can say I small R1 + I small R2 = to Epsilon 1 plus Epson 2 simple very simple right or we can say that I is basically Epsilon 1 + Epson 2 -
55:47 I R1 + R2 this is the equation number one okay this is the potential this is the terminal potential this is called this term IR is also called terminal potential okay sometimes we represent the same equation using V not required now for equivalent circuit this is from the given circuit from the given circuit and remember that to for this derivation the image both images both diagrams are compulsory okay without drawing the diagram if you try to solve you will not get any marks be careful the this derivation is very simple now for the uh uh equivalent for equivalent
56:38 circuit equivalent circuit the current Remains the Same right that is the condition so therefore what we can say is that what is the current in this case current is I uh current is I basically so voltage drop across R is I voltage drop across uh R equivalent is I equivalent then that should be equals to the potential right that is the cutos loop rule or cutos voltage rule right if you if you go in a loop okay if uh you are going in the direction of current then definitely uh every uh voltage drop across every component that should be
57:24 equals to the uh total potential right now I equal to epon equivalent minus I RQ now this is equation two now we have to do a small comparison what is the comparison see I I on our left hand side both are equal in equation one and equation two but in equation one the potential or the EMF is given by Epsilon 1 plus Epson 2 whereas in this case it is given Epsilon equivalent so comparing 1 and two we can write filon equivalent is equal to Epson 1 + Epson 2 and uh
58:09 what else see here again minus I R1 + R2 here minus i r equivalent so again we can say R equivalent is equal to R1 + R2 so this is the basic comparison very easy derivation and if you do it you will get five marks only thing that you have to remember a small small description is required and you have to mention the variables Epsilon 1 and Epsilon 2 epsilon1 and R1 and Epsilon 2 comma R2 are connected in CD so this this the description of the diagram is also important because the derivation part is very very simple and easy okay mostly I have seen that coming as a three mark question but they have given it as a five mark question in the model paper which if that comes in your board exam then you'll be lucky okay what I
58:55 have seen most cases parallel one is coming as a five mark question because that is little tricky derive the expression for the magnetic field at the point on the axis of a circular current Loop very very uh important derivation this is a very very very very very important derivation okay this derivation okay this one and the parallel obviously k Sorry wion bridge is also there from this chapter uh no sorry this is from Magnetic Moment right no anyway uh from this chapter force between two parall current carrying wires and this uh loop one is very very important okay most asked question one of the most asked question so we are trying to find magnetic uh field at this point P so we have to consider a small
59:40 element say DL the radius of the uh loop is capital r okay the distance from the point is X and if this temporary distance where what we are drawing okay my drawing uh skill is not super good you already know that right so I without using scale it is difficult to join two points also right so what I will do I will do some cheating I will remove this point from here and we will put this point on that right so this is point P we trying to find uh the magnetic field at this point this this distance a
1:00:26 r now if you apply Pythagoras Theorem you can see this is R this is X so small R definitely will be square root of capital r² + x² okay you can add everything in this diagram itself that will save some time for you to write okay now the current element is DL if you consider the current element you know that this is the element right this is the element now you are trying to find a point somewhere here on the axis so it is on the axis is somewhere here and if you consider this is the element then the line that you have you are joining that line and the tangent basically they are at 90° angle okay they are perpendicular so what you can do you can say that okay they are perpendicular so this angle is
1:01:15 90° okay now if this angle the internal angle the internal angle is five then the magnetic field will also make an angle uh five uh with this okay this is the thing that you have to remember if the internal angle is five the magnetic field will be making an angle five with this with this axis so this is say x- axis x-axis so it will make an angle five with the x-axis now you can divide this into two components uh so say this is the magnetic field DB Okay small component now you can make two components one component on y AIS say DB Y and another component along x-axis
1:02:05 say DB X right now about this circular uh coil the important thing is that if you consider a diametrically opposite Point diametrically opposite Point means you just need to construct the diameter and and the opposite point this is you don't need to draw I'm just expl exping it okay I'm just trying to explain it so that any doubt that you might have will be cleared so any if you consider any diametrically opposite point because of that uh whatever magnetic field you will have that will be exactly opposite to that five this is five means this will be also five right angle triangle right so this will be also five and you will get a component which is exactly opposite to DB okay so
