Tuesday, September 22, 2015

Physics 9702 Doubts | Help Page 201

  • Physics 9702 Doubts | Help Page 201



Question 972: [Measurement + Current of Electricity]
A power supply of electromotive force (e.m.f.) 50 V and negligible internal resistance is connected in series with resistors of resistance 100 Ω and 5 Ω, as shown.

A voltmeter measures the potential difference (p.d.) across the 5 Ω resistor and an ammeter measures the current in the circuit.
What are suitable ranges for the ammeter and for the voltmeter?

ammeter range / A       voltmeter range / V
A                     0 – 0.1                         0 – 1
B                     0 – 0.1                         0 – 3
C                     0 – 1.0                         0 – 1
D                     0 – 1.0                         0 – 3

Reference: Past Exam Paper – June 2015 Paper 12 Q5



Solution 972:
Answer: D.
Ohm’s law: V = IR
Current I in the circuit = 50 / (100+5) = 0.476.. = 0.5A

Since the current is 0.5A, the ammeter range cannot be less than this value. [A and B are incorrect]

From the potential divider equation,
p.d. across the 5 Ω resistor = [5 / (100+5)] × 50 = 2.38 V

The voltmeter range should be greater than this p.d. of 2.38 V. [C is incorrect]










Question 973: [Applications > Ultrasound]
(a) Explain the main principles behind the use of ultrasound to obtain diagnostic information about internal body structures.

(b) Data for the acoustic impedances and absorption (attenuation) coefficients of muscle and bone are given in Fig. 11.1.

acoustic impedance / kg m–2 s–1                      absorption coefficient / m–1
muscle                         1.7 × 106                                                          23
bone                            6.3 × 106                                                          130
Fig. 11.1

The intensity reflection coefficient is given by the expression
(Z2 – Z1)2 / (Z2 + Z1)2 .
The attenuation of ultrasound in muscle follows a similar relation to the attenuation of X-rays in matter.
A parallel beam of ultrasound of intensity I enters the surface of a layer of muscle of thickness 4.1 cm as shown in Fig. 11.2.

The ultrasound is reflected at a muscle-bone boundary and returns to the surface of the muscle.
Calculate
(i) the intensity reflection coefficient at the muscle-bone boundary,
(ii) the fraction of the incident intensity that is transmitted from the surface of the muscle to the surface of the bone,
(iii) the intensity, in terms of I, that is received back at the surface of the muscle.

Reference: Past Exam Paper – June 2009 Paper 4 Q11



Solution 973:
(a) A pulse of ultrasound is reflected at the boundaries / boundary and is received / detected (at surface) by a transducer. This signal is processed and displayed. The time between transmission and receipt of pulse gives (information about) the depth of the boundary. The reflected intensity gives information as to the nature of the boundary.

(b)
(i) Intensity reflection coefficient = (Z2 – Z1)2 / (Z2 + Z1)2 = (6.3 – 1.7)2 / (6.3 + 1.7)2 = 0.33

(ii)
{I = I0 exp(–μx). Fraction = I / I0 = exp(–μx).
Thickness x = 4.1cm = 4.1×10–2m}
Fraction = exp(–μx) = exp(–23 × 4.1×10–2) = 0.39

(iii)
{The incident beam of ultrasound of intensity I first travels from the transmitter through the muscle. The intensity of the beam that reaches the bone is 0.39×I – as calculated above.
But, at the muscle-bone boundary, some of this beam is absorbed while the rest is reflected back. The intensity of the beam that is reflected is 0.33  intensity of beam that reached the boundary. Intensity of the beam that is reflected is 0.33×(0.39×I).
Once reflected, the beam needs to travel through the muscle again in order to reach the receiver. Intensity of beam that reaches the receiver = 0.39×[0.33×(0.39×I)].}
Intensity = 0.33 × 0.392 × I = 0.050 I










Question 974: [Dynamics > Newton’s laws of motion]
A ladder is positioned on icy (frictionless) ground and is leant against a rough wall. At the instant of release it begins to slide.
Which diagram correctly shows the directions of the forces P, W and R acting on the ladder as it slides?


Reference: Past Exam Paper – June 2012 Paper 12 Q14 & June 2015 Paper 12 Q14



Solution 974:
Answer: B.
W is the weight and always acts downwards.
R is the normal reaction to the weight and acts upwards (since ground is frictionless).

For this question, you need to imagine this as a real case – how the ladder would move when it slides.

Before the ladder begins to slide, there is a horizontal force acting to the right at the point of contact of the ladder with the rough wall since the ladder itself exerts a force to the left on the wall due to its mass.
(Diagram A corresponds to this [case where no friction acts and the ladder is not sliding] – The angle of the weight with the ladder is not zero, so there is a component of the weight acting against wall. From Newton’s 3rd law, there is a reaction from the wall – this is P before the ladder begins to slide).

As the ladder begins to slide, the point of contact of the ladder with the rough wall tends to move downwards. Since the wall is rough friction would act upwards, opposing the motion. So, at the point of contact, there is now a force to the right, along with a force acting upwards, the resultant of which is shown by P in diagram B.

The ground is frictionless, so there is no additional force opposing the motion.











Question 975: [Waves > Intensity]
Fig. 5.1 shows the variation with time t of the displacements xA and xB at a point P of two sound waves A and B.

(a) By reference to Fig. 5.1, state one similarity and one difference between these two waves.

(b) State, with a reason, whether the two waves are coherent.

(c) The intensity of wave A alone at point P is I.
(i) Show that the intensity of wave B alone at point P is (4/9) I.
(ii) Calculate the resultant intensity, in terms of I, of the two waves at point P.

(d) Determine the resultant displacement for the two waves at point P
(i) at time t = 3.0 ms,
(ii) at time t = 4.0ms.

Reference: Past Exam Paper – November 2005 Paper 2 Q5



Solution 975:
(a)
Similarity: example: same wavelength / frequency / period, constant phase difference        
Difference: example: different amplitudes / phase

(b) They are coherent since the phase difference is constant.

(c)
(i)
{The intensity of a wave is proportional to the square of its amplitude. The amplitude is the maximum displacement, and can be obtained from the graph.}
Intensity (amplitude)2
So, IA = I 32 = 9       and      IB 22 = 4       leading to
{For wave A, the amplitude is 3. Its intensity I is 32 = 9 units. So, 9 units [the square of the amplitude (32)] represents I. So, 1 unit would represent I/9. The amplitude of wave B is 2. So, its intensity is 22 = 4 units. In terms of I, this is 4 (I/9) = (4/9)I.}
IB = (4/9) I

(ii)
{Amplitude of wave A = (+)3×10-4 cm and the amplitude of wave B is (-)2×10-4 cm. The signs account for the waves being out of phase with each other.}
Resultant amplitude = (3 – 2)×10-4 = 1.0×10-4cm
{The resultant amplitude is 1 (let forget the ×10-4 cm for now). For the intensity of the resultant wave, we consider the square of the resultant amplitude = 12 = 1 unit.
As stated before, 9 units (amplitude squared) corresponds to I. The intensity of the resultant wave would correspond to [by proportion] (1/9) I}
So, resultant intensity = (1/9) I

(d)
(i) Resultant displacement = 0

(ii)
xA = -2.6×10-4cm         and      xB = +1.7×10-4cm
So, resultant displacement = (-) 0.9×10-4cm




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