9702 November 2009 Paper 42 Worked Solutions | A-Level Physics
Paper 42
SECTION A
Question 1
(a)
Earth considered to be uniform
sphere of radius 6.38x103km, with mass concentrated at its centre.
(i)
Gravitational field strength is
defined as the force per (unit) mass
(ii)
Considering the gravitational field
strength at surface of Earth, show that mass of Earth is 5.99x1024kg:
g = GM / R2
9.81 = (6.67x10-11)M /
(6.38x106)2
So, M = 5.99x1024kg
(b)
Global Positioning System (GPS) is a
navigation system that can be used anywhere on Earth. It uses a number of
satellites that orbit Earth in circular orbits at a distance of 2.22x104km
above its surface.
(i)
Use data from (a) to calculate
angular speed of a GPS satellite in its orbit:
(r = 2.22x107 + 6.38x106
= 2.86x107m)
Either
GM = ω2r3
(6.67x10-11) (5.99x1024)
= ω2 (2.86x107)3
ω = 1.3x10-4rads-1
Or
gR2 = ω2r3
9.81 x (6.38x106)2
= ω2 (2.86x107)3
ω = 1.3x10-4rads-1
(ii)
Use answer in (i) to show that the
satellites are not in geostationary orbits:
Period of orbit of satellites = 2π / ω = 4.8x104s (= 13.4 hours)
The period of orbit for a
geostationary satellite is 24 hours (= 8.6x104s). So, the satellites
are not in geostationary orbits.
(c)
Planes of orbits of GPS satellites
in (b) are inclined at angle of 55o to Equator. Why satellites are
not in equatorial orbits:
In this way, the GPS satellite can
then provide cover at the Poles
Question 2
{Detailed explanations for this question is available as Solution 906 at Physics 9702 Doubts | Help Page 185 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-185.html}
Question 3
Variation with displacement x of
acceleration a of centre of cone of loudspeaker shown.
(a)
2 features of Fig that show that
motion of cone is simple harmonic:
A straight line passing through the
origin
It has a negative gradient
(b)
Use data from fig to determine
frequency, in hertz, of vibration of cone:
a = – ω2x and ω = 2πf
750 = (2πf)2
x (0.3x10-3)
So, frequency f = 250Hz
(c)
Frequency of vibration of cone now
reduced to one half of that calculated in (b). Amplitude of vibration remains
unchanged. On axes of Fig, draw line to represent variation with displacement x
of acceleration a of centre of loudspeaker cone:
Graph is a straight line between
(-0.3, +190) and (+0.3, -190)
Question 4
(a)
Capacitance is defined as the ratio
of charge to potential.
(b)
Isolated metal sphere of radius R
has charge +Q on it. The charge considered point charge at centre of sphere.
Show that capacitance C of sphere is given by C = 4πϵoR where ϵo is permittivity
of free space:
Potential (at the surface of the
sphere) = Q / 4πϵoR
So, C = Q / V = 4πϵoR
(c)
In order to investigate electrical
discharge (lightning) in a laboratory, isolated metal sphere of radius 63cm is
charged to potential of 1.2x106V. at this potential, there is
electrical discharge in which sphere loses 75% of its energy.
(i)
Capacitance of sphere, stating unit
in which it is measured:
Capacitance, C = 4π (8.85x10-12) x 0.63 = 7.0x10-11
Unit: Farad / F
(ii)
Potential of sphere after discharge
has taken place:
Energy = ½ CV2
Remaining energy: 0.25 x ½ C x
(1.2x106)2 = ½ CV2
Potential of sphere after discharge,
V = 6.0x105V
Question 5
{Detailed explanations for this question is available as Solution 926 at Physics 9702 Doubts | Help Page 190 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-190.html}
Question 6
Ideal transformer illustrated.
(a)
(i)
Faraday’s law of electromagnetic
induction states that the e.m.f induced is proportional / equal to the rate of
change of (magnetic) flux linkage.
(ii)
Use law to explain why transformer
will not operate using direct current input:
An e.m.f is induced only when the
flux is changing / cut. A direct current gives a constant flux.