1:02:53 DB if you are considering this is plus dvy this will be minus DV so total component along Y axis will cancel each other so total component along XA y- axis will be zero whereas if you consider this x component X component dbx and dbx both will be in the same positive X Direction so they will add up so X component will add up whereas y component will be uh deleted right so similarly we can say that uh if we consider a diametric you have to write that you have if you consider a diametrically opposite component uh then only the Y AIS only y component will be cancelled and X component will remain okay so uh we will only consider the X
1:03:39 component that is very very important now what is DB first we will write DB from bio's law DB is equal to Mu by 4 Pi I DL sin Theta by r² now we have already seen that Theta is basically equals to 90° so Theta is 90° now many students and many uh I have seen many students they use same symbol Theta and five do not do that okay that is technically incorrect you cannot uh use same symbol for two different uh quantities right so you have to be careful uh you cannot use
1:04:25 same symbol for two different quantities here Theta equal to 90° not 5 5 is not 90° okay so DB becomes mu by 4i I DL ided by r² + x² okay why so because R small R square small R is basically the distance between the element and the point element and the point is small R which is square root of r sare + x² so R square is r capital r sare + x² right so this is from this Pythagoras Theorem and uh sin 90 is 1 so that's how we have
1:05:10 written it now as we have found out that total component out of the all the components only X component remains so we are only interested to calculate the dbx component so dbx is equal to DB what is DB component X component so this is the angle this is the component so this is an adjacent component for adjacent component it becomes cos right adjacent component is cos and the perpendicular component is s if you remember slightly about resolution of vectors right so DB Cosi so we have to find the value of cos pi as well so mu by 4 Pi I DL r² + x² square into cos 5 cos 5 is what
1:05:58 so if you consider this is the cos 5 this is five that cos 5 is basically Base by hypotenuse so base is your R whatever adjacent side that is your base so this is your base here R and hypotenuse is this term so R ided by small R small R is basically nothing but r² + x squ square root so this is the term so it will be mu not by 4 Pi i r by r² + x² to the^ 3x 2 okay into DL okay so what I have done I have taken
1:06:47 this to be to the^ half square root means 2 the^ half here we have 2 the^ 1 so 2 the^ half to the^ 1 if if you add them you will get 3x2 right uh I hope you have already practiced this multiple times this derivation if you have not you have to practice it at least three times to make it perfect right now total component total magnetic field BX is basically integral of dvx right dvx we have calculated so total magnetic field is basically integration so integration means if you see mu by 4i is constant for this given configuration current is constant current is the current that is flowing in this coil current is constant okay the radius uh
1:07:33 is constant the distance is given already so the distance is also constant even though it is given as X it is not varying it is not a variable only thing that is variable here is DL so DL you are taking here you can take here you can take here you can take DL for all small small parts along the circle so the DL is the only variable and if you add all the DLS basically DL means addition integration means addition so what you are doing you are taking small small parts of the circle and then you are adding them right small small parts of the circle and entire circle of complete circle right so this will be a closed integral so complete circle means basically it is the perimeter of the
1:08:19 circle so you are taking small small portion of the circle and you are adding them so you are getting the total uh perimeter of the circle so mu by 4 Pi okay I by r² + x² to the^ 3x 2 into square root uh close integral DL that is 2 pi r okay 2 pi r so you can cancel here so you will get mu by 2 pi okay let me write it here so you get mu by 2 I r² / r² + x² to the power
1:09:13 3x2 okay this is the formula this is the formula field at a point on the axis of the circular Loop so this is the formula this if you derive up to this point you'll get full marks but if they add one more small part that uh hence find the magnetic field at the center magnetic field at the center uh BX Center is when X = to 0 that is equals to Mu so at the center means you are finding here that means this x distance becomes zero right if you put xals to 0 you will get the magnetic field at the center mu by 2 into I r² okay uh X becomes zero so it will r²
1:10:00 to the^ 3x 2 r² to the^ 3x 2 so multiply so it will be R Cub right so r² by R cub and as you can see this will cancel so you'll get I by R mu by 2 I by R this is an important formula okay especially if you are preparing for K set exam then this formula has to be remembered also for derivation they might give you at one Mark extra so total derivation is four marks and they can add this extra derivation for one extra marks okay so be careful in case they ask you you have to know how to do that okay now let's move on for ta to coherent waves of constant phase difference undergo uh
1:10:45 interference obtain the expression for the resultant uh displacement write the conditions for constructive and destructive interference in terms of phase difference very interesting first of all 4 a 40 a not super common but we know uh we can derive that y1 equals to let us consider two uh waves y1 is equal to a sin Omega T and Y2 = to a sin Omega t + five these are the two waves so what is the meaning of that so two waves means we have we have to take two waves waves they are interfering waves uh you can take in any form this is the easiest form to take so y1 is equal to a sin
