 do this. Second one done. Somebody just created everything all of you try it. So electric field is given 2 i cap plus 3 j cap. So how the electric field is? Suppose this is your x. So 2 i cap plus 3 j cap. So how the field will be? Like this? Like that? So this is how the field will be? Yes or no? And how this line is? y is equal to minus 2 by 3 x plus 5 by 3. So this equation of this line is that. So it has negative slope. So is it perpendicular to this? Is this 90 degree? If it is 90 degree you are moving in the direction that is 90 degree to the field. And we know that E dot is equal to change in potential. So if E dot dr is 0, change in potential is 0 and hence change in potential energy is also 0. So if you move along these lines your potential energy won't change. If potential energy doesn't change then work done by u is 0. You are not doing any work because energy is not changing. Understood? These are called equipotential surface. You will learn more about it in class 12th, electrostatics. First one you want to know the answer. Minus 20 i cap plus j cap. So when you integrate that you get 2 x plus 3 y. Which one integrate? E dot dr. Take a dot product and then integrate. Yes so I will get 2 x plus 2 x i cap plus. Dot product is 0 because theta is 90 degree, cos theta is 90. I am just saying if I want the field, if I want the potential energy x, then I can say that that is 2 x i cap plus 3 y j. So they have given field. You want to know the potential, how much it is? 2 x i cap plus 3 y j cap. So E dot dr you have to integrate from the point which is known to you. You need to know potential at a point which could be infinity also. So from that point E dot dr you have to integrate r on to r2 like the way we have done the previous loop. So v1 minus v2 you will get always. You will not be able to get what is v1 and what is v2 because your limits can't be infinity to something. When you integrate you get change in potential. It is simple right? First one, what about? Yeah I know. This is normal at CRT I guess. 80,000 feet is how many meters? That many. That is a lot more than that. This is about 5.4 kilometers. 5,000 meters? Yeah 5,000 meters. So I have climbed 80,000 feet. At the point 2.56. Can't do 0.56 for the first one. 0.6. This is r. No it's r. It's r only right? Yeah. Now g effective at a height is g0 has to become g0 by 2. So h will be what? R2 minus r. Something root and all will come. R2 minus r. h is root 2 minus 1. h is root 2 minus 1. This is roughly 0.5 times r. These are all simple questions. But then this also is simple like this. This one. Try this second one. That will be the exact answer. Exactly. Yes. Whoever gets the right answer gets a break. Earn your break. 9.7 kg. 0.9 kg. Done this properly. g effective will be how much? What is the formula for that? G is equal to g0 minus omega square. Minus omega square r e cos square lambda which will become 1. This is what it is. Omega is how much? Y by 12. 24, 60, 60. Radius of perth 6000. This much meters. Have you done all these calculations? The answer is 0.997. This will give you effective g. The weight will be m into g and the mass is m into g divided by g0. So 1 minus omega square r e by g0. This is the mass because m is 1 kg. So please attempt these questions. It starts from question number 1. Second one. Second one. So you have done something similar to the first one. You have done at what height below? Yes. You should find it quickly then. This is the first one. First, first. What is the first one? Any of you got the answer already? Yes. How much? Minus per ounce. Okay. Let me check. What is the value? Sir, omega square. Should I do it? What exact answer? 0.0015 into radius over. How much is that? Should I do it now? Here g effective is how much? This is equator. Here g effective is g0 minus omega square r e. This is a g effective here. So basically I need to find out at what height over here. g effective becomes equal to that. Yes or no? Okay. So this is let's say h. So basically g0 divided by 1 plus h by r e the whole square. You have to find out where it will become equal to g0 minus omega square r e. This equation all of you got? What is this equation? Now assuming, omega is very small. So this effect of rotation is very less, which is a reasonable assumption. We can assume h is very less compared to radius. So I can write this down as g0 1 minus 2 h by r e like that. And then you can further simplify it. g0 g0 will be gone. So omega square r e will be equal to g0 2 g0 h by r e. So h will come out to be omega square r e square divided by 2 g0 which is roughly 10 kilometers. All of you understood this? So whatever, see the thing is that if you are at equator the effective g there is 10 kilometers above the north pole. What it means? There are two questions. So both of them. Calculations are there. But try to get it yourself. I think simple. The first one is simple Kepler's law. r 1 t square is proportional to r cube. t 1 square by t 2 square is r 1 cube by r 2 cube. r 1 and r 2 are their distances. Ratiovera is just a mass and the sun between the earth and the sun. So r 2 by r 1 is equal to t 2 by t 1 2 by 3. So t 2 is how much? t 2 is 1.88 years. So 1.88 raised to power 2 by 3. This is what? Roughly how much it is? 1.5. 1.5 2. Can you do the second one? This is from ncrd, similar to ncrd. What is the answer? 6.0 Yes, 6 into transfer 24. So time period of moon is given. So how is it related to the mass of the earth? 4 pi square into r cube. Using direct expression? You can just use mv square by r. mv square by r is equal to g m 1 and 2. So suppose this is moon and this is earth. This distance is given. Orbital radius is given. So mv square by r, where small m is the mass of the moon, is equal to g capital M is mass of the earth divided by r square. So from here you get velocity to be equal to root over g m by r. Then 2 pi r divided by v is the time period of revolution for the moon. From there you get the mass of the earth. Just by knowing the time period revolution of moon, you can get the mass of the earth. So you don't directly calculate the mass of the earth or distance of earth on the moon. So these are the indirect ways to find out the mass of earth or mass of moon like that. So these are all the questions of H.C. Varma. We are almost done with all the questions. First one is same. First one is same? Same experience. Okay, don't do the first one. Do the second one. Everything is given. Radius of earth. Radius of earth. 6400 kilometers. This is all the direct formula. So you should get the correct answer. This is a problem. How much? This is a square root of 84000. This is approximately 3.5 million. In pursuit scheme, it's not 3.5 million. This is the second question. The first part, what is the answer? root of 84000. 6.9 kilometers per second. Part A is... Can't you solve this? mv square by r is equal to gm1 m2 by r square. Same thing you have to do again and again. t will come out to be root over 2gm by r surface. r is what? Small r is 2000 plus 6400. You have to substitute here to get the value of velocity. Then, kinetic energy. Once you get velocity, kinetic energy is simply you can find out like this. 2.5 to 10. 2.38. 10 is for 10. This is the kinetic energy. c part is potential energy. It will be minus of 2 times of kinetic energy. minus of mv square is the answer. t part, time period of revolution. You know the direct formula, right? You can do it like this also. 2 pi r by v.