Saturday, September 13, 2014

9702 November 2009 Paper 42 Worked Solutions | A-Level Physics

  • 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 = (f)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
Amplifier incorporating an operational amplifier (op-amp) has 3 inputs A, B and C, as shown. Negative feedback provided by resistor RF of resistance 8.0kΩ. for each of the inputs A, B and C, amplifier may be considered as single input amplifier. That is, each input is independent of other 2.
When amplifier not saturated, output potential VOUT is given by
VOUT = – (4VA + GVB + VC) where VA, VB and VC are input potentials of inputs A, B and C respectively and G is a constant.
(a)
2 effects of negative feedback on amplifier:
Choose any 2:
It reduces the gain
It increases the bandwidth
There is less distortion
There is greater stability

(b)
In expression for output potential VOUT, constant G is gain associated with input B. Show that numerical value of G is 2:
Gain = - RF / RI = - 8.0 / 4.0 = 2
Numerical value is 2

(c)
Input potentials VA, VB and VC are either 0 or 1.0V. Magnitudes of some output potentials for different combinations of VA, VB and VC are shown.
(i)
Complete Fig for 3 remaining values of VOUT:
[VOUT = – (4VA + GVB + VC)]
2, 6 and 7

(ii)
Use for this circuit:
Examples:
Digital-to-analogue converter
Adding / mixing signals with ‘weighting’



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.




4 comments:

  1. can you help me on november 09 paper 41 question 7i and ii) ? your help will be appreciated

    ReplyDelete
    Replies
    1. See solution 675 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html

      Delete
  2. 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:
    pV=nRT
    1.0*10^5*(2.1-1.8)*10^-3=0.18*8.31*T

    ReplyDelete
    Replies
    1. 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.

      Delete

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