Physics: Post your doubts here!

[mehria]

Do some of them at least or by tonight.. :/ i know you have to prepare.

Question 3, 4 , 5 , 8 , 11 , 14 , 15 , 19 , 24 , 33 , 35 , 37 , 38 PLEASEEEE!http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Q3:- as Cd has no units the we won't mention it in our sol.
so ρ (V^n) A = F
V^n = F/ρA
now we will just put the base units of them...
(m/s)^n = (kgm/s^2) / (kg/m^3 x m^2)
(m/s)^n = ( m^2 / s^2 )
(m/s)^n = (m/s)^2
so n=2

Q4:-


Q5:- for ths question m gettng B :/ but still i'll try it once again n then tell u

Q8:-

Q11:- here the first we have to do is to change the cm^2 of A into m^2
so for that we will divide 1 by 10000 = 1 x 10^-4
then we will put all the values in the formula:- P= F/A
over here as we r given with the amount of a-particles colliding per second then by multiplying ths value with speed will give us the acceleration...
so P= ((6.6 x 10^-27) x (1.5 x 10^7) x (5.0 × 10^4)) / (1x10^-4)
= 4.95 x 10^ -11

Q14:- in ths question we will simply find the resultant force of both at X n Y and the subtract them frm eachother
the resultant force in X is :- 2(Tcosθ) = 2 ( 100 cos65) = 84.6
the resultant foce in Y:- 2(Tcosθ) = 2( 120 cos55) = 137.6
so the increase will be :- 137.6 - 84.6 = 53

Q15:-

Q19:- we r given with volume and density n we have to find the mass first..
so m=1000 x 340 = 340000 kg
time= 60 s ( 1 minute)
g= 9.81
h= 30m
put ths in the equation :- P= (mgh) /t to get the input power
our input power = 1667700 W

as efficiency= (output power / input power) x 100
then output power= 1500930 W = 1.5 MW

Q24:- F = ke
so e= F/k
e= 25/150
e=0.167
so original length= 0.55 - 0.167 = 0.383 m

Q33:- P=I^2R ( m using ths formula cuz current will be the same acroos both resistors as they r in series where as V will be dffrnt)
so P in internal resistor = I^2 R
in external resistor = (I^2) 2R
so their ration will be 2

Q35:-

Q37:- In ths question we have to check for the resistance... we know that when resistance is less then voltage will be also less
so there will be lesser voltage in resistor 2 Ω and will be greater in resistor 4 Ω
n as the voltage from X to Y is frm positive to negative so it will be decreasing passing each resistor

Q38:- ok so first we will find the total V across resistors R, 2R, 3R
V in R= 2 V
total resistance= 6R
so
Vr = (2/6) V (Vr is the V across resistor 2R)
so V=6
as the V remains the same in parallel circuits so V across 4R will be also 6V
n emf = 6V

 

[ashcull14]

 

[Dr.MMM]

First Question)

Okay, so what we need to find out is the percentage uncertainty in the value of the volume of the sphere with a radius measurement of r = 5.00 mm ± 0.01 mm.

The formula for volume of a sphere as a function of radius is 4πr³/3. Therefore, we are multiplying r into r into r to get r³, and this is the only value in the equation that has any uncertainty - there is no uncertainty in 4/3 and there is no uncertainty in π, so r is the only quantity we need to concern ourselves with.

The rule for uncertainties when any number of quantities are multiplied goes as follows:
"When one multiplies or divides several measurements together, one can often determine the fractional (or percentage) uncertainty in the final result simply by adding the uncertainties in the several quantities."

Let's do this by example. We are multiplying r by itself three times (to get r³) and so to get the percentage uncertainty in the final result, we resort to "adding the uncertainties in he several quantities". In other words, we add the (percentage uncertainty in r) to the (percentage uncertainty in r) to the
(percentage uncertainty in r).

Basically we multiply the percentage uncertainty in r by 3.
The percentage uncertainty in r = (0.01 mm/5.00 mm) * 100 = 0.2 %.
Multiplying this by 3, we get the uncertainty in the volume of the sphere with this radius to be 0.2 * 3 = 0.6% = C.

(more information here: http://spiff.rit.edu/classes/phys273/uncert/uncert.html)

Second Question)

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.

