Physics: Post your doubts here!

[Thought blocker]

this is a2 ppr

woops! Scarred.

 

[ZaqZainab]

MCQ QUESTIONS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf

Q2,7,14,16,22,23,25,26,27,28,29,31,32

Please help with even one of these questions

Q2 it can't be A and B as ohms and watts are not base units
now we have C and D
We know Power=Voltage*Current
Making Voltage the subject Voltage=Power/current
there is another formula saying power=energy/time
so now we have voltage=energy/time*current
another formula says energry=mgh,0.5mv^2.......... i will use mgh
so now we have voltage=mgh/time*current
g is ms^-2<-- from SI BASE units of acceleration=v/t
now we know the base units of
mass=kg
g=ms^-2
h=m
time=s
current=A

lets put it in voltage=mgh/time*current
Volts=kg*ms^-2*m/A*s
Volts=kg*m^2*s^-3*A^-1
7. P is 1/6 OF Earth
earth is 9.81 or whatever the value given in that year
so P=1/6 * 9.81 this will give g(The acceleration of free fall)on P
The acceleration of free fall on P=1.635
Weight=Mass*Gravity
Weight=30*1.635=49.05

 

[Thought blocker]

MCQ QUESTIONS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf

Q2,7,14,16,22,23,25,26,27,28,29,31,32

Please help with even one of these questions

2) I upload a file watch page 4
7) accn of free fall on moon is aboutt 1/6 that of earth so, accn of free fall on moon is 1.6 m/s
so W=mass * accn of free fall
= 30 * 1.6 = 48 so about 49 N
14,16,22 Suchal Riaz ZaqZainab midha.ch usama321
23) Energy = 0.5*F*change in length..
=0.5*6*(40-70) (Convert it to meter) that is 0.03 = 0.09 J
25,26,27,28,31 Suchal Riaz ZaqZainab midha.ch usama321

29) Sin(θ) (which is basically the spacing divided by the distance)

So since sin(θ) = nλ/d

If λ is 1/2 and d is doubled the sin(θ) is reduced by 1/4

So 3.0mm/4 = 0.75mm

32) P=I^2*R

Power in Y = 0.5^2 * 2 = 1/2
Power in X= 1

1/2/1 = 1/2

 

[Thought blocker]

Q2 it can't be A and B as ohms and watts are not base units
now we have C and D
We know Power=Voltage*Current
Making Voltage the subject Voltage=Power/current
there is another formula saying power=energy/time
so now we have voltage=energy/time*current
another formula says energry=mgh,0.5mv^2.......... i will use mgh
so now we have voltage=mgh/time*current
g is ms^-2<-- from SI BASE units of acceleration=v/t
now we know the base units of
mass=kg
g=ms^-2
h=m
time=s
current=A

lets put it in voltage=mgh/time*current
Volts=kg*ms^-2*m/A*s
Volts=kg*m^2*s^-3*A^-1
7. P is 1/6 OF Earth
earth is 9.81 or whatever the value given in that year
so P=1/6 * 9.81 this will give g(The acceleration of free fall)on P
The acceleration of free fall on P=1.635
Weight=Mass*Gravity
Weight=30*1.635=49.05

Check If I did any mistakes..

 

[xxxt]

There is a point on the line joining the centers of the earth and the moon where their combined gravitational field strength is zero.is this point closer to the earth or to the moon? calculate how far it is from the center of the earth.....( mass of the earth = 6.0*10^24)(mass the moon= 7.4*10^22) and G= 6.67 *10^-11....can anyone please tell me how to solve a qn like this???

 

[midha.ch]

they would give you the apparatus so you will decide like that. if you send me the confidential instructions of that year i can tell you how it has to be done. ok?

this is planning question! You have to plan how you would carry it out.

 

[ZaqZainab]

Check If I did any mistakes..

well i only just did 2

 

[Thought blocker]

well i only just did 2

But seeing my solutions.. Was I correct on which I took an attempt ?

 

[ZaqZainab]

But seeing my solutions.. Was I correct on which I took an attempt ?

totally correct

 

[Thought blocker]

totally correct

The rest I was not able to solve plz all of u help me and him

 

[talha-_-khan]

The Fly wheel has a small magnet of mass M mounted on it. Each time the magnet passes the coil, a voltage pulse is generated , which is passes to the c.r.o the display of the c.r.o is 10 cm wide . The flywheel is rotating at a rate of about 3000 revolutions per minute . Which time-base setting will display clearly separate pulses onscreen.