(b)
(i)
Lenz’s law states that the (induced)
e.m.f / current acts in such a direction to produce effects to oppose
the change causing it.
(ii)
Use Lenz’s law to explain why input
potential difference and output e.m.f are not in phase:
The (induced) current in the secondary
coil produces a magnetic field that opposes the (changing) field produced in
the primary coil. So, the input potential difference and output e.m.f
are not in phase.
(c)
Electrical energy usually
transmitted using alternating high voltages. 1 advantage, for transmission of
electrical energy, of using:
(i)
Alternating voltage:
An alternating voltage means that
the voltage / current is easy to change.
(ii)
High voltage:
A high voltage means that there is
less power / energy loss (during the transmission)
Question 7
{Detailed explanations for this question is available as Solution 958 at Physics 9702 Doubts | Help Page 198 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-198.html}
Question 8
(a)
The decay constant of a radioactive
isotope is the (constant) probability of decay per unit time.
(b)
Show decay constant λ related to half-life t½ by λt½
= 0.693:
[N = Noexp(-λt)]
EITHER when time = t½, N
= ½No OR ½No = Noexp(-λt½)
EITHER 2 = exp(λt½) OR
½ = exp(-λt½)
(taking logs), ln2 = 0.693 = λt½
(c)
Cobalt-60 is a radioactive isotope
with half-life of 5.26years (1.66x108s). Cobalt-60 source for use in
school laboratory has activity of 1.8x105Bq.
Mass of Cobalt-60 in source:
Activity, A = λN
1.8x105 = N x (0.693 /
{1.66x108})
N = 4.3x1013
Mass = 60 x (N / NA) or
60 x N x u
Mass = (60 x 4.3x1011) /
(6.02x1023) = 4.3x10-9g
SECTION B
Question 9
An amplifier incorporating an operational amplifier (op-amp) has three inputs A, B and C, as shown in Fig. 9.1.
Question 10
{Detailed explanations for this question is
available as Solution 448 at Physics 9702 Doubts | Help Page 86 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-86.html}
Question 11
Variation with time of signal
transmitted from an aerial shown.
(a)
Type of modulated transmission:
Amplitude modulation
(b)
Use Fig to determine frequency of
(i)
Carrier wave:
Frequency = 1/ period = 1 / (10x10-6)
= 100kHz
(ii)
Information signal:
Frequency = 1/ period = 1 / (100x10-6)
= 10kHz
(c)
(i)
On axes of Fig, draw frequency
spectrum (variation with frequency of signal voltage) of signal from aerial.
Mark relevant values on frequency axis:
Vertical line drawn at 100kHz along
with vertical lines at 90kHz and 110kHz. The lies at 90kHz and 110kHz are of
the same length and are both shorter than the line at 100kHz.
(ii)
Bandwidth of signal:
Bandwidth = 110 – 90 = 20kHz
Question 12
Block diagram representing part of
mobile phone network shown.
(a)
What is represented by
(i)
Blocks labeled X: base stations
Block labeled Y: cellular exchange
(b)
User of mobile phone is making a
call. Role of components in boxes labeled X and Y during the call:
The base station / X sends /
receives signal to / from the handset. The call is relayed to the cellular
exchange / Y (and on to the PSTN). A computer at the cellular exchange
monitors signal from the base stations and it selects the base station
with the strongest signal and allocates a (carrier) frequency / time slot for
the call.
can you help me on november 09 paper 41 question 7i and ii) ? your help will be appreciated
ReplyDeleteSee solution 675 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html
42/O/N/09 Q.2(c)(ii) Despite the fact that this particular section of a question was removed from the assessment, I think that the question provided all the essential information for the calculation to be done. Please check if this method is valid:
ReplyDeletepV=nRT
1.0*10^5*(2.1-1.8)*10^-3=0.18*8.31*T
This formula gives the temperature T at a specific volume V. V is not a change in volume here and T does not give the change in temperature. T is only a simple, specific temperature.
DeleteCan you help me with May/June 2009/04 question 6(c)
ReplyDeleteGo to
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