MJC 2010 H2 Physics Prelim Paper 2

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MERIDIAN JUNIOR COLLEGEPreliminary ExaminationHigher 2

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H2 Physics 9646/2

Paper 2 21 September 2010

1 hour 45 mins

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READ THESE INSTRUCTIONS FIRST

Class Reg Number

Candidate Name _____________________________

This booklet contains 7 questions.

Do not open this booklet until you are told to do so.

Answer all questions.

Write your answers on this question booklet in the blanks provided.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the endof each question or part question. Marks will bededucted if units are not stated where necessary or ifanswers are not quoted to the appropriate number ofsignificant figures.

All working for numerical answers must be shown. Youare reminded of the need for good English and clear

presentation of your answers.

Examiners Use

Section A

Q1 /15

Q2 /8

Q3 /8

Q4 /6

Q5 /8

Q6 /15

Section B

Q7 /12

Deductions

Total /72


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Preliminary Examination Meridian Junior College 21 September 2010JC2 H2 Physics 2010

3

DATA AND FORMULAE

Dataspeed of light in free space c = 3.00 x 108m s-1

permeability of free space o = 4 x 10-7 H m-1

permittivity of free space 0 = 8.85 x 10-12 F m-1

= (1/(36)) x 10-9 F m-1

elementary charge e = 1.60 x 10-19 C

the Planck constant h = 6.63 x 10-34 J s

unified atomic mass constant u = 1.66 x 10-27 kg

rest mass of electron me = 9.11 x 10-31 kg

rest mass of proton mp = 1.67 x 10-27 kg

molar gas constant R = 8.31 J K-1 mol-1

the Avogadro constant NA = 6.02 x 1023 mol-1

the Boltzmann constant k = 1.38 x 10-23 J K-1

gravitational constant G = 6.67 x 10-11 N m2 kg-2

acceleration of free fall g = 9.81 m s-2

Formulaeuniformly accelerated motion

s = ut +1

2at2

v2 = u2+2as

work done on/by a gas W = pV

hydrostatic pressure p = gh

gravitational potential = -Gm/r

displacement of particle in s.h.m. x = xo sin t

velocity of particle in s.h.m. v = vo cos t

= 2 2o

-x x

resistors in series R = R1 + R2 +

resistors in parallel 1/R = 1/R1 +1/R2+

electric potential V = Q/4or

alternating current/voltage x = xo sin t

transmission coefficient T = exp(-2kd)

where k =

2

2

8 ( )m U E

h

radioactive decay x = xo exp(-t)

decay constant =

1

2

0.693

t


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Answer all the questions in the spaces provided.

1 (a) State the Principle of Superposition.

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.............................................................................................................................................

.................................................................................................................................... [2]

(b) Two sinusoidal transverse waves W1 and W2, of the same type, are incidentsimultaneously on a point P. The amplitude of W2 is the sameas the amplitude of W1.The frequency of W2 is halfthe frequency of W1.

At a certain instant (time t= 0) at P, both waves have zero displacement and then bothdisplacements increase in the same direction.

Fig. 1.1 is a graph of displacement at P against time for wave W1.

(i) On Fig. 1.2, sketch a graph of displacement against time for wave W2. [3]

(ii) On Fig. 1.3, sketch a graph to show the resultant wave produced by thesuperposition at P of waves W1 and W2. [3]

Fig. 1.1

Fig. 1.2

Fig. 1.3

displacement(wave W1)

displacement(wave W2)

displacement(resultant

wave)

time

time

time


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(iii) The frequency of wave W1 is 4.2 x 1015 Hz. Determine the frequency of the

resultant wave produced by the superposition of waves W1 and W2.

frequency = . Hz [2]

(iv) Explain why it is incorrectto say that waves W1 and W2 are coherent.

...................................................................................................................................

.......................................................................................................................... [1]

(c) Blue light of wavelength 485.6 nm from a star is incident normally on a diffractiongrating. The light is diffracted into a number of beams as shown in Fig. 1.4.

The angular separation of the two second order beams is 45.7o. Calculate thenumber of lines per millimeter on the grating.

Lines per millimetre = . [4]

Fig. 1.4

first order

second order

first order

second order

zero order

grating


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6

2 (a) Air near the Earth conducts electricity to a small extent. Because there is an electric fielddirected towards the Earths surface, a small electric current exists in the atmosphere.Measurements of the electric field near the Earths surface give an average of about1.0 x 102 V m-1.

(i) Estimate the potential difference between the head and feet of an average adult.

potential difference = . V [1]

(ii) The electric field near the Earths surface is as large as 1.0 x 10 2 V m-1, explainwhy you do not experience an electric shock when you stand upright.

