# Physics 9702 Doubts | Help Page 28

__Question 163: [Current of Electricity > Potentiometer]__Diagrams show same cell, ammeter, potentiometer and fixed resistor connected in different ways.

Distance d between sliding contact and a particular end of potentiometer is varied.

Current measured is then plotted against distance d.

For which two circuits will graphs be identical?

A W and X B W and Y C X and Y D Y and Z

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q36*

__Solution 163:__**Answer: A.**

A potentiometer is different than a
regular resistor in that the potentiometer can allow for different values of
resistance.

As in this case, a potentiometer
usually consists of a long wire having some resistance (R = ρL /
A). In simple terms, we can imagine current to flow at one end of the wire and
up to a specific distance on the wire where is slider is placed into contact
with the wire. Then, instead of continuing along the rest of the wire, the
current would flow in the slider which is connected to another point in the
circuit. This is possible if, for example, the slider has a negligible
resistance. Current would chose the easiest way, that is, the path is the
smallest resistance.

W and X will give identical graphs. In
each of these 2 arrangements, the resistance of the length d of the
potentiometer is connected into the circuit but the right-hand end does not
contribute {current flows from positive terminal of the
cell – so the right-hand end does not contribute}.

In arrangement W, the right-hand end
is even not connected, so only the length d of the wire will contribute some
resistance, along with the resistance of the fixed resistor.

In arrangement X, the slider (head
of arrow in diagram) is connected the bold point shown. Since the slider has
negligible resistance, it is at the same potential as the point. So, the
potential difference between the slider and the point is zero. When the p.d.
between 2 points is zero, current does not flow through the component connected
between these 2 points. In this arrangement, the right part of the
potentiometer is connected between these 2 points, so no current flows through
the right part of the potentiometer. Instead, the current chooses the easiest
path – the path with no resistance – and this is the slider as shown.

For arrangement Y, the path other
than the distance d on the wire of the potentiometer is connected to the
circuit. So, this arrangement is different from those of W and X. Even if the
circuit is complete, the results obtained on a graph would not be identical to
that of W or X.

For arrangement Z, current flows from the positive terminal of the battery, then along a distance d on the potentiometer but here, the current does not flow through the fixed resistor. The reason is similar to the circuit in X. The bold point and the slider (head of arrow) are at the same potential and the fixed resistor is connected between these 2 points. So, current does not flow through it and instead, current flows through the slider. Thus, the total resistance in the circuit is not the same as in W or X.

__Question 164: [Waves > Superposition]__Noise reduction headphones actively produce their own sound waves in order to cancel out external sound waves.

Microphone in headphones receives waves of one frequency. Loudspeaker in headphones then produces wave of that frequency but of a different phase.

What is the phase difference between external sound wave and wave produced by the loudspeaker in the headphones?

A 90° B 180° C 270° D 360°

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q29*

__Solution 164:__**Answer: B.**

For the waves to cancel out (from the

**principle of superposition**), the waves should be out of phase (phase difference should be between 180° and 359°).
A phase difference of 360° means
that the waves are completely in phase (there is no
phase difference between them.). In this case, the noise is amplified to
twice its amplitude .

A phase difference of 90° also means
that sound becomes noisier.

To completely cancel the noises, the
ideal phase difference should be 180°.

__Question 165: [Matter > Stress]__
Diagram shows a large crane on
construction site lifting a cube-shaped load.

Model is made of crane, its load and
cable supporting the load.

Material used for each part of model
is the same as that in full-size crane, cable and load. Model is one tenth
full-size in all linear dimensions.

What is ratio

**of**stress in the cable on the full size crane**to**stress in the cable on the model crane?
A 10

^{0}B 10^{1}C10^{2}D 10^{3}**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q23*

__Solution 165:__**Answer: B.**

Stress = force per unit area = F / A

Let the force (weight) and the (cross-sectional)
area in the full-size crane be F and A respectively.

