9702/12/F/M/26 | AS PHYSICS PAPER 1 FEB MARCH 2026 | #alevel | #physics | #YTClassroom

Added: 2026-06-03

[00:00:00]Hello guys and welcome back to YT Classroom. I hope you all are doing well. Today we'll be solving AS level physics 9702 paper 1 2 from the Feb-March 2026 series. This is a 1-hour 15-minute MCQ paper consisting of 40 marks. And this is one of the three papers essential for the AS level physics qualification and one of the five papers for the A level qualification. If you're new to this channel, do not forget to like, share, and subscribe for more helpful content.

[00:00:32]Let's get started. Calculators are allowed in this paper, so do not forget to take your one right now before we start solving.

[00:00:39]And yep, let's get started.

[00:00:42]So, there is a certain data values and formulas that are given in this paper, so I would advise you to get familiarized with these formulas before the exam itself.

[00:00:51]Okay, first question. What is the value of 1 km over 1 nm?

[00:00:57]First of all, let's write 1 km in terms of the in terms of meters. So, 1 km is basically 1 into 10 raised to 3 m.

[00:01:06]And nanometers, nanometers is basically 1 into 10 raised to -9 m.

[00:01:13]Now, you can simply put this in the calculator. So, 1 into 10 raised to 3 over 1 into 10 raised to -9 will give you 1 into 10 raised to 1 into 10 raised to 12. Therefore, the value would be D. I hope that is clear and understood.

[00:01:27]Second question. What could reduce systematic errors? Now, there are two types of errors in AS physics. There is systematic errors and there is random errors. Systematic errors are caused by errors in the equipment or your measuring tools. Whereas, random errors do occur due to um you know, difference in the response times or taking incorrect measurements, etc. So, in order to reduce systematic errors, what you could do is you could ensure that your measuring equipments or instruments are calibrated carefully so that you don't get the wrong reading from them.

[00:01:58]All the other options, such as averaging a large number of measurements, or repeating the measurements, or reducing the sample size, well, this would not be possible, but averaging a large number of measurements and repeating the measurements could reduce your random errors.

[00:02:12]Moving on to question number three.

[00:02:14]A student measures the current and the potential difference for a resistor in a circuit.

[00:02:19]The measurements are used to calculate the resistance of the resistor. What is the percentage uncertainty in the calculated resistance? Okay. Now, we know the formula that potential difference V is equal to IR, and therefore the value of R is going to be V over I.

[00:02:34]So, V over I, the value we have 500.0 plus.1, plus or minus.1, divided by 50.00 plus or minus 0.01. Now, when you're dividing two values and you want the percentage uncertainty, what you could do is you could just add up the percentage uncertainty of both the values. That's going to be 0.1 over 500 * 100, which is for the percentage uncertainty of the numerator, plus 0.01 over 50 into a 100. Let's use our calculator to work this out. So, 0.1 over 500 into 500 into 100 plus 0.01 over 50 into 100.

[00:03:22]That would be 0.04 percentage, so the answer would be 3D.

[00:03:28]Graphs can be used to represent the motion of an object. Which relationship between the quantity of a motion and the property of a graph is correct?

[00:03:37]First option is acceleration is the area under the velocity-time graph. No.

[00:03:41]Area under the velocity-time graph is actually the displacement.

[00:03:45]Acceleration is the gradient of the velocity-time graph. Yes, this is indeed correct. Acceleration is the gradient.

[00:03:51]Okay.

[00:03:52]Moving on question number five.

[00:03:55]The water surface in a deep well is 78 m below the top of the well.

[00:04:00]A person at the top of the well drops a heavy stone down the well.

[00:04:05]Air resistance is negligible and the speed of the stone in the air is three The speed of sound in the air is 330 m per second.

[00:04:14]What is the time interval between the person dropping the stone and hearing it hit the water?

[00:04:19]So, we have a well and we have a person who's dropping a stone from this point over here. Now, in the well, there's obviously water and when the stone hits the water, it makes a sound back which travels up. They're asking us to find the time between when the person dropped the stone and when it actually when he heard the sound back.

[00:04:39]First, let's figure out that how much time will it take for the stone to reach the bottom of the well using SUVAT. So, SUVAT, S is for displacement, U is for initial velocity, V is for final velocity, A is for acceleration, and T is for time.

[00:04:53]The distance that the stone has to travel is 78 m. The initial speed is zero. Final speed is not given to us and acceleration is going to be 9.81 because it's the acceleration of free fall.

[00:05:03]Now, from this, what we're interested is in time T, so we can say that S is equal to ut + 1/2 at squared. Now, because U is zero, I could just completely cancel this out. S is equal to 1/2 at squared.

[00:05:16]t squared would be 2s over a.

[00:05:21]And the value of t would be the square root of 2s over a.

[00:05:26]All right. So, what I'm going to do is um let's write this down over here.

[00:05:30]Since two t is two times the displacement, displacement was 78 m.

[00:05:36]2s divided by the acceleration, that's going to be 9.81.

[00:05:40]Let's find this value out. Square root of two times 78 over 9.81.

[00:05:45]That's going to be um 3.

[00:05:49]3.987748823.

[00:05:55]All right. Now that this stone has actually hit the water, it must create the sound. How long does it take for the sound to reach the top of the well again? So that now we know the speed of the sound is 330 m/s. So for this, if I do the calculation on different pen, we know that speed is equal to the distance or the displacement over the time. Therefore, the time is equal to the displacement S over the speed V. That would be the displacement is 78 m. The speed is 330.

[00:06:26]So this would be 78 over 330. That's going to be 0.236 36 followed on. So when you add this value, 3.98, and to this value, you would get your final answer. This plus 3.987748823.

[00:06:44]The answer should be 4.22 seconds.

[00:06:46]Therefore, the correct answer for question number five is indeed D. And with that, we are done with the first five questions. The answers are D B D B D. D B D B D. Moving on to question number six.

[00:07:01]It says that a toy car moves in a straight line. The graph show the variation with the time of the displacement of the car from the initial position.

[00:07:09]So from zero until this point, it has not displaced. That means there is no velocity.

[00:07:14]From this point until this point, the graph the graph has a constant gradient.

[00:07:19]Hence, it has a constant velocity.

[00:07:23]Constant velocity.

[00:07:25]All right. Now in this particular section, the top section of the graph, it is stationary because there is no change in the displacement.

[00:07:34]And from this time, it is actually going to be again constant velocity because the gradient of the graph is constant. Let us read the question. What do they actually ask? A student investigates whether the forces acting on the car are in equilibrium at four different times, T1, T2, T3, T4.

[00:07:52]At which times are the forces acting on the car equilibrium?

[00:07:55]Now, I know that at time two, the forces would be equilibrium because the object is moving at a constant velocity. If there is a resultant force that, then it would not be a constant velocity. If there would be acceleration. But because there is constant velocity, T2 has a the the car is in equilibrium over here.

[00:08:12]And the same thing would apply at T4 as well because again constant speed, so the car has no resultant force.

[00:08:18]Therefore, it is in equilibrium. So, T2 and T4 are confirmed. Now, what about T1 and T3?

[00:08:24]To be honest, at T1 and T3 also the car is at equilibrium. Now, when an object is at stationary, there is no resultant force on the object. That is the only reason which is why it is a stationary.

[00:08:34]If there would be have been a resultant force, the object would start to move.

