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).
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 0.50B 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.
Calculate
(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:
See solution at
The thickness x of the layer of fat on an animal, as illustrated in Fig. 10.2, is to be investigated using ultrasound.
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?
ReplyDeleteI diagram has been added. Read the explanation again and see if you understand.
DeleteSir, for question 443, why is the thickness 2x X 10^-2?
ReplyDeletethe 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.
DeleteTotal distance travelled = x+x = 2x
hi for question 443, why Incident intensity is 0.018?
ReplyDeleteα = IR / I = 0.018
DeleteSo, 0.018 of the intensity has been reflected at the interface.
Now, this reflected intensity will move through the fat towards surface S.
Considering this intensity from the boundary towards the surface S, we can take this reflected intensity as the new incident intensity. The ultrasound will move from the boundary towards surface S.
After being transmitted in the fat, the intensity at surface S is given to be 0.012 I.
So, the intensity was 0.018 I at the boundary, and as the ultrasound moves through the fat, it got attenuated and the new intensity at the surface S is now 0.012 I.
Since we're taking the reflected intensity as the incident intensity, it means we're only considering the reflection of ultrasound from boundary to surface X. Then why double the distance x?
DeleteThe ultrasound moves through a distance x of fat to the boundary where some of waves are transmitted and others reflected. The reflected wave would also need to move a distance x in the fat to reach surface S where it is detected.
DeleteI still dont undertand this. so we say that instantly after the reflection t the fat muscle boundary, the intensity was .018I. and at S it was .012I. so basically it attenuated from .018I to .012I after travelling a distance x through the fat. it did NOT travel through x twice. it traveled straight from the interface to S. then why is the distance 2x?
ReplyDeletethe solution has been updated with much more details.
Deletesee the link above
Q443).
ReplyDeleteWhy did u take
10^-2. When considering the thickness of 2X. Because data is given in meters?
So my answer must also in metres and then I convert into cm but I don't get the answer when doing so.
see the updated link added above
Delete