Physics: Post your doubts here!

[Asad Moosvi]

What to explain in this yaar ?
Its C simple theory question, for which you need to read chapters, see my signature !!!!

Okay, ignore that question.

Can you help me with question 19 though? Same paper.

 

[Thought blocker]

Okay, ignore that question.

Can you help me with question 19 though? Same paper.

I said that's my doubt ._.

 

[Asad Moosvi]

Question 19, anyone??? Please!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf

 

[Ahmed H. Al-Neel]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf
questions 16 and 17 please!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf
questions 16 and 17 please!

16)
You know that the weight always acts at the centre.. so lets assume the distance XY to be 1m.
The horizontal component of force F is Fsin30.

Sum of clockwise moments = sum of anti-clockwise moments
0.5 x 10 = 1 x Fsin30
F = 10N

17)
The tension (4N) is the same on both sides; XY and XZ.
First, we have to find the angle YXP
Sin theta = 40/50
theta = 53

Next, find the horizontal component of the tension:
= 4cos53
= 2.4N

The same horizontal component will be experienced on XZ

Hence, 2.4 X 2

=> 4.8 N

 

[mahmuda akter]

could u plz explain me how to write the answer of the question about the justification of significant figures in q-2 0f p-3.
like the one in o/n10 variant 33 q-2 g(ii)
& the one in m/j 08 q--2c(iv)

 

[Asad Moosvi]

I said that's my doubt ._.

I understand it now. The mass of the model is 10 times smaller in each direction. So if the mass of the actual crane is 1000 kg, the mass of the model will be 10 kg. The young's modulus is the same in either case. We just have to solve for the extensions and then find the ratio. Simple as pie.

 

[Thought blocker]

I understand it now. The mass of the model is 10 times smaller in each direction. So if the mass of the actual crane is 1000 kg, the mass of the model will be 10 kg. The young's modulus is the same in either case. We just have to solve for the extensions and then find the ratio. Simple as pie.

Who taught you.
Still Where are the calculations man!

 

[sagar65265]

I found it very very hard.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf
Q36 38
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf
Q19

sagar65265

I think I did Question 19 here, just let me know if you've got it (if you still have doubts, no issues, let me know!).

I'll do the others after a few minutes, have a little work to do elsewhere .

 

[Thought blocker]

Wow, is this a good question! The trick here is to simply work through all the steps, and see what you get.
First step is in find the extension from the values we know.

The Young's Modulus Formula is

Y = Fl/Ae (where F is the tension, l is the original, unladen length of the sample, e is the extension, and A is the cross section perpendicular to the extension)

So, by multiplying both sides by e, we get

eY = Fl/A From here, suppose we divide both sides by Y, we get

e = Fl/AY which is the extension formula that we need. What we need to do is find out the values of this ratio for both the scale model and the full size version, and divide them to get the final answer.

A very important point to note here is that the question says that all linear dimensions are in a ratio of 1:10; this means that quantities such as length, radius, diameter, height, and so on are 10 times greater in the real version than in the scale model. Quantities such as Area, Volume, and Density are non-linear and so do not follow this ratio.

On reading the question again, there is something crazily interesting there - they have clearly mentioned that the load is cubic. Why? WHY?
It's important, that's why:

Suppose the material has some density ρ and it has a side length a. Then, for the scale model, the mass = ρV = ρa^3.
For the full size model, the same follows - however, the length has increased to 10 * a, so the mass of the load in the full-size crane = ρ(10a)^3 = 1000ρa.
Major importance there!
I'm pretty sure there is no more to be taken in consideration, since the extension of a wire requires you to consider only the dimension of the wire (going to do it), the load (done by above discussion) and the kind of material being stretched (Young's Modulus is the same since the material used is the same. Done!)

So, for the scale model:
i) The force can be written as "F" (let's say). This is also equal to ρa^3, as discussed above.
ii) The length of the cable without any strain applied can be written as "l".
iii) The Cross Sectional Area of the wire should be written as πr^2, where r is the radius of the cable.
iv) Finally, the Young Modulus can just be written as "Y".

Therefore, we can write

(extension of cable on the model crane) = [(ρa^3)gl]/Yπr^2

For the real crane:
i) The force can be written as "1000ρa^3", again as discussed above.
ii) The length of the cable without any strain can be written as "L". We know from the question that this is equal to 10 * l, so the length = "10l".
iii) The Cross Sectional Area of the wire should be written as "πR^2" where R is the radius of the full scale wire. We know this is equal to 10*r, so we can write that
the cross-sectional area of the real cable = π * (10r)^2 = 100πr^2.
iv) Again, the Young Modulus is just Y, since the material of the real cable is the same as the material of the scale model cable.