[00:08:38]So, it is stationary, hence there is no resultant force, hence it is also in equilibrium. So, it is in equilibrium at all four times. Therefore, the correct answer would be 6A.

[00:08:47]All right, I hope that's clear. Moving on to question number seven.

[00:08:51]A car travels anticlockwise along a horizontal circular road of radius 12 m as shown. The car takes a time of 4 seconds to move from position P to position Q.

[00:09:02]What is the magnitude of the average velocity of the car from of from the journey from P to Q?

[00:09:08]Now, velocity is actually displacement over time. Do not mistake this for distance over time.

[00:09:15]That is for speed. And remember, displacement is the shortest distance from your starting point to the ending point.

[00:09:21]Now, in this question, we know that the car is actually traveled along this path. But the shortest distance between these two parts is right over here.

[00:09:29]Okay, so first let's find the length of this particular distance.

[00:09:33]Now, this is going to be 12 m, so we We use Pythagoras. 12 squared plus 12 squared the whole root. This distance is 12 root 2.

[00:09:42]Okay? Now, velocity is again displacement over five times. So, displacement is 12 root 2. And the time is 4 seconds. So, it's going to be 12 root 2 divided by 4 seconds.

[00:09:52]That's going to be 4.24.

[00:09:54]So, we're going to select 7A, 4.24.

[00:09:58]I hope that is clear.

[00:09:59]Moving on to question number eight.

[00:10:02]A person is suspended from a rope vertically below a helicopter. The helicopter and the person are stationary above the ground.

[00:10:10]Which force forms a Newton's third law of force pair with the weight of the person?

[00:10:16]Okay. Now, the weight of the person, what exactly does it mean? It is the Earth's pull, the Earth's or maybe you can say the Earth's gravitational force or the Earth's pull on the object.

[00:10:31]Now, one thing that you need to know about the Newton's third law pair is, right? If a force acts on object A, if a force acts on object A by object B, the Newton's third law pair would be the object B experiencing the same force from object A.

[00:10:45]So, if I have one force, which is the Earth pulling on the person, the Newton's third law pair would be the person pulling on the Earth. The force is of a very It is of equal magnitude, but because the Earth is so huge, the person can't actually make any impact on the Earth.

[00:11:02]However, in Newton's third law, if A is applying a force on B, then B applies a force on A. So, Earth pulling on the person, the person pulling on the Earth is a Newton's third law pair. So, you would say that the gravitational force applied by the person on the Earth.

[00:11:18]Therefore, 8C is the right answer. You would learn more about this in the first topic, which is gravitation of A-level physics.

[00:11:27]Question number nine. Two small moving particles collide. After the collision, the particles are moving in opposite directions. What is essential for the total momentum to be conserved in the collision?

[00:11:38]Now, if you remember the law of law of conservation of momentum, which says that the total momentum before the collision is equal to total momentum after the collision, which is asked in your paper, too. It always mentions an isolated system, which is a system where there is no external force acting. And this is really important because if there's external force acting, then the momentum is unlikely to be conserved.

[00:11:59]So, it is very essential that there's no external force acting.

[00:12:03]No kinetic energy is lost. This is not correct because you could have an inelastic collision where the momentum is conserved, but the kinetic energy will be lost.

[00:12:12]Collision being inelastic, also the momentum is conserved.

[00:12:16]The relative speed of approach is equal to relative speed of separation. This is the case for elastic collisions, but here also momentum is conserved.

[00:12:23]So, what is essential is that no external force acts on the system of the two colliding particles.

[00:12:30]Okay.

[00:12:31]Moving on to question number the answer for nine is A. Let's move on to 10.

[00:12:36]Block one of mass 3 kg is on a horizontal frictionless table and a rope with negligible mass is attached to the block.

[00:12:45]The rope passes over a frictionless pulley. The block two of the mass 1 kg suspended from the other end of the rope as shown.

[00:12:52]The blocks are accelerating. The acceleration of free fall is G.

[00:12:56]The unit of G is meters per second squared and the and the tension the rope has unit Newtons.

[00:13:01]What is the tension in the rope in terms of G?

[00:13:05]Okay, let's understand this question carefully. You have two blocks that are connected together and the blocks are actually accelerating. So, it is likely to accelerate in this particular direction.

[00:13:14]Now, let's consider each of the blocks individually. If I consider the second block first, in this block, there is two forces acting. There's a tension force and there's a weight force.

[00:13:24]Now, because the mass of the block is 1 kg, the weight force would be W is equal to mg, but m is 1 so W is equal to g.

[00:13:33]So, I'm going to say that the weight force, which is g, minus the tension force is equal to the mass, which is just 1 kg, times the acceleration. I've just applied f is equal to ma.

[00:13:47]The resultant force is the gravitational pull, which is the weight, minus the tension force, is equal to acceleration a for the first object.

[00:13:55]Now, for the first object, the only pulling force is the tension force, which will actually pull it downwards towards the pulley. There's no weight force going to act over here because the object is moving in a horizontal direction.

[00:14:06]So, for the first object, I can say that my tension is equal to the mass. Now, this time the mass is 3 kg. So, I'm going to say 3 times whatever the acceleration is, which is a.

[00:14:18]Now, between these two equations, you need to understand that the acceleration will be constant because both of the blocks are attached together. This would be This would be the mechanic students, but again, it's very easy that if the block is accelerating from here, it would pull this block by the same acceleration.

[00:14:32]The tension force is also the same for both the blocks.

[00:14:35]Okay. Now, let's kind of work with the equations.

[00:14:38]The second equation, I can write it down as this.

[00:14:41]3g - 3t is equal to 3a.

[00:14:46]And the first equation I have is again t is equal to 3a.

[00:14:50]The reason I've multiplied the second equation is so that I can perform a substitution now.

[00:14:54]Now, rather than writing 3a over here, I can write t over there.

[00:14:58]So, I can write the equation for the second one as 3g - 3t is equal to t.

[00:15:05]And therefore, 3g will be equal to 4t.

[00:15:09]Remember what you were looking for is the tension in terms of g. So, tension will equal to 3 by 4g.

[00:15:15]Therefore, the correct answer is actually 10c. Let's take a look at question number 11 now.

[00:15:21]The mass of an object can be determined by its relationship to other physical quantities. The relationships are shown.

[00:15:27]Which relationship is always equal to the mass of the object? So, basically, they have given us three relationships, and they want us to check that which relationship will give us which will always give us the mass.

[00:15:38]The first relationship, resultant force on the object divided by the acceleration of the object. Now, this is referring to the formula which is called as F is equal to MA.

[00:15:48]So, if I do M is equal to F over A. And here, F is resultant force, and A is acceleration. Therefore, this formula would always work.

[00:15:57]The second one, linear momentum divided by the velocity. Now, I know that the momentum of an object is equal to the mass times the velocity. So, if I do the mass, it would equal to the momentum divided by the velocity. So, this formula would also work.

[00:16:15]What about the weight of the object divided by the acceleration?

[00:16:18]Now, for this particular thing, we know that W is equal to mg. So, W divided by g will give us m. But, this case only applies if g if the acceleration of the object is the acceleration of free fall.

[00:16:33]It will not always be the case. If the acceleration is something different, then this formula cannot be applied.

[00:16:39]Therefore, we cannot take three. And the correct answer would be one and two only. Therefore, 11 is B.

