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Physics: Post your doubts here!

D

Dr.MMM

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sagar65265

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huh said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Please Q20. It's urgent. Don't understand answer.
Click to expand...

This image should clear up things:

Image005.png


So for any stretchable material, we can say that there is a point of applied stress until which they will stretch in a linear manner, i.e. with a constant value of the Young's Modulus. So until this point, the stress will be proportional to the strain applied, which is why this point, the last point of proportionality, is called the limit of proportionality.

However, we have to take a close look at the definition of elasticity - a sample stretches elastically as long as it can return to it's original length by the same route it was stretched by. So the point to remember here is that elasticity does not mean proportionality.

Even after the limit of proportionality has been reached, the sample can still return to it's original length by the same route (initially non-linear, then linear).

But beyond a certain point, even that stops happening - in the diagram, we can see that at C', the sample returns by a different route than it took to reach there. It does not trace it's way back there, and thus has gone by the elastic limit. Therefore, the elastic limit is not the same as the limit of proportionality, and so we can say that the answer is B (A suggests that the limit of elasticity lies in the linear reason, which is wrong. C is wrong since S is not the limit of elasticity, it is the limit of proportionality, and D is wrong since the return of the sample from T is not by the same route).

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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Dr.MMM said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q:18 Ans: D

Q19: The Mariana Trench in the Pacific Ocean has a depth of about 10 km.
Assuming that sea water is incompressible and has a density of about 1020 kg m–3, what would
be the approximate pressure at that depth?
A 10^5 Pa B 10^6Pa C 10^7Pa D 10^8Pa
Ans D

HOW?
Click to expand...

The pressure of a liquid at a depth below it's surface is given by the formula

Pressure = ρgh

Where ρ is the density of the liquid, g is the average gravitational field strength along that liquid column and h is the depth of the liquid column from the surface till the point where the pressure has to be measured.

So, we can write here that the pressure at a depth of 10 kilometers (which is equal to 10,000 meters) when the density of the water is 1020 kg m⁻³ and the average gravitational field strength is 9.81 ms⁻² is:

Pressure = (1020 kg m⁻³) * (9.81 ms⁻²) * (10,000 m) = 10, 006, 200 Pascals. This is equal to 1.0006 * 10⁸ Pascals = about 10⁸ Pascals = D.

Hope this helped!
Good Luck for all your exams!
 
H

huh

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sagar65265 said:
This image should clear up things:

Image005.png


So for any stretchable material, we can say that there is a point of applied stress until which they will stretch in a linear manner, i.e. with a constant value of the Young's Modulus. So until this point, the stress will be proportional to the strain applied, which is why this point, the last point of proportionality, is called the limit of proportionality.

However, we have to take a close look at the definition of elasticity - a sample stretches elastically as long as it can return to it's original length by the same route it was stretched by. So the point to remember here is that elasticity does not mean proportionality.

Even after the limit of proportionality has been reached, the sample can still return to it's original length by the same route (initially non-linear, then linear).

But beyond a certain point, even that stops happening - in the diagram, we can see that at C', the sample returns by a different route than it took to reach there. It does not trace it's way back there, and thus has gone by the elastic limit. Therefore, the elastic limit is not the same as the limit of proportionality, and so we can say that the answer is B (A suggests that the limit of elasticity lies in the linear reason, which is wrong. C is wrong since S is not the limit of elasticity, it is the limit of proportionality, and D is wrong since the return of the sample from T is not by the same route).

Hope this helped!
Good Luck for all your exams!
Click to expand...
You sir are a living legend. I will remember you in my prayers. Thank you very much.
 
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mehria said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
Q15 and Q23
Click to expand...

Q15)
Forces On Ball.JPG

Since the ball is balanced and in equilibrium, we can say that the net force on the ball is equal to zero. We can also split this net force into two components, one in the up-down (vertical) direction and another in the left-right (horizontal) direction. Let's also say that the upwards direction is positive and the leftward direction is negative.

In the vertical direction:

i) T has a component of T * cos(30) = (√3/2) * T = √3T/2 = (+√3T/2) Newtons
ii) W has a component of 0.15 Newtons downwards, so it has a component of (-0.15) Newtons.
iii)The wind force is horizontal, so it has no component.

Summing these up should give us zero, so

(+√3T/2) + (-0.15) = 0
√3T/2 = 0.15
√3T = 0.3

T = 0.1732 Newtons

That is the magnitude of the Tension force. Now let's do the horizontal direction:

i) T has a component of T * sin(30) in the rightward direction = T * (1/2) = T/2. Since this is to the right, the sign is negative, i.e. (-T/2) Newtons.
ii) The weight has no horizontal component in this situation.
iii) The Wind force has a component F(air) to the left, and only the left. Therefore, it has a component of (+F(air)) Newtons.