**Stress in cable of full-size crane = F / A**

__As mentioned in the question, the model is one-tenth full-size in all linear dimensions.__

**For the model**,

The load, which has a
cube-shaped load, is in 3-dimension. So,
for each of the 3 dimensions, the length of the model is reduced by a factor of
1/10. Therefore, the volume of the load is reduced by a factor of (1/10)

^{3}= 1 / 1000.
Weight = mg and Mass =
density x volume. Since the same material is used, the density is the same. So,
the mass is proportional to the volume. A reduction in the volume causes the
mass to be reduced by the same factor. The weight, which depends on the mass (g
is constant), is also reduced by the same factor.

Force (weight) in model = F / 1000

Similarly, the
cross-sectional area (which depends on (diameter)

^{2}) will be reduced by a factor of (1/10)^{2}= 1 / 100.
(Cross-sectional) Area in model = A
/ 100

**Stress in cable of model crane = (F/1000) / (A/100) = 0.1 (F/A)**

Ratio = (F/A) / 0.1(F/A) = 10

__Question 166: [Waves > Phase difference]__
X and Y are two points on surface of
water in a ripple tank. Source of waves of constant frequency begins to
generate waves which then travel past X and Y, causing them to oscillate.

What is phase difference between X
and Y?

A 45° B 135° C
180° D 270°

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q28*

__Solution 166:__**Answer: D.**

A distance of 1 λ
represents a phase difference of 360

^{o}(that is, the points behave exactly in a similar way)
Consider point X to be the 1

^{st}node of a wavelength. Considering nodes and antinodes, X is followed by an antinode (minima), then a node and finally point Y which is an antinode (maxima). For 1 complete wavelength, we need to consider the node just after point Y.
So, distance between X and
Y is ¾ λ.

1 λ represents a phase difference of
360

^{o}
¾ λ represents a phase difference of
¾ x 360 = 270

^{o}

__Question 167: [Waves > Two-source interference]__Diagram shows experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S

_{1}and S

_{2}.

Detector is moved from point O in direction of the arrow. Signal detected decreases until the detector reaches point X, and then starts to increase again as detector moves beyond X.

Which equation correctly determines position of X?

A OX = λ

B OX = λ / 2

C S

_{2}X – S

_{1}X = λ

D S

_{2}X – S

_{1}X = λ / 2

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q27*

__Solution 167:__**Answer: D.**

A minima is found at X since the
signal keeps on decreasing until it reaches X (so at X,
the signal has the lowest amplitude). X is the first minima.

S

_{2}X and S_{1}X are distances.
Distance from point O to 1

^{st}minima = λ/2 (destructive interference) [This equation alone does not allow us to determine the position of X. So, B is not the answer here]
{The distance between
point O and the first MAXIMA would be λ}

This distance (λ/2) is also equal to
the path difference between S

_{2}X and S_{1}X which is (S_{2}X – S_{1}X)

__Question 168: [Matter > Kinetic Theory of Gases]__Graph shows distribution of speeds for molecules of a gas at a particular temperature.

Which statement is correct?

A All molecules have the same kinetic energy.

B Commonest value of speed is also the average speed.

C Graph shows that the molecules of a gas are widely spaced apart.

D Peak value of the graph would move to the right if the temperature is increased.

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q21*

__Solution 168:__**Answer: D.**

The speed varies, so the molecules
do not have the same kinetic energy. [A incorrect]

The commonest value of speed {this is the most frequent value of speed – it should be the
speed that the greatest number of molecules have} is the speed corresponding
to the peak number of molecules with speed v axis, not the average speed (the average speed gives an idea of the general speed of the
molecules as a whole). [B incorrect]

The graph does not show that the
molecules of a gas are widely spaced apart, but shows that most of the
molecules are concentrated to a specific range of energies (which is close to the peak number of molecules – the other
values of speed correspond to a relatively much smaller number of molecules).
[C incorrect]

The peak value of the graph would
move to the right if the temperature is increased (they
gain thermal energy which is converted into kinetic energy (= ½ mv

^{2}). So, their speed increases).

