 do this you want to do this or should we proceed to the next one you want to do this little different it is like this only in the question of draw your own diagram you can draw the circles right not as good as mine bio botany and zoology and both you came on your own your parents side both ways so it concludes your parent for case a i will write down the answer this is the answer try to take it anybody got this got it yes please help others okay param please please explain all of us all of us it is at a distance of x minus r from the radius of the small sphere okay so if it is at a distance r from here sorry x from here from center it will be x minus r all of you agree this from here this distance is x this distance is r so from center it is x minus r okay and we have already derived the gravitational forces how much if that is the case g m m dash divided by r cube into x minus r into x minus r into the solid and it is inside the shell we are using the direct expression which we have derived earlier the two questions back we are continuously using it the previous question as well as this one also and if it is here then x is here shell again does nothing but it is outside the solid sphere which will be like a point object so it will be g m m dash divided by x minus r whole square okay it was a simple question I thought six sentences are written x greater than two r then what will happen both will be a point mass this point mass will be located at the center its own center the shell's point mass is located here at the two forces all of you clear yes clear right okay in the c part you assume that the solid sphere is a point mass at the center on centers inside p2 this is p2 this is p1 you need to know where p1 is to know where p1 is p1 the uniform metal sphere of radius a clear please are a distance away from center no I was from the south what it wasn't funny you just said the same thing and he's like p is inside the circle it's funny he wasn't laughing he wasn't laughing he wasn't laughing I don't know a uniform metal sphere of radius a and mass m is surrounded by thin uniform spherical shell of equal mass m and radius 4a okay the center of the shell falls on the surface of the inner sphere okay so this is the center find the gradation there is a field at point p2 shown in the figure so from p1 to this this is a and this is also a from here to here are you getting it and everything lying along this line this is gm by 16 there it is gm by 900 61 p2 how much anybody else got the answer you are sitting next to him that's why you came he copied you iot equation before you i calculated i did half of work you did we did separate separately then simple simple questions these are all g means level so the at point p1 how you find it only for this one which is from center it is at a distance of how much this is 4a this one can't apply any force only the smaller one will apply you can find out easily okay and outside what will happen will apply from their own center okay i think straight forward and the earlier one was more difficult than this but same genre sir as master the third body is not required but then they want the potential what are you talking about sir they say that two masters the third they kept the experience of force extra information extra money what what is your answer 2 by root 2 plus 1 0.83 0.83 0.3 and what is the potential so what is the potential energy they are not asking potential they are asking potential energy sir what are you saying 0.83 0.83 from 2 kg you want me to solve this I got it it is not nice looking yeah it is not nice looking how do you define nice symmetrical see 2014 j vanes was very calculation oriented questions were very simple so those guys who have this habit of they just would not do anything but the concepts were very very simple you have to be good with numbers good with 14 you have to be good with numbers ok being good with numbers is an art it is not a you cannot look down upon someone who is good at numbers ok he is very few people are good with numbers fine so this one 2 kg 4 kg are kept separation is 2 meters where should a third mass no net gravitational force so it has to be between the 2 masses or not because if it is here both the masses will attract net force cannot be 0 here also it will be like that net force will get added up but if you keep it in between one force is like this other force will be like that so now they can be equal and opposite half the distance or more than half the distance less than half less than half the distance so it has to be closer to 2 kg so like if you have this basic sense you can eliminate the options also so 2 into 0.1 g into x square when this become equal to g into 0.1 into 4 stop talking total separation is 2 meters or 2 minus x whole square simplify it x will come out to be 0.83 0.83 0.82 0.82 minus 2 0.82 no because the you should focus here focus focus focus focus focus greater your potential energy of a system how do you find greater your potential energy x to mass at a time g m1 m2 by x minus of that minus of g this this divided by distance between them then minus of g this that divided by distance between these two simply you have to add it up don't care about the directions at all this one if you care about we have done it for square triangle is simpler than that so we will not do this you can just read it once we will move to next question the first question only but then I think this has multiple parts A B C what we do this what we do this 10 G 10 G 10 G 10 into capital G 10 into capital G 10 into capital G you can do u1 plus the work done is going to u2 plus k2 minus u1 plus k1 so change in potential energy final potential energy is 0 minus initial potential energy is the answer initial potential energy is minus of g m1 m2 by the radius okay coming around next one you have do the first two parts A and B A and B only I think it is 7 26 A part see field is what force per unit mass so if I multiply mass with the field I will get the force so 2 kg into this so 10 Newton 24 Newton J cap that is the force so total force is under root of x component of force that is y component of force simple field is a vector that vector multiplied by the mass will give you the force vector and then find the magnitude of the force vector B part potential minus 16 yeah potential at original is taken 0 not at infinity how are you approaching this how do you approach this take the work and moving it from what is the real thing what is given to his field do you have a relation between field and potential yes what it has E is equal to minus of daba V by daba R or you can also write down E dot dR is equal to minus of dV yes or no so when you integrate this the right hand side will give you change in potential stop talking this is V2 minus V1 final potential minus initial potential E dot is constant comes out here is what delta R so it will become E dot delta R delta R is what final position vector E dot R2 minus R1 will give you minus of V2 minus V1 your V1 is 0 because the first point is original that is what they are telling you to assume what is V2 R1 is 0 for the first point and R2 is 12 by cap for the second point R1 is 0 and R2 is 5 j cap I think you can do it now what is the answer this is 68 minus it is plus negative it is positive because they are taking 0 no minus 60 is the answer V at both just use this formula don't go by anything else V1 is 0 here because you have taken first point to be original R would have taken 0 if you assume V1 to be infinity R1 should be infinity then getting it but this formula is always valid whichever point you take as a reference is it clear so other parts we can do later as in you can do later