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ALL CIE PHYSICS DOUBTS HERE !!!
SkyPilotage:
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 :)
Rvel Zahid:
--- Quote from: SkyPilotage 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 :)
--- End quote ---
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
SkyPilotage:
--- Quote from: Rvel Zahid on April 08, 2012, 08:41:03 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
--- End quote ---
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
Rvel Zahid:
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s09_qp_1.pdf
mcq 13, 15
SkyPilotage:
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
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