(some one please answer this question with explanation and from where can i learn about c.r.o as i can not find anything useful in book )

 

[1357911]

MCQ QUESTIONS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf

Q2,7,14,16,22,23,25,26,27,28,29,31,32

Please help with even one of these questions

The rest of these questions answeers.
Q14- anticlocwise moment = clockwise moment
W*2a = W*a + F*h

Q-25 P is displacement. that is obvious. while finding q look at the axis.. the TIME is on the x axis. so its frequency

Q-27 Antinode. that word is specifically given. so its lamda/4.
so lamda/4 = x
4x = lamda
4x = speed/frequency
make f the subject
f = c/4x

Q-28 d sine theta = n*lamda
where d = 1/N
so 1/N sin theta = n*lamda
where n = 3
make sin theta the subject, ull get the answer.

 

[Thought blocker]

The rest of these questions answeers.
Q14- anticlocwise moment = clockwise moment
W*2a = W*a + F*h

Q-25 P is displacement. that is obvious. while finding q look at the axis.. the TIME is on the x axis. so its frequency

Q-27 Antinode. that word is specifically given. so its lamda/4.
so lamda/4 = x
4x = lamda
4x = speed/frequency
make f the subject
f = c/4x

Q-28 d sine theta = n*lamda
where d = 1/N
so 1/N sin theta = n*lamda
where n = 3
make sin theta the subject, ull get the answer.

ty

 

[usama321]

MCQ QUESTIONS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf

Q2,7,14,16,22,23,25,26,27,28,29,31,32

Please help with even one of these questions

Not quite sure about 14, though B makes no sense whatsoever, C doesnt either, because the weight of the ladder is acting clockwise, while the normal force is acting anticlockwise, so they should not be added. For D, I don't see how it can be 2FH. For A, it is taking moments from the diagonal of the wall. F would cause the ladder to turn clockwise, and so would the weight of the ladder. However, the contact force W, which is at a distance of 2a from the wall, would act counterclockwise, thus balancing out the two other forces.

16 Did this in terms of work done. Work is always done against some force. In this case, a force is acting on the particle towards the right. Now, we are not moving against the force but with it a distance of s meters. There is no force against which to act in the vertical direction, thus we don't take in account the distance moved y.

22 elastic deformation is when an object is extended and then the loads removed, the object returns back to its original shape. If the object returns back to its orginal length, then it is elastic deformation. One explaination for the curve in the graph could be a different scale being used, though i am not sure.

25: amplitude is the point of maximum displacement, so p is just displacement. The time axis explains it is frequence and not wavelength.

31: Five electrons so it is negative. Thus the force would be upwards. Calculate the electric field strength 5000/0.008 = 625000. Now charge on five electrons would be 5* 1.6*10^-19

Now 625000 = F/5*1.6*10^-19
F = 5^-13 Answer C

 

[usama321]

The Fly wheel has a small magnet of mass M mounted on it. Each time the magnet passes the coil, a voltage pulse is generated , which is passes to the c.r.o the display of the c.r.o is 10 cm wide . The flywheel is rotating at a rate of about 3000 revolutions per minute . Which time-base setting will display clearly separate pulses onscreen.

(some one please answer this question with explanation and from where can i learn about c.r.o as i can not find anything useful in book )

6000 revs per minute, that is 250 rev per second. As each revolution causes one pulse, just calculate the time for one pulse by 1/250 that is .004s. I can't give an exact answer but a time base of around 1 to 2ms should work.

 

[mehria]

MCQ QUESTIONS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf

Q2,7,14,16,22,23,25,26,27,28,29,31,32

Please help with even one of these questions

22) from the end point of the limit of proportionality to Y we have elastic limit...after Y it'll be permanently deformed so frm X to Y it'll be elastic but not plastic

25) it is a very simple n easy question... as d is not the maximum displacement frm the mean position so it won't be the amplitude. it's just a displacement frm the mean position... for h-t graph one wavelength represents a time period while for s-d graph it will be a wavelength...

26) intensity is directly proportional to (amplitude)2 and inversely proportional to (distance)2 so in this question we will take (amplitude)2 inversely proportional to (distance)2... so by increasing r, A will decrease i.e. 1/r=8 so 1/2r=4 by doubling the r, Amplitude will be halved.