__Question 169: [Waves > Stationary wave]__Sound from loudspeaker placed above a tube causes resonance of the air in the tube.

Stationary wave is formed with two nodes and two antinodes as shown.

Speed of sound in air is 330 m s

^{–1}.

What is the frequency of sound?

A 413 Hz B 550 Hz C 830 Hz D 1650 Hz

**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q27*

__Solution 169:__**Answer: A.**

From the diagram, it is seen that 0.75 λ (¾ of a wavelength) is formed in the 60.0cm
(0.60m) long tube.

0.75 λ
= 0.6m

Wavelength, λ = 0.8m

**Speed, v = f λ**

So, frequency, f = v/ λ = 330/0.8 =
412.5Hz

__Question 170: [Waves > Interference > Diffraction grating]__Diffraction grating has N lines per unit length and is placed at 90° to monochromatic light of wavelength λ.

What is the expression for θ, the angle to the normal to grating at which third order diffraction peak is observed?

A sin θ = 1 / 3Nλ B sin θ = 3N λ C sin θ = Nλ / 3 D sin θ = 3λ / N

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q28*

__Solution 170:__**Answer: B.**

For diffraction grating,

**d sin**

**θ**

_{n}= nλθ

_{n}is the angle made by the n

^{th}order diffraction peak

and λ is the wavelength of the monochromatic light

Slit separation, d = 1 / N

For 3

^{rd}order diffraction peak,

(1/N) sinθ = 3λ

sin θ = 3N λ

__Question 171: [Matter > Young modulus]__
The diagram shows a large crane on a
construction site lifting a cube-shaped load.

Model is made of crane, its load and
cable supporting the load.

Material used for each part of model
is the same as that in full-size crane, cable and load. Model is one tenth
full-size in all linear dimensions.

What is ratio

**of**extension of the cable on the full size crane**to**extension of the cable on the model crane?
A 10

^{0}B 10^{1}C10^{2}D 10^{3}**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q19*

__Solution 171:__**Answer: C.**

As explained in question
165 above, the full-size load has 1000 times the weight of the model and the
area of cross-section of the cable itself is 100 times that of the model.

Now, the length (which is 1
dimensional) of the cable should be 10 times that of the model.

**Young modulus, Y = Stress / strain**

Young modulus, Y = (Force, F / Area,
A) / (extension, e / original length, l)

Extension, e = force × length / (Y ×
area)

Since the same material is used, the
Young modulus, Y is the same.

Let all the quantities
involved in the determination of the extension be unity (1) for the model. The
extension in the model would be 1.

Then, for the full-size crane,

Extension = force × length / (Y ×
area) = 1000 × 10 / (1 × 100) = 100

Ratio = 100 / 1 = 100

Can you explain why in question 163, the right-hand of the potentiometer dosent contribute in all arrangements and also why in Y, the potentiometer dosent contribute at all even though the right hand end of it is connected in the circuit.

ReplyDeleteAs stated above, current flows form the positive terminal of the battery. The potentiometer is an apparatus such that the resistance can be varied - that is some resistance can contribute to the circuit and some can be prevented to do so.

DeleteIn the above diagram, the portion of resistance being contributed is that from the left end up to where the junction is placed. (Of course, this reference is just for explaining - the actual case may not be like this)

Like I said, the left end up to the junction contributes to the resistance. but since the left end is not even connected in the case of Y, it's as if it's not there. Place a component near a circuit and do not connect it. Will it affect the circuit? No.

But what about the right end of the potentiometer in Y. That end is certainly connected in the circuit. Will it contribute to the resistance?

DeleteSecondly, in X, the right hand end of the potentiometer is connected to a junction(shown by bold dot) so shouldnt that portion of the resistance be in parallel with the other portion?

Like I said, the left end up to the junction contributes. In all of them, the junction up to the right end are not contributing. So, even if it's connected, it is not contributing any resistances. Be sure to understand that a potentiometer is not like a fixed resistor.

DeleteFor the case in X.

If what you said is correct, then the potentiometer is contributing all its resistances or the potentiometer is just like a fixed resisor. By that is not correct.

