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Math 4024 Paper 1

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So guys today was another hay day..physics paper went brilliant..i hope all of you had a good paper..so next is maths..post any stuff or info you have regarding it eg topic patterns,notes,guesses,Q and Ans help..lets score an A star inshAllah
 
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Guys, how do we find the area of a shape in contact with water? Its a cylinder with diameter 60 and length 80. In diagram 2 angle AOB is 120 and OB is 30
 
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abc123 said:
its an old one and thats the best i could draw
I don't see any drawing. Type the stem of the question if you can, too.
 
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oh i forgot to attach. Here
 

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You are required to find the total surface area of the cylinder in contact with the water?
 
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I saw that question. (J91/2/6).
In part (c), you found the area of the segment ADB. The curved surface area of a cylinder is 2 pi r h.
So in this part, to find the total area in contact with water you have to calculate: 2(area of segment ADB) + (120/360 x 2 pi r h).
Substitute the values and you'll get the answer.
 
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Guys what would be the best way to prepare for Paper 1..as small but important things come in this paper...Calculator is not allowed....What topics are extremely important which always come in this paper????? :O:
 
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abc123 said:
But why the length of segment? :s
What length of segment? :S I said area of segment. Obviously, the water touches the circles at the ends of the cylinder, too as well as part of the curved surface area of the cylinder.
 
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i meant the circumference of the segment (120/360 x 2 pi r h). Can't we find the height? From diagram 2 we can find out OA and subtract that from radius to give the depth? Is that possible?
 
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@abcde bro it has also came in May/June 2010...

Okay so here is my try..note: i am not good at maths :p

First of all take the body in contact with the water as a body with a rectangular front
then find the arc length by using the sector formula
then use the length 80 and the length found using sector formula and multiply them...

hmmm..i cant think any further..sorry..maths killz me
 
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abc123 said:
i meant the circumference of the segment (120/360 x 2 pi r h). Can't we find the height? From diagram 2 we can find out OA and subtract that from radius to give the depth? Is that possible?
I think you meant OD. That is possible but you don't need to do that. What will you do with the depth then? I think the depth is useless here as you need to find the area of cylinder in contact with water.
 
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hey under root 5000
works out to be 50xunder root 2

so we have to learn the value
 
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i think here is your question..a bit similar like this its from summer 09..check the mark scheme and examiner report bro
 

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oh i had the volume in mind...listen if we had to calculate the volume we would have to calculate depth right? And how would the volume been calculated?
 
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