2013 J2 H2 Physics CT2 P1

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JJC 2013 9646/JC2 CT2 P1/2013 [Turn Over  JURONG JUNIOR COLLEGE 2013 JC2 Common Test 2 Name Class 12S PHYSICS Higher 2 Structured Questions Candidates answer on the Question Paper. No additional materials are required. 9646/1 2 Jul 2013 1 hour 15 minutes READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the space provided at the top of this page. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use highligh ters, glue or correction fluid. Section A  Answer every question. Section B  Answer any two questions.  At the end of the ex amination, fasten al l your work secu rely together. The number of ma rks is given i n brackets [ ] at the end of each question or part question. For Examiner’s Use 1 2 3 4 Total (This question paper consists of 17 printed pages)

Transcript of 2013 J2 H2 Physics CT2 P1

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JURONG JUNIOR COLLEGE2013 JC2 Common Test 2

Name Class 12S

PHYSICSHigher 2

Structured Questions

Candidates answer on the Question Paper.No additional materials are required.

9646/1

2 Jul 2013

1 hour 15 minutes

READ THESE INSTRUCTIONS FIRST

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

Write your name and class in the space provided at the top of thispage.

Write in dark blue or black pen.You may use a soft pencil for any diagrams, graphs or rough working.Do not use highlighters, glue or correction fluid.

Section A Answer every question.

Section B Answer any two questions.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of eachquestion or part question.

For Examiner’s Use

1

2

3

4

Total

(This question paper consists of 17 printed pages)

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Data

speed of light in free space, c = 3.00 10 8 m s 1 permeability of free space, o = 4 10 7 H m 1 permittivity of free space,

o = 8.85 1012

F m1

= (1/(36 )) 109

F m1

elementary charge, e = 1.60 10 19 C the Planck constant, h = 6.63 10 34 J s unified atomic mass constant, u = 1.66 10 27 kg rest mass of electron, m e = 9.11 10 31 kg rest mass of proton, m p = 1.67 10 27 kg molar gas constant, R = 8.31 J K 1 mol 1 the Avogadro constant, N A = 6.02 10 23 mol 1 the Boltzmann constant, k = 1.38 10 23 J K 1 gravitational constant, G = 6.67 10 11 N m 2 kg 2 acceleration of free fall, g = 9.81 m s 2

Formulae

uniformly accelerated motion, s = ut + 12 at 2

v 2 = u2 + 2 as work done on/by a gas, W = p V hydrostatic pressure, p = gh gravitational potential,

=Gm

r

displacement of particle in s.h.m., x = x o sin t velocity of particle in s.h.m., v = v o cos t

v = 2 2( )o x x

mean kinetic energy of a molecule of an idealgas E = 3

2 kT

resistors in series, R = R 1 + R 2 + . . . resistors in parallel, 1/R = 1/ R 1 + 1/ R 2 + . . .

electric potential, V =o

Q

ε r 4

alternating current / voltage, x = x o sin t transmission coefficient, T exp( 2kd )

where k =2

2

8 ( )m U E

h

radioactive decay x = x o exp(- λt )decay constant

λ =1/2

0.693t

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Section A

Answer every question in this section.

1 In a nuclear fusion reaction, two light nuclei combine into a heavier nucleus with the release

of energy. The most basic fusion reaction is the fusion of two hydrogen nuclei1

1H .

There are two possible outcomes to such a fusion reaction. In reaction (1), a helium isotope2

2He is formed. In reaction (2), a hydrogen isotope 2

1 H and an unknown elementary particle

( X ) are formed.

Reaction (1) 1 1 21 1 2H H He

Reaction (2) 1 1 2 a1 1 1 bH H H X

(a) Explain the term isotope .

[1]

(b) By reference to binding energy per nucleon, explain why energy is released in fusionreactions.

[2]

(c) State the values of a and b for the elementary particlea

bX .

a =

b = [1]

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(d) The hydrogen isotope 21 H is readily found on Earth but not the helium isotope

2

2He .

(i) Suggest a possible reason for this observation.

[1]

(ii) Hence deduce which reaction releases more energy.

[1]

(e) With the following information, show that the energy released in reaction (2) is1.49 x 10 -13 J. [2]

Rest mass of 11H = 1.007825 u

Rest mass of 21 H = 2.014102 u

Rest mass of X = 0.000549 u

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(b) Light of wavelength 590 nm is incident on a diffraction grating with6.25 x 10 5 lines per metre. The screen is placed 10.0 cm away from the grating.

(i) Determine the total number of images produced by the light transmittedthrough this grating.

number of images = [3]

(ii) Calculate the distance between the first-order maximum and the centralmaximum on the screen.

distance = m [3]

(iii) Another diffraction grating of the same slit separation is placed in front of theoriginal grating such that their slits are perpendicular to one another as shownin Fig. 2.1. A 2-dimensional pattern of bright spots is formed on the screen.

