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physics paper-6

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ajmanutd said:
and also wat is a resistance wire
Click to expand...

I guess you're referring to the /\/\/\/\/\/\ shaped symbol. That would be a resistor symbol or a "resistance wire".
 
SweetSomebody

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KareemXPF said:
Well i took a long time to solve this question,,
1st.. the mark will be given for 2 points
the calculation [1] and the answer [1]
the question says .. current will be 0.5 I. ( means the current will be half the value of I 0 .. which is 0.3 ohm.. in figure 3.2. ) if the total resistance in the circuit is twice the resistance of X

1. 0.5 x 0.3 = 0.15 Amps
2. If you look at the table at the top of the page .. you will find that when current is 0.15.. the resistance will be 10.1 or 10 ohms..
Therefore the answer is 10 Ω

I hope i answered well and you understood :D
Click to expand...

but shouldnt the value of X be 5 as it was mentioned that the current will be 0.5 I0 when the total resistance is double that of X??
 
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for 2005 a)




This is probably what he wanted

as for c) the resistance is 6 ohm because the resistance of a single one was approx 2 as was witnessed in circuit one.
 
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SweetSomebody said:
KareemXPF said:
Well i took a long time to solve this question,,
1st.. the mark will be given for 2 points
the calculation [1] and the answer [1]
the question says .. current will be 0.5 I. ( means the current will be half the value of I 0 .. which is 0.3 ohm.. in figure 3.2. ) if the total resistance in the circuit is twice the resistance of X

1. 0.5 x 0.3 = 0.15 Amps
2. If you look at the table at the top of the page .. you will find that when current is 0.15.. the resistance will be 10.1 or 10 ohms..
Therefore the answer is 10 Ω

I hope i answered well and you understood :D
Click to expand...

but shouldnt the value of X be 5 as it was mentioned that the current will be 0.5 I0 when the total resistance is double that of X??
Click to expand...

No because when the value of the current was 0.5. there were 2 resistors and the X resistor. and we know that when the current = 0.5 , the total resistance = 2x. since x is there already , the resistance of the other 2 must be equal to x as well.
 
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HorsePower said:
can anyone please solve question 1 of this paper.... http://www.xtremepapers.me/CIE/Cambridge IGCSE/0625 - Physics/0625_w06_qp_6.pdf
Click to expand...

a) you measure the length with your ruler
b)basically the rope was wound up around it 5 times and the start and the end were marked. then the rope was unwinded and the distance between the marks were measured. So the rope covered the circumference of the bundle 5 times. So you just need to divide by 5 to get the circumference.
Second part , you just sub values in the equation.

c) i) this one's a bit tricky , but basically you needed to realize that you couldn't calculate the air with any degree of accuracy and therefore you should have just given the original volume to less significant figures to indicate lack of accuracy. It's a pretty messy question but oh well.
The second part is straightforward. Even if you don't have the right value of part i) , you should still attempt it with any value as you get marks for the work.
 
HorsePower

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unplugged said:
HorsePower said:
can anyone please solve question 1 of this paper.... http://www.xtremepapers.me/CIE/Cambridge IGCSE/0625 - Physics/0625_w06_qp_6.pdf
Click to expand...

a) you measure the length with your ruler
b)basically the rope was wound up around it 5 times and the start and the end were marked. then the rope was unwinded and the distance between the marks were measured. So the rope covered the circumference of the bundle 5 times. So you just need to divide by 5 to get the circumference.
Second part , you just sub values in the equation.

c) i) this one's a bit tricky , but basically you needed to realize that you couldn't calculate the air with any degree of accuracy and therefore you should have just given the original volume to less significant figures to indicate lack of accuracy. It's a pretty messy question but oh well.
The second part is straightforward. Even if you don't have the right value of part i) , you should still attempt it with any value as you get marks for the work.
Click to expand...
thnx alot :friends:
 
KareemXPF

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unplugged said:
HorsePower said:
can anyone please solve question 1 of this paper.... http://www.xtremepapers.me/CIE/Cambridge IGCSE/0625 - Physics/0625_w06_qp_6.pdf
Click to expand...

a) you measure the length with your ruler
b)basically the rope was wound up around it 5 times and the start and the end were marked. then the rope was unwinded and the distance between the marks were measured. So the rope covered the circumference of the bundle 5 times. So you just need to divide by 5 to get the circumference.
Second part , you just sub values in the equation.

c) i) this one's a bit tricky , but basically you needed to realize that you couldn't calculate the air with any degree of accuracy and therefore you should have just given the original volume to less significant figures to indicate lack of accuracy. It's a pretty messy question but oh well.
The second part is straightforward. Even if you don't have the right value of part i) , you should still attempt it with any value as you get marks for the work.
Click to expand...

About (c)
I asked my teacher.. he said that the air gaps are 5-10% of the total volume
So just find the value for a 5 or 10 % and subtract it with the total volume
I hope i helped you :)
 
KareemXPF

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unplugged said:
ajmanutd said:
and also wat is a resistance wire
Click to expand...

I guess you're referring to the /\/\/\/\/\/\ shaped symbol. That would be a resistor symbol or a "resistance wire".
Click to expand...

I think both are the same rectangular symbol
 
KareemXPF

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No no bro
thats completely enough
have you solved the papers?
You must solve all the years + what i just said above
Good luck
Any help just tell me
 
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