Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 171652 times)

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #990 on: April 04, 2012, 03:09:06 pm »
Q1.Can someone shed light on the virtual earth approximatio in operatioal amplifiers?

-Is it a rule for all op-amps?
-why does the non-inv input have to be the same as the inv input? i.e why does the diff have to be zero?

Q2.How does reducing the gain cause less distortion, increased bandwidth and stability ?

Thank you in advance.

Offline Physics09

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #991 on: April 04, 2012, 06:56:30 pm »
Thankyou so much!!
But just one thing, how is the total resistance 1.5?
« Last Edit: April 04, 2012, 06:58:58 pm by Physics09 »

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #992 on: April 05, 2012, 12:31:27 pm »
Resistors Q and R are parallel with each other! Since they have the same value, the total resistance is halved so its 1/2 R .
Resistor P is in series with the equivalent of both resistors Q and R so The total resistance of the circuit is 1R + 0.5R = 1.5R

Hope it cleared it up.

Offline Physics09

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #993 on: April 05, 2012, 01:16:19 pm »
Oh right!
Thank you!

Offline Rvel Zahid

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #994 on: April 08, 2012, 04:09:58 am »

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #995 on: April 08, 2012, 06:30:59 am »
Q-25 Alright for this question, you need to identify water waves as transverse waves. For transverse waves, point particles move up and down and NOT left and right.
So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.
First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.
To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.

Q-33 Resistance R is proportional to L and 1/A.
If L is doubled, then R is doubled ----> 2R
If the radius is doubled, then the area (Pie)(2r)^2 = 4*pie*r^2, Resistance is divided by 4 ----> R/4
Therefore, The total resistance is 2R/4 = 0.5 R

Hope that helped, and feel free to post any more Qs :)


Offline Rvel Zahid

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #996 on: April 08, 2012, 08:41:03 am »
Q-25 Alright for this question, you need to identify water waves as transverse waves. For transverse waves, point particles move up and down and NOT left and right.
So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.
First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.
To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.

Q-33 Resistance R is proportional to L and 1/A.
If L is doubled, then R is doubled ----> 2R
If the radius is doubled, then the area (Pie)(2r)^2 = 4*pie*r^2, Resistance is divided by 4 ----> R/4
Therefore, The total resistance is 2R/4 = 0.5 R

Hope that helped, and feel free to post any more Qs :)


Thankyou so much man. Hats off to your explanation specially in the 1st mcq, i was just not getting the point.
Yes! now you have triggered a bomb i will be coming up with more Qs. :D

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #997 on: April 08, 2012, 09:06:45 am »
Thankyou so much man. Hats off to your explanation specially in the 1st mcq, i was just not getting the point.
Yes! now you have triggered a bomb i will be coming up with more Qs. :D

You dont have to take off your hat *doubt you were wearing one* , I'm more than glad to help!
Plus its always a challenge to defuse bombs, So Bring it on! :D

Offline Rvel Zahid

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Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #999 on: April 08, 2012, 09:53:33 am »
Q-13 Torque due to the 2 forces F is force x perp distance = 1.2 x F  Nm
        Moment due to the weight of the block = 900x0.2 Nm
Therefore F = (900x0.2) / 1.2 = 150 N ---> B

Q-15 Usually in g.p.e problems they just give you the height and you just use the number given.
But you have to know it is the height the Centre of gravity of the object moved.
So in this problem p.e is mxgx(h/2) when the tap is closed because the centre of gravity is 1/2 h from the bottom.
When the tap is opened, the new p.e of the water is mxgx(h/4) because the centre of gravity is 1/4 h from the bottom.
Therefore the loss is m.g.h/4-----> B

Offline Rvel Zahid

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #1000 on: April 08, 2012, 07:20:12 pm »
lol yeah not wearing a hat. its summer time :D
still man thanks for prompt and comprehensive response.



i didn't really got the 15th mcq,
the gravitational force shouldn't be constant on the liquid? :/ or if we r talking bout liquids we always have to consider centre of gravity while calculating G.P.E

another mcq: http://www.xtremepapers.com/community/attachments/physix1-jpg.6355/

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #1001 on: April 08, 2012, 07:46:50 pm »
When calculating g.p.e h = is the height of the centre of gravity from the bottom.

For that other mcq, you have to draw out several lines from the origin to each of the points A/B/C/D.
So you will notice that the line is steepest i.e highest gradient when it connects the origin and point C. Gradient is determined by V/I.
Since R is I/V so at point C , it is the lowest resistance.
A common error is drawing tangents to the points on the graph so be careful.


Offline ashwinkandel

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #1003 on: April 09, 2012, 10:12:36 am »
Rvel Zahid I have attached solutions to your questions:

Offline SkyPilotage

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #1004 on: April 09, 2012, 10:18:44 am »
Q1.Can someone shed light on the virtual earth approximatio in operatioal amplifiers?

-Is it a rule for all op-amps?
-why does the non-inv input have to be the same as the inv input? i.e why does the diff have to be zero?

Q2.How does reducing the gain cause less distortion, increased bandwidth and stability ?

Q3.Are the guard bands included in the bandwidth or only the sidebands?

Thank you in advance.

If anyone would be kind enough :)