................

................

................

................

.............. [2]

(b) A photomultiplier tube is a device which has a common electrode (the photocathode) anda number of other electrodes (the dynodes), which must be maintained at definitepotentials relative to the common electrode. A potential divider circuit may be used tosupply these potentials.

Fig. 2.1 illustrates a photomultiplier tube with a photocathode and six dynodes,numbered 1 to 6. A potential divider of six resistors, each of resistance R, using a supplyvoltage of 1050 V, is connected to the electrodes in the tube.

Fig. 2.1

photocathode

1050 V

R R R R R R

1 2 3 4 5 6dynodes

potential divider

photomultiplier tube


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(i) For the situation where there is no electron current inside the photomultiplier tube,determine the potential difference between dynode 4 and the photocathode.

p.d. between dynode 4 and the photocathode = V [2]

(ii) A fault develops inside the photomultiplier tube causing a short circuit betweendynodes 3 and 5. Determine the new potential difference between dynode 4 andthe photocathode. Explain your reasoning.

new p.d. between dynode 4 and the photocathode = .. V

Explanation: ..

................

................

....... [3]


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3 From the power station, the voltage is stepped up to about 230 kV along the high voltagecables before a series of step down transformers near homes are used to reduce the voltageto 240 V at our homes. Fig. 3.1 shows the last iron-cored step down transformer before ourhomes with the actual turns ratio indicated. The home 240 V output has a frequency of 50 Hzis connected to a 58 resistor of a home appliance.

(a) Determine the number of similar turns-ratio transformers required to step down from thehigh voltage (230 kV) cables to the 240 V output used at homes. State one assumptionmade.

Number of transformers = ..................

Assumption: ........................................................................................................................

...................................................................................................................................... [4]

(b) Explain the purpose of the iron core in the transformer.

.............................................................................................................................................

.............................................................................................................................................

.................................................................................................................................... [1]

Fig. 3.1

home

secondary coilprimary coil

iron-core

58


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(c) Determine the average power dissipated in the 58 resistor of a home appliance.

Average power = .. W [2]

(d) Suggest a typical household appliance with such power rating.

Appliance is .... [1]

4 A p-n junction is formed between slices of p-type and n-type semiconductor material as shownin Fig. 4.1

Fig. 4.1

(a) On Fig. 4.1, draw an arrow indicating the direction of movement of holes when the twoslices are brought into contact. [1]

(b) Boron is used as a dopant in one of the semiconductor slices. State and explain whetherthe addition of Boron creates a p-type or a n-type semiconductor.

...................

...................

...................

...................

...................

...................

................. [3]

p-type n-type


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(c) State and explain what will happen to the depletion region when a battery is connected inas shown in Fig. 4.2.

Fig. 4.2

...................

...................

...................

...................

................. [2]

5 (a) Explain what is meant by binding energy.

...................

...................

................. [1]

(b) Calculate the binding energy of a thorium nucleus 22690

Th.

Given that

rest mass of 22690

Th = 226.0249 u

rest mass of proton = 1.0073 urest mass of neutron = 1.0087 u

Binding energy = MeV [2]


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(c) A thorium nucleus 22690

Th originally at rest decays and forms a radium nucleus 22288

Ra* and

an alpha particle as shown below. The radium nucleus 22288

Ra* is in an excited state.

226

90Th 222

88Ra* +

2He

Given that

rest mass of226

90 Th = 226.0249 urest mass of 222

88Ra = 222.0154 u

rest mass of2

He = 4.0026 u

(i) Calculate the kinetic energy of the radium nucleus if the alpha particle is emittedwith a kinetic energy of 2.38 MeV.

Kinetic energy of Ra = MeV [3]

(ii) The excited radium nucleus, 22288

Ra* further undergoes a gamma decay as shown below:

222

88Ra* 222

88Ra +

Calculate the energy of the gamma ray.

Energy of the gamma ray = MeV [2]


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6 Most Singapore buildings are built using a framework of concrete beams, slabs and columns.The concrete columns need to carry both the ultimate vertical load, Nand the ultimate bendingmoment, M induced from the attached beam/s as shown in the 3-D pictorial diagram ofFig. 6.1. In practice, the concrete columns are reinforced with steel bars.

To design for the steel bars in such columns, design charts are available from the BritishStandard Structural Use of Concrete, BS8110.

