Physics: Post your doubts here!

[Thought blocker]

Okhaaa ... Will tag ya .. Lemme just finish this paper

okok

 

[Mohammed salik]

okok

Quest

Question 4 as well please..

Guys , Here is My Way !
Speed (Avg) = Total distance/Time..
So 40/2.5=16 ms^-1
Now check How many SF is Time and Distance Given ? Its 3 SF.. So Speed Must be 3 SF !
So Ans would be 16.0 ms^-1 .. Hence Ans is C or B Possible..!
Then Add Fractional Uncertainties of Time and Distance (Like in Theory)!
For Time= (0.05/2.5) * 100 =2 %
For Distance=(0.1/40)*100=0.25%
Total Percent Uncertainty in Speed= 2+0.25=2.25%

So Uncertainty = (2.25/100) * 16=0.36.. Which Rounds upto 0.4!
Ans is C!

Hope i Helped!

 

[Thought blocker]

Guys , Here is My Way !
Speed (Avg) = Total distance/Time..
So 40/2.5=16 ms^-1
Now check How many SF is Time and Distance Given ? Its 3 SF.. So Speed Must be 3 SF !
So Ans would be 16.0 ms^-1 .. Hence Ans is C or B Possible..!
Then Add Fractional Uncertainties of Time and Distance (Like in Theory)!
For Time= (0.05/2.5) * 100 =2 %
For Distance=(0.1/40)*100=0.25%
Total Percent Uncertainty in Speed= 2+0.25=2.25%

So Uncertainty = (2.25/100) * 16=0.36.. Which Rounds upto 0.4!
Ans is C!

Hope i Helped!

I do the same way

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf qs 33, 35
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf why is 36 B and not C? Why can't we just find the potebtial difference and powers directly without using current...why does that give a wrong answer?

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf q10

 

[Snowysangel]

Um could someone explain the concept of weight, upthrust and viscous drag to me? Which is the largest in air? And in water? And what us the result force acting on a body under water?
Also does an object accelerate or decelerate as it falls through air...in both the absence and presence of air resistance
These are AS level questions btw!

 

[Fatima Iftikhar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf q10

kinetic energy remains conserved,
initial = final
since its a stationary body, initial kinetic energy = zero
if velocity of m is v, then velocity of 2m has to be such that sum of momentum remains zero.
therefore,
so velocity of 2m is -v/2 ( negative 'cause its in the opposite direction)
you can check, mv + (2m)(-v/2) = 0
now, [ 1/2(m)(v^2) ] / [ 1/2(2m)(-v/2)^2]
it becomes 2/1.
so its C.

 

[sagar65265]

26) C is just a random value I included; we know that the value of A changes by a factor; from A it becomes (something) * A, not (something) + A. So, I just decided to make that (something) C - it isn't part of the question, it is just something I put it. You will get the same answer by replacing (CA) with A(final). It doesn't matter, you can put whatever you want there.

27) The harmonics do not matter here;

I'll try posting the rest, I have to go right now.

27) Think about it this way - for an electromagnetic wave (or ANY transverse wave, or even a longitudinal wave) regardless of the wavelength and frequency, regardless of all other properties of the wave, as long as it is periodic (as long as it repeats itself in a certain time period), the distance from one crest to the next crest is equal to the wavelength. Similarly, the distance from one trough to the next is equal to the wavelength.

Similarly, on a standing wave, regardless of any other properties of the wave, the distance between two nodes or the distance between two maxima (antinodes) is equal to half the wavelength of that wave. The number of harmonics is dependent on the frequency of the wave, and this does not affect the fact that the distance between two antinodes is half the wavelength.

What is the doubt you're having with 33, exactly? I'll reply as soon as I can.

 

[sagar65265]

How do you get symbols like, lambda, rho.. etc etc ?

Google Search, then copy-paste

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf qs 33, 35
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf why is 36 B and not C? Why can't we just find the potebtial difference and powers directly without using current...why does that give a wrong answer?

