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Ok,so i'll get the ball rolling.
How to do Q1,b.ii?
 

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Ok,so i'll get the ball rolling.
How to do Q1,b.ii?
Radius = Diameter / 2

Area = Pi(r^2)

Percentage error = Pi (r^2) +/- (2*(0.02/0.50)) %

= 8%

NOTE: 8% IS DERIVED FROM THE EQUATION ON THE RIGHT. FOR EXPLANATORY PURPOSES, I WROTE THE WHOLE EQUATION.
 
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Radius = Diameter / 2

Area = Pi(r^2)

Percentage error = Pi (r^2) +/- (2*(0.02/0.50)) %

= 8%

NOTE: 8% IS DERIVED FROM THE EQUATION ON THE RIGHT. FOR EXPLANATORY PURPOSES, I WROTE THE WHOLE EQUATION.
Could you solve on a copy and send a shot?I cant understand whats going on.What about the DeltaA/A stuff?
 
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Could you solve on a copy and send a shot?I cant understand whats going on.What about the DeltaA/A stuff?

What do you even mean by Delta A? You sure you on the right paper?

All this is just a calculating the percentage of how much the estimated cross-section area can be away from the actual area.
 
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Could you solve on a copy and send a shot?I cant understand whats going on.What about the DeltaA/A stuff?

Since we have squared the radius, the error will double (you will get an error twice).

We already know that the percentage error can be found by

(Error/ Reading) * 100

Error = 0.02

Reading = 0.50

Both of the above are given.




Now read the first line of this comment.

Thus we will double the expression (It is not 'ExpressionS'. It is the quotient of what is in the brackets.)

2* (Error/ Reading)

Now find the percentage of the above expression (simply multiply it 100)

Did you get it??
 
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Since we have squared the radius, the error will double (you will get an error twice).

We already know that the percentage error can be found by

(Error/ Reading) * 100

Error = 0.02

Reading = 0.50

Both of the above are given.




Now read the first line of this comment.

Thus we will double the expression (It is not 'ExpressionS'. It is the quotient of what is in the brackets.)

2* (Error/ Reading)

Now find the percentage of the above expression (simply multiply it 100)

Did you get it??
But what about dividing by 2?Radius is to be 0.25 by dividing the diameter.Wont we take that into account?I've understood the answer now ,but shouldn't it be like
δA/A = δPir^2/Pir^2
By eliminating the constant Pi we get δr^2/r^2 which will give 2(0.01/0.25)=8%.I've got it now but i was just pointing out that it should be δr not δd.
What is DeltaA?

δ is the lowercase symbol for ,which is used to show change.like in mc∆θ=H
 
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But what about dividing by 2?Radius is to be 0.25 by dividing the diameter.Wont we take that into account?I've understood the answer now ,but shouldn't it be like
δA/A = δPir^2/Pir^2
By eliminating the constant Pi we get δr^2/r^2 which will give 2(0.01/0.25)=8%.I've got it now but i was just pointing out that it should be δr not δd.


δ is the lowercase symbol for ,which is used to show change.like in mc∆θ=H
Yeah you could do so; either way you will get the correct answer, and since this is a one mark question (1 mark for part 2), then only the answer is required.
 
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The new owner of the best profile shows the traits of it, modesty.

Anyway, we need a treat. You live in Lahore right?

I regret not meeting Zain_Rocks. What an incredible guy he was :'(
If you want to talk about this, my profile or those threads are the right place, not this thread.
 
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I need some advice or just tell me whether my combination is fine
Economics, Accounting, Maths and English Language
 
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