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Bhai busy... gimme some time
I'll solve it... they seem difficult... hope koi karley... warna me sikha dunga -
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Physics: Post your doubts here!
- Thread starter XPFMember
- Start date
I'll solve it... they seem difficult... hope koi karley... warna me sikha dunga 
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This seems easy but i messed up. Any other concepts rather than P=hdg?
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Konse year ka hain
?
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9702/12/M/J/13
the answer is D
it drove me cray!
the answer is D
it drove me cray!
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2)
3000 revolutions per minute means 50 revolutions in one second. And therefor this the frequency.
Frequency = 50Hz
Now, using the frequency, we find the time taken for one revolution -----> T = 1/f = 1/50 -----> 0.02 seconds.
Time is o.o2 seconds and so, in other words, 20 milliseconds (0.02 x 1000).
So you see, out of all the options given, 10 ms cm^-1 is the closed. So, the answer is B.
If you use 1 s cm^-1 as time base, there would be too many oscillations on the screen and wouldn't give you a good display.
Similarly, 100 microseconds or 1 microsecond cm^-1 will give you a very extended display, you wouldn't be able to see proper oscillations.
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33)
The total resistance of the cable is 8 ohms. The cable is 800 meters long since that's the distance between the power supply and the relay and there are 2 wires in the cable. 0.005 ohms per m means in one wire the resistance is 0.005 ohms x 800 = 4 ohms. So total resistance is 2 x 4 = 8 ohms.
Add the resistance of the wires to the resistance of the relay. Relay's resistance is R = V/I = 16/0.6 = 26.7 ohms. (26.7 + 8)
To find the output voltage = IR = 0.6 x (26.7 + 8) = 20.82 and the answer is C.
We are treating the relay and wires as one unit. The current through wires does not change and so we can plug the 0.6 amps of the relay into the V=IR formula.
It just seems like it's complicated but it really isn't. Try looking at it in a simplified way.
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35)
This is another tricky question. The voltmeter reading will be constant. Why? Because we are not measuring the voltage of the potentiometer P only but we are measuring the voltage for the whole circuit(see the connection of the voltmeter) so even though you slide the potentiometer along x to y it will not have any effect. However, if we are measuring the voltage of the potentiometer only, then the voltage will change when you slide the slider along x to y. So the answer will be A.
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
Guys question 15 please !! And 12 as well!!!! 23 as well !
Thank you
12)
23)
the area under graph is the elastic PE, so we can c that the answer is obviously (i hope its obvious for u too) B. not much that i could explain here
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74-63-224-172.rotatemyad.com/click.php?aid=861&said=06b_05281514p46&v=15&id=303f1e95923419f4c3082dbd549d4fc0&url=http%3A%2F%2F216.144.227.221%2Fclick.php%3Fkey%3Dc32f16f1fe5e9acbf3c3c0421f8d41318b078834ea2&btn=r&bUrl=http%3A%2F%2Fpapers.xtremepapers.com%2FCIE%2FCambridge%2520International%2520A%2520and%2520AS%2520Level%2FPhysics%2520(9702)%2F9702_s11_qp_11.pdf quest 13 ?
74-63-224-172.rotatemyad.com/click.php?aid=861&said=06b_05281514p46&v=15&id=303f1e95923419f4c3082dbd549d4fc0&url=http%3A%2F%2F216.144.227.221%2Fclick.php%3Fkey%3Dc32f16f1fe5e9acbf3c3c0421f8d41318b078834ea2&btn=r&bUrl=http%3A%2F%2Fpapers.xtremepapers.com%2FCIE%2FCambridge%2520International%2520A%2520and%2520AS%2520Level%2FPhysics%2520(9702)%2F9702_w10_qp_11.pdf
quest 5
74-63-224-172.rotatemyad.com/click.php?aid=861&said=06b_05281514p46&v=15&id=303f1e95923419f4c3082dbd549d4fc0&url=http%3A%2F%2F216.144.227.221%2Fclick.php%3Fkey%3Dc32f16f1fe5e9acbf3c3c0421f8d41318b078834ea2&btn=r&bUrl=http%3A%2F%2Fpapers.xtremepapers.com%2FCIE%2FCambridge%2520International%2520A%2520and%2520AS%2520Level%2FPhysics%2520(9702)%2F9702_w11_qp_11.pdf
ques 16
help please
Q13)
To balance the rule, the net moment on it about any point should be equal to zero.
Furthermore, since the tension in the string is trying to turn the rule anti-clockwise, we have to place a weight in such a position that it will try to rotate the rule clockwise. Intuitively, we can say that the weight should be on the right side of the pivot, to ensure that it tries to rotate the rule clockwise. So suppose we place the weight at some point "x" from the pivot. Note also, however, that the rule's mass plays a part here, so it depends on which moment is greater, the moment of the 0.02 kg mass of the ruler's mass.
Let's see what forces we get - if the 20 gram mass has to remain in equilibrium, using Newton's Law in the vertical direction,
Tension = mg = (20/1000) * g = 0.02g Newtons.
