 All right, friends, so today I have taken questions from J-Mains 2019. So these questions have recently been asked in J-Mains. And also these questions are taken from different sources. These are not the official source from which I have taken. So they are probably memory-based. So let us take one by one these questions. So you can see the first question on your screen. So there is a block of mass 10 kg which is kept on a rough incline as shown in the figure. Coefficient of friction is given between the block and the surface, that is 0.6. Two forces of magnitude 3 Newton and P Newton are acting on the block. This is P. If the friction on the block is acting upward, then minimum value of P for which block remains at rest is what is asked. So let's say friction is this. Now the block is at rest. So we can balance all this. Now there are only just these forces. P is there, friction is there, this 3 Newton external force. And there will be a component of gravity, which is mg sine theta. That will be acting like this. Another component will be perpendicular to the incline. That is mg cos theta. And then here will be normal reaction. So along the incline if we balance the forces, we will get P plus FR minus mg sine theta minus 3 is equal to 0. So the value of P will be equal to 3 plus mg sine theta minus FR. Now the value of FR is what we need to find. So the more the value of FR, the lesser the value of P is. For minimum value, friction should be maximum. So FR maximum is mu times normal reaction. Now how to get the normal reaction? Just balance the force perpendicular to the incline. So that will be n minus mg cos theta. This will be equal to 0, isn't it? So normal reaction will come out to be equal to mg cos theta. So friction force will be mu times n. So the minimum value of P will be equal to 3 plus m into g which is 10 into 10 let's say 100 sine theta is 1 by root 2 as theta is 45 degree minus mu times which is 0.6 times normal reaction which is mg cos theta. That is again 100 by root 2. So the minimum value of P will come out to be equal to 3 plus this is 40 divided by root 2. So how much this is? This you can say that this is equal to 3 plus 20 root 2. Now root 2 we know that this is 1.4, 414. So I will approximate it as 1.4. So that into 20 will be 28 plus 3, 31. So I can safely say that option 2 is correct for this particular question. Let us move to the next one. So here is another question. This is you can make out that this is from the thermodynamics section of the physics. So for path ABC heat given to the system is 60 joule and work done by the system is 30 joule. So ABC this is the scenario. For path ADC work done by the system is 10 joule heat given to the system for path ADC is asked. So for solving this question you need to be very good at the sign convention and also first law of thermodynamics you should be very comfortable with. First law of thermodynamics says delta Q is equal to delta U plus W. Now one thing you will notice for path ABC and path ADC is that initial and final points are same. So because of that the change in internal energy for both the processes will be same. So for process 1 change in internal energy you can find from that. So that is equal to delta Q minus W. So delta Q is what? Delta Q is heat given to the system. Now that is 60 and work done by the system is 30. So 60 minus 30 so that is 30 joules. And for process 2 again I will be using first law of thermodynamics delta Q is equal to delta U plus W. So for process 2 work done by the system is 10 joules. So delta Q is asked there. Now delta U will be same as the process 1 because initial and final points are same. Internal energy is a state function. So delta U is 30 plus work done by the system is 10. So 30 plus 10 so that will be 40 joules. So that is our option number 3 is correct over here. So here is a question on optics. Initially an object is kept at a distance of 10 centimeter from the convex lens. Sharp image is formed 10 centimeter ahead of the lens on this screen. So it must be a real image. It is formed on this screen. So using this information you can get the value of focal length. So object is kept at a 10 centimeter from the convex lens. So U is equal to minus of 10 centimeter and V is equal to plus 10 centimeter. So if I use 1 by V minus 1 by U is equal to 1 by F I will get the value of focal length. So this is 1 by 10 minus of 1 divided by minus 10 is equal to focal length 1 by F. So F is equal to 5 centimeter. So this is the value of focal length for this particular lens. Now the glass slab of thickness 1.5 centimeter this is thickness that is equal to refractive index in fact. So this is mu glass slab of refractive index 1.5 and thickness 1.5 centimeter is placed between object and the lens. The distance by which screen is to be shifted to get sharp image on this screen will be what? Okay. Now because of this glass slab there will be a shift of the object itself. So if you place an slab like this then what will happen is that the light will appear to be shifted slightly. Okay. It will come closer by this distance. So this lens will feel as if the object has shifted towards the lens. Okay. That shift that shift will be equal to 1 minus 1 by mu times the thickness. Okay. Now this is equal to 1 minus mu is 3 by 2 which is 1.5 so this into 1.5 centimeters so that. Okay. 1.5 divided by 3 so 0.5 centimeter the object is appear to be shifted. Okay. So the new objects image will be equal to what? It will be it was initially 10 centimeter from the lens. Now it has shifted towards the lens by 0.5 so you will be equal to minus of 9.5 centimeter. Okay. And focal length is 5 centimeter. Okay. All I have to do is to find out where is the location of final image to get the location of the screen. Okay. Wherever is the image that is where you need to put the screen to get the image. Okay. All right. So let us use this formula only again. So 1 by V minus 1 divided by 1 by minus of 9.5 is equal to 1 divided by 5. Okay. So 1 by V will be equal to 1 by 5 minus 100 divided by 95. All right. So from here 1 by V will become equal to let us take 95 common. Okay. So you can check that 5 ones of 5 then is 19 times. Right. So 19. So this will be 10. So it will be 19 minus 10. Okay. So from here you will get V to be equal to 95 by 9. Okay. 95 by 9 is what you are getting V. The new location of the image. Okay. So earlier it was 10 centimeter. Okay. So the shift. Shift will be equal to 10 minus or let us say it is 95 by 9. Right. So 9 minus 95 by 9. Okay. So the shift will come out to be in magnitude terms it will be coming out to be 5 by 9 meter. Okay. So that is why option 2 is correct over here. All right. Okay. Here is another question. A planet of mass m having angular momentum L is revolving around the sun. Aerial velocity of the planet will be what? Now aerial velocity is simply dA by dt. Right. Which we know that it is equal to L divided by 2 m. Okay. These things you must know as a fact. You should not stuck in derivation of it how it is coming because this is directly taken from NCRT. Okay. So answer is option number 2. Okay. All right. This one. The velocity of the particle at any instant is given by this the equation of trajectory is asked. Okay. So v is let us say vx is y and vy velocity in y direction is equal to x. Okay. So using this you can get vy divided by vx to be equal to x by y. Okay. Now vy is dy by dt and vx is dx by dt. So you can write it like this dy by dt divided by dx by dt. Okay. So this is equal to dy divided by dx. All right. So using this equation you will get x dx to be equal to y dy. All right. So let us integrate this. Okay. So from here you will get x square by 2 to be equal to y square by 2 plus constant. Okay. Or you can say that x square is equal to y square plus some constant k. Okay. And if you take k on the left hand side you will get y square is equal to x square plus some constant k. Okay. So that is why option 2 is correct over here. Okay. So these are the first 5 questions from J mains paper 1. All right. We will come back with another video on next 5 questions. Okay.