1:11:32 Omega t uh this is the first disturbance and the second disturbance is y1 Y2 is equal to a sin Omega t + 5 okay so this is the constant phase difference so what is the total phase total uh wave will be after the interfere total wave will be simply addition right so a will be common sin Omega t plus sin Omega t + 5 okay I think it is 3 + 2 so this derivation is three this derivation is 2 this much is two okay so now s uh C + sin D what is the formula 2
1:12:17 into cos Omega T minus Omega t + 5 by 2 into cos so sin uh C + sin D that is 2 sin Omega t + Omega t + 5 by 2 into cos Omega T minus Omega t + 5 / 2 right this is the formula we all know that now let's simplify a 2
1:13:03 a let's take this two outside right so s it will be Omega t plus Omega 2 Omega t plus so it will be Omega t + 5/ 2 okay into cos Omega T Omega T will cancel so it will be 5 by 2 or you can say it will be 2 a cos 5 by 2 into sin Omega t + 5 by 2 right so this is the new amplitude so whatever is after the waves interfere the new amplitude of the wave is given by this so in initially the amplitude has no C or sign term in it but after the interfere the
1:13:51 uh amplitude now has a cos 5 why cos 5 ter why we are not including this term because it has a Time dependent term whereas cos 5 by 2 is a because phase difference is fixed it is basically a constant okay cos 5 by2 is a constant depending upon whatever is the phase difference it can have different different values but for a given phase difference this is a constant now they have said that expression for the resultant displacement so this is the expression for the resultant displacement as simple as that now if they ask you the expression for result in intensity so intensity is basically proportional to amplitude Square so here amplitude is basically 2 a cos 5 by 2 so intensity will be equals to 2
1:14:39 a cos 5 by 2 whole Square okay so or you can write this as 4 4 a² cos 5 by 2 now it is all all of them are proportional relation not equal there is a proportionality relation because intensity is not exactly equal so there is a proportional value constant we do not know but what we know that initially both the waves have same intensity what was that both waves had same intensity say I not I not was both
1:15:25 for both the waves was proportional to a square right so this wave was also amplitud a this was also amplitud so both cases had proportional to a square now proportionality constant will be same whatever proportionality constant here the same proportionality constant will be here so you can say that I is equal to 4 I cos² 5 by2 this is the expression for intensity this is extra this is not asked in this question so you can uh but if they ask you and hence find the expression for intensity you have to know what to do okay you can directly Do I by I so if you do that I by I directly you can the a square will be cancelled okay so that is one of the way anyway
1:16:12 now let's uh see what is the condition for uh maximum constructive interference so phase difference in terms of phase difference constructive interference is when it is equals to n Lambda okay where so the phase difference uh for constructive interference is 2 pi n okay uh n is uh equals to 1 uh sorry 0 1 2 dot dot dot okay so this is the uh condition for constructive interference that is even multiple of n so if it is 0 2 pi 3 Pi then it is basically uh giving you constructive interface and if it is a odd multiple of Pi then
1:16:58 you are getting uh destructive interference so you can uh write it like this two uh it will be uh odd multiple right so it will be n + 2 n + 1 into Pi so if it is 2 n + 1 into Pi where n equals to again 0 1 2 3 then you will get a destructive interference this is the condition for constructive interference this is the condition for destructive interference in terms of phase difference now let's move on to the uh next question what is a rectifier explain the working of a full wave rectifier using a neat circuit
1:17:45 diagram draw this uh input output wave froms very very very very important uh question okay let's see a rectifier is an electronic device that converts alter alting current AC into direct current DC okay positive half cycle of AC input during the positive half cycle of the AC input the voltage uh the upper half of the Transformer secondary winding becomes positive diode D1 becomes so let's see the diagram so this is the diagram this is the easiest diagram given in your textbook so in the positive half cycle positive half cycle means when the current is increasing okay in that case your uh first part becomes positive and the bottom part becom negative right so this is how you can find the uh uh uh uh AC circuit right so
1:18:33 uh first diode D1 is conducting when this is positive the diode D2 will conduct when the uh circuit is becoming negative okay so first half cycle that is first half cycle means you can divide the circuit into two different halves okay first half cycle as it is a AC AC can be divided into two halves like this okay so first half it is positive second half it is uh negative so first half cycle is this one is called first half cycle this one is called second half cycle okay in the first half cycle diode D1 becomes forward biased and conduct so at this point this is positive this is negative and again uh if you consider this one then this one is positive this
1:19:19 one is negative so diode D1 conducts diode D2 does not conduct so only one way to flow the current right so only one way to flow the current now because of this you can see that uh the diode D1 is conducting di D2 is not conducting but when it is conducting when it is conducting the current is flowing through uh this RL right the current is flowing through RL now in the negative half cycle in the negative half cycle this part becomes positive this part becomes negative again the current can only flow in this Direction only so again the current is flowing through this XEL in the same direction no matter the current direction changes but uh the current flowing through RL Remains the Same
1:20:04 Direction right let's see so uh in the
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