JEE (Advanced) 2026 Physics Video Solutions of Official Paper 1 (Q. 1 - Q. 4)

Ajouté : 2026-05-19

[00:00:08]We are starting with paper one first question. So I just want to make sure you get these logics properly. Okay. So we have this first question. Come on sir. Consider a large disc of radius capital R and two smaller disc each of radius small R lying on its circumference as shown in the figure.

[00:00:28]See this is how it is. We have a disc like this. Okay. And there is a smaller disc here. Another smaller disc. These two are identical. They're all small RDS only. Yeah. Both of them. And this is bigger RDS. They're asking from here those two right the small are initially in contact with each other with angular separation of delta theta. this point to this point. This angular strength of the direction. Okay. They are made to roll without slipping in opposite directions with constant angular speeds omega and 2 omega shown in the figure. So this is omega. This is 2 omega. So right while the large disc is held stationary this they are fixing it. Okay. This is fixed disc. They're asking the time taken time to at which the smaller disc are again in contact is the question right. So they have given four options and all.

[00:01:28]Yes sir.

[00:01:28]>> When you saw this one what went in your mind? What was the logic?

[00:01:33]>> Sir this is this is a fairly common question you had given in class.

[00:01:37]>> Okay.

[00:01:38]>> But I was saved by the options.

[00:01:39]>> Okay.

[00:01:40]>> Because I forgot that when it comes here again we have the same delta theta.

[00:01:44]>> Yes.

[00:01:44]>> And I forgot about it until I saw the options.

[00:01:46]>> Oh nice. What was in your mind?

[00:01:49]>> Same thing. Same thing. Yeah. But see uh they have how are you bringing in this uh rolling motion this one like see the concept wise you have rolling in a flat surface here we have a curved surface.

[00:02:03]How are you bringing that?

[00:02:06]>> So I mean as this we just consider the center of mass with some other omega then we observe this omega.

[00:02:13]>> Okay.

[00:02:14]>> Then we just observe this point velocity and use our similar logics. So even though we might not have explicitly done rolling or curved surfaces, it's still doable with whatever we know.

[00:02:24]>> Yes.

[00:02:25]>> So don't you think it is similar to having rolling motion on a surface like a trough and we have rolling and oscillations in it and similar to Yeah.

[00:02:36]Okay. Anyways, here they're asking us to ignore gravity also. So let's ignore and let's try and solve and we'll give our viewers also like how to solve this one.

[00:02:44]Okay. Yeah. Let's do it. So everybody this problem see we know this this is a this let's say we have this and as we have learned this if there is a surface if this is uh this polar disc if this radius is r let's say if if it starts rotating with omega with respect to the surface we are observing that the bottom pointer which is in contact with this particular bigger disc that point velocity should be addressed the point which belongs to the disc right it should be addressed so because of rotation you'll have omega r speed right and because because of translation if it has speed v then the bottommost point will have the speed v because of translation we know that the bottommost point will have the velocity because we have learned this idea with respect to our logics in relative to motion. See velocity of point P which is at the bottommost point of the disc right that velocity is with respect to ground frame is equal to velocity of point P with respect to center with respect to center plus velocity of the center. Okay. So velocity with respect to the center would be omega R because of rotation because of transmission you have velocity V right. So bottommost point if it is at rest then d should be equal to omega. Got it? So I think this this idea you are familiar and this we know we have done this right. So this point will be going with a speed of omega r that let me as v and on the same logic this point will be moving with a speed v ddash and v ddash on the same lines v dash on the same lines we can write this one as 2 * of omega r why 2 * of omega r because we have 2 omega speed the bottommost point we'll have the speed because of translation vdash and because of rotation 2 omega r in this way 2 omega Right? So we know that this is move with V speed this is movement V dash speed. So it is more like now we have solved almost this problem only thing is we have to bring in little bit of chics and complete it right. So this particular disc as it starts rolling and reaching here let's say since it is moving bit slower and this is faster let's say it goes faster and reaches somewhere here. I'm just guessing some location. All right. And this is also the location of the other disc. Now, no matter where they meet, still the angular separation will be the angular separation will be dela theta only.

[00:05:30]Delta theta will be the angular separation. Right? So if this is the case then we know that keeping one point at stationary condition looking at other sitting at one of the center of this particular sphere and looking at the other one. So we call this one as relative motion and we keep one body at virtual rest and looking at the another body and we know that effectively they have to travel a distance. So our logic is distance by time taken would be our speed. That's our general logic. Right? I'm just going to observe that the time taken would be distance by speed. Time taken would be distance by speed. Now we work with relative distance and relative speed. The time would be same. Got it? Now using this logic we have the distance that it has to travel relatively if you keep one and look at the other. So you know that this has to cover all the way and reach the other one. Right? So overall we have 2 pi radius effective radius. If it was like really really small size we know that the distance would be 2 pi r for you know circumference. But from this center if you observe the radius is effectively r plus smaller right? So we can write it as 2 pi small r okay capital r plus small r. This is the effective distance as we consider from this point all the way back to the same point. But there is an angle of separation. So meaning there is some separation right. We'll bring that separation also. And also on top of it they're saying that delta theta right is very small. So sin delta theta is delta theta which means delta theta is small.

