Saturday, August 8, 2015

Physics 9702 Doubts | Help Page 185

  • Physics 9702 Doubts | Help Page 185

Question 904: [Options > Medical Physics > Ultrasound]
(a) Briefly explain the principles of generation of ultrasound.

(b) An ultrasound pulse is transmitted into the body of a patient. The pulse is partially reflected at fat/muscle boundary and then at a muscle/bone boundary. The reflected pulses are received back at the transmitter and are displayed, after processing, on the screen of a cathode-ray oscilloscope (c.r.o.). The trace is shown in Fig.1.

Time-base of the c.r.o. is 10 μs cm–1.
Use Fig.1 to determine
(i) thickness of the layer of fat given that the speed of ultrasound in fat is 1450ms–1,
(ii) muscle thickness given that the speed of ultrasound in muscle is 1590ms–1.

Reference: Past Exam Paper – June 2004 Paper 6 Q8

Solution 904:
(a) It consists of a piezo-electric/quartz crystal across which an alternating voltage is applied. This causes the crystal to vibrate at its resonant frequency.

{There are 2 reflected pulses – one at the fat/muscle boundary and one at the muscle/bone boundary. To determine the thickness of the layer of fat, we need to consider the first reflected pulse, which is a distance of 4.0 mm from the transmitted pulse, as seen from the graph.
5 squares represent 1cm. 2 squares represent 2 (1/5) = 0.4cm = 4.0mm}
Trace length = 4.0 mm
{1cm represents a time of 10 μs. 0.4cm represents 0.4 (10 μs) = 0.4 × 10 × 10-6 s}
Distance = speed × time = 1450 × (0.4 × 10×10-6) = 5.8 × 10-3 m
{The above distance represents the distance travelled by the pulse as it leaves the transmitter, reaches the fat/muscle boundary and then it is reflected at that boundary and thus travels back to the transmitter where it is received. So, it travels twice the thickness of the fat.}
Thickness {= 0.58 / 2} = 0.29 cm

{This is the length between the transmitted pulse and the 2nd reflected pulse in the graph.}
Trace length = 5.2 cm
{1cm represents a time of 10 μs. 5.2cm represents 5.2 (10 μs) = 5.2 × 10 × 10-6 s
Distance = speed × time = 1590 × (5.2 × 10×10-6) = 0.0827m = 8.27cm}
Thickness {= 8.27 / 2} = 4.1 cm

Question 905: [Measurement]
The first definition of the metre was one ten-millionth of the distance between the north pole and the equator of the Earth. Use this information to estimate the radius of the Earth. State one assumption which you have made in your estimation.
[Answer: 6.37 × 106 m]

Reference: ???

Solution 905:
One ten-millionth = 1 / (10 × 106) = 1×10-7

Let distance between the north pole and the equator of the Earth = y
So, from the definition of a metre,
(1×10-7) y = 1 m
Giving y = 1×107 m

Assuming the Earth is a perfect sphere,
The distance between the north pole and the equator of the Earth is the length of arc formed by the radii of the Earth which are at 90o to each other (considering a vertical plane on the sphere, the angle between the north pole and the equator is 90o).

Length of arc, y = rθ
So, rθ = 1×107 m where r is the radius of Earth and θ = 90o = (π/2) rad
R (π/2) = 1×107 m
Radius, r = 2 (1×107) / π = 6.37×106 m

Question 906: [Thermodynamics > First Law of Thermodynamics]
(a) State what is meant by internal energy of a gas.

(b) First law of thermodynamics may be represented by the equation
ΔU = q + w.
State what is meant by each of the following symbols.

(c) An amount of 0.18 mol of an ideal gas is held in an insulated cylinder fitted with a piston, as shown in Fig.1.

Atmospheric pressure is 1.0 × 105 Pa.
Volume of the gas is suddenly increased from 1.8 × 103 cm3 to 2.1 × 103 cm3.
For the expansion of the gas,
(i) calculate work done by the gas and show that the internal energy changes by 30J
(ii) determine the temperature change of the gas and state whether the change is an increase or a decrease.

Reference: Past Exam Paper – November 2009 Paper 42 Q2

Solution 906:
(a) The internal energy of a gas is the sum of kinetic and potential energies of molecules / particles / atoms that are in a random distribution.

+ ΔU: increase in the internal energy 
+ q: heating of / heat supplied to the system 
+ w: work done on the system

(1cm = 0.01m. So, 1cm3 = (0.01)3m3 = 1×10-6m3. So, 1.8×103cm3 = 1.8×10-3m3.)
Work done by gas = pΔV = (1.0×105) × (2.1 – 1.8)×10-3 = 30J
w = 30J and q = 0. So, ΔU = 30J

(ii) [These 3 marks were removed, as insufficient data was given in question for a correct calculation of the required temperature change]

Question 907: [Electric field]
Isolated conducting sphere of radius r is given a charge +Q. This charge may be assumed to act as a point charge situated at the centre of the sphere, as shown in Fig.1.

Fig.2. shows variation with distance x from the centre of the sphere of the potential V due to the charge +Q.

(a) State the relation between electric field and potential.

(b) Using the relation in (a), on Fig.3 sketch a graph to show the variation with distance x of the electric field E due to the charge +Q.

Reference: Past Exam Paper – June 2005 Paper 4 Q5

Solution 907:
(a) Electric field strength = potential gradient [- sign not required]

(b) For the graph, there is no field for x < r. For x > r, the graph is a curve in the correct direction, not going to zero. There is a discontinuity at x = r (vertical line required).


  1. Please consider answering ALL of the following questions before October:
    4/O/N/02 Q.3(b)(i),Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.4(c),Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.8(b),Q.9(b)(iii),Q.11(b)
    04/O/N/04 Q.3(c)(ii)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.7(b)(i),(c),Q.10(c)
    04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)
    04/O/N/08 Q.7(c)
    04/M/J/09 Q.10(c)(ii),Q.11(b)(iii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    42/O/N/09 Q.5(a),Q.7(b)(ii)

  2. For 04/O/N/04 Q.3(c)(ii), see solution 908 at


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