 this question came in 1983 here when Simon here got born okay can you tell me one thing their top surface do they remain at the same level yes or no they'll remain at the same level because at the top it is atmospheric pressure right the pressure is same so along horizontal line only pressure will remain same if horizontally the system is not accelerating okay so they both have to be in the same level if they are not let us say right-hand side liquid is slightly more then pressure over here will be atmospheric pressure plus rho gh of this liquid but that is not the thing pressure both side has to be atmospheric pressure now let us say second liquid is up till here so from here I can draw a line like this okay then since it is horizontal line pressure over here let's say is p1 pressure here let's say is p2 p1 has to be equal to p2 okay now p1 is atmospheric pressure plus rho of the second liquid g into h where this distance is h okay and p2 is what p2 is atmospheric pressure plus density of the first liquid g into h right and both of them are equal because they are along the same horizontal line so since p1 is equal to p2 this implies density of first liquid has to be equal to density of the second liquid so if specific gravity of first one is 1.1 for the second one also it has to be 1.1 okay so like this you have to do you know this question any doubts on this okay I'll move to the next question so you can see there are two questions on your screen try attempting both of them it is written that it's a large open tank so you can consider that the area of process of tank is much more than the area of the holes so you can use Toshery's law what is Toshery's law I'll quickly introduce Toshery's law so is it option A wait if this is area of procession A and the hole is very small and at a depth y then the velocity of the liquid that is coming out will be equal to root 2 g y assuming A is very large compared to smaller so this is Toshery's law remember this or you can derive it also but it is good to remember now quantities of water flowing out is what quantity of water that is flowing out per second is nothing but dm by dt this is mass flow rate which is rho a into v okay so if density is same I can say that a1 v1 is a2 v2 this is not continuity equation but what is given here is that if this is square hole and this is circular hole a1 v1 here will be equal to a2 v2 there that is what it means okay if mass flow rate coming out both sides are same okay now the quantities of water coming out is same circular hole has radius r and depth 4 y okay so okay circular is at a depth of 4 y so this will be a square which is at y and this will be let's say circular which has a depth of 4 y okay so let's say this is v1 and this is v2 okay so v1 is root of 2 g y 2 g into 4 y is v2 okay so this will come out to be two times under root 2 g y there is v2 now a1 is what a1 is the area of the square hole which has side length l so l square is the area this into v1 under root 2 g y1 this should be equal to a1 of the square hole which has radius r is pi r square this is area of this circular hole this into velocity of the water coming out from there okay so when you equate it root 2 g y get cancelled out okay and r will come out to be equal to l divided by under root 2 pi okay so many of you have solved these questions before so others those who do who are not able to get it from the first go because you are seeing it for the first time don't get unnecessarily intimidated by that fact okay there will be many people who have solved this question four or five times also at times anyways sixth one what is the answer okay Vaishnav is asking how you arrive dm by dt is rho av Vaishnavi velocity is what meter per second okay when you multiply with area of cross section you get volume per second area into meter per second volume per second okay now that volume per second when you multiply with density so density into volume becomes mass but then it become mass per second so rho av it comes so I am telling you in a very crude way in case you want to you know get into the details of it you can refer the notes where we have derived the Bernoulli's theorem there we have discussed in greater detail okay sixth one any answer no one should I do it okay let me do this now there is a hemispherical portion of radius r removed from the bottom the cylinder of radius r okay so this is a cylinder hemispherical portion is removed the volume of the remaining cylinder is v and the mass is m so volume of the cylinder is given okay it is suspended by the string in the liquid of density rho it stays vertical the upper surface is at a depth h the force on the bottom of the cylinder by the liquid is what okay first of all we know that net force due to liquid in this case is what bow and force okay which is what v rho g isn't it this is the net force due to the liquid now if you draw free by a diagram of this cylinder due to the liquid there will be a force from the top and there will be a force from the bottom okay there will be sideways force also but they will get cancelled away okay but these two forces will not get cancelled all right so we can say that pressure over his is p1 so p1 into a is the force from the top but here the problem is that this hemispherical portion every point is at different depth so let's just keep it as f only okay we know that net force is v rho g so f minus p1 into a is v rho g okay and of course there will be you know this you might be wondering well what about the force due to the tension which will be from here let's say this is tension okay and what about the mg force so first of all you need to understand I'm not doing the force balance here okay I am not writing force balance equation what I'm writing here is the force due to the liquid equation okay the force due to the liquid the net force is in upward direction that is the buoyant force which is v rho g okay now this v rho g you can split it into two components the force from the top and force from the below okay so this is not force balance so f will be equal to p1 a plus v rho g okay now p1 is not the total pressure here p1 is just the pressure due to the fluid okay so what I'm trying to say here is that you know there will be a atmospheric pressure also here p a plus rho g h is p1 okay so this is p1 okay similarly the atmospheric pressure will be acting from below as well okay but we are trying to find out the force due to the liquid not because of the atmospheric pressure okay so since when you subtract the pressure due to the atmosphere will anyway get cancelled away so I'll just consider the rho g h pressure due to the liquid okay that is that is simply rho g h only rho g h so f will become equal to p1 into a now a is what pi r square so rho g h into pi r square plus v rho g this is the answer for the force from the bottom option number d so if you understand discussion properly many concepts will be clear because here we have actually utilize the fact that buoyant force is the outcome of all the forces because of the fluid any doubts on this question anything fine let's move to the next one these two okay so back so for the seventh see you you have seen a similar question before also isn't it so when the coin is on top of the wooden block when the coin is on top of wooden block then it is displacing more volume than its own volume isn't it so when the coin is in the is at the top of the wooden block it is displacing more volume than its own volume okay but when the coin is fall fell when the coin fell down then it is just displacing you know it is just displacing volume equal to its own volume are you getting it so earlier it was displacing more volume okay and now it is displacing lesser volume so h will decrease h will go down getting it because the water is getting more volume to occupy so h will reduce okay so what about l see l is nothing but this lambda only which is shown over here so even this will reduce because now the wooden block doesn't have to displace more volume earlier it was getting compressed below because of coin okay so both h and l will decrease why this will increase both will decrease right what about eighth one what is the answer for eight seventh answer d is correct d for deli all of you understood why seventh d is correct type in yes or no okay eighth okay eighth one so this is showing a graph between delta l and the load that is connected or to the other end the cross section of the area of the wire is this calculate the Young's modulus of the material so Young's modulus is simply given as sigma divided by epsilon right sigma is what force per unit area divided by this is delta l by l okay so you will get f l divided by a delta l now here we have l given as one meter so it is simply f divided by a times delta l is Young's modulus okay and area of cross section is also given so Young's modulus is simply f by delta l multiplied by 10 this power 6 so I think this is a straightforward question where f is 20 20 newtons okay and delta l is 10 is a power minus 4 this into 10 is power 6 so you'll get 2 into 10s for 11 a is the correct answer okay at times they will give you just rather than weight they can give you mass so you need to multiply that with g also all right so we'll move to the next one any doubt guys please type in in case you have any doubts no doubts