27) f=V/2l
V= speed and l= wavelength (x)
as from the node to antinode it is a half wavelength so full wavelength will be 2l
AND f=c/(2x2)l = c/4l

28) sinθ=nλ/d
for the third order of diffraction, wavelength = 3λ
d=1/N where N is the number of lines
so: sinθ=3λ / 1/N => sinθ=3λN

31) F=eV/d
as it's gaining 5 electrons so; e= (5) x (1.6 x 10^-19) = 8 x 10^-19
so F= (8 x 10^-19) x (5000) / (0.8 x 10^-2)= 5 x 10^-13 N
it is negatively charged so it will move upwards
remember to chnge cm into m........

 

[sudeep1097]

2) I upload a file watch page 4
7) accn of free fall on moon is aboutt 1/6 that of earth so, accn of free fall on moon is 1.6 m/s
so W=mass * accn of free fall
= 30 * 1.6 = 48 so about 49 N
14,16,22 Suchal Riaz ZaqZainab midha.ch usama321
23) Energy = 0.5*F*change in length..
=0.5*6*(40-70) (Convert it to meter) that is 0.03 = 0.09 J
25,26,27,28,31 Suchal Riaz ZaqZainab midha.ch usama321

29) Sin(θ) (which is basically the spacing divided by the distance)

So since sin(θ) = nλ/d

If λ is 1/2 and d is doubled the sin(θ) is reduced by 1/4

So 3.0mm/4 = 0.75mm

32) P=I^2*R

Power in Y = 0.5^2 * 2 = 1/2
Power in X= 1

1/2/1 = 1/2

16 is A:
because R is moving upwards and rightwards, we only need to consider rightwards as that makes the difference, because force is experienced towards right by the positive charge so moving upwards at 90 degrees doesn't make any difference in a uniform magnetic field. As positive charge moves towards direction of force the potential decreases as charges attract each other ( farther charges are more potential energy is present).

25 is B as p is not amplitude but just displacement and q is time period because graph is height against time.

26: D
We have I is directly proportional to A^2 and inversely proportional to x^2 so A^2 is inversely proportional to x ^ 2.
Hence, at P 8^2 is directly proportional to 1 / x^2 at Q A^2 is directly proportional to 1 / (2x)^2
Solving and removing proportionality sign we get (64/A^2) = 4 so A is 4.0 at Q.

31 is C :
E=V/D = 5000 / (0.8/100)
Now, electrostatic force (F) = q x E where q is charge of one electron.
But since the oil droplet has gained 5 electrons F = 5 x 1.6 x 10^-19 x E
and since it is negatively charged it is attracted towards positive that is upwards.

 

[Thought blocker]

16 is A:
because R is moving upwards and rightwards, we only need to consider rightwards as that makes the difference, because force is experienced towards right by the positive charge so moving upwards at 90 degrees doesn't make any difference in a uniform magnetic field. As positive charge moves towards direction of force the potential decreases as charges attract each other ( farther charges are more potential energy is present).

25 is B as p is not amplitude but just displacement and q is time period because graph is height against time.

26: D
We have I is directly proportional to A^2 and inversely proportional to x^2 so A^2 is inversely proportional to x ^ 2.
Hence, at P 8^2 is directly proportional to 1 / x^2 at Q A^2 is directly proportional to 1 / (2x)^2
Solving and removing proportionality sign we get (64/A^2) = 4 so A is 4.0 at Q.

31 is C :
E=V/D = 5000 / (0.8/100)
Now, electrostatic force (F) = q x E where q is charge of one electron.
But since the oil droplet has gained 5 electrons F = 5 x 1.6 x 10^-19 x E
and since it is negatively charged it is attracted towards positive that is upwards.

Genius mode activated ?

 

[yousef]

Hi ,
mass of gamma to be used is mass of electron , but how ? it should be as i guess 2* mass of electron as there are 2 moles

 

[Thought blocker]

The Fly wheel has a small magnet of mass M mounted on it. Each time the magnet passes the coil, a voltage pulse is generated , which is passes to the c.r.o the display of the c.r.o is 10 cm wide . The flywheel is rotating at a rate of about 3000 revolutions per minute . Which time-base setting will display clearly separate pulses onscreen.

It is been replied.

6000 revs per minute, that is 250 rev per second. As each revolution causes one pulse, just calculate the time for one pulse by 1/250 that is .004s. I can't give an exact answer but a time base of around 1 to 2ms should work.