00:02
Hello, so in this question we have been told that the density of ice which we will take as row 1 is equal to 917 kg per meter cube and the density of sea water which we will take as row 2 is equal to 1 ,025 kg per meter cube.
00:18
Now for the heaviest weight that the ice can support, we have to equate the density of the system of ice and the density of, sorry, the density of seawater, those should be equal for the limiting value.
00:37
So the density, sorry, the density of the system of ice and bear should be equal to the density of seawater, which means if we take the density of this system to be row 3, so it should be equal to the density of seawater.
01:29
So it is, what is row 3? so it is the density of beer and ice system.
01:37
So it can be written as the mass of beer plus the mass of ice divided by the mass of the volume of ice.
01:58
So let us write it in symbols.
02:01
So here we have the mass of beer can be written as mv, the mass of ice as m i, and the volume of ice as v i.
02:10
So this is equal to row 3.
02:13
And now we know that this should also be equal to row 2 in order for this to float.
02:21
Now, row 2, which is the density of seawater, is given to us as 1 ,025.
02:32
So here we have mv plus m i would be equal to 1 ,025 times v...