Fig. 6.1

Beam

M

Column

N


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In order to choose the correct design chart, the values of the following must match the designdetails:

fcu = Ultimate crushing pressure limit of concretefy = Ultimate characteristic strength of steel bar

dhwhere d is the distance from the edge of the column to the centre of the steel bar that is

furthest from that edge and his the larger dimension of the column

The cross section of a column is shown in Fig. 6.2:

Dimensions of the column are:

h= 450 mm (larger dimension of the column)b= 200 mmd= distance from the edge of the column to the centre of the steel bar that is furthest from that

edge.

(a) The chosen design details for the column above are:

fcu = 50 N mm-2

fy = 460 N mm-2

cover = 40 mm (from edge of column to edge of steel bar)

Assuming that 4 numbers of 32 mm diameter steel bars are to be used. Determine the

value of

d

h and hence explain why Chart No. 49 is appropriate to be used.

d

h=

[1]

Explanation:

..............................................................................................................................................

..................................................................................................................................... [1]

Fig. 6.2

32 mmdiameter steelbar

cover= 40 mm

b= 200 mm

dh= 450 mm

Fig. 6.3

h= 450 mm

b= 200mm

Height ofcolumn


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(b) The coordinates of the chart derived from the values ofN

bhand

2

M

bhcan be used to

establish the corresponding value of SC100A

bhsuch that ASC (the total required

cross-sectional area of steel bars) can be calculated. It is necessary to provide enoughsteel bars, with a total area more than ASC.

An example of reading off the SC100A

bhvalue is:

Value ofN

bh= 32.5 N mm-2

Value of2

M

bh= 3.20 N mm-2

From Chart No. 49, the corresponding coordinate is marked with a cross and labelled

A. This coordinate corresponds to a value of SC100

4 5A

bh

< < . More specifically, the

value of SC100A

bhis 4.5. Using this value, ASC can then be calculated.

Based on the same design details in (a), the loads carried by the column in Fig 6.2 are:M= Ultimate bending moment = 91.2 kN mN= Ultimate vertical load = 2460 kN

(i) Determine the values ofN

bhand

2

M

bhfor the column in Fig 6.2.

N

bh= N mm-2

2

M

bh= N mm-2 [2]


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(ii) Using Chart No. 49, determine whether 4 steel bars of diameter 32 mm with atotal area of 3220 mm2 is sufficient. [4]

(c) A young engineer designed the same column except that he used a different concretewith an ultimate crushing pressure limit, fcu of 25 N mm

-2. Compare your calculated value

ofN

bhin (b)(i) with this value of fcu = 25 N mm

-2. Comment what would happen to the

concrete.

.............................................................................................................................................

.............................................................................................................................................

.................................................................................................................................... [2]

(d) The height of the column designed is 3.8 m. Suggest one possible problem with another12.0 m height column of similar size subjected to a similar vertical load and bendingmoments.

.............................................................................................................................................

.................................................................................................................................... [1]


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(e) The design of another column has the values of SC100

2.0A

bh= and

22.0

M

bh= .

Using Charts No. 39 and 49, determine the percentage decrease in ultimate verticalload, Nif fcu = 50 N mm

-2 changes to fcu = 40 N mm-2.

Percentage decrease of N= .....................................% [4]


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Section B

It is recommended that you spend about 30 minutes on this question

7 The piezoelectric effect describes an effect that converts a mechanical effect to an electricalsignal or vice versa. For example, piezoelectric sensors such as quartz, silicon, manmadecrystals or even flexible polymer sheets can produce electricity when squeezed, moved orbent. On the other hand, piezoelectric transducers, such as those used in speakers, rapidlychange shape when subjected to an electrical current.

Many modern devices, like the Nintendo Wii remote as well as smartphones such as the AppleiPhone have tiny inbuilt piezoelectric accelerometersto help sense motion and other actions.

Although called an accelerometer, it is actually the inertial force on a known mass that ismeasured by the piezoelectric material when the device is moved, which is then converted intoan electrical signal and interpreted accordingly.

Design an experiment using a sample of piezoelectric material to find how the strength of theelectrical signal depends on how much force it is subjected to.

The equipment available includes the following, besides common apparatus found in thelaboratory:

An A4 size flexible sheet of piezoelectric material (you may assume that the surfaces areconducting)

Digital Multimeters

10 slotted masses

You should draw diagrams to show the arrangement of your apparatus. In your account youshould pay particular attention to

a) The equipment you would use for the investigation,b) The procedure to be followed,c) The control of variables,d) Any safety precautions,e) Any precautions that you would take to improve the accuracy of the experiment

Electrical signal developed acrosshere when piezoelectric material iscompressed

Piezoelectricsample

Compressive force


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Diagram:

Please make use of this page and next two pages to write your answers to Q7:

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MJC 2010 H2 Physics Prelim Paper 2