If you look carefully, you will see that no component is connected between the bold dot and the right end of the potentiometer. So, it's as if that wire is not even there. It does not make any difference. That's why I said W and X can be considered to be the same.

Note that I'm saying the left end up to the junction contributes because the left end is connected to the positive terminal from which current flows. So, the current should go through the left end. If the right end was connected to the positive terminal of the battery, then the part contributing resistance would be the right end up to the junction.

Try to look for some good notes on potentiometer. I have not yet prepared on it.

I still have one misconception . In Y , aren't the cell, the slider, the right hand end of the potentiometer , the fixed resistor and the ammeter all making one complete loop. In that case, shouldn't the current flow from the positive terminal of the cell through the sliding contact and then through the right hand end of the potentiometer all the way back to the negative terminal?

DeleteBasically yes. But as I have said before, the slider up to the right end does not contribute any resistance. You COULD say that the current flows from + terminal to slider and the slider is connected directly to the wire that goes to the fixed resistor. You could say that the part: slider to right end of potentiometer is just like a normal wire used to connected the components. Current flowing through this part can be considered to be just like current flowing in the connecting wires.

DeleteI understand that but isnt the slider connected to a portion of the potentiometer in Y, i mean how can we just consider that it is like a normal wire when it actually is a resistor?

DeleteA potentiometer is a variable resistor to be more precise. So, while part of its resistance is being contributed to the circuit, the other part is not. Now, in the case we are discussing, the part that would contribute resistance is not even connected to the circuit (there's no wire on the left end). So, even if the 'part that do not contribute resistance' is connected, it's still no contributing any resistance.

DeleteWell, if you consider a variable resistor, we can choose the resistance to be zero. So, there's no problem to the fact the the potentiometer is not contributing any resistance in some cases.

So in the case of Y, irrespective of the slider position, the effective resistance will be the resistance of the fixed resistor only and so current will not change with any change in d?

DeleteBasically, yes.

DeleteHmm....i just want to understand why the right hand end dosent contribute even if it is connected to the fixed resistor in Y....the current from the positive terminal flows though the slider and then through a portion of the potentiometer's resistance....so why dosent right hand end contribute even if it is connected by a wire to the fixed resistor.

DeleteFor this, you need to look how a potentiometer / variable resistor is built. There are 3 outlets and what has been explained above can be done,. Ask your teacher to explain it to you in the lab, it would be better.

DeleteTo have a better idea, you need to see and use it by yourself, so i understand that it can be difficult to properly understand what I said here,

Ok thankyou for your help

DeleteThe explanations have been updated. The previous one contained some mistakes. So, I recommend you don't consider the above set of comments as it was based on the previous explanations.

Deletefor question 165, there was an exact same question in another variant which is variant 13 but they're asking for ratio of extension of the cable on the full size crane/extension of the cable on the model crane. how would we do that

ReplyDeleteLook at question 171 on the same page here.

Deletewhere is the november 2014 worked solution?

ReplyDeleteA list of worked solutions is available at

Deletehttp://physics-ref.blogspot.com/2014/05/physics-9702-notes-worked-solutions-for.html

i need november 2014 paper 22

ReplyDeleteI'll try to add it later

Deletejune q23 and q5 paper 1

ReplyDeleteWhich year, paper, question number???

Deleteq18 nov 2007 paper 1

ReplyDeleteSee solution 957 at

Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-197.html

in question 166

ReplyDeleteif the distance between x and a point after y equals half wavelength

then shouldn't the distance between x and y be less then half wavelength?

As stated above, distance between X and Y is ¾ λ.

DeleteHi admin, can u do worked solutions for paper 13, m/j 2015???? I faced a lots of problems there. Thanks!!!

ReplyDeleteState which questions you are having problems

DeleteFor Question 165: [Matter > Stress] (MJ13 paper 13)...ur answer is B..but the answer in the CIE ms says C...which one is right?

ReplyDeleteIt's from P12.

DeleteWhat you are referring to is probably solution 171 on this same page