00:01
Here in this question we are given a scow.
00:07
Here this is the dimensions of the given scow 6 meter, 15 meter and 8 meter and this is the given angle theta and weight w bb dash.
00:20
Here we must also know that 1 ton is equal to 9806 .65 newton.
00:28
So this is the relation to be known here.
00:34
First we need to calculate the displaced weight here.
00:39
Displaced weight will be equal to the equation is row into g into the volume of scow.
00:48
Here displaced weight is already given which is 164 ton.
00:52
So 164 into 9806 .65 newton is equal to density raw will be equal to 1 ,000 inch 10 .7 .7 .7 .7 .000.
01:00
G is 9 .81 into here the volume will be equal to 15 into 8 into h.
01:10
Upon solving this we will get the immersion height h is equal to 1 .3662 meter.
01:24
So this is the value of immersion height here.
01:29
Now we need to calculate metacentric height which is represent.
01:35
Represented as gm will be equal to bm minus bg or this is equal to instead of bm we can write minimum value of moment of inertia divided by displaced volume minus b into sorry minus bg here in this question bg means the distance between buoyancy and center of gravity here, value of g is equal to 6x by 2 which is equal to 3 meter.
02:17
B is equal to depth of immersion divided by 2 which becomes depth of immersion is 1 .366 divided by 2 and this turns out to be b is equal to 0 .683 meter.
02:36
Therefore value of bg will be equal to 3 minus 0 .683 this is equal to 2 .317 meter.
02:51
Now we have to calculate the value of gm here.
02:58
We know that gm is equal to substituting the values 8 cube into 15 divided by 12.
03:10
All divided by 15 into 8 into 1 .366 minus 2 .317.
03:20
Upon solving we will get the metacentric height to be gm represents metacentric height...