Problems From JEE Adv 2026 | Paper 1

Added: 2026-05-18

[00:00:00]Hello guys, welcome back. So in this video, we will cover some JEE 2026 problems. Okay, so this is the first question. We have a large disk whose radius is R and we have two smaller disk. Okay, the smaller disk rolling on another disk. This question was just a previous year question itself. So small R radius is given. They're lying on the circumference. They are rolling basically. Okay. Initial separation between their centers is given to be delta theta. They are made to roll with constant angular velocities omega and two omega. Okay, it's shown in the diagram. The question is when will they meet? So here the first thing is let's draw the circle in which the center of mass is moving. So that circle will look something like this. Okay. This is the center of mass of the first disk and this is the center of mass of the second disk. And this pink circle is the circle in which they are moving. Let's say they meet somewhere over here. Something like this. Let's say this is where they meet. This angle will still be delta theta. So this is the situation when they meet. Okay. So now the thing is the center-to-center distance, this particular distance we can say it is approximately 2R, right? Okay. And the reason we can say this is because this this angle is a very small angle, delta theta. Okay. So similarly, this distance also we can approximately say it is 2R.

[00:01:09]Okay. So this distance is covered and this distance is also covered. So now the thing is observe the center of mass of the first disk. It moves it covers this particular distance. Okay. And if I call this S1, this I can just write as speed into time. So the speed of the first center of mass, so this disk which was rolling with omega, the speed of this guy's center of mass will be small R omega. Okay, just the simple rolling condition. So S1 I can write it as R omega, speed of the center of mass, into time, tau. Similarly, the other distance which is S2, the distance covered by the center of mass of the other disk, this S2 we can write it as 2R omega into tau.

[00:01:46]Now the total circumference of this outer disk, outer circle, is 2 pi R, right? So 2 pi capital R plus small R.

[00:01:54]Because this is a circle, this circle has a radius of capital R plus small r.

[00:01:59]So 2 pi capital R plus small r we can say it is equal to 2 r plus the other 2 r plus S1 plus S2. Okay, S1 plus S2 is 3 r omega tau. Okay, and that is it. So this will be 2 pi and so capital R plus small r is 51 divided by 50 times capital R minus 4 times small r and this is equal to 3 capital R by 50 omega tau.

[00:02:24]So capital R by 50 you can just cancel out.

[00:02:27]And from here tau turns out to be take 51 common because the options are telling you to and divided by 3 omega times 2 pi minus 4 by 51. Okay.

[00:02:38]So I think this matches with option C.

[00:02:41]Okay, so C option should be the correct answer.

[00:02:44]So that is it.

[00:02:45]Okay, so in this question we just had to equate the distances.

[00:02:49]Okay, so now in the next problem we have this situation. This is a we have an AC voltage applied across a capacitor and this particular double solenoid type of a circuit.

[00:03:00]Okay, and the question is the number of turns per unit length is capital N. Cross sectional area is S and D for the bigger solenoid. Okay, or bigger coil. And inside there is another coil whose length is half and cross sectional area is half. So basically volume is 1/4 and the number of turns per unit length is double. Okay, and the ends of the smaller coil are connected with each other by an insulating conducting wire. Okay, so this is the hint. So this inside particular coil that you're observing, this is a superconducting coil. Okay, so it is connected with a conducting wire. So basically this is a resistance less circuit. Okay, they're telling you to neglect the ohmic resistances. So which means this is this inside particular small conducting loop or solenoid, this is a superconducting solenoid. So which means the flux through it is going to be constant throughout. Okay, so before let's say when the current in the circuit is zero, initially the flux through both of these coils will be zero. So the right, so the initial flux is going to be zero for the smaller coil. So if we are talking about the smaller coil, initial flux would be zero, so the final flux will also be zero, okay? So this is a simple idea of superconducting coils. And I mean the proof for this is quite easy.

[00:04:14]So if you apply like KVL in this smaller loop, so if we apply KVL equation for this loop, it will be minus d phi net by dt. This will be the net EMF induced would be equal to zero because there is no IR term, right?

[00:04:29]So which means the net flux through this superconducting loop, it will be constant, right? So so what that means is like in short this question you can just instantly solve it.

[00:04:42]So if we send a current I through the bigger solenoid, we know that the net flux, and let's say this is the smaller solenoid, the net flux through the blue region, which is basically inside the smaller solenoid, this will be zero always, okay?