[00:16:46]Question number 12. A basket is sliding on a horizontal frictionless surface with a velocity of 30 cm per second.

[00:16:54]The mass of the basket is m.

[00:16:57]A block of mass 2m walls falls vertically onto the basket.

[00:17:02]The block and the basket then move together with the same horizontal speed V. What is V?

[00:17:07]Now, initially, what what happened is we had the basket the original basket of mass m moving at 30 m per second. So, the momentum was m times 30 m/s.

[00:17:19]I'm not going to convert the units because they've used the same units everywhere.

[00:17:23]Now, this shall equal to the final momentum. The final momentum is 2m + m * v.

[00:17:31]Now, a lot of you might be wondering that why have we not, you know, used this particular object's momentum?

[00:17:37]Because this object has also fallen vertically. That means it was in motion when it hit the other object. But, what you need to understand is the object fell vertically. And therefore, it was having vertical momentum. Now, for any sort of object or for any sort of system, we keep the vertical and the horizontal momentum distinct. That means separate from each other.

[00:18:00]That means that if an object's, you know, if the vertical momentum when it falls you see that the block and the basket they are moving in the same horizontal direction. They do not move in the vertical direction anymore. That means the vertical momentum came to zero and now they started in the horizontal direction. So, hence we do not have to consider the momentum of this because it was only having vertical momentum. But, in the final scenario, we only had horizontal momentum. Let's work out this equation. So, m * 30 is equal to 3m * v.

[00:18:32]So, m and m get cancelled out. 3v is equal to 30 and therefore, the value of v is equal to 10.

[00:18:40]So, 12a would be the right answer.

[00:18:44]Two children are balanced horizontally in equilibrium on a plank. So, we have a plank and then we have it is of a length of 3 m. You got two children sitting, 15 kg and 20 kg. And the plank itself has a mass of 7 kg.

[00:18:58]The plank is uniform with a mass of 7 kg and a length of 3 m. There is a pivot at the midpoint. So, the pivot is exactly at the midpoint. Now, when they say that the pivot is the midpoint, you do not need to take into consideration the moment caused by the mass of the plank itself. This is not really needed now.

[00:19:15]Child X has a mass of 20 kg and is at distance X. Child Y has a mass of 15 kg at a distance of Y. What is the ratio of X over Y?

[00:19:23]Now, because this is in equilibrium, what happens is the clockwise momentum the clockwise momentum about the pivot, obviously, should equal to the anticlockwise momentum.

[00:19:42]Right?

[00:19:43]So, let's consider the momentum due to object child X. It has a mass of 20 kg and sitting at a distance of about X, which we do not know, shall be equal to the other child of 15 times the distance Y. So, 20X is equal to 15Y. If I want the ratio of X over Y, I need to do X over Y, which will be 15 over 20. I'm just going to swap everything. So, if you're not sure how that is done, 20X is equal to 15Y. I'm going to bring the Y over here. So, 20 and I'm going to bring the 20 over here. So, that is Y.

[00:20:18]Oh my god, just give me a sec.

[00:20:20]This Y comes over here and this 20 goes over here. So, X over Y is equal to 15 over 20, which is 0.75. You can check that using the calculator if you're not aware of it. 13 C would be the right answer.

[00:20:35]Question number 14. The diagram shows two ways of hanging the same picture. In both cases, the string is attached to the same points on the picture and looped symmetrically over a nail in the wall.

[00:20:47]The forces shown are those that act on the nail. The weight of a string is negligible.

[00:20:53]In diagram one, the string loop is shorter than in diagram two.

[00:20:57]What information is correct about the magnitude of the forces is correct? Now, one thing is this question is very, very, very much repeating. This is really my third or the fourth paper that I've done and I've seen the same question. So, do take up these type of questions. You might find the explanation already on YouTube. Now, here is the explanation to this. First up, let's talk about R1 and R2.

[00:21:17]Now, here's the thing. The painting used in both of the scenarios is exactly the same. So, it has the same weight, right?

[00:21:25]The weight is same. And R2 and R1 are the only balancing forces in this particular case. That means that R1 should be equal to the weight and R2 should be equal to the weight. And because the weight is same, R1 and R2 are equal to each other. That is why this option and this option is neglected. Now, let's talk about the tension forces.

[00:21:45]Okay.

[00:21:46]Now, I'm going to often try to use certain values of angles which are not true, but they would look as per the scale diagrams.

[00:21:54]So, let's say that this angle is 60° or I'm going to keep it as 59° and this angle is going to be 29°.

[00:22:04]Okay. Now, we know that they both are in equilibrium. So, we could say that R1, which is the upward force about the nail, shall be equal to the the vertical component of T1 and the vertical component of this T1.

[00:22:19]Now, how do we find the vertical component? This angle is 59, so it's going to be T1 cos 59 and T1 cos 59.

[00:22:29]Right? So, I could say that my R1 is equal to 2 * T1 into cos of 59.

[00:22:35]And the value of T1 could be R1 over cos of 59, 2 * cos of 59.

[00:22:42]Okay. Now, the same thing would apply for T2 as well. I could say that my T2 would be R2 over 2 cos of 29. Now, what I want you to do is try to work out this. Use the same values for R1 and R2 because we already told that R1 and R2 are equal to each other. So, let's say I'm going to use my the value of R1 as 15. 15 over 2 cos 59 shall give me T1 as 14.5.

[00:23:13]And T2 would be a 2 cos of 29 is 8.57.

[00:23:20]So, we just found out that T2 is actually less than T1 or T1 being greater than T2. That means the correct option would be 14 B. I hope this is very much clear.

[00:23:31]Okay, so what I've done is first of all, I figured out that R1 and R2 are equal because they should be equal to the weight. Then I took the vertical components of the tension forces and equaled that to R2. With that, I was able to form an equation of R1 and R2 in terms of like the tension in terms of R1 and R2.

[00:23:49]Now, because I know R1 and R2 are equal, I could take any possible values for them and substitute them in this equation and I got the values for T1 and T2. This would be different for you if you guys use different values for R1 and R2, but the example or the explanation is valid. I hope this clears out your doubt in this question.

[00:24:08]Question number 15. The diagram shows four metal cylinders submerged inside a container of water. Which cylinder has the greatest pressure on its top surface?

[00:24:17]Now, pressure in the water is given by rho gh. Rho is the density of the water, g is the acceleration due to gravity, and h is the depth. Depth and density. That means the deeper you go, the greater the pressure will be on the top surface. So, we will need to find the object or the metal cylinder with the with the deepest top surface.

[00:24:38]That is C because it has the deepest top surface. So, C has the greatest pressure. 15's answer is C.

[00:24:45]16.

[00:24:46]Three solid objects cube P, cube Q, cube piece, cuboid Q, and cylinder are shown.

[00:24:53]The dimensions are shown in terms of X and Y. Which two objects have the same density?

[00:24:58]Now, density rho is given by mass over volume.

[00:25:02]So, this one, the mass is M, the volume is X, so it because it's a cube, I'm just going to say X into X into X as the volume. That would be X cubed.

[00:25:11]What about this one? Mass of 4M.

[00:25:14]Now, um the dimensions are 2X * X * X, so it's going to be 4M over 2X cubed.

[00:25:22]Okay, which can be simplified as 2M over X cubed.

[00:25:27]The third one.

