14_15_h2_j2_ct2_p2

JJC 2015 9646/JC2 Common Test 2 P2/2015

JURONG JUNIOR COLLEGE JC2 Common Test 2 2015

Name Class 15S

PHYSICS Higher 2

Structured Questions

Candidates answer on the Question Paper. No Additional Materials are required.

9646/2

30 June 2015

1hour 45 min

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name and class in the spaces 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 paper clips, highlighters, glue or correction fluid. Answer all 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 each question or part question.

For Examiners Use

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Total

(This question paper consists of 22 printed pages)

JJC 2015 9646/JC2 Common Test 2 P2/2015 [Turn Over

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Data

speed of light in free space, c = 3.00 108 m s1

permeability of free space, o = 4 107 H m1

permittivity of free space, o = 8.85 1012 F m1 = (1/(36)) 109 F m1

elementary charge, e = 1.60 1019 C

the Planck constant, h = 6.63 1034 J s

unified atomic mass constant, u = 1.66 1027 kg

rest mass of electron, me = 9.11 1031 kg

rest mass of proton, mp = 1.67 1027 kg

molar gas constant, R = 8.31 J K1 mol1

the Avogadro constant, NA = 6.02 1023 mol1

the Boltzmann constant, k = 1.38 1023 J K1

gravitational constant, G = 6.67 1011 N m2 kg2

acceleration of free fall, g = 9.81 m s2

Formulae

uniformly accelerated motion, s = ut + 12

at2

v2 = u2 + 2as

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 = xo sin t

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

v = 2 2( )ox x

mean kinetic energy of a molecule of an ideal gas

E = 32 kT

resistors in series, R = R1 + R2 + . . .

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

electric potential, V = o

Q

r4

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


3

1 (a) (i) What is meant by the term internal energy of a system?

[1]

(ii) State the first law of thermodynamics.

[2]

(b) A monatomic ideal gas undergoes a cycle of changes A B C A, as shown in Fig. 1.1.

Fig. 1.1

(i) Calculate the work done by the gas during the change C A.

work done by gas = J [2]

volume V / cm3

pressure p / 105 Pa

A

B

C

20 5

1

7


4

(ii) Show that the internal energy of the gas at A is 3.0 J. [1]

(iii) Fig 1.2 is a table of energy changes during one cycle. Complete Fig. 1.2.[3]

Fig. 1.2

(iv) Determine the net work done on the gas over one cycle.

work done on gas = J [1]

section of cycle heating supplied to

gas / J work done on gas / J

increase in internal energy of gas / J

A B 0 +2.25 +2.25

B C -4.5

C A


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2. A pendulum bob in a particular clock oscillates so that its displacement from a fixed point varies with time t as shown in Fig. 2.1. The mass of the pendulum bob is 0.100 kg.

Fig. 2.1

(a) Determine

(i) the angular frequency of the oscillations,

angular frequency = rad s-1 [1]

(ii) the magnitude of the maximum velocity, and

maximum velocity = m s-1 [2]

(iii) the total energy of the system.

total energy = J [1]


6

(b) Sketch on Fig. 2.2, with values on both axes, the graph of velocity against displacement. [2]

Fig. 2.2

v / m s-1

x / m


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3. (a) A cable used for the transmission of electrical energy has a circular cross-section of radius 0.011 m. Fig. 3.1 is a full-scale drawing showing the electric field surrounding the cable together with lines of equal potential at an instant when the potential of the cable is +564 000 V.

Fig. 3.1

(i) State the equation relating field strength to potential gradient.

[1]

475 000 V

450 000 V

Cable at potential

+564 000 V

550 000 V

525 000 V

500 000 V


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(ii) Use Fig. 3.1 to estimate the potential gradient near the surface of the cable.

potential gradient = V m-1 [2]

(iii) Explain why a cable of larger radius but the same potential, will have a smaller electric field at its surface.

[1]

(b) Fig. 3.2 shows two horizontal metal plates, each of length 100 mm, separated by a vertical distance of 7.0 mm. The upper plate is at a potential of - 2.0 V. An electron placed between the two plates experiences an electric force of 3.2 x 10-16 N acting upwards.

Fig. 3.2

(i) Determine the electric field strength between the two metal plates.

electric field strength = V m-1 [2]


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(ii) Hence, or otherwise, determine the potential of the bottom plate.

potential = V [2]

4 (a) Charged particles from the Sun, on approaching the Earth, may become trapped in the Earths magnetic field near the poles, as shown in Fig. 4.1. This can cause the sky to glow. The phenomenon is called aurora borealis.

Fig. 4.1

Some of the charged particles travel in a circle of radius 50 km in a region where the magnetic flux density is 6.0 x 10-5 T.

For a charged particle of charge to mass ratio e/m, show that the expression for its speed v when travelling in a circle of radius r within a magnetic field of flux density B is given by

ev Br

m

[2]


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(b) Use your answer to (a) and the information about the path of the particles to show that the charged particles causing the aurora cannot be electrons.

