• 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 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 = – 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 = 4πϵ_oR where ϵ_o is permittivity of free space:

Potential (at the surface of the sphere) = Q / 4πϵ_oR

So, C = Q / V = 4πϵ_oR

(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 t_½ by λt_½ = 0.693:

[N = N_oexp(-λt)]

EITHER when time = t_½, N = ½N_o    OR ½N_o = N_oexp(-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 (1.66x10⁸s). 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 / N_A) 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.