[00:25:28]Now, this one, um the mass is M, and what about the the volume? For certain objects, when you do not know the volume, what you want to do is find out the area of the cross section and multiply it with the height. So, pi X squared multiplied with X over pi. So, this pi and this pi gets cancelled out, it's going to be M over X cubed.

[00:25:51]So, I can see that my object P, which is over here, and object R, which is over here, they both have M over X cubed and M over X cubed, so hence they both have equal densities, P and R.

[00:26:03]16's answer would be C.

[00:26:06]17. A small ball is projected into the air from horizontal ground. The diagram shows part of the path of the ball.

[00:26:13]The ball leaves the ground at point X with a kinetic energy of 8 J.

[00:26:17]A short time later, the ball is at a point Y, where it has a kinetic energy of 3.5 J.

[00:26:23]Point Y is at a vertical height of 8.8 m above the ground, and air resistance is negligible.

[00:26:29]What is the mass of the ball?

[00:26:31]Wonderful. Now, in one case, this is about um this is about, you know, motion. What what is this? Parabolic motion or whatever this motion is. It's a two-dimensional motion, as you know, horizontal motion and vertical motion.

[00:26:46]And one thing you need to understand is that in this case of motion, the momentum or the total energy, in fact is always conserved. Right? So, here I might have kinetic energy because I'm being projected. Now, when I go up, I lose that kinetic energy and gain potential energy. When I come back down, I'm losing potential energy. I'm going to gain kinetic energy, but the total energy remains constant.

[00:27:08]Initially to start off, you had 8 J.

[00:27:11]Okay, 8 J at the this particular point.

[00:27:14]Then, when you reach to the top, you had 3.5 J of kinetic energy. That means 8 - 3.5, which is going to be, let's see the calculator, 8 - 3.5 is 4.5. You had 4.5 J of gravitational potential energy.

[00:27:31]Gravitational potential energy is 4.5 J because energy is either going to kinetic energy or gravitational potential energy. Air resistance is negligible. Hence, you're not doing any work against the air.

[00:27:42]Now, if I have 4.5 J of gravitational potential energy, which is equal to m * the value of g * the value of h, which is 8.8.

[00:27:52]All right, so I can say that 4.5 is equal to m * 9.81 * 8.8.

[00:27:59]Therefore, the value of m would be 4.5 / 9.81 * 8.8.

[00:28:07]That would be giving us the mass as 0.052 kg. So, 17's answer would be B. I hope that is clear.

[00:28:17]Question number 18.

[00:28:19]Now, question number 18 says that a boat requires a power p to move through still water at a constant velocity, and the drag force f acting on the boat. What will be the velocity of the boat?

[00:28:29]There's a very easy formula for this.

[00:28:30]Power is equal to force * velocity. So, therefore, the velocity will be equal to power * over the force. So, the correct answer would be 18 C.

[00:28:39]Now, let me give you a quick recap on how do you actually get the formula for power.

[00:28:44]We know that power is basically the rate at which work is being done.

[00:28:49]And work done is force times the displacement in the direction of the force.

[00:28:53]And what about displacement?

[00:28:55]Displacement is equal to velocity into time. So I can write this equation work done as F times VT.

[00:29:02]And therefore I can write power as FVT over time. These both get cancelled and I'm left with power is equal to FV.

[00:29:11]I hope this is clear.

[00:29:13]Question number 19. When the speed of an object is increased from 10 to 20 m/s, the energy increases by E.

[00:29:20]When the increase is in the kinetic energy, what is the increase in the kinetic energy when the speed increases from 70 to 80? So for a certain object, when the speed is going from 10 to 20, the kinetic energy increases by E.

[00:29:32]And when the speed increases from 70 to 80, the kinetic energy increases by what factor? Let's find that out.

[00:29:38]Now, for these types of questions, I like to use a certain value for mass which should be constant everywhere. Let's say my mass is 15 kg. So initially, let's try to find the value of E.

[00:29:49]Half times 15 kg into the velocity squared, which is 20 squared minus half times 15 over the velocity of 10 squared.

[00:30:00]Half times 15 times 20 squared minus half times 10 half times 15 times 10 squared.

[00:30:08]That would give you 2250.

[00:30:12]Okay?

[00:30:13]Now, if I do the same thing, half times 15 times 80 squared minus half times 15 times 70 squared.

[00:30:25]So, I'm going to replace the 20 with my 80 and going to replace this with my 70. This is actually going to give me 11250.

[00:30:36]Now, when you do 11250 divided by 2250, it's actually equal to five. So if you take this as the value of E, then this is actually 5 E.

[00:30:46]So, it's very simple. What I've done is I know that they're only asking about the kinetic energy. So, I just assume that I have a mass of 15, found the value of E by finding the kinetic energy at 20 m/s and 15 m/s and and subtracting them, then found the value of 5 E by first finding the difference in kinetic energy between 80 and 70, and then dividing it.

[00:31:06]So, I got the answer as 5 E.

[00:31:08]Okay? And you could use any mass for this. The answer would still be the same because it Just ensure that you use the same mass across all the questions.

[00:31:17]Train supply a power coal. Train supply coal to a power station. The The table shows quantities describing the operation.

[00:31:24]So, the power station output is P in terms of watts. That means um per second.

[00:31:30]The number of trains that are coming into the power station is N, and each train has M kg of coal, and the energy is actually E in joules from 1 kg of coal.

[00:31:40]The number of seconds in 1 day. Now, efficiency is basically going to be efficiency is power output over power input. Now, output power. I know that the power station's average output is P, and if you multiply that with S, you will get the output per with number of seconds in 1 day.

[00:32:05]So, it's just P into S.

[00:32:07]Okay?

[00:32:08]Divided by Now, the number of trains that are arriving per day is actually N, multiplied by the mass of the train mass of the coal in each train. So, I'll get the total mass, and multiplied by E, I'll actually get the total energy.

[00:32:20]Right? So, this thing, power in 1 second multiplied by the number of seconds in 1 day, gives me the power output in 1 day. The number of trains per day multiplied by the mass of 1 train into the energy of 1 kg of coal will give me the energy input. And this is how I can actually find the um efficiency. So, it's going to be PS over NME.

[00:32:42]PS over NME, that's going to be 20A.

[00:32:45]All right, I hope this is clear.

[00:32:47]This paper seems to be pretty easy for now. Let's check question number 21.

[00:32:52]Which descriptions of object undergoing elastic and plastic deformations are correct? Now, elastic deformation, what happens is that the object is not permanently deformed and once you remove the force, it returns back to its originally original shape. Whereas in plastic deformation, it does not return back and is permanently deformed. Therefore, the answer would be 21D, which says that in elastic deformation, it is not permanently deformed, whereas in plastic deformation, it is permanently deformed.

[00:33:18]A force F extends a length L of a uniform wire by an extension E. The wire has a diameter D and obeys Hooke's law.

[00:33:25]What is the Young's modulus of the material of the wire?

[00:33:28]Now, the Young's modulus is given by the stress divided by the strain. It's the ratio of the stress over the strain.

[00:33:41]Now, stress is actually given by the force over the area. And the strain is given by the extension over the original length.

[00:33:50]So, I have applied a force F, so I'm just going to call that F. On what area though? The wire has a diameter of D.

[00:33:57]That means the area of the wire will be pi pi times r squared. Now, here's one thing that you need to understand. If the diameter is D then the radius radius would be 1/2 D. And if I do r squared, that would be 1/2 D squared or 1/2 D the whole squared.