Therefore, we can write

(extension of cable on the full-size crane) = [(1000ρa^3 ) * 10l]/[100Yπr^2] = [(100ρa^3)gl]/Yπr^2

Dividing those two, we get (extension of cable on the full-size crane)/(extension of cable on the model crane) = 100 = 10^2 = C.

Hope this helped!
Good Luck for all your exams!

"On reading the question again, there is something crazily interesting there - they have clearly mentioned that the load is cubic. Why? WHY?
It's important, that's why" This line was Just ^_^
Secondly, I get it TY!

 

[Asad Moosvi]

Who taught you.
Still Where are the calculations man!

Assume the young's modulus is 1 and then arrange the equation so that the extension is on one side and everything else is on the other side. Then just assume values for F, the original length, x, and area, A. Then find the extension. This will be the extension for the crane. Do the same for the model using values that are ten times less than the values used before, but F should be 10 times less in each direction. So if F was 1000 before, now it's 1. If x was 10 before, now it's 1. If Area was 1 before, the radius must've been one, so now the radius should be 0.1 and then the area could be assumed to be 0.1^2. Then solve for the extension of the model. This is what I did and I'm getting a ratio of 100, which is the correct answer.

 

[Thought blocker]

Assume the young's modulus is 1 and then arrange the equation so that the extension is on one side and everything else is on the other side. Then just assume values for F, the original length, x, and area, A. Then find the extension. This will be the extension for the crane. Do the same for the model using values that are ten times less than the values used before, but F should be 10 times less in each direction. So if F was 1000 before, now it's 10. If x was 10 before, now it's 1. If Area was 1 before, the radius must've been one, so now the radius should be 0.1 and then the area could be assumed to be 0.1^2. Then solve for the extension of the model. This is what I did and I'm getting a ratio of 100, which is the correct answer.

Now this is how I got it from sagar! And I will remember this way only. Thanks pal

 

[sagar65265]

could u plz explain me how to write the answer of the question about the justification of significant figures in q-2 0f p-3.
like the one in o/n10 variant 33 q-2 g(ii)
& the one in m/j 08 q--2c(iv)

Hopefully this link helps, and the file attached should also make matters clear - if not, there's this little summary here:

"When you multiply or divide a set of number to obtain a product, the number of significant figures quoted in the product should be the same as that of the number with the least number of significant figures."

As an example, suppose you have the following numbers:

i) 27.35 centimeters - this is to 4 significant figures, and was obtained using a ruler with a millimeter scale on it.
ii) 2.50798 seconds - this is to 6 significant figures, and was obtained using a light gate.
iii)0.25 (no units) - this is to 2 significant figures (zeroes before a nonzero digit do not count)

So we have one value to 4 sig. figures, and another value to 6 sig. figures and a third to 2 significant figures.

Suppose we want to find out the value of (27.35/2.50798) * 0.25. Let's work the steps out, here.

We divide the first by the second to give us

27.35/2.50798 = 10.90519063 centimeters/second.

Since we have one number to 4 significant figures and another to 6 significant figures, you might feel like saying "27.35 has 4 sig. figures, 2.50798 has 6 sig. figures, so we have to write this 10.90519063 to 4 sig. figures before multiplying by 0.25", but don't do this. Since our calculation is not yet over, we should not round it yet based on sig. figures or decimal points. Rounding it off now just gives us more uncertainties, so do not do that.

We take the number as it is in the calculator result of 27.35/2.50798, and multiply it by 0.25. Our answer to this is 2.7262976.

Now we can round off. We have used one number with 2 sig. figures, another with 4 sig. figures, and another with 6 sig. figures for this calculation. The least number, therefore, is 2 significant figures. Therefore, we round 2.7262976 to 2 significant figures, which gives us 2.7. This is our final answer, as we should give it. It seems like very little for the calculation, but do take a look at the video above to see another example, and it should give you a good idea of the process.

In the question, we can rearrange the equation to give us

k = Vd²/l

So here, you carry out the calculation on a calculator (multiply V and d², and divide the result as it is, without rounding off, by l). After that, take a look at the values of V, d and l - which one has the least number of significant figures? Take that number of significant figures and quote it in your answer.

So, you can say for g(ii) that you quoted it to so-and-so significant figures since that is the number of significant figures of the least precise value used in the calculation (the more sig. figures you have, the more sure you are of the value - therefore, the least precise value is the one with the least significant figures).

Hope this helped!

Good Luck for all your exams!

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf

27.

 

[sagar65265]

Can anyone help me?
M/J/13- (12) -3,9,19,20,23,27
M/J/13-(11) - 8,11,17,18,19
M/J/13-(13)- 10,15,16,27

For M/J/13 - 12:

3) The point where the net force can be assumed to act is the point where the net force will exert the same torque as the single forces taken one by one - suppose I have 3 forces A, B and C acting on a door.
The apparent line of action of the net force will be the place where the net force exerts the same torque as the sum of the torques exerted by A, B and C.