Q33)

The potential difference across the Upper branch is equal to the potential difference across the Lower branch; another assumption we can make is that the potential at P is 12 Volts, and the potential at Q is 0 Volts. So, we can take each branch separately as follows:

Potential Difference across Upper Branch = 12 Volts.
Net Resistance in Upper Branch = 500Ω + 1000Ω = 1500 Ω

So the current through the Upper Branch = I = V/R = 12/1500 = 0.008 Amperes.
Therefore the drop in potential across the 500Ω resistor = IR = 500Ω * 0.008 A = 4 Volts.

Since the potential at P is 12 Volts and the drop in potential energy across the 500Ω resistor is 4 Volts, the Potential at X = 12 - 4 = 8 Volts.

For the lower branch, we repeat the analysis:

Potential Difference across Lower Branch = 12 Volts.
Net Resistance in Lower Branch = 2000Ω + 1000Ω = 3000 Ω

So the current through the Lower Branch = I = V/R = 12/3000 = 0.004 Amperes.
Therefore the drop in potential across the 2000Ω resistor = IR = 2000Ω * 0.004 A = 8 Volts.

Furthermore, if the potential at P is 12 Volts and the drop across the 2000Ω resistor is 8 Volts, the Potential at Y is = 12 - 8 = 4 Volts.

So the difference between the two is 8 - 4 = 4 Volts = B.

Q35)

I remember getting this wrong during a practice test; my classmates had corrected me later with the following argument:

"If you take the resistors on the top and bottom of the diagram and move them towards the resistor on the right so that they kiss (no kidding, that's what they said! awesome guys ) each other, they will be in series; then you add them up to give 10 + 10 + 10 = 30Ω total.
Now, they'll be parallel to the 10Ω resistor on the left side, so the resistance is calculated with

1/R(equiv) = 1/30 + 1/10 = 4/30 So R(equiv) should be equal to 30/4 Ω. This is between one and ten, so the answer should be B."
Thanks, guys!

2007 Paper:
Q36)

I'm not sure what the problem is here - the answer is C, not B - either ways, just a quick little rundown in any case:

Any battery (except an ideal one, but the one in this problem is not an ideal one) is made of two components as far as circuit diagram components are concerned - the cells that produce a potential difference in the circuit and convert chemical energy to electrical energy, and the internal resistance that practically any real component will have.

Therefore, when the phrase "terminal p.d." is used, they mean the potential difference across both of these components taken together. It is not the potential across the cells alone, or the potential across the internal resistance alone.

However, potential energy is only lost in any kind of resistance (internal resistance is no exception to this) when a current passes through it. So, that is why we have to consider the current as well - a fraction of the energy produced by the cells is lost even before the current leaves the battery, and this loss is in the internal resistance. Not only that, this loss depends intimately on the current produced, so we need to know the current to find out the loss of energy in the internal resistance.

Applying Kirchoff's Second Law going anti-clockwise through the circuit (this is the direction of the current, and this makes it easiest to calculate the values required. We can go in any direction, but this choice makes our job a little easier. Also, the internal resistance can be treated as a separate resistor for this equation):

-4.0I - 2.0I + 3.0 = 0
3.0 = 6.0I
I = 0.5 Amperes.

Therefore, the potential energy lost in the internal resistance is equal to V = IR = 0.5 * 2 = 1 Volt lost.
So, putting this together, we see that the potential difference across the internal resistance is -1.0 Volt and the difference across the cells is +3.0 Volts. Summing them algebraically, we get Terminal PD = 2.0 Volts.

Further, the output power is the power expended by components outside the battery; so this power is given by I²R = 0.5² * 4Ω = 1 Watt
Putting this together we get the answer C.

Hope this helped!

Good Luck for all your exams!

 

[Snowysangel]

Q33)

The potential difference across the Upper branch is equal to the potential difference across the Lower branch; another assumption we can make is that the potential at P is 12 Volts, and the potential at Q is 0 Volts. So, we can take each branch separately as follows:

Potential Difference across Upper Branch = 12 Volts.
Net Resistance in Upper Branch = 500Ω + 1000Ω = 1500 Ω

So the current through the Upper Branch = I = V/R = 12/1500 = 0.008 Amperes.
Therefore the drop in potential across the 500Ω resistor = IR = 500Ω * 0.008 A = 4 Volts.