[00:34:21]This can be written as 1 by 4 times the value of D squared. So, I have 1 by 4 times the value of D squared over here.

[00:34:30]Divided by E is the extension, which is already given over here and l is the original length. Okay. Now we got to simplify this. First of all, because there's a division symbol and there's a fraction, I'm going to put that into multiplication symbol and put the l on the top. So it would be f over pi d squared divided by 4 into l over e.

[00:34:54]Okay. Now what happens is because this 4 is being divided in the denominator, I could put this back into the numerator over here. That becomes 4 f l divided by pi d squared e.

[00:35:07]So do I have an option similar to that?

[00:35:10]4 f l divided by pi d squared e. So therefore the correct answer would be 22d.

[00:35:15]I hope this is clear.

[00:35:18]Let's take a look at question number 23 now. Question 23 says an 8 Newton weight is attached to a lower end of a spring which is fixed at its upper end. The weight is initially held at rested position x and the spring is unstretched. The weight is then released and falls to position y which is 4 cm below x. The weight oscillates and then eventually comes to rest at position 0 which is 2 cm below x. How much energy is lost from the system?

[00:35:48]Now the formula to calculate energy in a spring is basically e is equal to half of f of x or half of kx squared where f is the force or the load and x is the extension. Now in the first case there is no extension. The spring is unstretched. So we will start taking energy from this point and this point.

[00:36:12]Okay, now let's consider the energy in this particular case which is number one.

[00:36:16]So half times the force which is 8 Newtons time extension. Now extension is given in centimeters. However, you should take it in meters so that your final answer is in joules. 0.04 Let's see what value does this give us?

[00:36:31]Half times eight times 0.04 That's going to be 0.16 joules.

[00:36:38]And what about the second case? Final position after many oscillations. Half times eight, which is the load, times 0.02.

[00:36:47]So half times eight times 0.02 is 0.08 joules.

[00:36:54]All right, so You had 0.16 and at the finally you have 0.08. That means you have lost 0.08 joules. Because 0.16 minus 0.08 is equal to 0.08 joules. Therefore, the answer for question number 23 is going to be B. I hope that is clear. Moving on to question number 24. An electromagnetic wave has a frequency of 3.3 into 10 raised to 14 hertz. Which principal region of the electromagnetic spectrum is this wave in?

[00:37:25]Okay. Now, obviously because you guys are taught the table of wavelengths, it's better to find the wavelength in this particular case. V is equal to F lambda. Therefore, the value of lambda will be V over F.

[00:37:37]Now, V as we know, it's 3.0 into 10 raised to eight. That's a constant value. Divide by the frequency, that's 3.3 into 10 raised to 14. So 3.0 into 10 raised to eight over 3.3 into 10 raised to 14 That's going to be 9.09 um continues on into 10 raised to -7 meters. Okay? And this can be written as about 900 nanometers.

[00:38:07]Now, if you remember the fact that 400 to 700 nanometers is your visible light.

[00:38:14]That means 900 is a greater wavelength than that. That means 900 has to be either radio waves, microwaves, or infrared waves. Because the spectrum says radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and then gamma rays. So, visible light over here, it's between 400 400 to 700 nm. So, 900 should be at least one of these. And in the options, we can see it is infrared. So, infrared will be the right answer, because visible light, it can't be that. It has to be between 400 to 700.

[00:38:46]>> [snorts] >> UV and x-ray will have a smaller wavelength than 400 nm. So, that is why the right answer is 24A. I hope that is clear.

[00:38:55]Question number 25. A wave on a stretched spring causes the particles on the string to vibrate. The diagram shows the part of the string at one instant.

[00:39:03]So, there are three points that are given.

[00:39:05]Which statement about two of the particles in the wave is correct?

[00:39:09]Okay, so first of all, the options are differentiating between progressive wave and a stationary wave.

[00:39:14]Okay, if it is a progressive wave, that means we do have both the options.

[00:39:19]All right, then what else do we have to look at?

[00:39:21]Particles P and Q vibrate with a phase difference of 90.

[00:39:25]And if it's a stationary wave. Now, there's one thing, right? In a progressive wave, if you want to talk about the phase difference of two particular points, you should either know the wavelengths or you should be given the angles or where exactly is this point along the entire wavelength.

[00:39:39]Without that, only the diagram, we cannot judge the phase difference of two points. On the other hand, if you talk about a stationary wave, we know that points which are between two nodes, that means this is a node and this is a node, the points between two adjacent nodes are in phase, that means these both are in phase. And Q and R or P and R are out of phase, because they're not between two adjacent nodes. So, I'm going to say that if it is a stationary wave, then particles Q and R vibrate with a phase difference of 180°, which is exactly over here. Right? So, what you need to remember here is that if you have a stationary wave and if you have one particle over here and the other one over here, then they have a phase difference of 180 regardless of where they are. However, for a progressive wave, it does not work in the same way. You should know at least one of the values such as their wavelength or their speed at that particular point. Wavelength or frequency frequency will be the same, but wavelength can tell you a lot about it or the angle itself directly.

[00:40:37]Okay, now moving on to question number 26. 25 is D. 26. What describes the displacement of a particle on a wave?

[00:40:45]Now, we know the displacement of a particle is basically the distance that the particles travel in a um specified direction from the equilibrium position.

[00:40:53]And displacement is a vector quantity.

[00:40:55]That is why it has to be in a specified direction. So, the answer for that would be 26 A. The distance in a specified direction from the equilibrium position.

[00:41:04]Okay? If you look at option B, the distance moved in a specified direction by the particle per unit time. Now, here they did not um they're not, you know, considering the fact the equilibrium position, which is very important because whenever you consider displacement, it has to be from the equilibrium position. Not from, let's say, from the entire thing from here to here. No. It has to be from the equilibrium position.

[00:41:26]Okay. So, 26 is also 26 is A. Moving on to question number 27.

[00:41:32]Now, a beam of vertically polarized light of intensity I0 is incident normally on a polarizing filter.

[00:41:39]The transmission axis of the filter is at an angle of 60° to the vertical. So, you have vertically polarized incident light, and you have a filter that is 60° to the light. The light is vertically polarized as you can see, and the filter is 60°.

[00:41:55]The transmitted light beam is then incident on normally on a second polarizing filter. So, after it goes through the first one, it goes to the second one as well.

[00:42:03]The transmission axis of the second filter is perpendicular to the vertical as shown.

[00:42:08]So, the transmission axis of the second one, right? It's It's over here. It's perpendicular to the vertical.

[00:42:14]The intensity of the light that emerges from the second filter is IT. What is the ratio of IT is to I0?

[00:42:20]Okay. Now, initially you had I0.

[00:42:23]After passing through the first filter, after passing the first filter, the it will be basically we know the formula that the transmitted intensity is equal to original intensity into cos squared theta, where theta is the angle between the direction of the light, which is vertically polarized over here, and the transmission axis of the filter. So, between these two in the first filter, it is 60°. So, it's going to be that after passing through the first filter, I T1, first filter, is equal to I0 into cos of 60 the whole squared.

[00:43:00]Cos of 60 the whole squared.

[00:43:02]Right? Now, when it passes through the second filter, what happens is that IT will equal to IT1, that was the transmitted intensity from the first filter, multiplied by cos. Now, here you have to be very careful. What angle should you consider? When the light came out from this filter, it was at a 60° angle. It was at a 60° angle.

