 okay first one all of you are giving the wrong answer draw the free by diagram yourself you will see okay so if this is the object alright this is the object it is falling under the gravity so its acceleration will be g alright then it will of course have mg force because of the earth and let's say up thrust is given as b so mg minus buoyant force should be equal to mass time acceleration which is mg only a is g so b comes out to be 0 buoyant force will not be applicable when it is falling under the gravity okay so vertically if things are at rest then only buoyant forces m into g otherwise otherwise you have to take g effective just like you know the weight of the object changes when it is inside the lift just draw the free by diagram you will be able to see what could be the effective g so buoyant force is rho v g effective when it is freely falling g effective is 0 anyways question number 2 the spring balance a reads 2 kg when the block m suspended from it a balance b reads 5 kg when the beaker with liquid is put on the ramcha you're there ramcha and kondi why you're on skype you're not talking anything just keeping silent you have to say okay every two minutes say something interact more the spring balance a reads 2 kg with the block I am suspended from it a balance b reads 5 kg when a beaker with a liquid is put on the balance the two balances are now so arranged that the hanging mass is inside the liquid in this situation okay second one you're saying see the balance a will read less than 2 kg of course it will read less than 2 kg because now the buoyant force is acting and we will read more than 5 kg that is correct because if liquid apply buoyant force on mass m upwards then mass m will apply equal and opposite force on the water downwards and hence according to Newton's third law option c should be correct okay let me so this is just one of the option there can be more than one option correct so can you see whether other options are also just evaluate let me quickly check the answer for you c is correct one more option is correct of course if c is correct b is also correct because b is subset of c so don't forget to mark options which are subset of each other okay anyways we'll move to next question see in case you have already seen these questions no point attending the session and you know you can as well tell me I can send you few new questions otherwise if you have solved this question couple of times you're wasting time okay now do this sir that is ridiculous yes amok that is ridiculous I can understand your emotions but J is the exam where you try to reject students based on their concept based on how careful they are how careless they are how much attention to detail they have but yes I can understand third one c amok is getting okay almost not almost everyone four people have got c as the option a vessel contains oil over mercury a homogeneous sphere floats with half of its volume immersed in mercury and half in oil the density of material of the sphere is what so just you need to find out the weight of the fluid that is displaced and equate that to the weight of the sphere okay so here let's say this is the vessel so we have the lower portion of course the upper one will be oil right and the lower one have to be mercury so if it is half submerged then v by 2 of the mercury is displaced so rho mercury v by 2g is the buoyant force because of the weight of the mercury and rho oil into v by 2 into g is the buoyant force because of the displacement of oil this has to be equal to the mg force okay mg is what the density of the material into v into g right the density of the material is rho mercury plus rho oil divided by 2 okay which is what 13.6 plus 0.8 which is 14.4 divided by 2 that is 7.2 okay so option c is correct over here what about fourth okay fourth also you are saying c let's see two rods of different material having coefficient of thermal expansion alpha 1 alpha 2 Young's modulus y1 by 2 they are fixed between two rigid massive walls the rods are heated such that they undergo same increase in temperature there is no bending and alpha 1 divided by alpha 2 is equal to 2 by 3 thermal stresses two rods are equal provided y1 and y2 is what now see this thermal stress sorry thermal strain is given as alpha delta t isn't it because delta l is what delta l is l alpha delta t so delta l by l which is strain is alpha delta t okay so epsilon 1 is alpha 1 delta t and epsilon 2 is alpha 2 delta t okay then a stress by strain sigma 1 by epsilon 1 is given as y1 so basically sigma 1 is y1 into epsilon 1 which is y1 alpha 1 delta t okay this should be equal to y2 alpha 2 delta t right because thermal stresses are equal now alpha 1 by alpha 2 is 2 by 3 so if I remove delta t so from here I will get y1 by y2 to be equal to alpha 2 by alpha 1 which is 3 by 2 okay so option c is correct over here fine we will move to next question now quickly solve this one I think this you can do straight away okay again see the first of all the velocity will increase because of gravity right so in order to keep a1 v1 equal to a2 v2 the area will decrease isn't it so it cannot be a so we have a2 to be equal to a1 times v1 by v2 right v2 is v1 plus sorry v2 square is v1 square plus 2 into g g you can take as 10 to g into s which is 0.15 okay so this is v1 is 1 plus 3 so this is 4 so from here I will get v2 to be equal to 2 meter per second fine so a2 will be half of the earlier area so half of the earlier area is c okay so that is how you have to solve this question I will move to the next one these two questions they are from j advance 2015 Saimir will explain how he has got that answer a and b is the correct answer both a and b see the curve will end the curve plotting will end where the material breaks okay so q can take this much stress okay whereas p can take maximum stress of slightly more than q all right so this is the maximum stress p can take and this is the maximum stress q can take and hence p is more p has more tensile strength than q so in order to understand this you need to have clear understanding of what is tensile stress and what does ductility and all means okay and also the strain of p is more than q okay for the same amount of stress if you have same amount of stress the strain of p is more than q fine so that is the reason why p is more ductile than q okay so this is how you have to solve this question now question number eight this is again a very different kind of question see if you know how to approach it okay spherical body of radius r consists of a fluid of constant density as and it is in equilibrium under its own gravity okay if p r is a pressure at a distance r which is less than the radius of the sphere then which of the options are correct so suppose you have let us say a dr width you are imagining a strip okay you are imagining a shell at a distance r so let's say the thickness is dr okay so there will be pressure right so let's say pressure from outside is p okay and pressure from inside is let's say p plus dp fine so if you consider that thin strip only consider that thin strip so from inside p plus dp is applied and from outside p is applied so pressure will increase as you go down go towards the center okay at the same time there is a gravitational attraction that is pulling this shell okay so let's say that gravitational attraction force is f g all right so if I write the force balance equation over here just taking the shell into consideration I can write you know since the shell is stationary so net force should be zero so p plus dp minus p okay this into this into 4 pi r square this is a net force due to the pressure okay minus the gravitational force should be equal to zero getting it so dp into 4 pi r square minus f g is equal to zero now you need to find out force of gravity on the shell fine so force of gravity on the shell because of the outer material will be zero so force of gravity will be only because of the inner material all right so you just find out force of gravity due to the inner material let us say mass of the inner material is small m okay so the force due to the gravity on that entire shell will be g m into rho into 4 pi r square dr divided by r square okay where m can be you know m can be written as small m will be written as rho into 4 pi sorry 4 by 3 pi r cube okay so like that when you substitute you will get dp and dr in the equation and then when you integrate pressure okay when you integrate the pressure from zero to p the r will vary from capital r to small r like that you will get pressure as a function of r and then you can you know simply evaluate which of these options are correct no aditya that is not correct pressure at r equal to zero is very high it tends to infinity actually no weight weight yeah pressure at r equal to zero is very high okay so option so I will just tell you the options which are correct you can try out from here I think it is just differential equation and I think you should try out after the class and let me know if you don't get it so b and c are the correct options so you'll get a differential equation just solve it and after this it becomes straightforward this came in j advance 2015