AS Physics P1 MCQs Preparation Thread.

[NokiaN95638]

Can any1 xplain the answer to these questions
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf
Q no 26,28,29,33
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
Q no 30, 31

 

[Soldier313]

please can someone give a detailed explanation for this, i just don't get it....a diagram would be appreciated too!.... thanx soo much and God bless

 

[Sana101]

M/J 2004 Q14) Q16)
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

thankyou!

 

[Sana101]

I need help with the following questions.
26 AND 28
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

 

[Sana101]

Q) 14 15 19 AND 27!

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

 

[Most_UniQue]

Someone help me with this . When Light intensity increases , resistance decreases across LDR and so the potential difference increases right? So shudnt it be either C or D? The answer's A in Markscheme!

 

[Sana101]

Someone help me with this . When Light intensity increases , resistance decreases across LDR and so the potential difference increases right? So shudnt it be either C or D? The answer's A in Markscheme!

View attachment 12426

When the light intensity is increased the resistance of the LDR decreases. this mean voltage across the LDR decreases as well. Since the highest voltmeter reading across it is 4, common sense would say the answer should be anything below 4 once resistance decreases!

 

[Sana101]

When the light intensity is increased the resistance of the LDR decreases. this mean voltage across the LDR decreases as well. Since the highest voltmeter reading across it is 4, common sense would say the answer should be anything below 4 once resistance decreases!

for potentiometers when the R decreases V decreases.

 

[Most_UniQue]

When the light intensity is increased the resistance of the LDR decreases. this mean voltage across the LDR decreases as well. Since the highest voltmeter reading across it is 4, common sense would say the answer should be anything below 4 once resistance decreases!

In semiconductors isnt potential difference inversely proportional to the resistance across it?

 

[NokiaN95638]

M/J 2004 Q14) Q16)
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

thankyou!

Q14- Since the system is in equillibrium all forces must act in the same direction and this is only possible in B (where all the forces act in the anticlockwise direction)
Q16- A- It is not possible since at max height V=0 son it cannot be true.
B- This is true since energy can neither be created or destroyed so total energy remains constant>
C- This is not true. Since POCM is Total m before collision= Total m after collison ( i don't know whether this is the correct explanation)
D- This is not true since after a certain time h would start to decrease when the ball starts to fall down so G.p.e does not always increase

Hope it helps.
Cheers

 

[NokiaN95638]

I need help with the following questions.
26 AND 28
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q 26- using dsin(theta)= n(lamda). Find the value of max value of n. n is max when sin(theta) is max (i.e. sin(theta)=1). n is calculated as 7.407. which means that 7 maxima are formed on one side. 7 more maxima will be produced on the other side of the central maxima since the pattern in a diffraction grating is symmetrical. so that means 14 maxima are produced but there is also the central maxima. But the total no. of maxima is 15 so the answer is D.

Q 28- Since the drop is stationary the upward force must be equal to downward force. The downward force is the weight of the charge therefore the upward force must be electrostatic force on the charge. So for the electrostatic force to act in the upward direction the charge must be negative.
now we know F=EQ and F=mg. therefore mg=EQ. so Q/m=g/E so the answer is B

 

[ibadsiddiqi]

In semiconductors isnt potential difference inversely proportional to the resistance across it?

Resistance is inversely proportional to current

 

[KurayamiKimmi]

Pwease help !!
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
Q no 12 --why is the answer A ?

 

[ibadsiddiqi]

Pwease help !!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf
Q no 12 --why is the answer A ?

This is an inelastic collision but momentum would remain conserved.
so the total momentum before collision equals momentum after collision
so (3m x v) - (2m x v) (- sign is because they are moving in the opposite direction)= (3m + 2m) vfinal
mv=(4m)vfinal
hence vfinal=v/4

 

[JD REBORN]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
Q 7

 

[ibadsiddiqi]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
Q 7

The main thing u should notice is that this is a velocity time graph, not the displacement time graph..so displacement is maximum when velocity is zero, so thats poits 5 and 15..acceleration at the point is zero. So the answer is D

 

[Soldier313]

can someone please help me with this? ms says A?! Shouldn't it be C????

and can someone please ans my prev qn too?....thanx

 

[00tanveer]

June 11 session:
16. Total anticlockwise moment about the hinge = total clockwise moment about the hinge
=> 2Fsin30*x = 10x
=> F= 10
19. Assuming, no energy is lost as heat, work done on the nail= 10J (kinetic energy is lost = work done on the nail by the wood )
so, force =work done by wood/ distance
=> force = 10/ 0.005 = 2000
34. You can fortunately do this making use of the eliminatio method.
A can't be true since R is changing so pd across r (the potential acroos cell, same thing yeah) can't be constant.
B can't be true since R s being increased which leads to an increase in combined resistance.
C can't be true since emf is ALWAYS constant.
SO D is the correct answer.

November 11 session:
.....Havent done em yet sorry!

 

[Pals_1010]

Help with this question plz?

 

[confused123]

What I don't like about this question is that they assume you know the units of specific heat capacity (which isn't in the syllabus, to my knowledge). From Chemistry, I learnt that q = m * c * ∆T

c = q / m * ∆T

Units of c = J . kg^-1 . K^-1
Which can be further be made by changing J into kg . m^2 . s^-2,

kg . m^2 . s^-2 . kg^-1 . K^-1

kg and kg^-1 cancel out, giving:

m^2 . s^-2 . K^-1

And in the question, b = c / T^3
b = m^2 . s^-2 . K^-1 / K^3

b = m^2 . s^-2 . K^-4

I hope this is the right answer. If not, I must have made a careless mistake somewhere.

well we know about specific heat capacity from Olevels. nothing new. but still it should be clearly mentioned in the syllabus. Thanks for the answer.