[00:43:25]Okay?

[00:43:26]So, now it means that if the light is at this particular angle, and the new transmission axis is at this particular angle, the angle between these two is 30°.

[00:43:35]Again, the angle between these two is 30°. So, you have to take the angle as 30°. 30°. So, that means the final answer will be cos of 60 the whole square, which will be the intensity after passing through the first one, multiplied by cos of 30 the whole square, which is the intensity after passing through the second one. So, cos of 60 the whole square times cos of 30 the whole square will give you 0.1875, which is 0.19.

[00:44:01]That is why 27 will be B. The only careful thing here is that which angle you're taking. The angle has to be between the direction of the polarization of the light which is passing through it multiplied with and the transmission axis of the filter. In the first case, it was 60°. Why? Because the light was vertically polarized and the transmission axis was like this. So, this angle was 60°. In the second case, your light was something like this and your filter was something like this. So, that is why this angle was 30°. And then you apply the same formula, cos of 60 squared or cos of theta the whole squared.

[00:44:36]I hope that is clear. Now, moving on to question number 28.

[00:44:39]A car is driving along a straight road a straight road while emitting a sound of a constant frequency.

[00:44:45]There's a stationary observer at each location X, Y, and Z.

[00:44:50]From X to Y, the car accelerates uniformly from rest to a high speed.

[00:44:54]From Y to Z, the car continues at this constant high speed. Which observer will hear the highest frequency of the sound?

[00:45:01]And what is the position of the car when the frequency is heard by the observer?

[00:45:05]Now, the formula to actually calculate this is basically um that observed frequency again observed frequency is equal to the actual frequency into the speed of the um sound in medium over the speed of the sound in medium plus or minus um we will often call it as VS, which is going to be the speed of the source.

[00:45:30]Okay, so this is the formula.

[00:45:33]Now, what I'm what I'm thinking is um it's likely to be very high at point Z. Let's understand that why.

[00:45:41]Okay. Now, if you want the highest frequency for for any of these points, the original frequency remains the same because it's a constant frequency. And the speed of the sound in this particular medium is also same because the medium is not changing. So, everything over here these all are same.

[00:45:57]Only thing that's changing is VS, the speed of the source.

[00:46:00]Okay. Now, when the car is approaching Z, if I want to use this particular formula, I will have to use it with a negative symbol, the minus symbol.

[00:46:09]Because when you're approaching something, you use the minus one, and when you're moving away from something, then you use the plus. Okay. Now, if the car is at the highest speed between these two sections, it's at a constant high speed, that means this value will be very high. Okay? And if this value is very high, that means my denominator overall is a very small value.

[00:46:29]And if I have a small denominator, that means that this frequency will be the larger one. Okay? Just like if I write eight Okay, let's take a larger value.

[00:46:38]If I say 60. 60 over two is 30, but 60 over 60 over four is 15. So, when I reduce my value from four to two, this 15 went to this 30. That means if I have a denominator which is smaller, my frequency observed frequency will be larger. So, the highest frequency will be observed at point Z when the position of the car is between point Y and Z.

[00:47:04]Now, we don't actually consider the region XY because the car is still accelerating. It has not reached the high speed. It is accelerating from rest to the high speed. So, the speed will be variable, and therefore, the frequency will be increasing, but it will not be as high as this particular region.

[00:47:21]Okay? I hope that is very clear, and that's why the answer is 28D.

[00:47:26]Moving on to question number 29.

[00:47:29]Each diagram shows a stationary wave on a stretch spring. Which diagram is labeled correctly?

[00:47:34]Now, node is a point with zero displacement, antinode has a maximum displacement, and Okay, now first of all, I'm going to, you know, completely reject these two ones because we know that the wavelength is from one node the entire wave completed. So, it goes starts from here and goes like this, like this, like this. The wavelength has to be over here. So, this and this are out. A and B are out.

[00:47:54]In C, they have not labeled node and antinode correctly. Only in D, they have labeled it rightly. That is anti-node is over here and node is over here and the wavelength is also shown rightly. That's why 29 would be D.

[00:48:06]30.

[00:48:07]Two coherent electromagnetic waves are traveling in a vacuum.

[00:48:12]The two waves meet at a point. At this point the two waves have different intensities. Which statement is not correct? Now, when they say coherent, one criteria for a wave to be coherent is that they have constant phase difference. So, this statement is definitely correct. So, hence it cannot be our option because we want to select the wrong statement.

[00:48:32]Okay. Now, if something has a constant phase difference, that means it should have the same frequency if same frequency.

[00:48:39]So, in order for two waves to be coherent, they should have a constant phase difference and they should have the same frequency. This is a theory part of the AS syllabus and these are the two conditions which are required to be coherent.

[00:48:50]Also, now if they have the same frequency and they're traveling in in a vacuum, it's electromagnetic wave which is traveling in a vacuum. That means they both will obviously have the same speed as well. That means this is statement is correct. Hence it cannot be the right answer. Therefore, the correct answer will be 30 B. They have the same amplitude at that point. Now, this is not necessary that two waves could have different amplitudes. It says the phase difference has to be true. The amplitude is not regarded to the phase difference in any order.

[00:49:19]Okay, but phase difference tells you that how much part of the wave has traveled in comparison to the whole wave. Whereas amplitude is the maximum displacement. So, amplitude is not related and we've also seen that if we check all the other options, they all can't be the right answer. Therefore, 30 B is the right answer.

[00:49:36]31.

[00:49:37]Laser light of a single frequency is incident normally on a diffraction grating with lines 5.00 * 10^5 lines per meter.

[00:49:45]A diffraction pattern is formed on a flat screen which is parallel to the grating.

[00:49:50]The screen is a distance of 33.5 cm from the grating. The distance between the first two order intensity maxima is 15 cm. What is the wavelength of the light? Okay. Now, first of all, in these types of questions, what you're allowed to do is you write the formula, which is going to be d sin theta d sin theta is equal to n lambda. d sin theta is equal to n lambda. Okay. Now, we are we given the value of d for this particular diffraction grating? Yes, we have 5.00 * 10 raised to 5 lines per meter. That means if we do reciprocal of this value, which is 1 over 5.00 * 10 raised to 5, we will be able to find the distance between two lines. So, d is secured.

[00:50:34]sin theta, we do not know sin theta yet.

[00:50:37]n, n would be 1. Now, here's what they're saying. If your diffraction grating is over here and if the screen is over here so, this is our grating and this is our screen.

[00:50:50]The distance between these two is 33.5 cm, 33.5 cm.

[00:50:58]Okay. And one more thing which is that the first order intent the distance between the two first order intensity maximum. We know that first one will be over here and the second one will be over here. Both have n values one. It is 15 cm. So, the distance between these two is 15 cm.

[00:51:20]15 cm. Okay. Now, with this data, we will actually be able to work out the our or the distance that we're actually looking for. Let me draw a larger diagram so that you guys can see it much more clearly.

[00:51:34]grating.

[00:51:37]The distance from here to here is 33.5 cm.

[00:51:43]I have the n is equal to 1 over here and I have n is equal to one over here as well. Okay, the distance between these two being about 15 cm.

[00:51:54]All right.

[00:51:55]Now, in this particular formula, if I use n value as one, then the theta value is the going to be this one. This is the theta value that I'm supposed to use.

