AS Physics P1 MCQs Preparation Thread.

[Capricedcapri]

Any one plz explain me 21 and 35 McQ of oct nov 2002.
help will b appreciated....( extremely desperate!!)

 

[ibadsiddiqi]

please?

okay so this is simple
for question 10, the rate of change of momentum is the resultant force ( newtons second law) so u find the gradient of the graph
so u apply the formula (p1-p2)/(t2-t1) which gives u B.
now question 34. The light intensity is increasing. Hence the resistance of the LDR falls. Pd is directly proportional to the resistance, so if the resistance falls, so does the pd across it. Now the new pd can be anything lower than 4V and for that there is only one option that is A.
Question 36..u find the current in the whole circuit 3=I(4+2), I=0.5
Terminal potential difference is the pd the cell provides after taking in account the pd being lost in the internal resistance..that is V=0.5x2=1. 1 volt is being lost, hence terminal pd is 3-1=2
to find power output just use the formula V2/R= 2^2/4=1
hence the answer is C. Hope this helped and GOOD LUCK for tomorrow

 

[pearl angel]

could someone pleaseee explain me uncertainty

 

[malik234]

o/n/2010/11
mcq no. 9... is anyone there to xplain plz ???????

 

[A.ELWY 7]

QUESTION 28, YOUR ANSWER IS CORRECT BUT YOUR EXPLANATION FOR C IS WRONG, LIGHT CAAAAAAN! BE POLARISED!! IT IS A TRANSVERSE WAVE... LOOK AT ANY OF YOUR SUNGLASSES, THEY POLARISE LIGHT, AND THUS PROTECT UR EYES!!!

really sry..looks like i mis-explained or got confused with another question..thnx very much

 

[umarashraf]

could someone pleaseee explain me uncertainty

post the question mate...

 

[umarashraf]

o/n/2010/11
mcq no. 9... is anyone there to xplain plz ???????

mcq 9 of both n 10 an d 11..??

 

[bloody_mary]

21 C and 35 D

Q21 i tried and failed

Q35 answer is D because:
you have to remember the formula v/E = R1/R1 + R2

as voltage across one resistor increases the voltage across the other decreases
and R= V/I in the wire
so in order to balance resistance loss you move the contact towards Y

 

[bloody_mary]

Any one plz explain me 21 and 35 McQ of oct nov 2002.
help will b appreciated....( extremely desperate!!)

sorry for late reply

 

[Soldier313]

Any one plz explain me 21 and 35 McQ of oct nov 2002.
help will b appreciated....( extremely desperate!!)

oct nov 2002.....qn 21:
find what the pressure (P) for liquid 1 is by using formula of pressure in liquids = density x gravity x height
1800 x 9.81 x 0.2 = 3531.6
find pressure of liquid 2 is = 1200 x 9.81 x 0.6 = 7063.2
hence if P = 3531.6
? = 7063.2
cross multiply, u get 2p, so i reckon C is the answer

 

[ullahabd]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
guys i need the explanation for Q17, Q33, Q34
THANKS

 

[perkypearl]

anyone can help me with collisions part ??
i always get confused there :/

 

[umarashraf]

anyone can help me with collisions part ??
i always get confused there :/

upload any particular question..???

 

[N.M]

i think this is a really simple question
but i dont know i m not able to do it right now
question # 14
pg no 10
from the topic work energy
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

the answer is D

And also question # 26
my answer is not matching the ones in the options
for q26 the answer is C
May Alah give you reward for this
plz some one
jzkAllah

 

[Capricedcapri]

sorry for late reply

ITS OK BUT THANKS FOR HELPING!!:d

 

[Soldier313]

i think this is a really simple question
but i dont know i m not able to do it right now
question # 14
pg no 10
from the topic work energy
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

the answer is D

And also question # 26
my answer is not matching the ones in the options
for q26 the answer is C
May Alah give you reward for this
plz some one
jzkAllah

qn 14:
do this: force applied by person - mgsintheta - friction = 0
force applied by person = 200sin30 - 150 = 250
work done by person = force x distance
= 250 x 3 = 750 J
so i guess answer is D

 

[leadingguy]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
guys i need the explanation for Q17, Q33, Q34
THANKS

17 )

force is 4n wil be same at both sides of the hinge, X as x is the centre.
now find the angle between the line XY and XP

finding angle..
sin(theta = 40/50 (40 is the half of the 80mm )
theta = sin-1 (0/8
theta is 53

now find the horizontal component of tension in the streched elastic XY
it wil be 4cos(53 = 2.4N
now same wil be the component frm tension in XZ

means 2.4N *2 = 4.8 ans is C i guess... (I have not seen the m.s yet cud be wrong but dont think so...)


34)

A... the terminal pottential wil be the Ir= E - IR so if R is varing it cannot be constant.

B.... I = v/R so increasing R wil reduce current

c.... the source is a battery and wil always provide a fixed amount of energy.. no means of varing R can increase the battery power...

d.... remains so wil be rite... as P= VI and same resistance wil provide same voltage and highest current , I .

