DIRECT 70 MARKS | RE-NEET 2026 | TRUST BLINDLY | Yawar Manzoor sir

[00:00:00]First one.

[00:00:02]First one is simple.

[00:00:03]This is a question of average velocity.

[00:00:06]Let's suppose an object covers this distance X with velocity V1.

[00:00:11]Then it covers next X with velocity V2.

[00:00:15]Same X, same X. If they ask you to find the average velocity or average speed, that will be 2 upon 1 by V1, 1 by V2.

[00:00:22]Directly remember.

[00:00:24]Now, if you have three equal distances, X X X. Here it is moving with V1, V2, and V3. So, here it will be 3 upon 1 by V1, 1 by V2, 1 by V3. Let's take a look at the question. This is the question which was asked in, I would say, JEE 2023.

[00:00:41]An object covers X distance with V1, next X distance with V2. What will be the average velocity? Okay. Okay. And in this particular case, it will be 2 upon 1 by V1 plus, I would say, 1 by V2. Now, average velocity in this question is given from there only. So, you'll be putting over here V. So, this will be 2 upon 1 by V1 plus 1 by V2. Take it over here. So, it will be 1 by V1 plus 1 by V2 will be over Take this V over here.

[00:01:11]So, 2 divided by V. I think option four is matching in this question. Let's take a look at the next question. In this particular case, an object covers AB with X, BC with X, and CD with X. You can read the question. It will be like this only. They ask you to find the average velocity in this particular case. So, it will be simply 3 upon 1 by V1 plus 1 by V2 plus 1 divided by V3.

[00:01:35]Now, if you solve this, okay, it will come out to be option three.

[00:01:40]Four.

[00:01:41]Then, my dear NEET aspirants, one more important concept is there. That is time of flight. Let's suppose you have launched an object with velocity U, okay, at an angle theta.

[00:01:52]This is the horizontal component of the velocity. That is called U cos theta. We also call it UX.

[00:01:57]This is what we call vertical component of the velocity U sin theta, it is also called UY. Now, this projectile reaches here. This is called the maximum height, how high it can go. This is called the range, how far it can go. And time of flight is how much time it is going to take from here to here. Time of flight, you can write this, 2 U sin theta by G.

[00:02:17]U sin theta also means UY, so 2 UY by G.

[00:02:21]Maximum height, you can write U squared sin squared theta by 2G. So, U sin theta means UY, so UY squared by 2G. Range is 2 U cos theta U sin theta by G. So, U cos is UX, U sin is UY, so this will be the range. Now, see, so many times frequently they have asked the question from this particular topic, okay? Now, take a look. In this particular case, an object is projected with velocity U is equal to alpha I cap plus beta J cap.

[00:02:51]When object is given this initial velocity, it means this is the X component, which is with I cap, this is Y component. So, X component of initial velocity is alpha, Y component is beta.

[00:03:02]This is the question of JEE 2024, JEE 2024.

[00:03:07]Now, take a look at this. You are supposed to tell me what is the value of time of flight, 2 UY by G. UY is beta, so 2 beta divided by G. You are supposed to tell me what is the value of maximum height. So, UY, UY is beta squared divided by 2G. You are supposed to tell me what is the value of range, so it will be 2 UX is alpha, UY is beta divided by G. Done and dusted with the question.

[00:03:32]Next, relation between horizontal range and maximum height. This is the relation which is between horizontal range and maximum height. H is equal to R tan theta by 4, where theta is the angle at which you have this launched. So, H is height and R is this. Now, see, JEE 2024, the angle of projection of a projectile for the same horizontal range is we have to find what is the value of theta if a projectile is launched which is having the same horizontal range as maximum height. Basically, range and height is given as same. We'll be using H is equal to R tan of theta divided by 4. We have to find the angle of projection. At what angle you have launched? So, you'll be putting over here R is equal to R tan of theta divided by 4. R and R will cancel. Take it over here. So, I can say tan theta will be equals to 4. Theta will be equals to if you take it over here, tan inverse of 4. This is the answer to the question.

