Problems From JEE Adv 2026 | Paper 2 | Q15-Q18

Hinzugefügt: 2026-05-21

[00:00:00]Okay, so hello everyone, welcome back.

[00:00:01]So, in this video we will cover the last four problems of of the paper. So, yeah, in this question we have a container whose height is two and length is two.

[00:00:10]Okay, so this is two by two and into the page the breadth is 1 m.

[00:00:14]It is made of an insulating vertical walls and two large area horizontal metal plates which extend beyond the vertical walls in all all directions.

[00:00:23]Okay, so this these metal plates are considered to be much larger than the dimensions of these containers.

[00:00:29]Uh and these containers are made of insulating vertical walls. Okay.

[00:00:35]It is partitioned into two equal chambers with a thin insulating vertical wall. The partition wall contains a small hole of area of cross-section square root of 10 cm squared uh near the bottom edge. Initially, the hole is closed and the left chamber of the container is completely filled with a liquid of dielectric constant 15.

[00:00:54]The right chamber is empty. Okay, so initially the So, basically the total height of this water is actually two 2 m.

[00:01:03]Um so, initially this left container was full.

[00:01:06]So, from this we can actually um observe something. So, this area of cross-section and this area of cross-section are both identical.

[00:01:12]Right? So, which means if um which means by volume conservation, if the total height of the liquid initially was actually 2 m, it will always be 2 m.

[00:01:24]So, if this height is h1, let's say, and if this height is h2, then we can also say h1 + h2 the sum will always be 2 m, okay, by volume conservation. Uh if the areas were different, then the then that equation will also change. But, yeah. At time t equal to zero, the hole is opened and the liquid flows from the left chamber to the right chamber.

[00:01:45]If in both the chambers the space above the liquid has epsilon r equals one and it is maintained at atmospheric pressure. Okay, so the air above is at atmospheric pressure. Acceleration due to gravity is 10. So, the first question is asking us the height of the left liquid. Uh so, this is going to be So, the height is going to vary now with respect to time. So, we need to find the height H1 at 500 seconds. And the second option is asking us to find the capacitance of the system. Right? So, uh of course, we need to begin by solving the fluid dynamics problem. So, in this uh particular problem, there will be a few assumptions that we are going to use. Okay? Okay, so first of all, it's obvious that uh fluid from the left container will flow into the the right container through the small hole, and the height of the right container will start to increase. Okay, so let's just say at some instant of time, the height of the liquid in the left container is H1, and the height of the liquid in the right container we'll say it is H2. And we have a small orifice at the bottom over here.

[00:02:50]And uh fluid from the left container will enter the right through this hole.

[00:02:56]So, the first thing is if you observe something, the area of cross section of this orifice, let's call it as A, uh and if the area of this container is capital A, this area A is significantly smaller than capital A. Right?

[00:03:11]So, capital A is in fact 1 m squared, and uh small A is given to be root 10 cm squared. Okay? So, the thing is So, the as a consequence of that, when fluid from the left container enters the right container, the volume of the fluid that is entering every second, if we call it as DV, this will be a very small volume.

[00:03:32]Okay? So, so if this is the fluid on the right, okay? We have some DV amount of fluid entering the fluid on the right at some speed V. So, now the thing is this is This we can consider as a large a significantly large compared to the volume entering. We can say that this DV volume is not going to disturb this V volume by that much, okay? Which means we can consider this fluid of volume V to be in static situation, okay? But if this hole was like very large, okay?

[00:04:06]Then we cannot say that. Then a large amount of fluid will enter per second.

[00:04:11]Okay? And then the flow here will be turbulent, okay? And the fluid on this container will be completely disturbed.

[00:04:18]And we'll have like waves on the surface of this fluid and all, okay? But because this orifice is is very tiny compared to this area of cross section, we can say that the tiny DV volume entering every second is not going to disturb the fluid on the right, and we can consider this fluid to be approximately in a static situation.

[00:04:39]Even though strictly speaking, the moment the fluid enters this vessel, it will come to rest after traveling some distance, okay? And you can think of the fluid particles that are incoming with a velocity V colliding with colliding with the fluid particles that are at rest in this location. And there is some sort of an elastic collision happening here, okay? Okay, so actually what will happen is this is the fluid volume DV entering with some velocity V. It will come to rest after traveling some distance, okay? And and now as we are considering this um body on the right, the fluid on the right to be approximately in a static situation, we can consider the pressure very close to this orifice to be the static pressure due to the liquid on the right. So, if the pressure here is P naught, we can consider the pressure here to be P naught plus rho g h two, okay? Uh once again, this is strictly because we're assuming the situation on the right to be static, okay? And that assumption is more correct the smaller the orifice becomes. So, now what we'll do is we'll consider the flow in the left container to be a laminar flow. So, we'll consider a streamline something like this connecting the free surface of the first liquid and a point very close to the orifice. Okay? So, near the orifice we'll consider the pressure to be the static pressure due to uh the right fluid. Okay? And this is kind of an approximation because actually, you know, very close to the the situation is not a static situation because the because the fluid is still kind of moving. But, because the volume DV is very small of the entering fluid, it's not going to disturb the large fluid body that much. So, we can consider the situation to be approximately static.