[00:52:05]Okay, theta value that I'm supposed to use. Now, let's understand that how exactly will we get this theta value.

[00:52:10]If I have 15 cm over here, I can say that this distance will be 7.5 cm, right? And then, in order to work out theta, tan theta, which is opposite over adjacent, it's 7.5 over 33.5.

[00:52:29]Okay, so the value of theta is going to be tan inverse of 7.5 over 33.5, which is the value of theta is 12.

[00:52:39]619 322 29. Exact value just ensure that our answer is correct. Okay, now we can use the formula. Now, the formula says, let me do the working over here in different color.

[00:52:53]The formula says d sin theta is equal to n lambda. So, we want to work out lambda. Lambda will be d sin theta over n. d sin theta over n. Now, the value of d is 1 over 5.00 into 10 raised to 5.

[00:53:11]The value of sin theta. Now, sin of theta, theta is 12.6193229.

[00:53:18]Okay, divided by n. n, we are looking at n is equal to one, so you can just let it be or you can write over here. Now, what you want to do is simply put this in a calculator. 5.00 into 10 raised to 5 multiplied by sin of 12.619300029.

[00:53:35]That will give you 4.36 into 10 raised to -7. Now, all the answers are given in Newton meters over here, I mean in nanometers. So, what you do is you divide your answer by 10 raised to -9, which will give you 436 nanometers. So, that's the correct answer will be 437 nanometers.

[00:53:53]I hope that is very much clear for you.

[00:53:55]Try to solve this question on yourself.

[00:53:56]It's not that hard. Just requires a bit of understanding. And if you can crack this diagram, you understand the question very well.

[00:54:04]Which graph represents the variation with the current and the potential difference for a filament lamp?

[00:54:08]Okay. Now, for a filament lamp, what happen what happens is normally the after a time or after the current starts to increase, the temperature is increases itself, right? The temperature of the filament inside the lamp increases. And when the temperature increases, because it's a metal, the resistance also increases. So, we should see a pattern which shows increasing resistance.

[00:54:30]Now, for this particular graph, we know a formula which is V is equal to IR. So, R is equal to V over I.

[00:54:37]The gradient of this graph is V over I.

[00:54:40]In all cases, the gradient is V over I.

[00:54:47]That means that the gradient of the graph should increase as the current increases. Now, let's look at the first case. In the first case, we can see that the gradient of the graph is actually decreasing because here, now, how do we identify it's decreasing?

[00:54:59]You can see this part is has a more gradient, like it's more steep, and this is less steep.

[00:55:04]Second case, this is less steep, this is more steep. That's why the gradient of the graph is increasing. And therefore, we would choose this particular pattern.

[00:55:11]And we can see that this obeys on the other side as well because no matter in which direction the current is flowing, the resistance will increase because the wire or the filament does get hotter.

[00:55:22]Okay. So, the 32 B would be the right answer.

[00:55:25]Moving on to question number 33 now.

[00:55:28]What is the definition of potential difference across the component? It is the energy transferred per unit time.

[00:55:34]So, the 30 to 33 A would be the answer.

[00:55:37]If you have a component, let's say, which is a bulb, okay? And if you attach a voltmeter inside this, so the reading on the voltmeter is actually giving the potential difference, which is how much work did a charge have to do to get from this point to this point. Basically, to pass the resistance of this particular component, how much work did it have to do or how much energy has been transferred by the unit charge.

[00:56:02]Okay, so 33A is the right answer. Moving on to question number 34.

[00:56:07]Now, 34 says that conducting putty is a substance that conducts electricity and can be molded into various shapes.

[00:56:14]Can be molded into various shapes. A cuboid made from this putty has a length of L and a width of X. The resistance between the square faces is 15 uh omega.

[00:56:25]So, the resistance between these faces are they talking about? These are the square faces here, because they have X and X as the length.

[00:56:31]Is 15 omega.

[00:56:32]Some more of the putty is molded into a cylinder that has a length of L by 2.

[00:56:37]The resistance between the two circular faces of the cylinder is also 15. What is the radius of the cylinder?

[00:56:44]Interesting. So, they have this particular um in a cuboid made of the putty and it had the resistance of 15 with one square face.

[00:56:52]They also used some of the putty to make another um cylinder and the resistance of that cylinder is also 15.

[00:57:00]So, we have to work out that what is the radius of the cylinder in terms of X by X and by.

[00:57:05]Okay. Now, one thing we know is resistance is equal to um rho times L, which is the length, over the area. And rho is a constant, which is called as the constant of resistivity.

[00:57:17]Now, the constant of resistivity would be would be the same for both the materials, because would be the same for both the cases.

[00:57:24]Now, first case, let's investigate with this cuboid. So, if resistance is equal to resistivity into length over area, I can say that 15, which is the resistance, is equal to resistivity into the length. The length is L and divided by the area, the cross-sectional area, which will be x squared.

[00:57:42]The cross-sectional area basically refers to this area because they said that the resistance between the square faces. So, x squared is this area.

[00:57:50]Now, let's talk about um the cylinder now. Now, cylinder 15, it has the same resistance is equal to rho times the length. The length is L by 2, so I'm just going to say rho times L by 2.

[00:58:04]multiplied by 1 over area. multiplied by 1 over area. So, the area will be pi times r squared. We don't know what's r squared till now. We just held as variable.

[00:58:15]Okay, so if you're confused, what I actually did is I split this formula into rho L into 1 over area, which is the same thing. So, I have rho over here. I have L by 2, so rho into L by 2 into 1 over pi pi r squared.

[00:58:30]Okay. Now, because both of them both of these equations give us 15, we can equate them together.

[00:58:36]So, I have that rho L over x squared is equal to rho L over 2 pi r squared. Now, this rho L and this rho L gets cancelled, and what we have is that x squared should be equal to 2 pi r squared.

[00:58:54]Okay. They both are the denominators, but you can equate the denominators together because they're equal. So, if I have r squared will be x squared divided by 2 pi, and therefore r will be the square root of x squared over 2 pi.

[00:59:10]Now, this can be written as r is equal to x over the square root of 2 pi alone.

[00:59:16]So, x over the square root of 2 pi, the correct answer is actually 34B.

[00:59:21]I hope that is understood.

[00:59:23]Now, moving on to question number 35, each of Kirchhoff's laws is linked to a conservation of a physical quantity.

[00:59:30]What conserved physical quantities are used in the derivation of Kirchhoff's first law and Kirchhoff's second law?

[00:59:36]Now, this is a very repeated question and it's a simple knowledge question.

[00:59:39]Kirchhoff's first law is used for the conservation of charge and Kirchhoff's second law is used for the conservation of energy. That is why 35 would be C. I hope that is clear.

[00:59:53]Moving on to question number 36.

[00:59:55]Eight identical resistors are connected as shown. So, they're connected in a series circuit, but inside the series circuit, you have one component as over here, which is a parallel combination of resistors.

[01:00:06]Another component over here, which is a parallel combination of resistors.

[01:00:10]Ammeter one reads 1.50 A amperes. What is the reading on ampere ammeter two?

[01:00:17]So, this one over here has a reading of 1.50 amperes.

[01:00:22]What is going to be this reading?

[01:00:24]Now, before we start solving this question, let us quickly revise the concepts of series and parallel circuits. Now, in a series circuit, what happens is every single component gets the same amount of current. So, this entire component gets the same amount of current and this component also gets the same amount of current because current is not shared.