Take the center of mass of the disc into consideration here - what is the net torque of the forces about this point?

The torque of the 4 Newton force is zero, since it acts at the point we are considering. So the distance "d" in the formula Torque = τ = Fdsinθ is zero, and that's why the torque due to this force is zero.

The torque of the 3 newtons force is also zero - the angle between the force and the line of action is 0 degrees, so the θ in the same formula as above is zero. This means that sin θ is also equal to zero, so the torque due to this force is zero.

Therefore, there is no net torque on the disc. Therefore, the net force should also not exert a net torque on the disc.

This happens in B and D - if the net force acted at the points shown in B or D, then they would exert a net torque on the disc. We have shown that the net torque is zero, so this cannot be the answer.

This leaves A and C. But we know from the magnitudes of the force that the left-right force (of 4 Newtons) is stronger and has a greater magnitude than the up=down force of 3 Newtons. So the net force should have a greater left-right component than the up-down component.

The option that shows a stronger left-right component than an up-down component is option A, which is our answer.

Q9)
This is an earlier answer - if you still have any doubts, be sure to put them up on the forums.

Q19)

When any sample of any material is at a particular temperature, the kinetic energy of 1 mole of that material will be the same as the kinetic energy of any other sample of any other material at the same temperature. This kinetic energy is related to the molecular movements - vibrational movement in a solid, sliding movement of molecules in liquid and linear movement of molecules in a gas.

The potential energy, on the average, of a molecule, is related to the average distance between it and the other molecules around it. Molecules very close together have low potential energies, while molecules that are far apart have large potential energies.

In the case of water molecules, they are actually farther apart in the solid state (ice) than in the liquid state (normal water) but this doesn't happen in any other common material. In other materials, the atoms/molecules are closer in the solid state than in the liquid and gas state.

But since the distance between molecules in the solid and gas state is practically independent of temperature, the mean potential energy does not depend on the temperature, but on the state. Therefore, the mean potential energy in a solid state is lower than the mean potential energy in a liquid state. So the potential energies are not the same, and we can eliminate A and B.

We have also established above (in the first paragraph of this question) that the mean kinetic energy only depends on the temperature of the sample concerned, so we can say that C is the correct answer, but to understand why D is wrong, we have to see that since the kinetic energy of both ice and water is the same but the potential energy of each sample is different, the sum can definitely not be the same. Therefore our answer, for sure, is C.

Q20)

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.

Q23)

There are quite a few discussions on this, so here, here and here are some explanations.

Q27)

If the frequency of waves is 1 GHz, this is equal to 1 * 10^9 GHz = 10⁹ Hertz.
The speed of these waves are is 3 * 10⁸ ms^-1 since they are waves in the electromagnetic spectrum, and all the waves in the spectrum travel at this speed in a vacuum/ in open air.
Therefore, the wavelength of these waves is given by v = fλ; rearranging, we get λ = v/f = (3 * 10⁸)/10⁹ = 0.3 meters

The distance we are concerned with is 45 centimeters = 45/100 meters = 0.45 meters.

Since the question says standing waves are formed, the number of wavelengths produced have to be a multiple of 0.5 (0.5, 1, 1.5, 2.0, 2.5, etc).

We see that is true by noting that there are 0.45/0.3 = 1.5 wavelengths in this length.

Taking a look at the image of a standing wave, 1.5 standing waves have 3 antinodes (each 0.5 wavelength has 2 nodes and 1 antinode; a node, an antinode, and then again a node). Therefore, our answer is C.
In case you have any doubts, just reply on the forums.

Hope this helped!
Good Luck for all your exams!

 

[Ahmed H. Al-Neel]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
questions 10,17,23,26,35 please!

 

[unkidd]

7)
Acceleration upwards is always -9.81 and downwards 9.81. As upward motion is positive, + multiplied ny - gives you -9.81. Vice versa with downward motion. - multiplied with +9.81. The magnitude stays the same as long as air resistance is neglected, and the sign depends on which direction you take as positive.

9)
initial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it will lose some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.

22)
because since k is low for the first spring, the box will prevent it from breaking by over extending. Hence A

26)
We know that distance between two minima is given by λ/2 (example, distance b/w T and S is λ/2)
We also know that Microwave travels with the speed of light c i.e is 3 * 10^8 m/s
Given distance b/w each minima is 15 mm so total distance from T to P would be 15 * 4 = 60 mm

Now distance b/w:
T and S = λ/2,
S and R = λ/2
R and Q = λ/2
Q and P = λ/2
and them you get 2λ
so 2λ = 60*10^-3
hence λ = 0.03 m

We know that c = f * λ
So f = c / λ
f = ( 3 * 10^8 ) / ( 0.03 )
f=10GHz

27)
Let the distance between the double slit and the screen be 1m initally.
When the distance is increasd BY 2m the NEW distance is 3m.
Using the formula:
Wavelength= (fringe seperationx distance between the double slits)/ distance between the screen and the slits.
Wavelenght=600nm=600x10^-9m.
Fringe seperation= 3mm= 3x10^-3m
Distance between screen and slits= 3m (1+2)
Distance between the double slit=?
Put these values in the formula, the final answer is 6x10^-4m =0.6mm
Hence the answer is B.

