Friday, June 12, 2015

Physics 9702 Doubts | Help Page 171

  • Physics 9702 Doubts | Help Page 171



Question 855: [Waves > Stationary waves]
A stationary wave is set up a stretched string, as shown.

Which statement about points on the string is correct?
A Point Q vibrates with the largest amplitude.
B Points P and R vibrate in phase.
C Point S is an antinode.
D The horizontal distance between R and S is half the wavelength.

Reference: Past Exam Paper – June 2014 Paper 13 Q29



Solution 855:
Answer: B.
Points P and R (they both move up) vibrate in phase while (P and Q) or (Q and R) are out of phase.

Point R vibrates with the largest amplitude (R is an antinode).
Point S is a node (zero amplitude).

The horizontal distance between R and S is a quarter of the wavelength.











Question 856: [Current of Electricity]
Diagram shows a fixed resistor and a light-dependent resistor (LDR) in series with a constant low-voltage supply.

When LDR is in the dark, the fixed resistor and the LDR have the same value of resistance.
Light is shone on the LDR.
What happens to the potential differences across the two components?
p.d. across resistor       p.d. across LDR
A         decreased                    increased
B         increased                     decreased
C         no change                    increased
D         no change                    decreased

Reference: Past Exam Paper – June 2011 Paper 12 Q37



Solution 856:
Answer: B.
The resistance of the LDR decreases as the light intensity on it increases. Thus, in the dark, the resistance of the LDR was greater.

Ohm’s law: V = IR
When light is shone on it, the resistance of the LDR decreases. So, the p.d. across the LDR decreases.

From Kirchhoff’s law, the sum of p.d.’s in a circuit is equal to the e.m.f. in the circuit. Thus, the p.d. across the fixed resistor increases such that the sum of p.d.’s is still equal to the e.m.f.











Question 857: [Radioactivity]
The isotope 22286Rn decays in a sequence of emissions to form isotope 20682Pb. At each stage of decay sequence, it emits either an α-particle or a β-particle.
What is the number of stages in the decay sequence?
A 4                              B 8                              C 16                            D 20

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q39



Solution 857:
Answer: B.
An α-particle is 42α (mass number = 4 and atomic number = 2) and a β-particle is 0-1β (mass number = 0 and atomic number = -1).

Difference in mass number A between 22286Rn and 20682Pb = 222 – 206 = 16
Since a β-particle does not contribute anything to the mass number, this difference in mass number is solely due to the α-particles.

Number of α-particles = 16 / 4 = 4
Number of stages liberating an α-particle = 4

Increase in atomic number due to α-particles = 4 × 2 = 8

Let the number of β-particles be x.
Consider the initial and final atomic numbers, which should be equal.
86 = 82 +8 + x(-1)
x = -4 / -1 = 4

Thus, 4 β-particles are liberated, each one in a stage. There are 4 stages here too.

Total number of stages = 4 + 4 = 8











Question 858: [Waves > Diffraction]
Diagram 1 shows ripple tank experiment in which plane waves are diffracted through a narrow slit in a metal sheet.
Diagram 2 shows same tank with a slit of greater width.
In each case, the pattern of the waves incident on slit and the emergent pattern are shown.

Which action would cause the waves in diagram 1 to be diffracted less and so produce an emergent pattern closer to that shown in diagram 2?
A increasing the frequency of vibration of the bar
B increasing the speed of the waves by making the water in the tank deeper
C reducing the amplitude of vibration of the bar
D reducing the length of the vibrating bar

Reference: Past Exam Paper – June 2010 Paper 11 Q24 & Paper 12 Q23 & Paper 13 Q25



Solution 858:
Answer: A.
The amount of diffraction is less as the curvature of the diffracted wavefront decreases (diffraction is more in diagram 1).

During the diffraction of plane wavefronts through an opening, neither the frequency nor wavelength (given by the distance between the wavefronts) changes. The frequency only depends on the source of the vibrating bar (e.g. a motor).

The amount of diffraction (the sharpness of the bending) depends on the wavelength. Diffraction increases with increasing wavelength and vice versa.

The closer the obstacle/opening/slit is to the wave's wavelength, the greater the amount of diffraction. When the slit is equal to the wavelength, maximum diffraction is seen and the waves spread out greatly, with the wavefronts being almost semicircular.

If the slit is larger than the wavelength, diffraction is less. The water wave passes through the gap and does not spread out much on the other side. See diagram 2. Actually, when the wavelength is smaller than the slit, diffraction may not be noticeable.

We want the waves in diagram 1 (with a smaller slit) to be diffracted less. So, the difference between the slit and the wavelength should be large, resulting in the diffraction to be less. Since the slit is itself small, we should be producing waves of even smaller wavelength.
Speed v = fλ
By increasing the frequency f, the wavelength is decreased (while the speed is constant), causing the relative size of the opening/slit to be larger which results in less diffraction. [A is correct]

As the depth of water in the tank is increased, the wavelength and thus the speed increases. A wavelength smaller than the slit causes less diffraction while a wavelength larger than the slit would result in relatively more diffraction. [B is incorrect]

The amplitude is the maximum displacement of the water molecule from its equilibrium position. The amplitude gives us an idea of the energy of the wave, but does not affect the wavelength and diffraction. [C is incorrect]

Reducing the length of the vibrating bar does not affect the experiment since it is the wave (in the water) that moves. [D is incorrect]




5 comments:

  1. Hi
    For question 858
    Why B is incorrect? The speed of the water wave when the depth of the tank increases. So therefore wavelength increases as frequency constant. This means more diffraction. So therefore B is not correct. Is my understanding correct?

    ReplyDelete
    Replies
    1. I'm confused now. In my O Levels, we learn that the speed of wave increases when the depth of the water tank increases.

      Delete
    2. Sorry what I mean is the speed of wave increases as the depth of the water in the water tank increases.

      Delete
    3. Sorry, you are right. There may have been some mistakes in the explanation. It is now been corrected. Read through it again, it should be clear now.

      Delete
    4. I understand the explanation now. Thanks admin :)

      Delete

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