[01:00:41]Okay? Now, but when current enters over here, it gets divided between these two branches in terms of their resistances.

[01:00:49]That means the branch which has less resistance gets more current and the branch which has more resistance gets less current.

[01:00:57]The same thing happens over here. Let's say the circuit over here, we had 10 amperes. Let's just say that. So, over here also you have 10 amperes, but then it gets divided between these two.

[01:01:07]This one will also get 10 amperes, but it gets to divide between these two.

[01:01:12]Okay.

[01:01:13]So, this part is clear. Now, if this branch got 1.50 ampere with one resistor, so the other branch has two resistors, double the resistance means half the current. So, the above branch will get 0.75 amperes of current. That means in this circuit in total there is 2.25 amperes of current. Okay, 2.25.

[01:01:36]That means over here you had 2.25 as well and over here also you had 2.25.

[01:01:42]Okay. Now, this 2.25 amperes of current, it gets divided between these two branches. But, how does it get divided?

[01:01:49]It doesn't get divided equally. The branch which will get the which has the lower resistance gets more current. So, if I do 2.25 into 3 by 5 because one total there are five resistors and one branch has three resistors. So, 2.25 into 3 by 5 is actually 1.35.

[01:02:08]That means the above branch will actually get 1.35 amperes.

[01:02:13]And the remaining one which is 2.25 minus 1.35 is actually 0.9. So, the below branch gets 0.9 amperes. Now, you can check the math. 1.35 + 0.9 is actually 2.25.

[01:02:27]So, this branch, it has one the lower branch has one more resistor one more resistor, that's why it got less current. And the above branch has one less resistor, that's why it got more current.

[01:02:38]So, the final answer as we can see, it will be 1.35 amperes. 36 C is the right answer.

[01:02:46]Okay, moving on to question number 37.

[01:02:48]A battery of electromotive force 6 volts and a negligible internal resistance is connected in series with a resistor of 6 ohms and a variable resistor of resistance from 0 to 4 ohms.

[01:02:59]So, you have a 6-volt battery and which is connected in series with a 6-ohm resistor and a resistor whose resistance can be adjusted from 0 to 4 ohms.

[01:03:10]A voltmeter is connected across the variable resistor. The resistance of the variable resistor is changed. What is the range of the voltmeter reading? Now, the range of the voltmeter reading depends on the the resistance of this particular resistor. Let's say it is actually 0 ohms. That means all of the potential will be used up in this particular resistor. That means this voltmeter gets 0 volts. So, the starting the minimum is 0.

[01:03:36]Now, let's say maximum. When does it get the most voltage? When the resistance is 4 ohms. 4 ohms.

[01:03:44]In that case, how will we determine how much voltage does this get? That will be 6 into or because let's it will be shared, yeah? So, we have 4 ohms and the total resistance is 6 plus 4, which will be 10, multiplied by 6.

[01:04:01]So, 4 over 10 multiplied by 6 means 2.4.

[01:04:06]That means that the reading on the voltmeter can range from 0 to 2.4 volts.

[01:04:12]Therefore, the value is 37 A. I hope that is clear.

[01:04:19]Question number 38. In the alpha particle scattering experiment, a beam of alpha particles is aimed at a thin gold foil. Most of the alpha particles go straight through or are deflected by a very small angle.

[01:04:30]A very small proportion of the alpha particles are deflected by more than 90°, effectively rebounding towards the source of the alpha particles.

[01:04:38]Which conclusion about the structure of an atom cannot be drawn from this experiment alone?

[01:04:44]Okay, most of the atom is empty space.

[01:04:46]Now, this one we can actually understand from this experiment because it says that a very small number of alpha particles are deflected. That means most of them just went straight through. That is why the most of the atom is empty space. So, this can be understood from this particular experiment, from the first statement. Because most of them went straight through, that is why the atom is just empty space.

[01:05:07]Most of the mass of the atom is concentrated in the nucleus. This statement can also be understood because it says that a very small number of alpha particles are deflected by more than 90°. You are deflected by more than 90° when you hit the center of the in the center of the atom, the nucleus itself. It has all the mass, hence you cannot pass it. So, that is why this is also understood from this.

[01:05:30]The nucleus contains both protons and neutrons. Now, this is not very clear.

[01:05:34]Okay? The reason is it tells us that there is a nucleus which is probably positively charged, but it cannot, you know, it cannot it does not say that it has protons and neutrons, which was determined from other experiments.

[01:05:47]So, this is something that we cannot understand from this experiment. 38C.

[01:05:51]38D, the nucleus is charged. Yes, this is true because they're deflected by a small angle.

[01:05:57]All right, then. So, 38C.

[01:06:00]Now, moving on to 39. In a type of alpha alpha decay, an up quark changes to a down quark. Which row shows the new hadron created in this beta decay and the sign of the charge on the emitted alpha particle?

[01:06:12]Now, you have two types of beta decay.

[01:06:15]You have beta minus beta minus and you have beta plus.

[01:06:23]Now, in a beta minus decay, what happens is a neutron gets changed to a electron. A neutron is basically converted to a proton. Along with that, an electron is released and an antineutrino is released, antineutrino.

[01:06:42]In a beta plus decay, however, you have a proton that changes to a neutron plus an positron and a neutrino. And you're supposed to remember this by heart.

[01:06:56]Okay?

[01:06:57]So, if an up quark changes to a down quark, this will probably be beta plus because proton has um up up down And a neutron has down, down, up. So, when a proton changes to neutron, all it needs to do is change one up to one down. Okay? So, this is a beta plus decay. The hadron created is a neutrino in this particular case, and the particle which is released, the emitted alpha particle is a positron. So, it has a positive charge. Therefore, the right answer will be 39 A.

[01:07:30]Okay, moving on to question number 14, which is the last question. Uranium 238 92 decays to radium 226 88 after a sequence of alpha and beta minus decays.

[01:07:40]Which row gives the information about the correct number of alpha and beta minus decays?

[01:07:45]Now, alpha decay is represented Alpha is represented by this. Beta minus is 0 1.

[01:07:50]So, one thing you need to understand is that the mass number, which is at the top over here, is only impacted by the alpha decays. So, if I have 238 92, after one alpha decay, what happens is this becomes 234 90.

[01:08:05]Plus one alpha particle. Plus one alpha particle.

[01:08:11]Okay, so I started off with 238 92.

[01:08:15]It then became 234 9 234 90, because the mass the mass number reduces by four, the atomic number reduces by two.

[01:08:23]And then it becomes 230 88. And then it becomes 226 86. 226 86. So, now we have reached the desired amount of mass number. After how many decays? After one, two, and three.

[01:08:40]Three alpha decays. Okay.

[01:08:43]Now, that means the right option has to be It can't be this one. It can't be this one. How many beta decays? Now, when you are at 226 86, if you if you have one beta minus decay, because beta is 0 -1, so, you will be at 226 87.

[01:08:59]Okay, the mass number increases by one.

[01:09:02]Okay? but we wanted 88, so we'll have to do one more beta decay to 226 88. That means two beta decays. Therefore, 40 B is the right answer.

[01:09:13]Okay, and with that, I've reached the end of this paper. So, thank you so much for watching. I hope this video was really helpful. And if it is, drop a like and make sure to subscribe to the channel for more helpful content and good luck for your exam. Bye-bye.