 two questions trust me this is the only time in the next whatever in class till class 12 gets over this is the only time where you are focusing only on gravitation problem solving you will never get time like this ok so value it and do it learn something and go 4.2 times or minus 5 meter per second the first answer again u1 plus k1 equal to u2 plus k2 have you understood the first one but the velocities of both of them will not be same that what you will do the momentum conservation of velocity will use right because net external forces 0 ok so u1 plus k1 u1 is minus of g 10 into 20 m1 m2 divided by 1 k1 is 0 separation decreases from g into 10 into 20 divided by 0.5 and k2 there will be two kinetic energies half m1 v1 square plus half m2 into v2 square both will acquire velocities the momentum will also be conserved so in each momentum 0 will be equal to 10 into v1 minus 20 into v2 there are two equations v1 equal to 2 v2 substitute here we will get the v2 clear no doubts right second one second one involves integration little bit same is over wire of length l and mass m total mass is m mass per unit length will be a small mass is kept over here. Where do you think the net force will be vertically downwards why because if you draw a dotted line like this every horizontal line you draw you will find two symmetrical points on this wire will attract this in this way and that way so net force will come out to be like this the component in hundred are going to get cancelled away due to symmetry the net force will be down. So I just need to integrate the downward component of the force because horizontal component of force integration is 0 summation is 0 due to symmetry. So let us see that so whenever you have a circular symmetry like this what you do you assume angle and d theta like this this is theta it is a standard process d theta so this is my dm why I am using dm because the formula known to me is only for dm I do not know the formula for the entire semicircle right Newton has given us the formula for the point masses so assuming this is a point mass m this is dm it is given a we can find out r square okay is the force in this direction and which component I should integrate cos theta theta is like this so this is the force so I need to integrate cos theta component of this gm r square cos theta so this if I integrate I get the total force the dm and cos theta 2 variables are there so I will write dm in terms of d theta how much it will be m by l is mass per unit length into length of that is I d theta this one so I will substitute there gm g small m by r square dm is m by l it will be cos theta d theta what will be the limits minus pi by 2 to pi by 2 minus pi by 2 to pi by 2 because I am first going in this way and then going in that way clockwise and interclockwise I am moving so my orientation of angle is changing so if one orientation is positive or there will be negative like this what is the answer g small m capital m pi by l by r l this will be interior cos theta that is sin theta sin pi by 2 minus of sin of minus pi by 2 it will be 2 so 2 will come here and r is l by 2 pi this is the answer direction also tell this downwards m by l is mass per unit length if I multiply that with a length I will get mass of that length so if this is my dm the length of this dm is how much this d theta r d theta is the length so r d theta if I multiply with mass per unit length I will get mass of that length this kind of standard process we have followed everywhere be it center of mass moment of inertia and to find a circle or field every time we do like this only whenever we see circle or a sphere d theta all that we should get used to of these kinds the first one we have already done second one already this one see the outer shell mass is inside the outer shell outer shell will not exert any force the inner shell will act like a point mass located at the center so only inner shell will exert force this is scenario this is r1 this is r2 listen where will be r2 by 2 it will be outside r1 but less than r2 it will be somewhere here so this mass won't get affected by the outer shell the inner shell will act as if it is located at the point entire mass so only this mass have to consider both this is m1 right so g m1 m divided by the distance from the center this is the answer outer shell will have no effect first one just do the first one right now so only the gravitational force the normal force also it is through the force and the normal force tunnel is dug find the force on a particle of mass and place in a tunnel at a distance gravitational force only gravity net force will be anyway 0 right it is hanging so if we consider earth rotation no it is x1 x from the center anyone has any answer mass of earth and radius of earth can be in the answer because you know gmb by re square is mgx by r okay i will do it now suppose we have got this you can attempt the second one so you are at a distance x from the center you are here okay if you are here this portion this portion will not exert any force on you which is outside that because if you are inside a spherical shell the force due to the spherical shell of the mass is 0 so if you consider that entire earth is made of a spherical shell of different radius so after this point whatever are else they are not exerting any force on this mass so only this portion is exerting the force and this portion will act as if located at the center so mass of that portion is how much mass of earth divided by mass of the density into the volume of this portion 1 by 3 pi x cube so you will get me x cube by re cube as the mass of the portion which is this so the force is g mass of that portion which is me x cube by re cube divided by what x square distance from the center that is the answer second one so now you have to find the normal reaction in the second question do the force balancing it is a very very small length and surprisingly it does not depend on x mg by mg by 2 just draw a free word diagram the earth perpendicular distance r by 2 yes mg by 2 now at a distance x from the center of tunnel you have a mass we need to find the force exerted by the earth on this mass okay mg by 2 try mg by 2 how how do you get it sir i took a confidence normal force it balances the gravity force you stick f normal force will be in which direction this one no the theta doesn't matter it will be into sin theta exactly you don't need the theta and you can't know theta either because it's like you'll have to express theta is tan inverse but then it cancels theta is known now is x if x is known you can find theta what is the problem yeah it's a tan inverse theta right now along this direction you can balance the force okay there'll be an acceleration also in this direction because there is a net force in that direction friction is upset one component of gravity will be acting this way and another component of gravity will be acting this way this component of gravity balance of the normal reaction that component of gravity creates an acceleration okay so gravity force is how much how much is the gravity force on this it's going to be g massive sir g into massive object divide by root over r we just derived which is derived you can use that okay fine r e cube into x into the distance of the center from here to here which is how much r by 2 by cos theta yeah yes or no so r by 2 divided by cos theta this is what we have derived the previous question yes or no yes or no yes right so this is the gravitational force and if this is normal reaction the gravitational force this is theta normal reaction is gravitational force cos theta so normal reaction is this times cos theta so cos theta is gone so this is r e this is also r e right yes sir so you will have g m e m by r e square 2 r e square by 1 by cos theta into cos theta so cos theta cos theta cancels and balancing force in this direction so gravity force cos theta comes okay and the gravity force sin theta is the acceleration this way so it will come out to be tan theta when you put sin theta over here you will get a force which is equal to g m e m u r e square 1 by cos theta into sin theta so that will be tan theta this is the force which is equal to mass time acceleration okay and if your x is very small you displace little bit here it will do shm because tan theta is equal to x by r roughly so if theta is one edge to the other edge it won't do shm oscillation okay is it clear clear but oscillation and shm later on don't worry