Since the potential at P is 12 Volts and the drop in potential energy across the 500Ω resistor is 4 Volts, the Potential at X = 12 - 4 = 8 Volts.

For the lower branch, we repeat the analysis:

Potential Difference across Lower Branch = 12 Volts.
Net Resistance in Lower Branch = 2000Ω + 1000Ω = 3000 Ω

So the current through the Lower Branch = I = V/R = 12/3000 = 0.004 Amperes.
Therefore the drop in potential across the 2000Ω resistor = IR = 2000Ω * 0.004 A = 8 Volts.

Furthermore, if the potential at P is 12 Volts and the drop across the 2000Ω resistor is 8 Volts, the Potential at Y is = 12 - 8 = 4 Volts.

So the difference between the two is 8 - 4 = 4 Volts = B.

Q35)

I remember getting this wrong during a practice test; my classmates had corrected me later with the following argument:

"If you take the resistors on the top and bottom of the diagram and move them towards the resistor on the right so that they kiss (no kidding, that's what they said! awesome guys ) each other, they will be in series; then you add them up to give 10 + 10 + 10 = 30Ω total.
Now, they'll be parallel to the 10Ω resistor on the left side, so the resistance is calculated with

1/R(equiv) = 1/30 + 1/10 = 4/30 So R(equiv) should be equal to 30/4 Ω. This is between one and ten, so the answer should be B."
Thanks, guys!

2007 Paper:
Q36)

I'm not sure what the problem is here - the answer is C, not B - either ways, just a quick little rundown in any case:

Any battery (except an ideal one, but the one in this problem is not an ideal one) is made of two components as far as circuit diagram components are concerned - the cells that produce a potential difference in the circuit and convert chemical energy to electrical energy, and the internal resistance that practically any real component will have.

Therefore, when the phrase "terminal p.d." is used, they mean the potential difference across both of these components taken together. It is not the potential across the cells alone, or the potential across the internal resistance alone.

However, potential energy is only lost in any kind of resistance (internal resistance is no exception to this) when a current passes through it. So, that is why we have to consider the current as well - a fraction of the energy produced by the cells is lost even before the current leaves the battery, and this loss is in the internal resistance. Not only that, this loss depends intimately on the current produced, so we need to know the current to find out the loss of energy in the internal resistance.

Applying Kirchoff's Second Law going anti-clockwise through the circuit (this is the direction of the current, and this makes it easiest to calculate the values required. We can go in any direction, but this choice makes our job a little easier. Also, the internal resistance can be treated as a separate resistor for this equation):

-4.0I - 2.0I + 3.0 = 0
3.0 = 6.0I
I = 0.5 Amperes.

Therefore, the potential energy lost in the internal resistance is equal to V = IR = 0.5 * 2 = 1 Volt lost.
So, putting this together, we see that the potential difference across the internal resistance is -1.0 Volt and the difference across the cells is +3.0 Volts. Summing them algebraically, we get Terminal PD = 2.0 Volts.

Further, the output power is the power expended by components outside the battery; so this power is given by I²R = 0.5² * 4Ω = 1 Watt
Putting this together we get the answer C.

Hope this helped!

Good Luck for all your exams!

I did all that except the last part where you subtract 4 from 8...why do u do that? The pd on either side of X and Y is 8+4=12 so shouldnt the total difference in pd be 12-12=O

 

[sadiaali]

Please anyone explain me Q:14 . Thank a lot in advance
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf

 

[Snowysangel]

Q33)

The potential difference across the Upper branch is equal to the potential difference across the Lower branch; another assumption we can make is that the potential at P is 12 Volts, and the potential at Q is 0 Volts. So, we can take each branch separately as follows:

Potential Difference across Upper Branch = 12 Volts.
Net Resistance in Upper Branch = 500Ω + 1000Ω = 1500 Ω

So the current through the Upper Branch = I = V/R = 12/1500 = 0.008 Amperes.
Therefore the drop in potential across the 500Ω resistor = IR = 500Ω * 0.008 A = 4 Volts.

