00:02
Hi, here this given problem it includes three different problems which are based upon archimedes principle.
00:09
In the first one there is a wood log whose weight in the air that is given as w in air it is equal to 1 ,540 newton.
00:37
So its mass will be given by let it be capital m that is its weight in the air divided by acceleration due to gravity means this is one 540 divided by 9 .8 kilogram then mass of the lead which is attached with it small m that is given as 34 kilogram its density density of the lead that is given as 11 ,300 kilogram per meter cube so volume of the lead v lead will be given by mass upon density means this is m small m by density of the lead now suppose volume of the woodland log, suppose this is v, capital v, then the volume of water displaced, that will be total volume, volume of the wood log plus volume of the lead.
02:29
So upthrust acting on it, upthrust acting on the system, that will be equal to the weight of the water displaced and as the system is in equilibrium so it will be equal to the total weight of wood log plus lead so we can see total volume of the water displaced into density of the water into g that is equal to weight of the wood log 1 ,540 plus weight of the lead m into g 34 into 9 .8 or this v plus v lead that is equal to 1540 plus 34 into 9 .8 and and divided by density of water in mcs 1000 kilogram per meter cube into g which is 9 .8.
03:57
So density of wood log will be sorry volume of the wood log will be equal to 1540 plus 34 into 9 .8 divided by 980 980 minus volume of the lead which we have already seen here that is m mass of the lead divided by density of the lead small m by row lead so plugging in all the known values here this volume actually it will be given by volume equals to mass by density so mass of the wood log divided by density of the wood log which is missing we have to find it and that is equal to for the first term it comes out to be equal to 0 .191 and for the second term means volume of the lead this is 0 .003 so it comes out to be equal to 0 .188 or finally density of the wood log will be equal to 1540 divided by 9 .8, this is the mass of the wood log, into 0 .188.
05:24
And here it comes out to be equal to 835 .9 kilogram per meter cube, which is the answer for the first problem here.
05:35
Then there is one more problem in which average density of iceberg that is given as row of ice is equal to 917 kilogram per meter cube and average density of sea water that is row of water is equal to 1 ,0 and 402 kilogram per meter cube.
06:17
So fraction of the volume of the iceberg under sea water that is given by an expression v by v.
06:48
Volume small v is the volume which is dipped into the water capital is the total volume of the iceberg and that is given by density of the iceberg divided by density of the sea water.
07:01
So here it will be 917 divided by 1042.
07:06
Therefore, percentage of the volume which is emerged in the water, that will be given by 917 into 100 divided by 1042 and it comes out to be equal to 88 percent, which is answer for this second problem here.
07:26
Then in the third problem diameter of the balloon here that is given as 10 meter so its radius will be r equals to half of diameter means 5 meter mass of the people there are two people each having 70 kg so the combined mass will be two times of 70 means this is 140 kilogram density of air that is given as row air is equal to 1 .16 kilogram per meter cube...