[00:07:20]If it is small then you can treat this length right it is almost as straight line but actually you might think sir that distance straight line or curved you can think like this is a curve okay where we have arc length arc length equals radius into plane angle so I can write 2 r equals 2 r equ= r + r * of delta theta this relation probably you know that Arc length is equal to radius into angle of angle plain angle that we have right. So here this one right and in our formula it is a curved line right but here you can treat this curved line or straight line are almost same because delta theta is small right so using that logic here we can go ahead and say that here the initial separation was * of delta theta pi * of delta theta here we have right so let me write not* let me write as r + r * beta see because from here to here it is r r r r r r r r r r r r r r r r r r r r r + r r r r r r r r r r r r r r r r r r r r r + r into data is this distance right so that's the distance we have and when it reaches back again the final separation will also be involving the same distance of separation between the sectors so it is two times of it okay so this is overall length minus of this this effective relative separation between them so we got effective relative separation divide by effective speed. So this is moving this way. This is moving the opposite way. The speeds gets added up. So we can write it as V + V dash. That's the time taken. All we have to do now is to just solve this mathematics. Nothing much. So we can write to = 2 pi * say r + r by 50 that they have given minus 2 * delta theta. Delta theta we can write as let me substitute here R by 50 and R * 1 + 1 by 50 times of delta. So from this we can simplify and say 2 R1. So 50 so 2x 51 would be our diameter. Okay. So 2x 51 2x 51 here and r + r which I'll write as r + r by 50. Okay but whole thing divided by what do we have here? So v + v dash v is omega v dash is 2 omega. So omega r by 50 plus 2 omega r by 50. This is what we have. Now let's simplify this. to equals here 2 pi we'll have so 51 and r we have so let me write r here and there is 50 here in the denominator minus 4 * of there is 50 here so let me write let me separate this so that we don't mix up things okay this is how it looks like yep so 4 by 51 And there is 50. Let me write 50. 50 here. And there is R here and 51 by 50.

[00:10:54]Just checking with the math. 4 * R 51 51 here 50 here whole thing divided by 3 omega R by 50. Right? This is what we have. So let's simplify this.

[00:11:11]Okay. Now this 51 and 51 canceling up.

[00:11:15]This 50 this 50 this 50 will cancel up.

[00:11:18]Right. So we can now write our final answer. Okay. So I will erase a bit of it. I will erase a bit of it.

[00:11:30]So here this concept we have understood right. So I'll just take this one out.

[00:11:41]So we have to = 2 pi * r. So we can just simplify that and write 2<unk>i * 51 r.

[00:11:53]So 2 pi r now = 2 pi * 51 r 51 r - of 4 r 50 4 r but 50 is already cancel right so 4 r / 3 r so we can just get rid of all this r dependency Right. And let's look at the options.

[00:12:27]We need this 51 outside. So let me take 51 outside and let's have 3 omega also outside and we'll see what is there inside the bracket. So 2 pi will be there and 3 omega I have taken outside. 51 I have taken outside. So 2 pi minus of. So I think we have here. So one of the term uh 51 by 50. So when we take 51 outside right we are taking 51 outside. So I multiply by 51 here divide. Okay. So I'm taking 51 outside. So 2 pi we have here minus of 4 by 51.

[00:13:13]So we are just checking out the options again. 51 by 3 omega 2 p<unk> - 4 by 51.

[00:13:19]So we got bunch of other options also right I think this is very simple enough I think you can easily observe that it is option C and that's our answer we will discuss question number one question number two paper one first we read the question consider a circuit consisting of a capacitor of capacitance C and a coil with n turns per unit length Cross-section area S and length D where D² must greater than S. There is another coil of length DX2. Cross-section area S by two and two end turns per unit length completely inside the larger coil as shown in figure. The ends of the smaller coil are connected with each other by an insulated conducting wire. The self inductance of larger coil is L.

[00:14:15]Neglecting S effects and all the omic resistances. The unit frequency of the circuit is one coil is this one and other coil inside this. Other coil inside this and it is short circuited and it is capacitor and this is a source AC source.

[00:14:40]First we see this concentrate on this part.

[00:14:45]Suppose two coils like this current in this one is I1 and this is I2.