Summing these up should give zero, so

(+F(air)) + (-T/2) = 0
F(air) = T/2 = 0.1732/2 = 0.866 Newtons = 0.87 Newtons = B.

23)

When the force W is applied on R, the spring R gets the full load of the force. In other words, the force extending R is equal to the force W. Therefore, the extension of spring R is equal to (using the formula |F| = kx):

x = |F|/k = W/k = W/3k (since the spring constant of R is 3k).

Now, the spring R will exert the same force at both ends - it will exert a force W on the lower end, and it will exert a force W on the upper end, i.e. on the bar.
For the bar to be in equilibrium, the force on it have to balance out, so

(Force from spring P) + (Force from spring Q) = (Force from spring R) (The bar has negligible weight)

For moments to be zero about the center of mass, the (Force from spring P) has to be equal to the (Force from spring Q). Therefore,

2(Force from spring P) = 2(Force from spring Q) = W

Therefore, the (Force from spring P) = W/2 = (Force from spring Q).

The extension of spring P if it has to exert a force of W/2 is equal to

x = |F|/k = (W/2)/k = W/2k

The extension of spring Q is the same, so that the system is stable and balanced.

Therefore, the overall extension is

W/2k + W/3k = 3W/6k + 2W/6k = 5W/6k meters = A.
If you still have a doubt concerning either of these questions, just post it on the forums and i'll see if I can answer.

Hope this helped!
Good Luck for all your exams!
 
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crazytaylorfanXD said:
yeah :)
Click to expand...

Alright, then:

31)

When a charged particle is placed anywhere in space, it extends an electric field around it, that spreads in all possible directions, and exists whether or not there is any other particle present to experience the force. Therefore, the particle Q creates an electric field around it in all directions.
The important point to note here is that the particle cannot exert a force on itself.
Another important point to note, therefore, is that when the question asks for the field strength of charge Q, we do not need to consider the force on charge Q but we need to consider the force exerted by the field on charge q.

The definition of electric field strength at a point is "...the force experienced per unit charge at a point in an electric field gives us the strength of the electric field at that point in N/C". So, if a force "F" is experienced by the particle "q", then at the point where q is located the strength of the field is given by "F/q" = D.

One more point worth noting is that we can also say that the electric field strength at the location of "Q" is given by the ratio of the force exerted on it divided by the charge - thus, the electric field strength at the point where Q is located will be "F/Q", instead of the above "F/q". This is not the same electric field strength, but it is such that it will exert the same force.

32)

Interesting question here - the good part for us is that the current is reduced steadily, so we can take an average value and still get a correct answer.
The average current that flows in this time duration of 8.0 seconds is

(100 mA + 20 mA)/2 = 120 mA/2 = 60 mA.

This is the average current flowing past that point over the 8.0 seconds.

Since Q = It (Charge flowing past a point in a time interval = Average current over time interval * length of time interval), we can write

Q = 60 mA * 8.0 Seconds = 480 mC = C.

I'll reply for the others later, just that Q36 has a gigantic explanation (and deserves one, in any case).

Hope this helped!
Good Luck for all your exams!
 
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crazytaylorfanXD said:
yeah :)
Click to expand...

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.

Q37)

The potential difference between points P and Q is the same as the potential difference across the fixed resistor.

Suppose the Voltage of the source is some value V. Also assume that the resistance of the fixed resistor is R.
If we write the resistance of the thermistor as R(t), then we can find that the current in the circuit is equal to

I = V/(ΣR) = V/(R + R(t))

Since the potential difference across the terminals of a resistor can be given by V(across) = I * R, we can see that the potential difference across the resistor R is given by

V(across R) = V(between P and Q) = IR = VR/(R + R(t))

So this equation can be used to draw a graph of V(between P and Q) on the y-axis and R(t) on the x-axis.

But we note that they have plotted θ on the graph instead of R(t). No problem, we just note that as temperature increases, the resistance R(t) of the thermistor decreases. Therefore, as we go along the x-axis, we have an increasing temperature and a consequently decreasing resistance.

From the formula above [V(between P and Q) = IR = VR/(R + R(t))] we can see that since V and R are assumed to be constant, decreasing R(t) increases the Voltage and increasing R(t) increases the denominator and decreases the voltage. So the voltage is increasing as resistance R(t) decreases.

Therefore, the Voltage increases as temperature θ decreases. So we have C or D.

The last thing to note is that this is not a linear formula - to prove this, we can say that R(t) = k/θ (k is some constant).

