# 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 6.38x10

^{3}km, 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 5.99x10

^{24}kg:
g = GM / R

^{2}
9.81 = (6.67x10

^{-11})M / (6.38x10^{6})^{2}
So, M = 5.99x10

^{24}kg**(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 2.22x10

^{4}km above its surface.
(i)

Use data from (a) to calculate
angular speed of a GPS satellite in its orbit:

(r = 2.22x10

^{7}+ 6.38x10^{6}= 2.86x10^{7}m)
Either

GM = ω

^{2}r^{3}
(6.67x10

^{-11}) (5.99x10^{24}) = ω^{2}(2.86x10^{7})^{3}
ω = 1.3x10

^{-4}rads^{-1}
Or

gR

^{2}= ω^{2}r^{3}
9.81 x (6.38x10

^{6})^{2}= ω^{2}(2.86x10^{7})^{3}
ω = 1.3x10

^{-4}rads^{-1}
(ii)

Use answer in (i) to show that the
satellites are not in geostationary orbits:

Period of orbit of satellites = 2π / ω = 4.8x10

^{4}s (= 13.4 hours)
The period of orbit for a
geostationary satellite is 24 hours (= 8.6x10

^{4}s). So, the satellites are not in geostationary orbits.**(c)**

Planes of orbits of GPS satellites
in (b) are inclined at angle of 55

^{o}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 = – ω

^{2}x 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 = 4πϵ

_{o}R where ϵ_{o}is permittivity of free space:
Potential (at the surface of the
sphere) = Q / 4πϵ

_{o}R
So, C = Q / V = 4πϵ

_{o}R**(c)**

In order to investigate electrical
discharge (lightning) in a laboratory, isolated metal sphere of radius 63cm is
charged to potential of 1.2x10

^{6}V. 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 CV

^{2}
Remaining energy: 0.25 x ½ C x
(1.2x10

^{6})^{2}= ½ CV^{2}
Potential of sphere after discharge,
V = 6.0x10

^{5}V

__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 t

_{½}by λt_{½}= 0.693:
[N = N

_{o}exp(-λt)]
EITHER when time = t

_{½}, N = ½N_{o}OR ½N_{o}= N_{o}exp(-λt_{½})
EITHER 2 = exp(λt

_{½}) OR ½ = exp(-λt_{½})
(taking logs),

__ln2 = 0.693 = λt___{½}**(c)**

Cobalt-60 is a radioactive isotope
with half-life of 5.26years (1.66x10

^{8}s). Cobalt-60 source for use in school laboratory has activity of 1.8x10^{5}Bq.
Mass of Cobalt-60 in source:

Activity, A = λN

1.8x10

^{5}= N x (0.693 / {1.66x10^{8}})
N = 4.3x10

^{13}
Mass = 60 x (N / N

_{A}) or 60 x N x u
Mass = (60 x 4.3x10

^{11}) / (6.02x10^{23}) = 4.3x10^{-9}g**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 R

_{F}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 V

_{OUT}is given by
V

_{OUT}= – (4V_{A}+ GV_{B}+ V_{C}) where V_{A}, V_{B}and V_{C}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 V

_{OUT}, constant G is gain associated with input B. Show that numerical value of G is 2:
Gain = - R

_{F}/ R_{I}= - 8.0 / 4.0 = 2
Numerical value is 2

**(c)**

Input potentials V

_{A}, V_{B}and V_{C}are either 0 or 1.0V. Magnitudes of some output potentials for different combinations of V_{A}, V_{B}and V_{C}are shown.
(i)

Complete Fig for 3 remaining values
of V

_{OUT}:
[V

_{OUT}= – (4V_{A}+ GV_{B}+ V_{C})]
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 station__s__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.

Delete