[00:06:23]Okay? So, hence we can write the pressure here as P naught plus rho g h2.

[00:06:28]So, now we can apply Bernoulli's theorem in this streamline between points 1 and 2. So, at location 1 the pressure is P naught.

[00:06:36]And if I consider the zero PE level to be this horizontal level, the base, then the potential energy term will be rho g h1. And because the area and because this area is significantly larger than this area, we can consider V1 to be significantly small compared to the velocity of efflux at this location.

[00:06:55]So, the velocity of efflux I'm considering it as V here. Okay? So, the kinetic energy term will be zero. And for the right hand side, the pressure we'll use the static pressure due to the right fluid. So, that'll be P naught plus rho g h2. And the kinetic energy term will be half rho V squared and the potential energy term will be zero. So, from here P naught cancels out and we get V squared as 2 g h1 minus h2. Now, h1 minus h2 is actually the difference the height delta of the two water surfaces. Okay? This height difference between the two water surfaces is h1 minus h2. This we'll call it as some h.

[00:07:30]Okay? So, height h is h1 minus h2.

[00:07:34]Earlier we also noted the fact that h1 plus h2 was 2 m. We can use these two relations. So, the V squared term will become 2 g h.

[00:07:42]We can create some differential equation on how H varies with time. Even though the the original question is what is height on the left chamber, so that will be H1. So we need to find H1, but you know, the differential equation H is going to be a little bit more cleaner.

[00:08:00]So now what we can do is we can use continuity equation. So we said the height of the first surface is at H1. So the rate of change of this height uh so this height is going to decrease.

[00:08:12]So we can call this magnitude as H1.

[00:08:14]Right? Okay, which is like DH1 by DT.

[00:08:17]So H1. is a rate of change of height in the first container.

[00:08:22]If you multiply that with the area of the first first container, right? Both of their areas are the same.

[00:08:27]So if you multiply it with area, this is the rate of change of volume of the fluid in the first container.

[00:08:33]And this will be equal to Okay, so fluid is exiting the first container, let's say at a flow rate of Q. So Q we can write it as area of the orifice times the efflux velocity. So water in the first container is exiting at a rate of Q per Q meter cubed every second.

[00:08:51]So this H1. into A will be minus Q.

[00:08:54]Okay. Similarly, water is entering the right container at a rate of plus Q. So its volume change, so that will be H2.

[00:09:02]times area.

[00:09:03]This will be equal to plus Q. Okay. So now Q we can write it as So if I take the first equation, we can say H1. into area equals Q is small A into efflux velocity, which is root 2G H. Now H1 we need to replace with H, so we can use this relation. Right? So H. is equal to H1. minus H2.

[00:09:26]Uh and H2. from here is simply minus H1.

[00:09:29]Right? So this will be minus H1.

[00:09:32]So this is actually 2H1.

[00:09:34]So instead of H1. we can use H. by 2. H.

[00:09:37]I'm going to write it as dh/dt divided by 2. So area by 2 is equal to small a times root 2gh. So there was a negative sign here as well. Now So now we can do variable separation. So dh divided by root h equals minus 2 small a by capital A root 2g times dt. So now we can integrate. So at t equal to 0 the height h was 2 m, right? The separation between the two surfaces was 2 m.

[00:10:07]And at t equal to t we need the height h. So this will be 2 root h after integration. So this is 2 root h minus root 2 and the right hand side is 2 a by capital A times root 2g into t. So 2 cancels out and we have square root of height h equals root 2 minus small a by capital A root 2g into t. Okay, so now small A is root 10 cm squared and g is 10 divided by A was 1 m squared into t. Okay, so now we need the height at t equal to 500 seconds. Okay, so root h is root 2 minus So this will become 10 into 10 to the power minus 4. So root 2 t by 1000.

[00:10:52]Okay. Now t we will set it as 500. This will become half. So this is root 2 by 2 on the right hand side.

[00:11:01]So the height h, if you square it will become half meters.

[00:11:04]Okay, now if h is half meters, we needed h1. So we'll add these two equations. So 2h1 is equal to 2 plus h and h was half meters. So this will become 2.5 meters.

[00:11:17]So h1 is 1.25 meters at the 500 second mark. Okay. So this answer will be 1.25 meters. Now once we have that, we need to calculate the capacitance in the next case. So for that in the first situation, this this first container was fully filled with the dielectric liquid. Okay?

