 Alright friends, so here is a question, question number 6, J mains 2019, paper 1, day 1. So initially block of mass capital M is at rest on frictionless floor and the spring is in its relaxed condition. A constant force is applied on the block S showing the figure, the maximum velocity of the block is what? Now figure is this, this is the figure, this is mass M. So let us say this is the constant force F. So this is the force that is applied on the block, horizontally we need to find the maximum velocity of the block. Now the block will accelerate forward. So slowly what will happen, spring will get extended. So spring force will increase in this direction. And if at any moment the extension in the spring is X, so K times X will be the force due to the spring. Now till this value of force F is more than K times X, till that time the block will accelerate in the forward direction as in right hand side. And because it is accelerating forward as if F is greater than K X, it will accelerate forward and since it is accelerating forward, its velocity will keep on increasing. And till what time it will increase, till this force is greater than K X. And after the point when F becomes equal to K X, after that point in time, the value of spring force becomes more than the applied force and hence during after that point onward, the velocity will start to decrease. So if I want to find the maximum velocity, it will be at a point where all the forces horizontally are first time balanced. So F minus K X naught will be equal to 0. So from here you will get the extension in the spring to be equal to F divided by K. So when spring get extended this much, the mass will gain the maximum velocity. Now all I have to do is now to apply work energy theorem. So W is equal to U2 plus K2 minus U1 plus K1. Now this force which is applied over here, the point of application will be moving by a distance of X naught. So the work done by the force will be what? It will be simply force into X naught. So force into X naught, it will be like this. This is the work done by the force. This is equal to final potential energy which is half K times X naught square. So F by K whole square plus K2. Now K2 is kinetic energy of mass at that point. So that is half m v square. Then minus of U1 is what? U1 is 0. Spring is not extended initially and also I can say this horizontal line is gravitational potential energy is 0. So since height is not changing, gravitational potential energy is not coming in the picture. U1 is 0 and K1 is also 0. Now let us simplify it further. So you can cancel out half and 2 will go that side. So you will have 2 F square by K minus F square by K which is m into v square equals to F square by K. Or the velocity will become equal to F divided by under root m into K. So this is what you are getting the maximum velocity to be. So that is option number 1. Let us move to the next question. This is I guess looking at the diagram you can easily make out it is a standard type of question. I think you might have already solved many these kind of questions. So this time in J, many routine questions were asked where students have felt that something similar they have already practiced. So I would say that this paper is straightforward if I talk about physics section. Now magnetic field because of the circular arc is mu naught i divided by 2 times the radius. And due to an arc of angle theta, it will be theta divided by 2 pi of that. This is the magnetic field due to an arc of angle theta. Theta should be in radiance. Take care of that. Now here you will see that 2 arcs are there, this one and that one. And this straight section will not contribute to any magnetic field. So this and that they will not have any magnetic field at point O because the line of this current is passing through the point where you are trying to find the magnetic field. So that is why no magnetic field because of the straight sections. Only is because of this curved section. Now if you look at the curved one, because of this if you curl your right hand in the direction of current, you will see that the magnetic field is coming out of the screen. And because of the outer curve, the magnetic field is going into the plane. So total magnetic field will be a subtraction of the magnetic field due to the inner one and magnetic field due to the outer one. So current is same. Current is 10 ampere only. So you can take a mu naught i divided by 2 and in fact you can pi theta is same for both of them. Only radius is different. This divided by 3 centimeter for the inner one 3 into 10 to the power minus 2 minus 2 into 10 to the power minus 2 like this. So just I have to substitute the value. So mu naught by 4 pi I know is 10 ratio power minus 7. So mu naught into 4 pi is 10 to the power minus 7. Current is 10. Theta is 45 degrees. So that is pi by 4. So this into 10 to the power 2. Then the second radius is not 2 actually. This is 2 plus 3. It is 5. So this is 5. So 1 by 3 minus 1 by 5. So here we will have 5 minus 3 divided by 15. So it is 2 by 15. So all I have to do is simplify this. This is 10 to the power 3. So this will become 10 to the power minus 4 and then I will get pi divided by 30 into 10 to the power minus 4. Or I can say pi by 3 into 10 to the power minus 5. So I can safely say that it is close to 10 to the power minus 5 because pi is close to 3. So I would mark option 2 to be correct over here. Let us move to the next one. So here is