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AS Physics P1 MCQs Preparation Thread.

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Ok my problems in m/j 2007 paper. Hope someone can help me out here,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q10) the answer is B, all i dont get is shouldnt it be p2-(-p1) in the numerator, and then becomes p2+p1? why is it p2-p1?
Q13) The answer is B, the only problem i have is, how do we find the perpendicular distance?
Q20) the answer is B, i dont get it completely, whats tension and compression is such scenarios?
Q37) The answer is C, how do they have equal resistances?
PLEASE HELP, I WOULD TRULY APPRECIATE IT, THANKYOU :)

i have the same questions too :(
 
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q 19 and 37 for 37 i am sure that its B then why its C :S pls help needed :(
pls help needed any one plsssss :( and these 2 mcqs too no 4 and no 33 pls friends :( :( :(
 

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pls help needed any one plsssss :( and these 2 mcqs too no 4 and no 33 pls friends :( :( :(
Q4) I varies with 1/d² (given in the question) and the further you are from the light, the less light there is, so they are proportional, hence I α 1/d². This results in:

I = k * 1/d²

So it's a straight line graph from the origin with I on the y-axis and d² on the x-axis.


19 is just kinetic model theory. Water at 50°C is all liquid, so the have a range of speeds. Since the molecules are held together, they do have attractive forces. And at 50°C some molecules can definitely leave the liquid due to evaporation.

I did 33 by calculating the current using V/(total resistance). Then find the voltage of X and Y using V = IR and calculating the difference in their values. Long, but I got the right answer.

Total V = 12
Total R = 1000 Ohm

So Total I = V/R = 0.012 A.

Current going into X = 0.008 A, so voltage at X = 0.008 * 500 = 4V.
Current going into Y = 0.004 A, so voltage at Y = 0.004 * 2000 = 8 V.

Difference = 8-4 = 4V.

Q37 is a little tricky. The voltmeter is connected to the thermistor, NOT the LDR. That means if you increase the resistance of the thermistor, you will get a higher voltage reading. Since a thermistor has a NTC (negative temperature co-efficient), high resistance is at a lower temperature.

Now, for the LDR. If you want the voltmeter of the thermistor to have a higher reading, you want the resistance of the LDR to be lower. You can verify this using the potential divider formula if you want. And a lower resistance of the LDR means more brightness, i.e. high illumination.
 
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What is best definition for Power?
1. Work done per unit time
2. Force x velocity?
 
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Q4) I varies with 1/d² (given in the question) and the further you are from the light, the less light there is, so they are proportional, hence I α 1/d². This results in:

I = k * 1/d²

So it's a straight line graph from the origin with I on the y-axis and d² on the x-axis.


19 is just kinetic model theory. Water at 50°C is all liquid, so the have a range of speeds. Since the molecules are held together, they do have attractive forces. And at 50°C some molecules can definitely leave the liquid due to evaporation.

I did 33 by calculating the current using V/(total resistance). Then find the voltage of X and Y using V = IR and calculating the difference in their values. Long, but I got the right answer.

Total V = 12
Total R = 1000 Ohm

So Total I = V/R = 0.012 A.

Current going into X = 0.008 A, so voltage at X = 0.008 * 500 = 4V.
Current going into Y = 0.004 A, so voltage at Y = 0.004 * 2000 = 8 V.

Difference = 8-4 = 4V.

Q37 is a little tricky. The voltmeter is connected to the thermistor, NOT the LDR. That means if you increase the resistance of the thermistor, you will get a higher voltage reading. Since a thermistor has a NTC (negative temperature co-efficient), high resistance is at a lower temperature.

Now, for the LDR. If you want the voltmeter of the thermistor to have a higher reading, you want the resistance of the LDR to be lower. You can verify this using the potential divider formula if you want. And a lower resistance of the LDR means more brightness, i.e. high illumination.
mate 19 and 37 are of other years pls do them i have given the link on previous pages :) pls
 
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q 19 and 37 for 37 i am sure that its B then why its C :S pls help needed :(
OK for 19)

Initially, the white sugar granules are small crystals, so they have a crystalline sort of structure. That gets rid of options A and D. From B and C, an amorphous solid is formed when something is supercooled and their shape isn't geometrical (like wax, toffee, candy floss etc.). This is because when they're supercooled, the atom arrangement is disordered. Polymeric substances are sort of elastic and 'rubbery', so it can't be C. So the answer is B.

37) This question has the same sort of concept for my explanation of the other Q37 (from W06), they are almost the same question. Please do give that one a read. If you still don't get it, then do let me know.
 
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OK for 19)

Initially, the white sugar granules are small crystals, so they have a crystalline sort of structure. That gets rid of options A and D. From B and C, an amorphous solid is formed when something is supercooled and their shape isn't geometrical (like wax, toffee, candy floss etc.). This is because when they're supercooled, the atom arrangement is disordered. Polymeric substances are sort of elastic and 'rubbery', so it can't be C. So the answer is B.

37) This question has the same sort of concept for my explanation of the other Q37 (from W06), they are almost the same question. Please do give that one a read. If you still don't get it, then do let me know.
thanks alot for replying may Allah bless u :) but i have doubts in second question pls expalin me tht one i am again and again getting B as my choice :S ?
 
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thanks alot for replying may Allah bless u :) but i have doubts in second question pls expalin me tht one i am again and again getting B as my choice :S ?
You're welcome. :)

For the reading on VL to be high, the resistance of the thermistor must be high (since V = IR, as R increases so will V), which means the LDR must be in the dark. So low illumination.

This is not the case for VT, because the voltmeter isn't connected with the thermistor but with a fixed resistor instead. So the V = IR relationship applies for the fixed resistor only. If you use the potential divider formula here, you will notice that as you increase the resistance of the thermistor, the voltmeter reading will decrease. So you want the thermistor to have a low resistance, i.e. high temperature.
 
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Ok my problems in m/j 2007 paper. Hope someone can help me out here,
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q10) the answer is B, all i dont get is shouldnt it be p2-(-p1) in the numerator, and then becomes p2+p1? why is it p2-p1?
Q13) The answer is B, the only problem i have is, how do we find the perpendicular distance?
Q20) the answer is B, i dont get it completely, whats tension and compression is such scenarios?
Q37) The answer is C, how do they have equal resistances?
PLEASE HELP, I WOULD TRULY APPRECIATE IT, THANKYOU :)

10... p2 is negative.. so overall change in momentum is -p2-p1.. which gives same numerical value as p2+p1... the answer is B
 
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You're welcome. :)

For the reading on VL to be high, the resistance of the thermistor must be high (since V = IR, as R increases so will V), which means the LDR must be in the dark. So low illumination.

This is not the case for VT, because the voltmeter isn't connected with the thermistor but with a fixed resistor instead. So the V = IR relationship applies for the fixed resistor only. If you use the potential divider formula here, you will notice that as you increase the resistance of the thermistor, the voltmeter reading will decrease. So you want the thermistor to have a low resistance, i.e. high temperature.
ARghhh miss the line of reading on both should be high :mad:@@@ btw thanks alot man
 
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