Since the string is assumed to be massless and the pulley is frictionless, the tension at all points in the rope is the same value. This value is 0.02g Newtons.
Therefore, the force trying to rotate the rule anti-clockwise is equal to 0.02g Newtons.
Now when you place the weight some distance x from the pivot, for it to be stationary, the normal force exerted on it by the rule has to be equal to it's weight. By Newton's Third Law, the weight exerts an equal and opposite Normal Force on the rule. The magnitude of this force = (mass of weight) * g = (50/1000) * g = 0.05g Newtons. This is the force that tries to rotate the rule clockwise.
Lastly, the force of gravity acts at the mid-point of the rule, and so that force has a magnitude of 0.1g Newtons. This acts at the 50 c mark, which is 10 cm away from the pivot, or 0.1 meters from the pivot.
Let's take moments about the pivot. Suppose Anti-Clockwise is positive, and Clockwise is negative, we get:
(0.02g) * (1 - 0.4) - (0.05g) * (x) - 0.1g * (0.1)
This is equal to zero, because the rule is balanced. So,
0.02 * 0.6 - 0.01 = 0.05 * x, and so x = 0.04 meters.
Since the distance of 0.04 meters is from the pivot, and the pivot is at the 0.4 meter mark on the rule, the marker where we put the weight should be 0.4 + 0.04 meters = 0.44 meters = C.
Q5)
Since the flywheel rotates 3000 times every 60 seconds, it rotates (3000/60) = 50 times every second. This means that the magnet passes the coil 50 times every second, since each revolution takes it past the coil once. Further, this means that the magnet passes the coil once every (1/50)th of a second, i.e. every 0.02 seconds.
Every time the magnet passes the coil, a pulse is generated on the screen. The time period between any two pulses is the time taken for the magnet to go around the circle and return to the coil (if the magnet starts at the coil, there is a pulse. It goes all the way around, and when it next reaches the coil another identical pulse is produced).
So, the time between any two pulses = 0.02 seconds.
Therefore, to see clearly separate pulses, the screen has to show one pulse in the beginning, a gap representing 0.02 seconds, and another pulse.
So the screen has to show at least the 0.02 second gap between pulses in 10 cm. Let's take a look at the options.
Suppose we take D. The base there is set as 1 µs / cm, i.e. each centimeter on the screen will represent 1 µs = 1 * 10^-6 seconds. We have 10 cm, so we can see
10 * 1 * 10^-6 seconds = 10^-5 seconds. But 0.02 seconds = 2 * 10^-2 seconds, so we will see only a small fraction of the time gap. This is wrong.
Suppose we take C. The base there is given the setting of 100 µs / cm , i.e. each centimeter on the screen will represent 100 µs = 100 *10^-6 = 10^-4 seconds. With 10 cm, we will see 10^-3 seconds. This is still very much smaller that the time gap, so we will again see only a small fraction of it. We won't even see any pulse! Next!
Let's take B. The base there is set as 10 ms / cm , i.e. each centimeter on the screen represents 10 * 10^-3 = 10^-2 seconds. In 10 seconds we can see what happens in 10^-1 = 0.1 seconds. In 0.1 seconds, we will see 5 pulses, but what will be the gap? Since we want to see a clear gap of 0.02 seconds, we can calculate that in this case, the distance on the screen representing the gap will be equal to 2 centimeters, which is easily visible. Therefore, our answer is B.
Suppose we take A, the time base is too large, and so the gap on the screen of 0.02 seconds will have a length of 0.02 cm, while is a fraction of a millimeter. We would see a mat of thorns without any gaps, there! So B is the answer.
Q16)
Suppose you have a graph of y versus x. Then the gradient is written as dy/dx , right? Similarly, in this case we have momentum on the y axis and time on the x axis. Therefore, the gradient at any point can be written using calculus as d(mv)/dt.
Since (mv) is commonly written as (p), we write that again as dp/dt.
From what we have learned of Newton's Laws, Resultant Force is defined as dp/dt, a special case being mdv/dt = ma. So the gradient is the resultant force on the car = A.
Hope this helped!
Good Luck for all your exams!
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How do you solve question 5 on this http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
Also question 20 of the same year
Please explain in full details
Also question 20 of the same year
Please explain in full details
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Any of you geniuses out there, please explain the following to me. PLSSS.
Answer: C
Answer: D
Answer: D
Answer: C
Answer: D
Answer: D
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15)
The diving board is in equilibrium - it is not accelerating, it is not revolving.
If an object is not accelerating, there is no net force on the object.
If an object is not rotating, there is no net torque on the object.
The boy is stationary, so the force exerted on him by the board that holds him up is equal in magnitude to his weight. By Newton's Third Law, he exerts the same force downwards on the board. Since his weight is mg, the force he exerts on the board is equal to mg, and acts downwards. This force acts 5.0 meters away from the hinge.
The spring, if deformed by a distance "x", will exert a force of magnitude "kx" upwards on the board. This force contacts and pushes the board 2.0 meters from the hinge.