[00:04:31]Okay. Let's move on to the next topic.

[00:04:33]That is this one. Area under velocity time graph. Very important. Let's suppose you'll be making the velocity time graph of an object. It is this. If you somebody asks you what is the area, area means displacement. Area of velocity time graph gives you the displacement covered by the object. Now, let's suppose this is a question which was asked in JEE 2025. This is the velocity time graph. You have to tell me what is the displacement from 0 to 4. 0 to 4 means this displacement. Means you have to come basically find this area.

[00:05:02]What will be this area? That is this area of triangle. 1 by 2 base into height. Base is 2. This base is 2. And how much is height? Height is 10.

[00:05:12]Plus area of this this rectangle. Its length is 4 minus 2 is 2.

[00:05:18]2. Area of rectangle is breadth length into breadth. So, breadth is I would say 10. Okay. So, 2 and 2 will cancel. 10 plus 20 area is 30. So, 30 m is the displacement in this particular case.

[00:05:31]Okay. Yes.

[00:05:33]Now, guys, one more important concept, Bernoulli's equation. What does Bernoulli's equation say? Bernoulli's equation say, if the liquid is flowing through the pipe, if you take the unit volume of the liquid at any point. Listen to me very carefully.

[00:05:48]It's potential energy unit volume potential energy of this unit volume and kinetic energy of this unit volume since it is flowing it has kinetic energy. It is at height H it will be having potential energy and pressure at that point. It will remain constant.

[00:06:03]Potential energy per unit volume, kinetic energy per unit volume and pressure will remain constant. Let's suppose you take the unit volume here, you take the unit volume here. Now, what is the potential energy at this point uh rho g h1 directly remember. What is the kinetic energy of the unit volume here?

[00:06:18]1 by 2 rho v squared where rho is the density, g is acceleration due to gravity, h1 is the height of this liquid at this point. V1 is the velocity of this liquid at this point, okay? Then P1. P1 is the pressure at this point.

[00:06:32]So, these three things will remain constant means what next point is this.

[00:06:37]So, whatever is the potential energy at this per unit volume at this point, kinetic energy per unit volume plus pressure, so it will remain the same. In case of horizontal pipe, we assume potential energy to be zero, okay?

[00:06:50]Because we consider this on the reference line. In case of horizontal pipe, this P1 plus this will be equals to this and this. So, here potential energy per unit volume will be zero, only kinetic energy per unit volume will be there. Now, take a look my dear NEET aspirants. This is a question which was asked in JEE 2020. You are supposed to tell me in this particular case, you are supposed to tell me in this particular case, what is the value of this velocity v?

[00:07:14]Okay, a fluid is flowing through this pipe. Velocity of the fluid here is v, pressure is p. Velocity here it is u.

[00:07:21]Here it is v, pressure is p by 2. We'll use the horizontal pipe concept. That is P1 plus 1 by 2 rho v1 squared plus is equal to is equal to P2 plus 1 by 2 rho v2 squared. You can say you can use this particular concept over here. Now, my dear NEET aspirants, this is p, Okay, plus 1 / 2 row V1 squared. Okay, so V1 is a row U squared is equal to P2 is P / 2. Pressure here is P / 2. Plus 1 / 2 row V squared. Take it over here. So, this P P minus P / 2 is P / 2 is equal Take it over there. So, it'll become 1 / 2 row is common V squared minus U squared. I'm doing the simple calculation. This and this will cancel.

[00:08:07]So, P / row and this plus U squared will be equal to V squared. Take this root over here. V will be under root of I would say P / row plus I would say U squared. This is the answer to this particular question. Okay?

[00:08:22]Yes. So, that's all for today. Thank you so much, my dear NEET aspirants. Make sure you smash the like button and you share it with your friends, also.