[00:04:57]Uh okay, so the net total flux through this entire solenoid will only be present on the rest of the volume, okay? There won't be any flux in this region.

[00:05:08]So the volume of the inner solenoid was 1 1/4 of the total volume.

[00:05:14]So the total flux will basically be 3/4 of LI, okay? If this inner solenoid was not present, then the total flux will be LI, okay? But because of the inner solenoid being present, the just the flux press just the flux through this region will be zero itself, okay? So the volume of the inner solenoid is 1/4 of the total volume. So 1/4 of the flux will vanish, and the total flux will just be 3/4 of LI, okay? So now if you So now if you look at the total picture, if you want to replace the entire inductor with a simple L effective, uh and if you send the same current I, the new flux will be L effective into I.

[00:06:00]And this will be equal to 3/4 of LI. So, the equivalent uh inductance is 3/4 L.

[00:06:06]So, the new resonant frequency is 1 by square root 3/4 LC.

[00:06:12]Okay? So, this is 2 by root 3 LC, basically.

[00:06:17]So, that is it.

[00:06:19]So, C option, I guess.

[00:06:21]Now, the next question.

[00:06:23]Okay, so the next question is also a previous year question. So, here we have a solid cylinder and it is rolling.

[00:06:29]And uh the moment when the cylinder loses contact with the surface due to rotation around the corner. Okay, so it is uh rotating about the corner.

[00:06:37]Uh and let's say this is the situation when it loses contact.

[00:06:42]So, losing contact just means the normal reaction will become zero uh at the contact point or the contact line. So, which means the only force in this situation is the force of gravity. Okay, so mg will be acting downwards. Uh the center of mass will have some speed and there will be some rotational velocity at this instant.

[00:07:00]This distance is R.

[00:07:01]Okay? So, uh if I say this angle is theta, then I can write uh the centripetal acceleration equation. So, that is mg R cos theta equals mv squared by R. So, this will be my first equation.

[00:07:15]So, v squared is rg cos theta.

[00:07:20]Now, So, now we can conserve energy. The center of mass has descended a bit.

[00:07:24]Initially, it was at a height of R.

[00:07:26]Now, it is at a height of R cos theta, right? So, uh the drop in height of the center of mass is uh or the drop in potential energy is mgr 1 minus cos theta. This will be equal to the change in kinetic energy.

[00:07:39]Okay, so for the change in kinetic energy, we can write it as 1/2 I about I A R times omega squared. So, this will be 1/2 moment of inertia about the instantaneous axis is 3/2 MR squared times Okay, so the over here the omega is V by R. So, we have V squared by R squared.

[00:07:58]Minus the initial omega is V naught divided by R. Okay, so V naught squared by R squared will be G by 3R. So, G divided by 3R.

[00:08:07]So, now we just have to solve for V squared.

[00:08:10]So, MGR can be canceled out. Right? So, if I substitute V squared as RG cos theta Um yeah, we can cancel out MGR and what we are left with is 1 minus cos theta equals 1/4.

[00:08:23]Not 1/4, cos theta by 4. Okay, there will be a three here. So, 3/4 cos theta minus 1/4. Okay, so from here we get So, 1 + 3/4 is 7/4 cos theta.

[00:08:37]Equals 5/4. So, cos theta will be equal to 5/7.

[00:08:41]So, if cos is 5/7, then V is square root of 5RG by 7. Okay.

[00:08:47]So, option B will be correct. Okay, so this is the fourth question. This is just a simple question. So, this is just based on the lens maker's formula for a double convex lens. Okay. So, they're asking the power versus you know NL function. So, basically the power formula will be the relative refractive index which is 1.5. 1.5 is the refractive index of glass.

[00:09:13]Divided by the refractive index of the liquid minus 1 times 2 by R. Okay, and R is given to be 20 cm. So, this will be 1/10.

[00:09:22]So, this is the function.

[00:09:24]So, yeah, to 1.5 the power will be zero.

[00:09:27]So, these two are wrong.

[00:09:28]And and it won't be a constant slope graph, so B is also wrong. So, the answer is A.

[00:09:33]Okay. So, that is it guys for this video. I'll bring the rest of the problem soon as well. That's it.

[00:09:39]Thank you.