Problems From JEE Adv 2026 | Paper 1 | Q13-Q16

Added: 2026-05-19

[00:00:00]Hello guys, welcome back. So in this video, we will cover problems 13 to 16.

[00:00:05]Okay, so basically the last four questions. Okay, so in this So the first question is basically we have we have four configurations and basically a sound wave whose wavelength is given enters these structures at point S and a detector is placed at D. So here we have the detector and here we have the source. The sound travels through only the tubes and list two contains the possible values possible smallest values of L for which the detector records maximum amplitude. So which means the path length So this is let's say if one of the path through which the sound wave travels and this is the second path. So the path difference will be lambda in this case, right? Because we have to consider the smallest value of L. So for the first case the longer path length is pi into radius, which is pi into 0.5 L. So that is pi by 2 into L and the other path length minus the second path length is L. So this is equal to path difference and that should be equal to lambda.

[00:01:00]Okay. So from here L is 2 lambda divided by >> [clears throat] >> pi minus 2. Okay, so this you know, after you calculate turns out to be 0.51. Okay, so the only challenging part in this question is calculation. So P matches up with three. Okay, so P3 and if you look at the options Q is just four. So let's just solve R now. So the first So this path length is just L. The other path length is L plus this pi r.

[00:01:27]Okay, so if these two are L the diameter is root 2 L. So this length is L by root 2, right? So the radius of the circular path is L by root 2. So the path length is L plus pi L by root 2 minus L. This is the path difference. This will be equal to lambda. So from here L is root 2 lambda divided by pi. So this turns out to be 0.13.

[00:01:50]Okay, once you calculate. So R matches up with five. P3 R5 P3 R5 only one option. D is the one that is matching. Okay. For this case, you'll have to use sine rule if you are calculating it. You'll have to use sine rule and figure out the other side lengths of the triangle. So, this is the next question. This is a match the list question. So, here phenomena physical phenomena of light are given and then they are talking about some theory that is explaining them. Okay.

[00:02:21]So, the first case, the colorful sky in the North Polar region. So, this particular phenomena, the aurora borealis. So, this is fundamentally like we can observe this event during the night time. Even during night time, right? So, this phenomena, of course, you know, even if you uh do it didn't study anything about it. Uh this has to do with some physical phenomena that results in light being generated, right? So, if you look at the options, option five is hinting at that fact, right? It's the emission of radiation from oxygen and nitrogen atoms excited by charged particles. So, here uh we have emission of radiation, right?

[00:03:00]So, this explanation is hinting at generation of light. So, so option P is basically match matched with five. And then rainbow is of course explained best as dispersion and reflection. So, it's basically like white light.

[00:03:16]Um so, first they internally reflect inside a rain inside raindrops and then they get dispersed, right? So, the best option is uh dispersion and reflection. And dark and bright fringes are observed in the diffraction pattern, right? And partially polar- So, there is only one option that matches, which is P5 S3. P5 S3 only A option matches. And of course, partially polarized sunlight is due to scattering of light by the atmosphere.

[00:03:44]Okay.

[00:03:46]All right. So, that is it for this one.

[00:03:49]So, here this is also a pretty standard question. We have to talk about the I versus T graph, right? And notice that the clockwise current, the clockwise current is what is depicted as the positive current. So, the in the graphs, the positive current is corresponding to the clockwise current. So, when the semicircular wire, this particular semicircular wire, when it enters, so so let's say after some time, a very tiny area of dA enters this particular magnetic field, so the extra flux through the semicircular loop of the magnetic field will be just B into dA, okay? So, the rate at which this area is entering the semicircle is entering the region is constant, right? So, basically dΦ by dt, if you observe, it is just B times dA by dt. Now, this dA by dt is a constant in this case because omega is constant, right? It is rotating at a constant rate, so every dt second, the same dA area enters. So, after another dt seconds, the new area that enters is also equal to dA itself. So, every dt seconds, the same dA area enters, so the rate of change of flux will also be constant, okay? This means the induced EMF is constant, uh hence the induced current is constant, okay? So, the graphs will be mainly horizontal lines when it comes to the current function, right? As you can see.

[00:05:18]So, now in the first case, so basically the in the first case, the graph is quite simple. T is the time period of the circular motion, it's the time in which the each of these loops cover 360°, right? So, after T by 2 time has passed, this semicircle will be fully inside the circular region. So, until this instant, um the flux through the semicircle will increase.

[00:05:42]So, flux increases T by 2, okay? And also the out the magnetic field lines in the outward direction, okay? So, by Lenz's law the induced EMF will try to oppose the outside electric flux. So, which means it will be in the clockwise sense, okay? So, when So, until T by 2 time when the semicircle is in this orientation, basically, the induced EMF will be clockwise or the induced current will be clockwise itself. So, which means the current will be positive.

[00:06:13]After T by 2 after T by 2, what will happen is the semicircle will start exiting, okay? So, now the flux will start to decrease, which means d5 by dt will become negative.

[00:06:26]So, now the induced current will be in the opposite direction. So, the graph in which the current is positive and constant until T by 2 and then negative is option three, okay? So, P matches with option three. So, if we look at the options, P matching with option three is only happening in these two options.

[00:06:46]So, Q is basically redundant. We don't have to solve it.

[00:06:50]So, let's solve for R.

[00:06:52]So, R is the 60° wire, right?

[00:06:57]So, same again, the logic is still the same. After dt time, dA area enters. So, So, the flux through the loop will increase and same clockwise current will exist, okay? So, initially So, until this 60° arc fully enters the region.