In this case, we can see that the board isn't rotating, so the net torque on it is zero. So, to find out our answer, we take moments about the hinge:
Moment due to the boy = mg * 5.0 meters = 5mg in the anti-clockwise direction.
Moment due to the spring = kx * 2.0 meters = 2kx in the clockwise direction.
So the net torque is zero, therefore 5mg = 2kx. Therefore, x = 5mg/2k
Putting in the values, we get x = 5 * 50 * 9.81/(2 * 10,000) = 1962/20000 = 0.0981 meters
This is equal to 0.0981 * 100 = 9.81 centimeters = C.
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Wow, no idea on 5. However, I did manage to find out something:
Between each level (100 to 1000, 1000 to 10000, etc) there is a block of height 0.6 cm - this does not correspond to a linear increment, i.e. this does not mean an increase of 500 ohms, or something like that - the distances on the scale depend on the level you are at.
In other words, when you go up one millimeter from 1000 Ω, the change in resistance corresponding to that increment is different from the change in resistance if you started at 100 Ω and went up the same distance - 1 millimeter.
So my guess is that the increments on the scale depend on the level below you - if you are between 100 and 1000, the scale is a function of 100. If you are between 1000 and 10000, the scale depends on 1000, etc.
So, suppose we have a logarithmic scale - the distance from 100 to 1000 (and any other level, but let's take this for an example) is 2.0 cm, and the resistance varies as
100(x)^d = R
Where "d" is the distance you've gone up from the level of 100, R is the resistance of the thermistor at that value d from the lower level and "x" is some number.
Suppose you put d = 2.0 cm, 100(x)^2.0 should be equal to 1000. In other words,
x^2 = 10
x = (10)^0.5 = 3.166
So, suppose you start at the level of 100 and you go up 3 mm = 0.3 cm, the resistance at that height should be equal to
100 * (3.166)^0.3 = 141 Ω.
Suppose you start at the level for 1000, the equation should be
1000 * 3.166^d = R.
Where d is the distance from 1000 level.
So, on measuring, we see that at 40 Ω, the line has a height co-ordinate of 1.2 centimeters. So we put this into the earlier formula to get
R = 100 * (3.166) ^ 1.2 = 100 * 3.98 = 398.1071 Ω
Which is approximately 400 Ω, answer C.
An absolutely, maddeningly, crazy question, but it has to have some sort of solution. This one appears to help, but i'm not sure.
20) Take the leftmost column of liquid.
There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."
What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)
So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.
Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.
When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write
P + ρgh1 = 16,000.
Repeating that calculation on the right side, we get
P + ρgh2 = 8,000
Eliminating P,
16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06
So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.
Hope this helped!
Good Luck for all your exams!
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Ammeter fluctuation is from 1.98 to 2.02, so the mean value would be 2.00 +/- 0.02, with a % uncertainty of (0.02/2.00 * 100)=1%. Since, question also says that there is a systematic uncertainty of 1%, overall uncertainty will be 2%. Therefore, the answer will be D because 2% uncertainty of 2.00 is 0.04.
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Q25)
Two ways of doing this - first, using gradients at a point, and second, using frame-shifts.
Let's do the second way first. You know the wave is going from left to right - the question tells you that. Now suppose you take the wave you have now, and mentally shift it very slightly to the right, in the direction of it's travel. That is where it will be in some very short time duration.
Take a look at point Q. If you shift the wave very slightly to the right, or put your pencil point on the point Q and drag it slightly to the left (only left, not along the wave), then you will note that the point on the wave straight down from your pencil is not very far down from Q itself; as this time internal becomes smaller and smaller, the difference becomes zero, i.e. in that time interval the particle at Q travels a distance of zero units, i.e. it's speed is zero.
However, at P, that isn't the case - even a very small displacement will result in a change of position for P, that position being lower than it originally was. Therefore, P is traveling downwards. Hopefully the diagram i'm attaching helps in some way or the other:
On the diagram you can see the purple horizontal line - that line is the same length at P and Q, but the vertical line (corresponding to their position after some time)
is almost non-existent for Q, while it is very, very visible for P. Usually the time interval we have to take is much smaller, but for this question the diagram below should serve well enough. So you can see that Q is stationary and P moves downwards, giving us an answer of A.
The next method is of gradients, and is very similar to the one above - basically, if you take the gradient at any point along the wave, the particle at that point will travel onto that position relative to the wave - the gradient will have a horizontal component, but you should ignore that, and the vertical component of the gradient will tell you how the particle moves.
E.g. at the zero-point, the gradient is maximum, and thus the vertical component of the gradient is also maximum - this follows with the observation that the fastest moving particles are the ones at the equilibrium position.
However, at the crest or trough of a particular wave/waveform, we see that the gradient is absolutely flat - this gradient has no vertical component, so the vertical velocity of the particle at the crest/trough is zero. This follows with out observation above that the particle at Q is stationary.
Hope this helped!
Good Luck for all your exams!
THANK YOU SO MUCH!