• 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}.

The flow of current in the 4 arrangements is shown below. This will allow us to better identify which circuit would give the same graph. Current flows from the positive terminal of the battery towards it negative terminal.

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:

Go to
Noise reduction headphones actively produce their own sound waves in order to cancel out external sound waves.

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 10⁰               B 10¹               C10²                D 10³

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 S₁ and S₂.

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 S₂X – S₁X = λ
D S₂X – S₁X = λ / 2

Reference: Past Exam Paper – June 2014 Paper 12 Q27



Solution 167:
Go to
The diagram shows an experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S1 and S2.








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 mv²). 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 sinθ_n = nλ

where d is the slit separation
θ_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 10⁰               B 10¹               C10²                D 10³

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