[00:07:16]So, how much angle did it cover?

[00:07:18]So, just look at the left radial line, okay? The left radius line, it covered 60° angle.

[00:07:24]So, which means this So, it enters the it fully enters the region after covering 60° angle. So, that happens after T by 6 time, okay? So, until T by 6 we will have a positive current. So, after T by 6, what will happen is the 60° arc is fully inside. Okay, fully inside the magnetic field region.

[00:07:46]Now, once it is fully inside, the flux is constant. Okay?

[00:07:51]So, after t by 6, flux is constant, which means there won't be any induced current now for some time.

[00:07:59]Uh until until the arc reaches this particular position. Okay? This particular position.

[00:08:05]After this position, it will start exiting, and now we will have a negative induced current, right?

[00:08:11]So, how much angle did it cover? Let's see. So, if you look at this radius line, so let me mark it with a different color. So, if you look at this radius line, it covered this much angle. Okay? This is the angle it covered, which is 120°.

[00:08:25]The flux remains constant from t by 6 till Okay, so it covered an extra 120° angle. 120° / 360 is 1 by 3. So, it stayed inside for another t by 3 time.

[00:08:39]So, if you add t by 3 to t by 6, this is basically t by 2.

[00:08:44]So, from t by 6 to t by 2, flux is constant. Till t by 6, there was a positive current. So, let's see.

[00:08:53]So, it looks like option one, right? So, let's say this is the t by 6 mark.

[00:08:59]So, until this t by 6 mark, current is positive, and then it suddenly became zero until t by 2.

[00:09:05]Okay, so for our option, it will be the first graph that matches. So, our matching with one, so the answer is C. Okay?

[00:09:14]Now, moving on to the final question.

[00:09:16]So, here we have to calculate the moment of inertias in these situations. Uh Mass of each rod and length of each rod is given. Okay?

[00:09:24]So, here the thing is the main pattern that is given is we have a we have an axis of rotation, and to this axis of rotation, we have some slanted line, or slanted rod, right? So, basically Let's say if this is the axis of rotation. All in all these situations, we basically kind of have a rod which is angled at some theta with respect to this angle with respect to this axis of rotation.

[00:09:51]So, let's So, the thing is you can quickly find out the moment of inertia of this rod in this situation.

[00:09:57]So, if I choose, let's say a tiny element, tiny mass dm on this rod, how do we calculate the moment of inertia?

[00:10:03]So, if this is my axis, I figure out the perpendicular distance from the axis.

[00:10:08]So, let's call this as So, in fact, I'm going to call this distance as x.

[00:10:13]Okay?

[00:10:14]Uh and this will be x sin theta, the perpendicular distance, right?

[00:10:19]So, the moment of inertia di due to this dm mass about this axis is going to be dm x squared sin squared theta.

[00:10:28]The sin theta, you can take it outside because it's just a constant, right? For all dm elements, it's going to be the same. So, if you want the total moment of inertia, it's simply going to be integral of dmx squared.

[00:10:39]So, if you ignore the sin squared theta, this is simply the moment of inertia about one of the end of the rod, okay? Moment of inertia about the end of the rod. So, this is just like calculating So, this formula is basically the same as this particular situation where we had a rod which was at a 90° angle to the axis, okay? And in this case, the moment of inertia was ml squared by 3.

[00:11:05]Okay?

[00:11:06]So, this particular dmx squared, if you evaluate it, it will simply come out to be um ml squared by 3.

[00:11:13]So, the only difference is if this is angled at a theta degree angle, uh we have an additional factor of sin squared theta being multiplied, okay? Uh you can put theta equal to 90° to verify it. If you put theta equal to 90, it will be ml squared by 3 itself. So, that is this case, right?

[00:11:31]So, ml squared by 3 sin squared theta, that is the formula. So, we are going to use that here.

[00:11:36]So, now the thing is in the first option the angle theta is just 45° and we have two such rods, okay? So, two for two of these rods times ml squared by 3 sin squared 45. Sin squared 45 will be half. So, this cancels out. So, the first option is just ml squared by 3. So, P matches up with five. So, P matching up with five happening in A, C and D. So, now let's say I try to solve Q. Okay, Q is this triangle.

[00:12:07]So, here the two rods are centered at 60°. So, for first let's write it for them.

[00:12:14]So, that will be two for both of these rods times ml squared by 3 sin squared 60. Sin 60 is root 3 by 2. So, sin squared 60 is 3 by 4. So, this is for the rods CA and CB. This rod is actually pretty easy to write because this is just parallel to the axis of rotation.

[00:12:32]So, each DM mass is at the same distance from the axis of rotation. So, the moment of inertia of the rod AB is simply going to be ml squared.

[00:12:42]Okay? So, what is this perpendicular distance?

[00:12:46]So, if the length of the equilateral triangle is L, this is going to be L root 3 by 2, right? L sin 60.

[00:12:53]So, that moment of inertia is m into L root 3 by 2 the whole square. So, ml squared into 3 by 4. So, three three cancels out. This becomes half.

[00:13:03]Half plus 3 by 4. That is 5 by 4.

[00:13:06]So, this becomes 5 by 4 ml squared.

[00:13:10]So, Q option matches up with one. Q one.

[00:13:14]So, P5Q1 is going to be option A. All right? So, that is it guys. So, that is it for paper one. So, these were all the questions. Now, we will take a look at paper two from the next video.