Note that increasing θ decreases R, decreasing θ increases R, and so on. Suppose we substitute for R(t) in the equation above, we get

VR/(R + k/θ) = VR/([Rθ + k]/θ) = VRθ/(Rθ + k)

This is not a direct proportionality, so the graph is not a linear graph. Therefore, you can see that the option D (which is a linear, straight line graph) is the wrong answer. The only remaining option is C.

Q38)

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 an increase 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.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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crazytaylorfanXD said:
Can someone please help me in JUN 2012 P11 Qs.( 5 , 8 , 15 , 37)
JUN 13 P12 Qs.( 6 , 20 , 32)
NOV 13 P13 Qs. ( 31 , 32 , 36 , 37 , 38 )

I know its a lot of questions but ive been so confused :(
Your help would be much appreciated. :)
Click to expand...

5)

There is a rule that when dividing or multiplying two or more numbers, the result should be stated to the same number of significant figures as in the multiplier with the same number of significant figures.

For example, suppose we have three numbers, A = 2.570, B = 5.0 and C = 5.89678.
Suppose we want to obtain the value of the product ABC, then we simply multiply the numbers to get

2.570 * 5.0 * 5.89678 = 75.773623

However, looking at the multipliers (A, B and C) we can see that A has 4 significant zeroes (trailing zeroes are always significant), B has 2 significant figures (same rule applies here) and C has 6 significant figures (all non-zero digits are significant). So, the least out of these is 2 significant figures from B. Therefore, we write our answer to 2 significant figures, as 76. Not 75.78, not 75.8, but 76.

Similarly, in this case we are dividing (distance) by (time) to get the speed of an object. So, our calculation is

(40.0 m)/(2.50 s) = 16 m/s.

But since both values, 40.0 and 2.50, are to 3 significant figures, we should give our answer as 16.0, which has 3 significant figures. This immediately points us to C, which is the right answer, but we can go further.

When we are multiplying or dividing two values, the % uncertainty in the final product is equal to the sum of the % uncertainties of the individual multipliers.
So, in this situation, the % uncertainty in 40.0 = 0.1/40.0 * 100 = 0.25 %. In 2.50, the % uncertainty is 0.05/2.50 * 100 = 2%.
Summing this up, we get 2 + 0.25 = 2.25 %.

This is the % uncertainty in our final value, 16.0. Therefore, the absolute uncertainty is 2.25/100 * 16 = 0.36.
However, from the Examiner's Report, a comment says that
"The very popular incorrect response was D. Candidates should realise that any uncertainty should be expressed to one significant figure."

So we need to write the uncertainty to 1 significant figure, which gives us 0.4 as the uncertainty, and 16.0 as the answer. So C.

8)

We know that since the tube is evacuated, a vacuum exists inside it. Therefore, no air resistance affects the motion of the feather and it falls freely with an acceleration of "g" ms⁻² (which we can assume is around 9.81 ms⁻²). So, it starts from rest and falls a distance "L" in time "T". Therefore, the parameters of motion are:

u = 0 ms⁻¹
v = ?
s = L m
a = g ms⁻²
t = T s

So, writing the equation s = ut + 0.5at², we can write (since the distance traveled in T seconds is L meters)

L = (0)T + 0.5gT²
L = gT²/2
2L/g = T²

So that T = √(2L/g) seconds.

Now we take a look at the situation where t = 0.50T = 0.5√(2L/g). Here, we again take a look at the parameters:

u = 0 ms⁻¹
v = ? ms⁻¹
s = ? m
a = g ms⁻²
t = (T/2) s

We again put this in the same equation to get

s = (0)(T/2) + 0.5 * g * (T/2)² = 0.5 * g * T²/4 = gT²/8
Putting the value of T we obtain earlier, we get

s= g/8 * 2L/g = L/4 = 0.25L = B.

15) Again, this is a little long so i'll post it later.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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crazytaylorfanXD said:
Can someone please help me in JUN 2012 P11 Qs.( 5 , 8 , 15 , 37)
JUN 13 P12 Qs.( 6 , 20 , 32)
NOV 13 P13 Qs. ( 31 , 32 , 36 , 37 , 38 )

I know its a lot of questions but ive been so confused :(
Your help would be much appreciated. :)
Click to expand...

Q15)

We have to consider the two elements or conditions of equilibrium:

i) For translational equilibrium, no net force must act on the system;
ii)For rotational equilibrium, not net torque must act on the system about any point.

Let's take the crane itself to be the system. Then forces are exerted on the crane by the Earth (Gravity), the weight W (this is stationary so it has to be balanced), the weight L (same justification as previous one) and the ground (Normal force, maybe frictional force).

Now the first thing to note is that moving the weight L to the right, the crane feels a tendency to rotate in the clockwise direction, since the net moment about the center (Let's take that point to be at the top of the vertical rod, and at it's middle) will try to rotate it in that direction.