another question. A charge q is uniformly distributed over a ring of radius r and the ring of radius r, the height h on the axis of the ring at which electric field is maximum is what? Now we know that electric field due to the ring is kqx divided by r square plus x square ratio power 3 by 2. Where the value of k is 1 by 4 pi epsilon 0. This is the electric field. We need to find the height h on the axis at which electric field is maximum. So here the value of x is h. So instead of x I will write it as h. This is the electric field or the magnitude of electric field. I need to find what height h it will be maximum. So for that I will use derivative of electric field and I will equate that to 0 value of h for the maximum electric field. So let us differentiate it quickly. So I will get kq divided by r square plus h square 3 by 2. Then plus kqh multiplied by, now I am taking derivative of r square plus h square ratio power minus 3 by 2. So minus 3 by 2 will come out and then you will have this multiplied by, let me write it below, kqh multiplied by minus 3 by 2. So r square plus h square minus 3 by 2 minus 1 which is minus 3 by 2 into 2h. So this I have to equate it to 0. So you will see that kq get cancelled and then I am getting here 1 divided by h square ratio power 3 by 2. Then minus 3h square divided by r square plus h square 5 by 2. So 2 and 2 get cancelled. So this will be equal to 0. And then I will cancel out ratio power 3 by 2 from here. So when 3 by 2 get subtracted from 5 by 2 you will get 1. So this will be off. Now I am getting 3h square divided by r square plus h square to be equal to 1. So from here I am getting r square to be equal to 2h square and r will be equal to root 2 times h or h is equal to r by root 2. From here you will get h is equal to r divided by root 2. So that is why option 1 is correct in this case. So you can see that in this particular question they have actually tested you on the fact that whether you remember the expression for electric field because if you do not remember you will take a lot of time to derive it and in the heat of the moment you may not get it right. So it is expected that you should remember the basic formulas. And also they have tested you on a mathematical concept of maxima and minima over here. So here is the next question. 2 radioactive elements A and B have initial activities 10 and 20 respectively. A has twice the number of moles as that of B. The decay constant lambda A and lambda B will be what? So it is lambda subscript A and lambda subscript B. This is what? So what is given is D and A by Dt. It is equal to 10 curie although curie is not the SI unit. So I am let us write it like this 10 into 3.7 into 10 is for 10. This is the SI unit now in terms of disintegrations per second. And D and B by Dt. This is equal to 20 into 3.7 into 10 is for 10. Now A has twice number of moles as B. So number of A is 2 times of number of B. So moles are 2 times so even number of nucleus will also be 2 times. So this you can write it as lambda A into Na and A is 2 times of NB. And this you can write it as lambda B into NB and just divide it. So when you divide it this multiplication factor goes off it becomes 1 by 2. So 1 by 2 will become equal to lambda 2 times of lambda A divided by lambda B. So from here you will get lambda A A divided by lambda B to be equal to 1 by 4. So lambda B should be 4 times lambda A. So if lambda A is 5 only this option makes sense because lambda B over here is 4 times lambda A. So let us move to the next one. Fine. Here is another question. A conducting loop of resistance 10 ohm and the area of 3.5 into 10 is per minus 3 meter square. It is placed in uniform and time varying magnetic field B why it is called uniform because with space it is uniform but with time it is varying. You need to find charge passing through the loop in t equal to 0 to t equal to 10 millisecond. So for that let us try to find out first the value of current because charge is nothing but integral of I dt. So for that I will first find the value of flux. Flux is B into A. So it will be simply like this and number of terms here. So emf will be equal to A times dB by dt area is not changing with time. So current will be emf divided by resistance. So A divided by R into dB by dt. This is the value of current. So if B is this dB by dt is what? dB by dt is 0.4 into 50 pi into cos of 50 pi t. So the value of current will be area that is 3.5 into 10 is so minus 3 divided by resistance that is 10 into 0.4 into 50 which is 20. So this is 20 pi cos of 50 pi t. So this is the value of current. Now to find the value of charge I just need to integrate current with respect to time. So here I will get as Q is integral I dt that is 7 pi into 10 is so minus 3 integral of cos 50 pi t. So 50 pi will come in the denominator like this and integral of cos is sin. So this is sin 50 pi t. The limits are from 0 to 10 millisecond. 10 millisecond is 10 is so minus 2 seconds. So pi fortunately is off. So charge is equal to 7 by 5. So this is 5 into 5 4 is 20. So 1.4. So this is 1.4 into 10 is power minus 4 when you put the limits when you put 10 is so minus 2 as t you will get it as pi by 2 sin of pi by 2 minus when you put t equal to 0 you get sin 0. So here you will get this as 1 sin of pi by 2 which is 1 minus sin of 0 which is 0. So this is what you will get as charge. This is also equal to 140 micro coulomb. So like this you have to solve this particular question.