 

[Jaf]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
i need help here plzzzzzz
in Q no 11,12,13,20,22,25....

I am sorry but I don't have any idea about how to do 11.

On 12, however, the answer is found as follows:

It gives us the torque. Now, torque is just the moment on an object moving about an axis and, so, the equation for moment (moment = force * perpendicular distance from pivot) will also apply here. They have given the diameter of Q to be 100 mm and, so, the radius is 0.05 m (50 mm). They also gave the torque as 3 Nm.

3 = Force (the tension in the belt) * 0.05
Force = 3/0.05
= 60.
This is found in D, so that is the right answer. You can also use the same equation, since you have found the tension, to find the torque on P by 0.075 m (the radius of P) * 60 and you also get 4.5.

---------------------------------
On 13, if I am not wrong, the answer (which I have seen in the marking scheme to be A) is found by considering the fact that if you throw something at 45 degrees to the horizontal, the kinetic energy is divided equally into the horizontal and vertical motions of the body, such that half of it is the vertical kinetic energy and half is the horizontal kinetic energy. Now, as the body starts to rise, the VERTICAL kinetic energy will start changing to gravitational potential energy, until it is reduced to zero, which will be the highest point it reaches. Although the VERTICAL kinetic energy is reduced to zero, the HORIZONTAL kinetic energy, however, remains the same (friction is negligible) and, therefore, the total kinetic energy will just be the HORIZONTAL kinetic energy, which, as I have explained above, is exactly half the original (0.5 E). Thus, the answer is 0.5E.

---------------------------------
For 20, you need to use the equation E = 1/2kx^2 (also E = 1/2final force * extension), know that the area under the curve represents the energy, and take into account the fact that the equation is for a STRAIGHT line, while this is a non-linear graph and, thus, compensate for this in your results. Assuming this was a straight line, the energy would be:

E = 1/2 * 100 * 0.002 m (the extension in the graph is in mm so divide by 1000 to give metres)
= 0.1 Joules

Now, like I have said, this would have been the case if the graph were a linear one. Since it isn't, all you need to do is make a straight line joining the two points. If you do that, you will see that the original, non-linear curve is above the straight line, meaning that the area under the original would be MORE than that under the straight line. Thus, your energy would be MORE than 0.1, but just by a tiny amount, which is 0.11 Joules. Thus, the answer is C

--------------------------------
22, just refer to a electromagnetic spectrum and you will find the answer to be 15.

25 is a bit hard...

 

[oldfashionedgirl]

I am sorry but I don't have any idea about how to do 11.

On 12, however, the answer is found as follows:

It gives us the torque. Now, torque is just the moment on an object moving about an axis and, so, the equation for moment (moment = force * perpendicular distance from pivot) will also apply here. They have given the diameter of Q to be 100 mm and, so, the radius is 0.05 m (50 mm). They also gave the torque as 3 Nm.

3 = Force (the tension in the belt) * 0.05
Force = 3/0.05
= 60.
This is found in D, so that is the right answer. You can also use the same equation, since you have found the tension, to find the torque on P by 0.075 m (the radius of P) * 60 and you also get 4.5.

---------------------------------
On 13, if I am not wrong, the answer (which I have seen in the marking scheme to be A) is found by considering the fact that if you throw something at 45 degrees to the horizontal, the kinetic energy is divided equally into the horizontal and vertical motions of the body, such that half of it is the vertical kinetic energy and half is the horizontal kinetic energy. Now, as the body starts to rise, the VERTICAL kinetic energy will start changing to gravitational potential energy, until it is reduced to zero, which will be the highest point it reaches. Although the VERTICAL kinetic energy is reduced to zero, the HORIZONTAL kinetic energy, however, remains the same (friction is negligible) and, therefore, the total kinetic energy will just be the HORIZONTAL kinetic energy, which, as I have explained above, is exactly half the original (0.5 E). Thus, the answer is 0.5E.

---------------------------------
For 20, you need to use the equation E = 1/2kx^2 (also E = 1/2final force * extension), know that the area under the curve represents the energy, and take into account the fact that the equation is for a STRAIGHT line, while this is a non-linear graph and, thus, compensate for this in your results. Assuming this was a straight line, the energy would be:

E = 1/2 * 100 * 0.002 m (the extension in the graph is in mm so divide by 1000 to give metres)
= 0.1 Joules



Now, like I have said, this would have been the case if the graph were a linear one. Since it isn't, all you need to do is make a straight line joining the two points. If you do that, you will see that the original, non-linear curve is above the straight line, meaning that the area under the original would be MORE than that under the straight line. Thus, your energy would be MORE than 0.1, but just by a tiny amount, which is 0.11 Joules. Thus, the answer is C

--------------------------------
22, just refer to a electromagnetic spectrum and you will find the answer to be 15.

25 is a bit hard...

Can you PLEASE help me out on question 15 and 26 of the following paper? If you have any idea? The answers are B for both:

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf

 

[Capricedcapri]

anyone here knows how to solve mcq 35 of o/n 2004??