34)
voltage decreases linearly so its supposed to be B : as rho increases, there would be more steepness in graph

Can you also explain q8 12 13 14 15 18 21 28 31 32 thanks

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
questions 10,17,23,26,35 please!

Q10)

This is a tricky one at first glance, but a little thought helps clear up things.

The sand is dropped horizontally into the cart, so it has no initial horizontal momentum. The cart has some momentum in the beginning, since it is moving initially. Suppose you take the sand and the cart together as the system, there is no net horizontal force on it. Therefore, the horizontal momentum component of the system remains constant.

So, in this case, when the sand falls in the cart both the sand and the cart have to travel at the same speed, so since the mass has increased the speed has to decrease.
This part is relatively clear - momentum has to be constant, but if mass increases, speed decreases to compensate. The momentum of the cart in the beginning is partially given to the sand, and partially remains with the cart, so that each one has some momentum of it's own. This narrows the answers down to A or B.

However, it is what happens at Y that can cause the confusion. What really happens when you let the sand fall?

Let's consider the sand along for a moment.
When the sand falls, it has some amount of momentum because it is moving along with the cart. No force from the trapdoor acts in the horizontal direction, since we are only dropping the sand, not ejecting it. Therefore, the horizontal velocity of the sand when it leaves the trapdoor (at the instant it leaves the trapdoor) is the same as the velocity of the cart+sand at Y, just before the sand is dropped.

So if the sand is moving with the same velocity, then the momentum it gained at X still remains with it, so the remaining momentum of the cart has to be the same, right? Since the momentum is constant, the speed remains the same at Y as it was just before reaching that point. So momentum remains the same.

The only option that agrees with both of these ideas is B, which is our answer.

17)

In any time duration, the only kinds of energy emitted by the bulb are Heat and Light energy. Usually, all the energy in the current is converted to these forms.
So, here 92 Joules of heat are evolved along with 8 Joules of light. The total energy being produced is therefore 92 + 8 = 100 Joules of energy.
This is also (by our assumptions) the amount of energy supplied to the bulb in the form of electrical energy. Therefore, the efficiency is
(energy produced in the form of light)/(energy supplied in electrical form) * 100 = 8 Joules/100 Joules * 100 = 8% = A.

23)

The energy of a deformed object is proportional to the area under a force-extension graph, since the stretching of any material can be modeled by the spring equation:

Spring equation: |F| = kx
Young Modulus Equation: Y = Fl/Ax
Rearranging, YAx/l = F and if we write (YA/l) = k (some constant) then we can write |F| = kx.

So the greater the area under the curve, the more the energy stored. Let's go option-by-option:

A seems to be very small - it has a large extension but a small force. This seems to be like a putty-type material, since it needs very little force to stretch.
B seems to have a good balance - it has a good extension for a rather large force. This is a much stiffer material that stretches in a linear manner.
C is the opposite of A - it has small extension but needs a large force for that extension. This is a very stiff material that also stretches in a linear manner.
D is a curve, and it is a very slim one at that - you can approximate it with a shallow line, but it doesn't look to display a large area due to a very weak force being needed to cause a large extension.

Therefore the best candidate seems to be B, our answer.

26)

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

35)

The total resistance of the circuit is equal to (Internal Resistance) + (External Resistance) = R(i) + R(e)

Therefore, the current flowing through the circuit (by V = IR) is equal to V/R = V/[R(i) + R(e)]

Further, the power dissipated as heat in a resistor is given by P = I²R = I²R(e) = V²R(e)/[R(i) + R(e)]²

This reaches a maximum value when R(i) = R(e):

P = V²R(e)/[R(i) + R(e)]² = V²R(i)/[R(i) + R(i)]² = V²R(i)/[2R(i)]² = V²R(i)/4R(i)² = V²/4R(i)

Thus, we get a maxima on the graph when the external resistance is equal to the internal resistance = 2Ω.

The only graph that shows a maximum value at 2Ω is A = our answer.

(This is a theorem, in fact, that states that the power dissipated in the load of a battery with an internal resistance is maximum when the value of the load resistance is equal to the value of the internal resistance).

Hope this helped!
Good Luck for all your exams!

 

[Ahmed H. Al-Neel]

thank you soooo sooo much bro that really helped alot! i just forogot to ask questions 12 and 15 also. sorry for all the troubles!
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf

 

[doremon]

15 B and 12 A . is it?