Third Question - Q22)

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-554#post-826497

Fourth Question - Q23)

From the earlier question I think you might see a pattern - the equivalent spring constant of springs in series (end-to-end attachment) is found in a similar manner to the way the equivalent resistance of resistors connected in parallel is found:

1/k(equivalent) = 1/k(1) + 1/k(2) + 1/k(3)..........+ 1/k (when n springs of spring constant k(1), k(2), k(3) until k are connected in series)

Similarly, the equivalent spring constant of springs connected in parallel (connected side to side at the same level) is found in a similar manner to the way the equivalent resistance of resistors connected is series is found:

k(equivalent) = k(1) + k(2) + k(3).............+ k (when n springs of spring constant k(1), k(2), k(3) until k are connected in parallel)

So, what we can do here is find the equivalent spring constant for each option, and see which one turns out to have the largest extension.

Let's take A first. There are two coils, which implies that there are two springs in that combination, connected in series. Also, both those springs have the same spring constant, let's say "k". Therefore, we can add their spring constants like so:

1/k(equivalent) = 1/k + 1/k = 2/k
Therefore, since 1/k(equivalent) = 2/k, k(equivalent) = k/2.
So the spring constant for the setup shown in option A is equal to k/2.

Onto option B. There are three coils, so we can assume that there are three springs connected in series, and each one of them has the same spring constant, "k" - the same spring constant as the one used for the previous option. So, their equivalent spring constant is given by

1/k(equivalent) = 1/k + 1/k + 1/k = 3/k
Since 1/k(equivalent) = 3/k, k(equivalent) = k/3.
So the spring constant for the setup shown in option B is equal to k/3.

Option C: We have two springs in parallel. Finally, an easier calculation!
Either ways, each one those springs has the spring constant "k", again the same as before. Since they are attached in parallel, we can find their equivalent spring constant as follows:

k(equivalent) = k + k = 2k.

Option D: Happy days, another simple calculation!
Okay, so there are three springs in parallel, and all of them have the spring constant k. So, their equivalent spring constant is given by

k(equivalent) = k + k + k = 3k.

Alright. Now for the extensions.
Each of those springs follows the formula |F| = kx, or (Magnitude of Stretching Force) = (Spring Constant) * (Extension). So, we can rewrite the equation as
x = |F|/k. Let's calculate these ratios for each of the options.

Option A: x = (2 Newtons)/(k/2) = (4/k) meters.
Option B: x = (1 Newton)/(k/3) = (3/k) meters.
Option C: x = (6 Newtons)/(2k) = (6/2k) = (3/k) meters.
Option D: x = (8 Newtons)/(3k) = (8/3k) = (2.667/k) meters.

Clearly, out of all these, the largest extension comes about for Option A, since the rest all have the same denominator but a smaller numerator. So, A.

Hope this helped!
Good Luck for all your exams!

Thanks A LOT.
It really helped

ans is B. How?

 

[danial 234]

A laser emits light of wavelength 600 nm.
What is the distance, expressed as a number of wavelengths, travelled by the light in one second?
A - 5 × 10^8
B 5 × 10^11
C 5 × 10^14
D 5 × 10^17
How do I solve this? What formula is being applied. :/

in one second light will travel 3 × 10^8 metres
1 wavelength= 600 nm
x wavelength= 3 × 10^8 metres
x= 3 × 10^8 metres/600 nm

 

[danial 234]

Thanks A LOT.
It really helped
View attachment 45269
ans is B. How?

cz you have to find the area ...!!

 

[Asad Moosvi]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Question 40

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Question 40

divide proton number by nucleon numbers.... Li will be the ans

 

[Asad Moosvi]

divide proton number by nucleon numbers.... Li will be the ans

Why do we do that?

 

[Thought blocker]

Why do we do that?

Idk that, but we do it... :/

 

[Laila39]

thankssss!

 

[danial 234]

Why do we do that?

i think it is the charge/mass ratios .. because they both affect the speed of particle... the greater the ratio.. the greater the speed..

 

[danial 234]

thankssss!

welcome

 

[Zepudee]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

no 36 (D) ,no 38 (C) thank youu!!

 

[Shaoli Hassan]

Can someone please help me out with October/November 2011/variant-12, Question Number 10? The correct option is A, but I do not understand why.

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf

no 36 (D) ,no 38 (C) thank youu!!

I already asked sagar
36)
Any two conductors connected to each other by another conductor will have the same potential.