So if the crane tries to rotate clockwise, for it to be stationary there has to be some other torque that tries to rotate it anti-clockwise, right?
This is the second part of the conditions, and yes, there has to be a torque that tries to rotate the system anticlockwise.

This eliminates A - the force by the ground on A will only rotate it further clockwise, not resist the clockwise motion (imagine the crane is a door hinged at the top, and apply that force - the door will rotate clockwise, right? This is the opposite of what we want happening).

B is possible - the force by the ground attempts to rotate the "door" anti-clockwise and could thus aid in stabilizing the system, but we'll get back to that later.

C is wrong - suppose the normal force R moves to the left, then it will still try to rotate the system in a clockwise manner (since it is not perfectly aligned with that point we are taking moments about, it will try to rotate the system clockwise, which is not what we want).

D is possible - the reaction force R attempts to turn the crane anticlockwise and thus could stabilize the system, but there's a little more work to do.

We have to apply the first condition of equilibrium - that no net force acts on the system.
In B, there is a vertical force of gravity, a vertical force from W, a vertical force from L, a vertical force from the ground and a horizontal force from the ground.
Hang on. This isn't balanced! There is not horizontal force counteracting the horizontal force from the ground, so how can the system remain in equilibrium?

Let's confirm with D. In D, there is no horizontal force that upsets the equilibrium, but there is a variable vertical force that ensures equilibrium both, for translation and rotation. Therefore, D is our answer.

Q37)

Let's put aside the light factor and deal exclusively with the resistance.
First things first - the output voltage will be equal to the potential difference across the resistor R, since the output terminals are connected across R, and a voltmeter connected in the same manner would give the potential difference across R which is what we get in the output circuit.

Since the same current passes through both the resistors, the net resistance will be equal to

(Resistance of R) + (Resistance of LDR)

And since we can write V = IR, we can say

V = 10 - 0 = 10 Volts = I (Resistance of R + Resistance of LDR)

and therefore, I = 10/([Resistance of R] + [Resistance of LDR])

This is the current flowing through both the resistors. Now, we can find out the potential difference across resistance R by again using V = IR to get

V(across R) = Output Voltage = 10 * [Resistance of R]/([Resistance of R] + [Resistance of LDR])

Right now, I suggest you write this down if possible, in whatever manner is easiest for you to understand, since the calculation we can do now is going to be difficult to display here in a post.

Once you're done, note that we have to maximize this value of output voltage. We can do this with calculus, or we can do this with a neat math trick, which i''ll describe below:

We can divide both the numerator and denominator of the Output Voltage equation by [Resistance of R] - this is allowed since while doing this we are essentially multiplying the equation by (1/[Resistance of R])/(1/[Resistance of R]) - this is equal to 1, and multiplying an equation by 1 doesn't change it.

In this post, i'll do it step by step. First let's divide the top half by [Resistance of R] and we get

10 * ([Resistance of R]/[Resistance of R])/([Resistance of R] + [Resistance of LDR]) = 10/([Resistance of R] + [Resistance of LDR])

Now let's divide the bottom half by [Resistance of R] to get

10/([Resistance of R]/[Resistance of R] + [Resistance of LDR]/[Resistance of R]) = 10/(1 + [Resistance of LDR]/[Resistance of R])

So the output voltage is also equal to 10/(1 + [Resistance of LDR]/[Resistance of R]). To maximize this, we have to minimize the denominator, since the numerator is constant and cannot be changed.
Furthermore, since the denominator is equal to (1 + [Resistance of LDR]/[Resistance of R]) we basically have to minimize the second half, since the first half is a constant (1) and cannot be changed.

So, to minimize [Resistance of LDR]/[Resistance of R], we have to maximize [Resistance of R] and minimize [Resistance of LDR], since that will give us the smallest fraction possible.
To minimize the [Resistance of LDR], we note that we have to maximize the intensity of the light falling on the LDR, and so the only option that shows both these options is A.

Hope this helped!
Good Luck for all your exams!
 
Hadi Murtaza

Hadi Murtaza

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Browny said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf

Sorry about that.
Click to expand...
24.
Intensity ∝ (Amplitude)²
Intensity ∝ (Frequency)²
Intensity = (Amplitude)² × (Frequency)²

Amplitude of wave P = x₀
Frequncy of wave P = F
So intensity of wave P = x₀² * F² = (Fx₀)²

Amplitude of wave Q = 2x₀
Frequency of wave Q = F/2 , since only half a wave passes at time t₀
So intensity of wave Q = (2x₀)² * (F/2)² = 4x₀² × F²/4 = F²x₀² = (Fx₀)²

Intensity of both is (Fx₀)² so intensity of Q will also be I₀
 
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