[00:11:40]And the in the other container, it was fully filled with basically air, right?

[00:11:44]epsilon r equal to 1, here epsilon r equals 50.

[00:11:47]So, these were connected across two metal plates.

[00:11:52]The potential difference across these two dielectrics will treated as equal. So, which means we'll consider these two as parallel, right? So, this is just the simple parallel formula. So, we can say the equivalent capacitance epsilon r epsilon naught area of cross-section was 1 m squared divided by d was 2 m, right?

[00:12:10]This height is 2 m. Plus for the other one, it is epsilon naught a by d.

[00:12:16]So, this is going to be 8 epsilon naught. Okay? Now, in the next situation, the height of the liquid on the left container was 1.25 m, so it is slightly greater than 50% and on the left container, it will be slightly less than 50%. So, basically this height was 1.25 m and this height will be 0.75 m. Similarly, over here this will be 0.75 m and this will be 1.25 m. So, this is going to be the fluid. This is going to be the air part. Similarly, over here this is going to be the air part and this is going to be the fluid part. And between them, we have the insulated wall. Okay?

[00:12:57]So, we will consider these two in series and calculate their equivalent capacitance. And then we'll consider these two in series, calculate their equivalent. And then just we'll come and then finally we'll add them together as a parallel. Okay? So, for the part on the left, we can say 1 over c1 is equal to 1 over the capacitance of this part, this dielectric is epsilon r 15 epsilon naught area of cross-section is area is all area is 1 m squared itself. Divided by D and D is 1.25. So, 1. 1.25 I am writing it as 5 [snorts] by 4 plus 1 upon epsilon naught.

[00:13:38]Uh and the other one is 1.75.

[00:13:41]The other is 0.75 which is 3 by 4, right? So, times 3 by 4. So, here 5 and 15 cancel out and we get 1 by 12 plus 9 by 12, so 10 by 12. So, C1 is 12 epsilon naught by 10, which is 1.2 epsilon naught. And in the other case, 3 by 4.

[00:14:05]This is the D times 15 epsilon naught plus the other D is going to be 5 by 4 times epsilon naught. So, this cancels out.

[00:14:15]So, 26 by 20 epsilon naught.

[00:14:19]So, C2 will be 10 by 13 epsilon naught.

[00:14:24]So, effectively the capacitance of the system is C1 plus C2.

[00:14:28]That is 10 by 13 plus 1.2 epsilon naught.

[00:14:34]Now, in the question what they asked was the difference in these two cases is written as 8 minus n epsilon naught.

[00:14:41]Now, initially if you observe the capacitance was 8 epsilon naught.

[00:14:45]So, basically this n epsilon naught is just the final capacitance, right? So, what they are asking essentially is the final capacitance. So, we just need to calculate this value because that will be the answer.

[00:14:59]So, this after calculation turns out to be 1.969.

[00:15:03]So, I guess rounded off this will be 1.97, okay? Epsilon naught. Okay, so the answer will be 1.97 I guess for the next one.

[00:15:13]>> [snorts] >> So, now coming to the last question. So, here we have a uniform circular disk whose radius is 0.2 m and mass is 1 kg.

[00:15:21]It's pivoted at the top point C such that it can rotate freely around C. So, this is the hinge. Initially, when the disk is at rest, it is a particle of mass 20 g traveling along the negative X direction in the XY plane. Initial speed was 100.

[00:15:36]It hits the circumference at the point P and after collision it moves along the negative Y direction. Okay? So, we can calculate its initial momentum and final momentum. So, using that we can get the impulse exerted by the force on the particle, right? By the disk.

[00:15:52]Uh so, we'll start with that. So, yeah, [snorts] the question and the main question is um after the collision, the disk starts to rotate about point C in the XY plane.

[00:16:01]What is the maximum change in height of its center? Okay? So, the max change of height is pretty simple. So, let's just say uh in the max height situation, the disk comes to rest something like this.

[00:16:12]Uh the disk is in some situation like this. So, this rise in the height is basically this maximum height change happens when the disk comes to rest. So, in the final situation, we'll just say the omega final is zero.

[00:16:24]That's when the height of the center of mass will be the greatest from the original location. Okay?

[00:16:30]And the second case, we have to find the amount of energy loss. So, that will be just the initial energy which is 1/2 mv 1/2 into m into 100 squared minus the total final kinetic energy of the system. Okay? So, let's do that. So, let's start.

[00:16:46]So, the first thing is if you observe the tiny mass, it has a leftward momentum initially and then and then of course the disk will exert some force which will have two components, one along the X direction and the other along the Y direction. And this force will exert some impulse on the small m mass and that impulse will change the momentum, right? And if you observe the initial momentum is along the left, that got completely killed. The final momentum is in the minus Y direction.

