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
But the ans to 14 is c on the mark scheme and yours is B (9702/12/m/j/15)
ReplyDeletethe ms says its B, may be you looked at the wrong variant
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