9702 November 2008 Paper 4 Worked Solutions | A-Level Physics
Paper 4
SECTION A
Question 1
{Detailed explanations for this question is available as Solution 709 at Physics 9702 Doubts | Help Page 143 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-143.html}
Question 2
{Detailed explanations for this question is available as Solution 1049 at Physics 9702 Doubts | Help Page 220 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-220.html}
Question 3
{Detailed explanations for this question is available as Solution 582 at Physics 9702 Doubts | Help Page 114 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-114.html}
Question 4
(a)
Equation to represent first law of
thermodynamics in terms of heating q of a system, work w done on system and
increase ΔU in internal energy:
ΔU = q + w
(b)
Pressure of ideal gas decreased at
constant temperature. What change, if any, occurs in internal energy of gas:
EITHER
The kinetic energy is constant
because the temperature is constant. The potential energy is also constant
because there are no intermolecular forces (in an ideal gas). So, there is no
change in the internal energy.
OR
The kinetic energy and potential
energy are both constant. So there is no change in the internal energy. Reason
for either constant KE or constant PE given
Question 5
{Detailed explanations for this question is available as Solution 559 at Physics 9702 Doubts | Help Page 109 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-109.html}
Question 6
Simple iron-cored transformer illustrated.
(a)
Why core is
(i)
a continuous loop:
EITHER To prevent loss of magnetic
flux
OR It improves flux linkage with the
secondary coil
(ii)
laminated:
It reduces eddy current (losses). It
also reduces losses of energy (in the core)
(b)
(i)
Faraday’s law of electromagnetic
induction states that the (induced) e.m.f. is proportional to / equal to the
rate of change of (magnetic) flux (linkage).
(ii)
Use law to explain the operation of
the transformer:
The changing current in the primary
coil gives rise to a changing flux in the core. The flux links with the
secondary coil, changing the flux in the secondary coil inducing an e.m.f.
(c)
2 advantages of use of alternating
voltages for transmission and use of electrical energy:
It can change the voltage easily /
efficiently
High voltage transmission reduces
power losses
Question 7
{Detailed explanations for this question is available as Solution 988 at Physics 9702 Doubts | Help Page 205 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-205.html}
Question 8
{Detailed explanations for this question is available as Solution 965 at Physics 9702 Doubts | Help Page 199 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-199.html}
SECTION B
Question 9
Different frequencies and
wavelengths used in different channels of communication.
Why
(a)
Infra-red radiation rather than
visible light usually used with optic fibres:
Infra-red radiation has less
attenuation (per unit length). So, fewer (repeater) amplifiers are needed /
There is a longer uninterrupted length.
(b)
Base stations in mobile networks
operate on UHF:
Either UHF has a limited range. So,
the cells do not overlap (appreciably)
OR UHF has a short wavelength. So, a
convenient length for the aerial can be used (on the mobile phone)
(c)
For satellite communication,
frequencies of order of GHz used, with uplink having different frequency to
downlink:
Frequencies are of the order of GHz
have a large bandwidth / large information carrying capacity. The frequencies
are different so that the uplink signal is not swamped by the downlink signal.
Question 10
(a)
Circuit for amplifier incorporating
an ideal operational amplifier (op-amp) shown.
(i)
1.
Name of this type of amplifier
circuit:
Inverting (amplifier)
2.
Why point P is referred to as
virtual earth:
The gain of the operational
amplifier (op-amp) is very large / infinite. The non-inverting input is at
earth / 0V. For the amplifier not to saturate, P must be about earth / 0V.
(ii)
Show that gain of amplifier circuit
given by expression G = - R2
/ R1:
The input resistance is very large.
So, the current in R1 = current in R2.
I = VIN / R1
And I = - VOUT / R2
Hence, gain, G = VOUT / VIN
= - R2 / R1
(b)
Circuit of Fig modified by
connecting light-dependent resistor (LDR) as shown. Resistance R1
and R2 are 5.0kΩ and 50kΩ
respectively. Input voltage VIN = +1.2V. High-resistance voltmeter
measures output VOUT. Circuit used to monitor low light intensities.
(i)
Voltmeter for light intensities such
that LDR has resistance of
1.
100kΩ:
Feedback resistance (= [1/50 +
1/100]-1) = 33.3kΩ
Gain (= (-) RF / RIN
= 33.3 / 5) = 6.66
VOUT (= Gain x VIN= 6.66 x 1.2) = 8.0V
2.
10kΩ:
Feedback resistance (= [1/50 + 1/10]-1)
= 8.33kΩ
(Gain = 8.33 / 5 = 1.67 )
VOUT (= 1.67 x 1.2) =
2.0V
(ii)
Light incident on LDR provided by
single lamp. Use answers in (i) to describe and explain qualitatively variation
of voltmeter reading as lamp is moved away from LDR:
(An increase in the lamp-LDR
distance gives) a decrease in intensity. The feedback / LDR
resistance increases. So, the voltmeter reading increases / becomes more
negative.
Question 11
(a)
Distinguish images produced by CT
scanning and X-ray imaging:
The CT image is a (thin) slice (through
the structure) + any further detail e.g. it is built from many ‘slices’ / 3-D
image
The X-ray image is a ‘shadow’ image
(of the whole structure) / 2-D image
(b)
Reference to principles of CT scanning;
suggest why CT scanning could not be developed before powerful computers were
available:
An X-ray image of a slice is
taken from many different angles. These images are combined (and processed).
This is repeated for many different slices in order to build up a 3-D image
which can be rotated. (Powerful) computers are required to store and process
huge quantity of data.
In Question 5 part (a), why do we multiply m by 2, and square the charge q?
ReplyDeleteI have updated the explanations.
Deletewhere can i find the question 5? when i press the link its display calibration chapter which is one of the AS topic
DeleteAs said above, its solution 559 on that link
Deletethank you very much.this blogspot is very useful.keep it up .cheers
DeletePlz explain qs 3 part c ii and in question 2 part b iii why can't we convert 64.7 g to 10 minutes?
ReplyDeleteSome details have been added for question 3.
DeleteI'll try to see the other one later
Can you explain question 1 a (iii) and b (i)?
ReplyDeleteThe details have been added
Deletecan you plz explain question no.8 last part
ReplyDeleteSome details have been added. See if it helps
Deletei got it alhamduillah
Deletejazakallah khair.
june 08 Q10, thanks!
ReplyDeleteCheck solution 1040 at
Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-218.html
question 2 biii, why did you times half of 16.6?
ReplyDeleteDetails have been added.
DeleteJune 2008 p4 Q6(@) please .
ReplyDeletego to
Deletehttp://physics-ref.blogspot.com/2019/10/a-small-rectangular-coil-abcd-contains_1.html