Friday, March 13, 2015

Physics 9702 Doubts | Help Page 84

  • Physics 9702 Doubts | Help Page 84

Question 440: [Waves]
Diagram shows a vertical cross-section through water wave moving from left to right.
At which point is water moving upwards with maximum speed?

Reference: Past Exam Paper – November 2010 Paper 11 Q24 & Paper 13 Q25 & June 2014 Paper 12 Q24

Solution 440:
Answer: C.
Here is a simple way to solving this problem:
We are asked to identify which point is moving upwards. So, we need to know at what position each point would be in the next instant. Since the water wave is travelling to the right, just shift the given wave to the right a bit and we are able to see how each point will result in the next instant of time.

At point A and B, the water is moving downwards while C and D move upward. [A and B are incorrect]

At D the motion is just slightly upwards but the answer is clearly C. [D is incorrect]
The motions of particles on a wave are periodic [in this case, each water particle will move up and down] and can be compared with the simple periodic motion of a pendulum bob.

At the maximum displacement, the bob is temporarily at rest and at the equilibrium position, the bob is moving with maximum speed. Point C is at equilibrium position while point D is at the maximum displacement.

Question 441: [Current of Electricity]
A 12 V battery is in series with ammeter, 2 Ω fixed resistor and a 0 – 10 Ω variable resistor. High-resistance voltmeter is connected across variable resistor.

Resistance of variable resistor is changed from zero to its maximum value.
Which graph shows how potential difference (p.d.) measured by the voltmeter varies with the current measured by the ammeter?

Reference: Past Exam Paper – June 2013 Paper 11 Q37

Solution 441:
Answer: D.
Note that the graph is that of the p.d. across the VARIABLE resistor against the current through it. This current is the same flowing in the whole circuit since this is a series connection.

When the variable resistor has value zero, the total resistance in the circuit is 2 Ω. So, there is 0V across the variable resistor and the current is a maximum (= 12/2 = 6.0A) in this case. On the graph, this would correspond to point (6.0, 0).

When the variable resistor is at 10Ω (maximum), the total resistance in the circuit is 12 Ω. So, the current is a minimum (= 12/12 = 1.0A) and the voltmeter reading is obtained from the potential divider equation.
p.d. across variable resistor = [10 / (10+2)] x 12 = 10V

This corresponds to point (1.0, 10).

So, if the variable resistance is zero the current will be large and the voltmeter reading will be zero. When the variable resistance is 10 Ω the current will be reduced, but not zero, and the voltmeter reading will be high.
The graph has a negative gradient since it passes through the points (1.0, 10), (6.0, 0).

Question 442: [Current of Electricity]
Battery has e.m.f. of 3.0 V and internal resistance of 2.0 Ω.

Battery is connected to load of 4.0 Ω.
What are terminal potential difference V and output power P?
V / V               P / W
A         1.0                   0.50
B         1.0                   1.5
C         2.0                   1.0
D         2.0                   1.5

Reference: Past Exam Paper – June 2007 Paper 1 Q36

Solution 442:
Answer: C.
The terminal potential difference V is the p.d. across the battery. Since the battery has some internal resistance, some volts is lost inside the battery – that is, the voltage available for the external circuit (the resistor) is not equal to the e.m.f.

Total resistance in the circuit = 2.0 + 4.0 = 6.0Ω
Current in circuit = 3.0V / 6.0Ω = 0.5A

p.d. lost due to internal resistance = Ir = 0.5 (2.0) = 1.0V
Terminal potential difference = 3.0V – 1.0V = 2.0V

The output power P is the power dissipated in the load.
Power dissipated = I2R = (0.5)2 (4) = 1.0W

Question 443: [Remote Sensing > Ultrasound]
(a) State what is meant by acoustic impedance Z of a medium.

(b) 2 media have acoustic impedances Z1 and Z2.
Intensity reflection coefficient α for boundary between the 2 media given by
α = (Z2 – Z1)2 / (Z2 + Z1)2
Describe effect on transmission of ultrasound through boundary where there is large difference between acoustic impedances of the 2 media.

(c) Data for acoustic impedance Z and absorption coefficient μ for fat and for muscle shown.
Z / kg m–2 s–1               μ / m–1
Fat                   1.3 × 106                      48
Muscle             1.7 × 106                      23
Thickness x of layer of fat on an animal, as illustrated in Fig, is to be investigated using ultrasound.

Intensity of parallel ultrasound beam entering surface S of layer of fat is I.
Beam is reflected from boundary between fat and muscle.
Intensity of reflected ultrasound detected at surface S of fat is 0.012I.
(i) Intensity reflection coefficient at boundary between fat and muscle
(ii) Thickness x of layer of fat

Reference: Past Exam Paper – June 2011 Paper 41 Q10

Solution 443:
(a) Acoustic impedance Z of a medium is the product of the density (of a medium) and the speed of sound (in the medium).

EITHER The reflected intensity would be nearly equal to the incident intensity
OR the coefficient for the transmitted intensity = (1 – α).
So, the transmitted intensity would be small.

(i) α = (1.7 – 1.3)2 / (1.7 + 1.3)2 = 0.018

{Thickness to consider = 2x × 10-2}
Attenuation in the fat = exp(-μx) = exp(- 4.8 (2x) X10-2)
[I = Ioexp(-μx)]
0.012 = 0.018exp(- 4.8 (2x) X10-2)
x = 0.42cm


  1. Sir I completely don't understand Question 440. It looks to me like A is going up and C is going down. I understand about the speed but I don't understand the upward/downward directions. It only makes sense that C is going down if we consider the wave to be the white part of the diagram, not the gray part. If a surfer were on this wave of water, and he was at point A, wouldn't he be going upwards, towards the crest of the wave?

    1. I diagram has been added. Read the explanation again and see if you understand.

  2. Sir, for question 443, why is the thickness 2x X 10^-2?

    1. the ultrasound is first incident in the fat. It travels a distance x. Upon reflection, it returns along the fat itself towards the detector. This is also a distance x.

      Total distance travelled = x+x = 2x


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