This is like having two separate tubes of water (each one represents one conductor), with different amounts and different heights in each (the height represents the potential, the amount represents the charge) - if they are connected by a tube or any other medium that allows free flow of water (this medium represents the conductor) then the heights will become equal (the potential will become equal).

Note that this does not mean "potential" flows from one to another - just like "height" does not flow, potential won't. The charges on an object make up it's electric potential, and these charges will flow (since all materials involved are conductors).
In the analogy, water (the charges) flows to equalize the levels.
In the end, the volume of water in each tube may be different (i.e. the charge on each conductor may be different) but the height (potential) will be the same.

However, this potential only becomes equal in a static situation.

So we can say the potential at point X is +24 Volts because it is connected to the positive terminal of the source, which is at a potential of +24 Volts itself.
Suppose current is flowing through the motor, energy is lost in the motor until the potential of the current becomes equal to the potential of the negative terminal. So the potential at Y is 0 Volts.

The problem arises when either wire is cut - suppose you cut the positive cable somewhere along it's length. What is the potential of X then? It is not connected to the 24 Volt terminal, so what else could the potential there be?

The answer is that the potential of X is 0 Volts, since it is connected to the negative terminal of the battery, which we have assigned to have a zero potential. When the wire is cut no current can flow through the circuit, so the resistance inside the motor stops acting like a resistance and acts like a normal conductor, which means the point X will be at 0 Volts.

Suppose we cut the negative cable. Following the same logic, since the motor no longer does anything (no current because circuit is incomplete), it simply acts as a conductor and equalizes the potential at X and Y, so that both have a potential of +24 Volts.

Suppose the connection in the motor breaks, again no current will flow through the circuit, and X will be at a potential of +24 Volts (since it is still connected by an unbroken connector to the positive terminal of the source) with Y remaining at 0 Volts (since it is connected to the negative terminal of the source).

So having seen all this, we can say that D is the only right answer among all the options.
Let me know if you have any doubts, since it was a little difficult to explain this one, and the concept is not so straightforward anyways.

38)
This is a neat little question, with the only law needed being Kirchoff's Second Law.

There is a very, very quick way of solving it, but first the long way:

i) Suppose you start at the bottom right corner of the circuit.
ii) You go left, across the battery, and find an increase in potential of 20 Volts.
iii) Okay, now you go up (no resistance), go right till the junction (again, no resistance) and take the upper branch.
iv) As you pass the resistor L, you see a drop of 7 Volts.
v) You continue across M where there is some unknown change in potential and then
vi) return to the bottom right corner.

So, the total change in potential should be zero. Therefore,

+20 Volts + (-7)Volts + (Change in potential across M) = 0
13 + (change in potential across M) = 0
Therefore, the change in potential across M = -13 Volts. The change is -13 Volts, the drop is 13 Volts. So B or C.

Let's narrow it down using Q.

i) Again, start at the bottom right corner of the circuit.
ii) Go left, cross the battery, and see an increase in 20 Volts.
iii) Continue until you reach the junction, and pick the upper branch.
iv) Cross P, seeing a drop of 7 Volts, and take the bridge between the branches.
v) Cross N, seeing a drop of 4 Volts, and go towards Q, i.e. to the right on the diagram..
vi) There is some unknown change in potential across Q, and
vii)you then return to the bottom right corner.

Again, the total change in potential should be zero. So,

+20 Volts + (-7) Volts + (-4) Volts + (Change in potential across Q) = 0
9 Volts + (Change in potential across Q) = 0
Change in potential across Q = -9 Volts. Therefore drop = 9 Volts, so the only option that has that is C.

The thing is, that you could have gotten the answer directly without the first part if you had just gone past Q - you would get 9 Volts, and the only option with a 9 Volts drop for Q would be C!

But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have anincrease in potential, not a decrease.

This is because current flows from a region of high potential to low potential - suppose you are going along the current, you are also going from the region of high potential to the region of lower potential. But if you go in the direction opposite to the current you are going from a region of low potential to a region of high potential, which is an increase in potential - you have to make sure you adjust the signs correctly.

 

[♣♠ Magnanimous ♣♠]

I already asked sagar
36)
Any two conductors connected to each other by another conductor will have the same potential.

This is like having two separate tubes of water (each one represents one conductor), with different amounts and different heights in each (the height represents the potential, the amount represents the charge) - if they are connected by a tube or any other medium that allows free flow of water (this medium represents the conductor) then the heights will become equal (the potential will become equal).