[00:17:16]So, we can just calculate the impulse on the x direction pretty easily. So, there was some nx force acting on the smaller mass.

[00:17:24]The impulse of this force jx will be just the change in momentum in the x direction.

[00:17:30]So, initial momentum was m into 100, right? So, that will be the impulse.

[00:17:35]So, 20 g * 100. So, this will be 2 N second. So, this is the x direction impulse that was exerted on the mass m.

[00:17:43]Similarly, there was a y direction impulse that was exerted in the downward direction and this y direction impulse gave it a y direction velocity. And that y direction impulse's magnitude, we can say it is So, the y direction speed is 90 m/s. So, it will be mass into 90. So, this is actually 1.8 N second.

[00:18:05]Okay, so So, on the m mass, the there was an x direction impulse jx and a y direction impulse jy uh that was exerted by the disk. Okay, by the force n by the disk.

[00:18:20]So, the impulse that was exerted by small m on the disk is just the opposite of this by Newton's third law, right? So, now if we take a look at the disk and let's also draw the location of the point at which the impulse is exerted. So, at this location, there will be an impulse to the left jx and an impulse along the plus y direction. Okay, so jx on the particle was on the right. So, on the disk, it will be on the left. jy on the particle was downwards. So, jx jy on the disk will be in the upward direction. Okay, so So, now this is the hinge.

[00:18:56]Uh this distance is actually r by root 2 because this angle is 45°.

[00:19:01]So, this distance will be r by root 2.

[00:19:04]And this vertical distance is r by root 2 plus r. Okay, so now we can calculate the angular impulse on this disk by this force, okay, about the hinge, okay? So, the net angular impulse about the point O.

[00:19:19]So, if you observe, due to this JX, the angular impulse is clockwise about O.

[00:19:25]And due to this JY, the angular impulse is counterclockwise. So, okay, so clearly JX's angular impulse component will be greater because JX is 2 N second and it is also at a larger distance from the axis.

[00:19:38]So, so the net angular impulse will be in the clockwise sense. So, the net angular velocity will also be in the clockwise sense.

[00:19:46]So, we can say JX, which was 2 N second, times R times 1 by root 2 plus 1. JY's component will be negative, right? So, minus JY was 1.8 times R by root 2. This will be equal to I omega, right? Change in angular momentum.

[00:20:05]Moment of inertia about O is 3 by 2.

[00:20:08]Mass of the disk was capital M, um, was 1 kg into R squared, right? So, into R squared times omega. So, from here we get Okay, so basically the right-hand side is the final angular momentum of the disk. So, from here, if we cancel out an R, omega becomes root 2 plus 2 minus 0.9 root 2 times 2 divided by 3.

[00:20:37]R, and R was 0.2 m. So, after the calculation, this becomes 7.138 rad/s.

[00:20:46]So, rounded off to two digits, this becomes 7.14 rad/s.

[00:20:51]>> [snorts] >> Okay, so now, what we'll do is we'll do energy conservation just for the disk.

[00:20:58]So, the disk for the disk, the initial kinetic energy, half I omega squared, will be completely converted into the mg delta h change in energy.

[00:21:09]When center of mass is at the highest point, I is 3 by 2, m is 1, radius is 0.2 whole square times omega square.

[00:21:19]And this thing divided by mg. Okay, m is actually 1 kg, g is going to be 10.

[00:21:25]This will give us the height h.

[00:21:27]Okay, and h comes out to be 0.1528 m.

[00:21:33]Which rounded off will be 0.15 m. And now the thing is uh So, in the next part we have to calculate the energy loss. So, the energy loss in collision will be the initial kinetic energy of the mass, which is half small m v squared minus the final kinetic energy of the system as a whole.

[00:21:54]Okay?

[00:21:55]So, this will be half 20 g times speed initially was 100, right?

[00:22:02]So, 100 squared is 10 to the power 4.

[00:22:05]Minus the final kinetic energy of the system. So, finally for the mass it is half into 20 into 10 to the power minus 3.

[00:22:13]And its speed was 90 m/s. So, 90 squared will be 81 into 10 squared.

[00:22:19]And for the disk it will be half I omega squared.

[00:22:23]Which is actually just mg delta h, right? So, we can use mg delta h instead.

[00:22:28]So, minus m is 1, g is 10. So, just multiply this with 10.

[00:22:34]So, it will be 1.528 minus 81 minus 1.528.

[00:22:43]So, this turns out to be around 17.

[00:22:46]Okay, rounded off, I guess, comes out to be 17.47 J.

[00:22:54]Okay, I guess they will give some range for this, but yeah.

[00:22:57]So, this is what I'm getting. So, yeah, that is it for this paper, guys. So, we have discussed all problems. If you have any doubts, you can ask below.

[00:23:06]And if you enjoyed, make sure to like, share, and subscribe. That's it. Thanks for watching.