 One small thing that was ignored earlier is change in density with the temperature. Suppose you have let us say liquid of volume V, you increase the temperature, will it mass change, will it volume change, so will it density change, extra change. So, how density will vary, that is what the next topic is. So, I come variation of, so density is nothing but mass divided by volume, original density let us say is rho naught, which is m naught by V naught. Now, if you increase the temperature, the density should be equal to mass remains in m naught and volume becomes V naught 1 plus gamma delta t, this is equal to m naught by V naught times 1 plus gamma delta t raise to power minus 1, do you remember binomial approximation, you remember binomial approximation goes like this, 1 plus x raise to power y is approximately equal to 1 plus x y, if mod of x is very less than 1, as in it is like 0.0001, I take it. So, similarly here, if you use binomial expansion, binomial approximation and naught by V naught is rho naught, so this is rho naught 1 minus gamma delta t, same volumetric expansion coefficient is coming, but density is original density into 1 minus gamma delta t, see here y is minus mass, here y can be anything, no y can be anything, y can be negative also, you do not know, so if y is negative it will be negative, head outs here, no let us quickly do a numerical on this, then I will proceed to next property, a solid floats in a liquid at 20 degree Celsius, with 75 percent of it immerse in the liquid, 75 percent is inside, when the liquid is heated to 100 degree Celsius, liquids temperature goes to 100 degree Celsius, fine, when this happens the same solid floats with 80 percent of it immerse in the liquid, this happens, then 80 percent is inside the liquid, fine, you need to find coefficient of expansion for the liquid, assume that volume of solid remains constant between 20 and 100 degree Celsius, volume will not expand, you will know that, fine, you need to find gamma for liquid, how much, see what will happen is that, what there is a weight of the liquid, that should be equal to weight of the object only, because buoyant force is balancing mg, where is my point, so if this is the solid, let us say this is the solid, if total volume is v, then 3 by 4 times v is inside the liquid, so buoyant force will be what, density of liquid into volume is placed which is 3 by 4 v into g, this is a buoyant force, any doubt here, this buoyant force is acting upwards, and then there will be gravity force, let us say which is mg, acting downwards, so initially rho times, this is initial density of the liquid, rho naught 3 by 4 v into g, this is equal to mg, any doubt here, no, this is your first equation, now after you heat it, density of liquid will change, fine, let us say new density is rho, so rho into, now 80 percent is inside the liquid, so 80 by 100 that is 8 by 10 times v into g, this should be equal to mg, any doubts, anything, now this v remains v only, repeating it, this volume is what, volume of the solid which does not expand, we are ignoring the expansion of solid, so it remains v, so just divide these 2 equations, you will get rho naught into 3 by 4 divided by rho into 4 by 5 is equal to 1, fine, so from here you will get rho is equal to or rho into 4 by 5 is equal to rho naught into 3 by 4, so rho will be equal to rho naught into 15 by 16, any doubts, so how will you get gamma from here, this should be equal to what, rho naught, what is delta T, 80 by 100 from 20, this is 80, rho naught and rho naught, so 80 times gamma is equal to 1 minus 15 by 60, so gamma is 1 by 80 times 1 by 16, did you get this kind of, did you follow this concept like this, so this is how you have to look this particular, this is what, 80 into 16, 1180 which is 8.47 into 10 so minus 4, 1 divided by this.