Fig. 2.1

Laser source

To Screen

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Sketch the pattern obtained, showing clearly the relative separation of thespots up to the 2 nd order maxima. [2]

(c) Fig. 2.2 shows an arrangement used to determine the wavelength of monochromatic light emitted by a laser.

Fig. 2.2

S 1 and S 2 are silts at right angles to the plane of this page. When the silts areilluminated by light from the laser, they form coherent sources of light. Aninterference pattern is formed on the screen from which measurements can betaken to determine the wavelength.

(i) Explain what is meant by monochromatic .

[1]

ScreenParallelbeam of

light fromlaser

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(ii) Describe how the concepts of diffraction and interference can be used toexplain the formation of the interference pattern.

[3]

(iii) If white light is used in the above experiment instead of monochromatic light,state how the interference pattern would change.

[2]

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3 A wire frame ABCD is supported on two knife-edges P and Q so that the section PBCQ onthe frame lies within a solenoid, as shown in Fig. 3.1. and Fig. 3.2.

Fig. 5.1

Electrical connections are made to the frame through the knife-edges so that the partPBCQ of the frame and the solenoid can be connected in series with a battery. When

there is no current in the circuit, the frame is horizontal.

(a) When the frame is horizontal and a current passes through the frame and solenoid,what can you say about the direction of the force, if any, due to the magnetic field of the solenoid acting on

(i) side BC,

(ii) side PB?

[2]

Fig. 3.1

Fig. 3.2

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(b) State two ways in which you could reverse the direction of the force on side BC.

1.

2.

[2]

(c) (i) The solenoid has 700 turns m -1 and carries a current of 3.5 A. Given that themagnetic flux density B on the axis of a long solenoid is

oB ni , where n is the number of turns per metre of the solenoid, and i isthe current in the solenoid,

calculate the magnetic flux density in the region of side BC of the frame.

magnetic flux density = T [1]

(ii) Side BC has length 5.0 cm.

Calculate the force acting on BC due to the magnetic field in the solenoid.

force = N [2]

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(iii) A small piece of paper of mass 0.10 g is placed on the side DQ andpositioned so as to keep the frame horizontal. Given that QC is of length 15.0cm, how far from the knife-edge must the paper be positioned?

distance = m [2]

(d) State Faraday’s and Lenz’s laws of electromagnetic induction.

[2]

(e) A pair of concentric coils is shown in Fig. 3.3.

Fig. 3.3

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The outer coil X has 2500 turns and is connected to a variable power supply by theterminals CD. The inner coil Y has 500 turns, a cross-sectional area of

7.25 x10 -4 m 2 and a resistance of 5.00 Ω. Coil Y is connected to a resistor R of resistance 5.00 Ω.

The variation with time t of the magnetic flux density B in coil Y is shown in Fig. 3.4.

(i) Calculate the maximum current in R.

maximum current = A [3]

(ii) On Fig. 3.5, sketch the variation with time t of current I in R. [3]

Fig. 3.4

Fig. 3.5

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(iii) Discuss the change in the answer to (e)(i), if any, when the number of turns ininner coil Y triples to 1500.

[3]

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4 (a) In a photoelectric experiment, metal G and metal F are illuminated with a lightsource in an evacuated photocell and photoelectrons are emitted.

Each metal occupies an equal surface area. Fig. 4.1 shows the variation of thestopping potential V

sfor metal G with frequency f of the light source.

(i) State what is meant by the work function of a metal.

[1]

(ii) Metal F has a work function of 0.125 eV.

Calculate the threshold frequency for photoelectrons to be emitted from metalF.

threshold frequency = Hz [2]

(iii) On Fig. 4.1, sketch the variation of stopping potential V s for metal F with thefrequency f of light source. Label your sketch F. [2]

V s / V

f / 10 13 Hz

metal G

2.60Fig. 4.1

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(iv) Electrons emitted from the surface of metal G and metal F are collected atplate D as shown in Fig. 4.2.

1. State whether metal G or metal F emits electrons with a higher maximum kinetic energy. Explain your answer.

[2]

2. The current detected in the ammeter is reduced to zero when thepotential at metal G and metal F is 8.00 V and the potential at plate D is5.00 V.

Determine the corresponding frequency of the light source.

frequency = Hz [3]

A

metal G andmetal F

Plate D

Light source

Fig. 4.2

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(b) Fig. 4.3 shows the variation of intensity I with wavelength λ of an X-ray spectrum.

Fig. 4.3

(i) Explain the origins of the features of the X-ray spectrum using quantumtheory.

[5]

I

λ 0 λ min

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(ii) The spectrum is obtained from an X-ray machine in which electrons areaccelerated through a potential difference of 40 kV.

Calculate the minimum wavelength λ min .

minimum wavelength λ min = m [2]

(iii) On Fig. 4.3, sketch the spectrum which would be obtained from the samemachine when the accelerating voltage is halved. [3]

End of p aper