Note that this does not mean "potential" flows from one to another - just like "height" does not flow, potential won't. The charges on an object make up it's electric potential, and these charges will flow (since all materials involved are conductors).
In the analogy, water (the charges) flows to equalize the levels.
In the end, the volume of water in each tube may be different (i.e. the charge on each conductor may be different) but the height (potential) will be the same.

However, this potential only becomes equal in a static situation.

So we can say the potential at point X is +24 Volts because it is connected to the positive terminal of the source, which is at a potential of +24 Volts itself.
Suppose current is flowing through the motor, energy is lost in the motor until the potential of the current becomes equal to the potential of the negative terminal. So the potential at Y is 0 Volts.

The problem arises when either wire is cut - suppose you cut the positive cable somewhere along it's length. What is the potential of X then? It is not connected to the 24 Volt terminal, so what else could the potential there be?

The answer is that the potential of X is 0 Volts, since it is connected to the negative terminal of the battery, which we have assigned to have a zero potential. When the wire is cut no current can flow through the circuit, so the resistance inside the motor stops acting like a resistance and acts like a normal conductor, which means the point X will be at 0 Volts.

Suppose we cut the negative cable. Following the same logic, since the motor no longer does anything (no current because circuit is incomplete), it simply acts as a conductor and equalizes the potential at X and Y, so that both have a potential of +24 Volts.

Suppose the connection in the motor breaks, again no current will flow through the circuit, and X will be at a potential of +24 Volts (since it is still connected by an unbroken connector to the positive terminal of the source) with Y remaining at 0 Volts (since it is connected to the negative terminal of the source).

So having seen all this, we can say that D is the only right answer among all the options.
Let me know if you have any doubts, since it was a little difficult to explain this one, and the concept is not so straightforward anyways.

38)
This is a neat little question, with the only law needed being Kirchoff's Second Law.

There is a very, very quick way of solving it, but first the long way:

i) Suppose you start at the bottom right corner of the circuit.
ii) You go left, across the battery, and find an increase in potential of 20 Volts.
iii) Okay, now you go up (no resistance), go right till the junction (again, no resistance) and take the upper branch.
iv) As you pass the resistor L, you see a drop of 7 Volts.
v) You continue across M where there is some unknown change in potential and then
vi) return to the bottom right corner.

So, the total change in potential should be zero. Therefore,

+20 Volts + (-7)Volts + (Change in potential across M) = 0
13 + (change in potential across M) = 0
Therefore, the change in potential across M = -13 Volts. The change is -13 Volts, the drop is 13 Volts. So B or C.

Let's narrow it down using Q.

i) Again, start at the bottom right corner of the circuit.
ii) Go left, cross the battery, and see an increase in 20 Volts.
iii) Continue until you reach the junction, and pick the upper branch.
iv) Cross P, seeing a drop of 7 Volts, and take the bridge between the branches.
v) Cross N, seeing a drop of 4 Volts, and go towards Q, i.e. to the right on the diagram..
vi) There is some unknown change in potential across Q, and
vii)you then return to the bottom right corner.

Again, the total change in potential should be zero. So,

+20 Volts + (-7) Volts + (-4) Volts + (Change in potential across Q) = 0
9 Volts + (Change in potential across Q) = 0
Change in potential across Q = -9 Volts. Therefore drop = 9 Volts, so the only option that has that is C.

The thing is, that you could have gotten the answer directly without the first part if you had just gone past Q - you would get 9 Volts, and the only option with a 9 Volts drop for Q would be C!

But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have anincrease in potential, not a decrease.

This is because current flows from a region of high potential to low potential - suppose you are going along the current, you are also going from the region of high potential to the region of lower potential. But if you go in the direction opposite to the current you are going from a region of low potential to a region of high potential, which is an increase in potential - you have to make sure you adjust the signs correctly.

 

[Thought blocker]

Can someone please help me out with October/November 2011/variant-12, Question Number 10? The correct option is A, but I do not understand why.

10)
At X, sand is stationary and truck is moving so when they stick together the speed decreases.
At Y, sand also leaves with some speed and as the total mass remains same, the speed will not change.

 

[mehria]

I already asked sagar
36)
Any two conductors connected to each other by another conductor will have the same potential.