Since the potential at P is 12 Volts and the drop in potential energy across the 500Ω resistor is 4 Volts, the Potential at X = 12 - 4 = 8 Volts.

For the lower branch, we repeat the analysis:

Potential Difference across Lower Branch = 12 Volts.
Net Resistance in Lower Branch = 2000Ω + 1000Ω = 3000 Ω

So the current through the Lower Branch = I = V/R = 12/3000 = 0.004 Amperes.
Therefore the drop in potential across the 2000Ω resistor = IR = 2000Ω * 0.004 A = 8 Volts.

Furthermore, if the potential at P is 12 Volts and the drop across the 2000Ω resistor is 8 Volts, the Potential at Y is = 12 - 8 = 4 Volts.

So the difference between the two is 8 - 4 = 4 Volts = B.

Q35)

I remember getting this wrong during a practice test; my classmates had corrected me later with the following argument:

"If you take the resistors on the top and bottom of the diagram and move them towards the resistor on the right so that they kiss (no kidding, that's what they said! awesome guys ) each other, they will be in series; then you add them up to give 10 + 10 + 10 = 30Ω total.
Now, they'll be parallel to the 10Ω resistor on the left side, so the resistance is calculated with

1/R(equiv) = 1/30 + 1/10 = 4/30 So R(equiv) should be equal to 30/4 Ω. This is between one and ten, so the answer should be B."
Thanks, guys!

2007 Paper:
Q36)

I'm not sure what the problem is here - the answer is C, not B - either ways, just a quick little rundown in any case:

Any battery (except an ideal one, but the one in this problem is not an ideal one) is made of two components as far as circuit diagram components are concerned - the cells that produce a potential difference in the circuit and convert chemical energy to electrical energy, and the internal resistance that practically any real component will have.

Therefore, when the phrase "terminal p.d." is used, they mean the potential difference across both of these components taken together. It is not the potential across the cells alone, or the potential across the internal resistance alone.

However, potential energy is only lost in any kind of resistance (internal resistance is no exception to this) when a current passes through it. So, that is why we have to consider the current as well - a fraction of the energy produced by the cells is lost even before the current leaves the battery, and this loss is in the internal resistance. Not only that, this loss depends intimately on the current produced, so we need to know the current to find out the loss of energy in the internal resistance.

Applying Kirchoff's Second Law going anti-clockwise through the circuit (this is the direction of the current, and this makes it easiest to calculate the values required. We can go in any direction, but this choice makes our job a little easier. Also, the internal resistance can be treated as a separate resistor for this equation):

-4.0I - 2.0I + 3.0 = 0
3.0 = 6.0I
I = 0.5 Amperes.

Therefore, the potential energy lost in the internal resistance is equal to V = IR = 0.5 * 2 = 1 Volt lost.
So, putting this together, we see that the potential difference across the internal resistance is -1.0 Volt and the difference across the cells is +3.0 Volts. Summing them algebraically, we get Terminal PD = 2.0 Volts.

Further, the output power is the power expended by components outside the battery; so this power is given by I²R = 0.5² * 4Ω = 1 Watt
Putting this together we get the answer C.

Hope this helped!

Good Luck for all your exams!

Wait so the terminal potential difference isn't the pd wcross the battery's internal resistance? I thought that's what it meant so I did 4/6 x 3 and got 1 V

 

[DeViL gURl B)]

Guys , Here is My Way !
Speed (Avg) = Total distance/Time..
So 40/2.5=16 ms^-1
Now check How many SF is Time and Distance Given ? Its 3 SF.. So Speed Must be 3 SF !
So Ans would be 16.0 ms^-1 .. Hence Ans is C or B Possible..!
Then Add Fractional Uncertainties of Time and Distance (Like in Theory)!
For Time= (0.05/2.5) * 100 =2 %
For Distance=(0.1/40)*100=0.25%
Total Percent Uncertainty in Speed= 2+0.25=2.25%

So Uncertainty = (2.25/100) * 16=0.36.. Which Rounds upto 0.4!
Ans is C!

Hope i Helped!

Thought blocker this is ma way of doin
Sorry for being late though!