[00:14:55]If we see about bigger coil its inductance L1 is mu n² area cross-section area S into D. It is for bigger coil. for solder coil and it is given to L in the question. Mu not 2 N is² S is S byX2 D is D /X2 so it it is also equal to L. So both coils have self inductance L right now second in the second case ABC in the smaller coil magnetic field due to only bigger coil.

[00:15:53]So if we find mutual inductance m is equal to magnetic field due to this is mu not n i 1 it is magnetic field into its area is sx2 and number of turns is 2n into dx2 divided by current I1 and from this we got lx2 it is second Now third step a smaller coil is short circuited like this one. So induced GMF net induced GMF should be zero. So L d2 by dt minus m di1 by dt is equal to zero is equal to zero. Here m is l by2. So you write di2 by dt is equal to lx2. d i1 by dt. So I2 is equal to i2 is equal to i1 by 2. Now we find in the next step equivalent inductance for first coil like this one. L equivalent di by dt di by dt is equal to l d1 by dt minus m di2 by dt. This is equal to L d1 by dt - lx 4 di by dt. So here we got l equivalent is 3 l by 4.

[00:17:49]So omega is 1 upon<unk> LC L equivalent into C that is L equivalent 3 L by4 3 LC and this one in the in this step second coil is not connected to first coil when for equivalent we consider only LDI1 by DT minus MDI2 by DT is the Also we discuss question number three in paper one.

[00:18:23]First we read the question. A solid cylinder of radius R rolls without slipping with a center of mass speed V2 grx by3 on a horizontal surface with a vertical as shown in figure. Here G is acceleration due to gravity. At the moment when the cylinder loses contact with the surface of the rotation around the corner the speed of center of mass is now I explain the solution. Just see it is a surface and this cylinder is rolling with velocity V suppose like this and the end when it reaches at this point it center is here like This but here surface end. So it starts touling about this. When it starts tling about this then suppose at this moment it loses the contact and its center is suppose this one initial center is here.

[00:19:37]Suppose this angle is theta and at this point velocity is v and at this point velocity is v.

[00:19:46]Till this instant rolling continue and we apply law of conservation of energy.

[00:19:52]So laws of potential energy mgz 1 - cos theta is equal to and energy of cylinder is/ mv² + i omega².

[00:20:09]So/ M r² omega is v² by r² that is equal to 3x4 mv² here v. So we write here 3x4 MV² minus V² right this loss in potential energy is equal to getting dynamic energy at this point contact so N becomes zero here mg this angle is theta so this angle is theta mg cos theta at this instant provide centrial force because up to this instant center of mass moving on a circuit. MG cos theta is equal to MV² by R.

[00:20:53]And after solving these two equations, you got cos theta is 5 by 7 and V is equal to<unk> 5 by 7 Z R.

[00:21:04]So it is the answer.

[00:21:08]Okay.

[00:21:10]Question number four.

[00:21:12]A double convex lens made up of glass of refractive index 1.5 length refractiveness 1.5 and radius of curvature 20 cm each. So it's a double convex lens immersed in liquid of refractive index NL correct plot showing the variation of power in the units of diapter. How to find power in diaper as a function of NL.

[00:21:52]So to finally get a graph of power versus N.

[00:21:59]So here we can just use uh lens makers formula 1x f= mu of length by surroundings - 1 into 1x r1 - 1x r2 this is the formula lens makers formula and 1x f is power so I'll just substitute the given values in the equation and you'll get the equation for power. So power is refractive index of lens is given as 1.5 by surrounding is nm -1 into 1x r1 for double convex lens r1 will be positive that is given as 20 1x 20 I'll convert that to m so that I'll get the power in diameters si unit so I have to substitute Radius in the meters 1x.2 minus of minus 1x.2 Here for second for R1 right side positive R2 will be negative so your minus will just simplify this 1 by L -1 2 by2 that is 1.5 by 10 L -1 into 10 I'll just substitute if you see the graphs NL value starts with one then I >> NL value if I take as one then power power I'm getting from this equation n value if it is 1 1.5 by 1 that is 0.5 sorry 1.5 minus 1 will be.5 into 10 five will come for NL equals 1 power getting five then if we substitute NL as 1.5 1.5 by 1.5 that is 1 1 - 1 0 power will get zero and you substitute NL as 2 1.5 by 2 - 1 into 10 that is 75 -1 into 10 that is - 2.5 will come. So if you look at the graph now it should start with five then 0 then minus 2.5.

[00:24:59]So you look at the graphs options A option is matching okay B also is similar but what's the difference in B there even a straight line and we know from this equation P is not directly proportional to NL is in the denominator therefore you'll not get a straight line so the only possibility is a a curved graph like this with starting value five when NL is zero five when NL is 1.5 Five power is zero and when I let this two power is negative. So option A is only matching.