This is like having two separate tubes of water (each one represents one conductor), with different amounts and different heights in each (the height represents the potential, the amount represents the charge) - if they are connected by a tube or any other medium that allows free flow of water (this medium represents the conductor) then the heights will become equal (the potential will become equal).

Note that this does not mean "potential" flows from one to another - just like "height" does not flow, potential won't. The charges on an object make up it's electric potential, and these charges will flow (since all materials involved are conductors).
In the analogy, water (the charges) flows to equalize the levels.
In the end, the volume of water in each tube may be different (i.e. the charge on each conductor may be different) but the height (potential) will be the same.

However, this potential only becomes equal in a static situation.

So we can say the potential at point X is +24 Volts because it is connected to the positive terminal of the source, which is at a potential of +24 Volts itself.
Suppose current is flowing through the motor, energy is lost in the motor until the potential of the current becomes equal to the potential of the negative terminal. So the potential at Y is 0 Volts.

The problem arises when either wire is cut - suppose you cut the positive cable somewhere along it's length. What is the potential of X then? It is not connected to the 24 Volt terminal, so what else could the potential there be?
y don't u just solve them in a much simpler way??
The answer is that the potential of X is 0 Volts, since it is connected to the negative terminal of the battery, which we have assigned to have a zero potential. When the wire is cut no current can flow through the circuit, so the resistance inside the motor stops acting like a resistance and acts like a normal conductor, which means the point X will be at 0 Volts.

Suppose we cut the negative cable. Following the same logic, since the motor no longer does anything (no current because circuit is incomplete), it simply acts as a conductor and equalizes the potential at X and Y, so that both have a potential of +24 Volts.

Suppose the connection in the motor breaks, again no current will flow through the circuit, and X will be at a potential of +24 Volts (since it is still connected by an unbroken connector to the positive terminal of the source) with Y remaining at 0 Volts (since it is connected to the negative terminal of the source).

So having seen all this, we can say that D is the only right answer among all the options.
Let me know if you have any doubts, since it was a little difficult to explain this one, and the concept is not so straightforward anyways.

38)
This is a neat little question, with the only law needed being Kirchoff's Second Law.

There is a very, very quick way of solving it, but first the long way:

i) Suppose you start at the bottom right corner of the circuit.
ii) You go left, across the battery, and find an increase in potential of 20 Volts.
iii) Okay, now you go up (no resistance), go right till the junction (again, no resistance) and take the upper branch.
iv) As you pass the resistor L, you see a drop of 7 Volts.
v) You continue across M where there is some unknown change in potential and then
vi) return to the bottom right corner.

So, the total change in potential should be zero. Therefore,

+20 Volts + (-7)Volts + (Change in potential across M) = 0
13 + (change in potential across M) = 0
Therefore, the change in potential across M = -13 Volts. The change is -13 Volts, the drop is 13 Volts. So B or C.

Let's narrow it down using Q.

i) Again, start at the bottom right corner of the circuit.
ii) Go left, cross the battery, and see an increase in 20 Volts.
iii) Continue until you reach the junction, and pick the upper branch.
iv) Cross P, seeing a drop of 7 Volts, and take the bridge between the branches.
v) Cross N, seeing a drop of 4 Volts, and go towards Q, i.e. to the right on the diagram..
vi) There is some unknown change in potential across Q, and
vii)you then return to the bottom right corner.

Again, the total change in potential should be zero. So,

+20 Volts + (-7) Volts + (-4) Volts + (Change in potential across Q) = 0
9 Volts + (Change in potential across Q) = 0
Change in potential across Q = -9 Volts. Therefore drop = 9 Volts, so the only option that has that is C.

The thing is, that you could have gotten the answer directly without the first part if you had just gone past Q - you would get 9 Volts, and the only option with a 9 Volts drop for Q would be C!

But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have anincrease in potential, not a decrease.

This is because current flows from a region of high potential to low potential - suppose you are going along the current, you are also going from the region of high potential to the region of lower potential. But if you go in the direction opposite to the current you are going from a region of low potential to a region of high potential, which is an increase in potential - you have to make sure you adjust the signs correctly.

y dnt u solve them in a mush simpler way? Thought blocker

 

[danial 234]

Can someone please help me out with October/November 2011/variant-12, Question Number 10? The correct option is A, but I do not understand why.

correct option is B

 

[Thought blocker]

y dnt u solve them in a mush simpler way? Thought blocker

Why you want me to solve if it is already solved ??
And its very simpler.... read it...