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Question 3 I dunno ., I don't get this!
Q.4
And q 31!!! Pleaseeeeeeeeeeeeeeeee!! D':
Thank you

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Question 3 I dunno ., I don't get this!
Q.4
And q 31!!! Pleaseeeeeeeeeeeeeeeee!! D':
Thank you

q.3
p = mv
p² = m²v²
p²/m = m²v²/m = mv² , dis is similar to da formula for KE except for da (1/2) , but becuz (1/2) is a constant n doesnt have any units, da base units of mv² n (1/2)mv² will b same, so answer is A: Energy

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Question 3 I dunno ., I don't get this!
Q.4
And q 31!!! Pleaseeeeeeeeeeeeeeeee!! D':
Thank you

q.4
% uncertainty = (uncertainty/reading) * 100
% uncertainty = (0.5/100-40)*100
% uncertainty = ± 0.83% <----- dis is da % uncertainty each time temperature is measured, so when it is measured twice (at 40 °c and 100°c) % uncertainty is added, so % uncertainty in % rise = 0.83% + 0.83% = 1.66% ≈ 1.7%
Answer: D

 

[sagar65265]

I did all that except the last part where you subtract 4 from 8...why do u do that? The pd on either side of X and Y is 8+4=12 so shouldnt the total difference in pd be 12-12=O

I can't understand what you mean by "on either side of X and Y is 8+4=12", could you explain?

Anyways, when they are asking for the potential difference between X and Y (which are points in the middle of the upper and lower branches respectively) they are asking for the (Potential at X) - (Potential at Y).

Think of it this way - every coulomb of current entering the combination at P possesses 12 Joules of electric potential energy (by the definition of potential).
Similarly, each coulomb of charge that enters the upper branch (after entering the branch, before entering the resistor) will have 12 Joules of electric potential energy. The same applies for the lower branch.

However, when that same coulomb of charge in the upper branch passes through the 500Ω resistor, it loses 4 Joules of energy (since each coulomb of charge loses 4 Joules of energy on passing through that resistor, by definition of potential difference). Therefore, each coulomb of charge has 8 Joules of electric potential energy remaining, when it is at point X.

For the lower branch, a coulomb of charge with 12 Joules of electric potential energy (before entering the 2000Ω resistor) loses 8 Joules of energy when it passes through the 2000Ω resistor, again by the definition of potential difference (we know the potential energy drops by IR when the current passes through the resistor; by definition of potential difference, V = W/Q, we know that each coulomb of charge also loses 8 Joules of energy in this process). Therefore, any coulomb of charge passing through the point Y will have 12 - 8 = 4 Joules of electric potential energy at point Y.

So, the difference between their potential is the difference between their energies - at X, each coulomb has 8 Joules of energy; at Y, each coulomb has 4 Joules of energy. Therefore, the difference in energy per coulomb between them (which is, again, equal to the potential difference between those two points) is 8-4 = 4 Joules of energy.

That is why the potential difference between the two points is 4 Joules.

Hope this helped!
Good Luck with all your exams!

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

Question 3 I dunno ., I don't get this!
Q.4
And q 31!!! Pleaseeeeeeeeeeeeeeeee!! D':
Thank you

31)
Total resistance in 6 and 3 ohm is 2 ohm, so the voltage will be 6V in d 2 ohm resistor. n 6V across d 6 n 3 ohm resistors respectively. we have the voltage n resistance. so V=I*R, 6=6*I. so current is 1A

 

[sagar65265]

Wait so the terminal potential difference isn't the pd wcross the battery's internal resistance? I thought that's what it meant so I did 4/6 x 3 and got 1 V

No, it isn't. The terminal P.D. is the potential difference across the terminals of any battery. If that battery has no internal resistance, then the terminal P.D. is equal to the EMF or the Voltage rating of the battery.
If the battery has an internal resistance, since that resistance is inside the battery itself, it is part of the terminal P.D. - you still have to remember that the cells are also part of the battery, so the potential difference supplied by the cells also count. Together, the two of these make up the terminal P.D. of any battery - in case of an ideal battery, there isn't any internal resistance, but it still applies.

Hope this helped!
Good Luck for all your exams!