Mistake in Allen AIOT 2026 Qn ? Best of Allen AIOT 2026 | JEE Advanced 2026

Added: 2026-05-11

[00:00:01]Let us look at some good questions from this year Allen All India test series.

[00:00:05]Okay. And also we'll see a small mistake in this question after we solve. Okay.

[00:00:12]So I'll just briefly about the question.

[00:00:15]There are two dumbbells here. Masses connected to springs.

[00:00:20]And they're coming in hitting each other. Okay. After hitting each other they compress. Okay. What is asked is what is the time at which you know the spring potential energy is maximum.

[00:00:34]Okay.

[00:00:35]Let us first understand what is happening.

[00:00:37]So this one is coming this side and this one is coming this side. Both are coming towards each other with velocity V0.

[00:00:45]Okay.

[00:00:47]When they strike each other this anyway there is no extension or compression in the spring. This velocity remains V0.

[00:00:56]Okay. And because both masses are striking they'll exchange their velocities. This velocity becomes V0.

[00:01:04]This velocity becomes V0 in this direction and this becomes V0 in this direction.

[00:01:09]And SHM motion starts.

[00:01:12]Okay. Now what happens? Both the springs compress. Okay. And the velocities final velocities become zero.

[00:01:22]And again they start you know expanding.

[00:01:25]When they expand the speed of this is in this direction V0.

[00:01:32]Okay. Speed of this is in V0 in this direction.

[00:01:37]Now the speed of this is in this direction V0 and this is towards left.

[00:01:43]Again what will happen because they collide again they exchange their velocities. Okay. So this becomes V0 and this becomes V0 in this direction.

[00:01:55]Okay.

[00:01:56]And finally what happens they move away.

[00:01:58]This goes away we know the direction.

[00:02:05]What he is asking is what is the time taken from here till the spring potential energy becomes maximum.

[00:02:13]So he has given us that this distance is 2 pi meters.

[00:02:21]Okay, so the time is first the time for them to come here. Plus the time for it to compress till here.

[00:02:29]Okay, so total time T naught equal to what is the time for these both to come here?

[00:02:38]So 2 pi divided by 2 V naught right? This is coming with V naught and this is coming with V naught.

[00:02:44]Relative velocity is V naught.

[00:02:47]2 V naught plus the time taken for this to go from here to here.

[00:02:54]Now just see this mass it's it has it is in the maximum velocity. So we can say it is in the mean position.

[00:03:01]And from here to here it has come to the extreme position.

[00:03:03]So from mean to the extreme if the time period of this oscillation was T.

[00:03:08]So from here to here it will take time T by 4. Okay.

[00:03:13]So T by 4 1 by 4 and we know that for two masses which are oscillating the time period of this oscillation is 2 pi under root mu by K. Mu here is both have same mass M.

[00:03:30]So M by 2 K.

[00:03:35]So this is the time. We just solve this so we get T naught equal to pi by V naught plus plus 2 pi by 2 under root this M by 2 k is 1 k m is 1 kg and the k he has given here as 0.5 Newton per meter.

[00:03:59]So, this comes out to be 1.

[00:04:02]And this v naught here is also given 1 meter per second.

[00:04:05]So, we get this t naught equal to 3 pi by 2.

[00:04:11]And yes, so from this we can say k equal to the value of k is 3.

[00:04:16]Okay.

[00:04:18]Now, what is the mistake in this question?

[00:04:21]He has said here that the spring potential energy will be maximum for the first time.

[00:04:30]The spring potential energy becomes maximum after that they both go away.

[00:04:35]Okay. So, there is no second time, third time that the spring potential energy is becoming maximum. There is only one time.

[00:04:43]So, when he's when he's saying here first time, okay, that means he he has given a hint that there could be the second time, there could be a third time.

[00:04:52]So, that is a small mistake in the framing of this question.

[00:04:57]Let us look at another question from this year's Allen All India Test Series paper.

[00:05:03]So, let me explain you the question.

[00:05:08]Okay.

[00:05:09]So, there are two plates metal plates here and this is connected to a positive terminal and that is connected to the negative terminal of the battery.

[00:05:17]And so, these both plates form a capacitor.

[00:05:20]There are um mini spheres, okay, which are going back and forth between the plates.

[00:05:27]So, when a sphere comes here, it gets stuck to this, takes positive charge.

[00:05:33]And because of the positive charge, it gets repelled and it goes here.

[00:05:39]Okay.

[00:05:40]After this, it gives away the positive charge, gets a negative charge and again is repelled by this and comes back here.

[00:05:47]And you know, so to and fro these spheres are moving in this direction in this direction. Okay.

[00:05:55]So, let us move on to what is the question that is asked here.

[00:05:58]So, the first question is the current between the plates in terms of the given parameters. Okay. So, what is current?

[00:06:06]Tell me. I equal to n a v d So, we know this already.

[00:06:16]And what is this n? n is the number of charges per unit volume.

[00:06:20]So, I can write this n as n naught by What is the volume of this? Let me The area of this is A. The separation is D.

[00:06:32]A into D.

[00:06:34]Okay.

[00:06:35]Now, this number of charges per unit area is given as n naught.

[00:06:42]I can just write this as n naught by We are just writing in terms of the quantities which are given in the question.

[00:06:52]Next, charge is given as Q. Area we have capital A.

[00:06:57]Now, what is the drift velocity?

[00:06:59]See, these spheres are getting repelled because of the electric force.

[00:07:04]So, here what happens is they are at rest.

[00:07:07]They get repelled. They go here.

[00:07:09]So, that means this is a uniformly accelerated motion. Okay.

[00:07:13]So, what we can write here is that this v d drift velocity is average velocity.

[00:07:21]But here it is going from zero to some maximum velocity.

[00:07:24]So, we can write this We'll write this maximum velocity as Q into V equal to half m v squared.

[00:07:32]The work done by electric force is getting converted to kinetic energy.

[00:07:36]So, what is V?

[00:07:38]This is speed, okay? This will write this as small v.

[00:07:43]2 Q V divided by This is the maximum speed.

[00:07:50]In a uniformly accelerated motion speed is going from 0 to V. What is the average speed?

[00:07:57]So, the V average equal to V1 + V2 by 2. So, this will be V average becomes QV by 2m.

[00:08:10]Okay?

[00:08:12]V average is nothing but the drift speed.

[00:08:15]So, we'll just put this in the drift speed. Okay?

[00:08:18]So, we'll get I equal to N is nothing but N0 by D times Q into area. The drift speed is under root QV by 2m.

[00:08:37]Okay?

[00:08:38]Now, here you see we have QV by 2m.

[00:08:41]And here we have N0 Q by epsilon not.

[00:08:44]So, basically he has given the capacitance as C.

[00:08:47]So, what can we write? Tell me. I can just write I equal to See, I can just divide by epsilon not and into epsilon not. This E epsilon not by D I can write as C.

[00:08:58]So, N0 QC divided by epsilon not under root QV by Okay?

[00:09:10]So, if you compare it with this, what is the value of K? Tell me. The value of K equal to equal to >> [clears throat] >> There is another question. He's asking, "What is the effective resistance of this device?"

[00:09:25]So, the effective resistance we can say R equal to V by I.

[00:09:30]Okay? We've already found out I. Okay?

[00:09:33]And just put V by I, you'll get what is the answer.

[00:09:37]So, for this you directly get K equal to root that is equal to 1.41 That is the value of K in this answer.