[2]

(c) Suggest a particle that could cause the aurora.

particle = [1]

(d) The atmosphere of the Earth is a mixture of nitrogen and oxygen atoms and molecules that surrounds Earth. With reference to the electronic transitions that can occur within the nitrogen and oxygen atoms and molecules, explain how an aurora can cause the sky to glow.

[3]

5 (a) State Faradays law of electromagnetic induction.

[1]


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(b) A small coil is positioned so that its axis lies along the axis of a large bar magnet as shown in Fig. 5.1.

The coil has a cross-sectional area of 0.40 cm2 and contains 150 turns of wire.

The average magnetic flux density B through the coil varies with the distance x

between the face of the magnet and the plane of the coil, as shown in Fig. 5.2.

Fig. 5.2

(i) The coil is 5.0 cm from the face of the magnet. With reference to Fig. 5.2, show that the magnetic flux linkage of the coil is about 3.0 x 10-4 Wb. [1]

Fig. 5.1


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(ii) The coil is moved along the axis of the magnet so that the distance x changes from x = 5.0 cm to x = 15.0 cm in a time of 0.30 s.

Calculate

1. the magnitude of the change in flux linkage of the coil,

change in flux linkage = Wb [2]

2. the mean e.m.f. induced in the coil.

mean e.m.f. = V [2]

(iii) State and explain the variation, if any, of the speed of the coil so that the induced e.m.f. remains constant during the movement in (b)(ii).

[2]

(iv) The magnet in Fig. 5.1 is now replaced with a solenoid connected to an

alternating voltage, with the axis of the solenoid aligned to the axis of the

small coil.

Suggest how such an arrangement may be applied in the wireless charging of mobile phones.

[2]


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6 An a.c. power supply is connected to a resistor R as shown in Fig. 6.1.

Fig. 6.1

A cathode ray oscilloscope (c.r.o.) is used to show the potential difference (p.d.) across R. The screen of the c.r.o. displays the variation with time of the p.d. across R power supply is connected to a resistor R, as shown in Fig. 6.2.

Fig. 6.2

On the vertical axis 1.0 cm represents 5.0 V. On the horizontal axis, 1.0 cm represents 10 ms.

(a) Use Fig. 6.2 to determine the frequency of the a.c. supply.

frequency = Hz [2]

1.0 cm

1.0 cm


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(b) The resistance of R is 500 .

Calculate

(i) the r.m.s. current in R,

r.m.s. current = A [2]

(ii) the mean power transformed in R.

mean power = W [2]


15

7 Ultrasonic sound waves (ultrasound) have frequencies outside the audible range of the human ear, that is, greater than about 20 kHz. As ultrasound passes through a medium, wave energy is absorbed.

The rate at which energy is absorbed by unit mass of the medium is known as dose-rate, measured in W kg-1.

The total energy absorbed by unit mass of the medium is known as the absorbed dose, measured in J kg-1.

Under certain circumstances, biological cells may be destroyed by ultrasound. The effect on a group of cells is measured in terms of the survival fraction SF, which is computed using the expression

SF = exposure before cells of number

exposure after surviving cells of number.

For any particular absorbed dose, it is found that the survival fraction changes as the dose-rate increases. Fig. 7.1 shows, for an absorbed dose of 240 kJ kg-1 for each sample of cells, the variation of survival fraction with dose-rate for samples of cells in a liquid.

Fig 7.1

(a) (i) Using Fig. 7.1, determine the survival fraction for a dose rate of 115 W kg-1.

survival fraction = [1]


16

(ii) Calculate the exposure time for an absorbed dose of 240 kJ kg-1 and at a dose-rate of 200 W kg-1.

time = s [2]

(b) Survival fraction depends not only on the dose-rate but also on the absorbed

dose. Fig. 7.2 shows the variation of 10log ( )SF with dose rate for different values

of absorbed dose.

Fig. 7.2

(i) Identify the line in Fig. 7.2 that corresponds to the data given in Fig. 7.1 and label it L. [1]


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(ii) Suggest why the survival fraction is plotted on a logarithmic scale.

[1]

(iii) Using Fig. 7.2, complete the table of Fig. 7.3 for a dose-rate of 200 W kg-1.

[1]

Absorbed dose / kJ kg-1 log10 (SF)

50 -0.650

100 -0.900

160

240 -1.575

340 -2.150

450 -2.850

560 -3.750

Fig. 7.3

(iv) Using the values in the table of Fig. 7.3, plot the points for absorbed doses 50 kJ kg-1, 160 kJ kg-1 and 240 kJ kg-1 into Fig. 7.4, the graph of

10log ( )SF against absorbed dose for the dose rate of 200 W kg-1. [2]

Fig. 7.4


18

(c) A researcher suggests a theory that at a high dose-rate of 200 W kg-1, two different effects cause cell destruction. The first effect, computed as (SF)1, is present for all absorbed doses. The second effect, computed as (SF)2, becomes significant only at high absorbed doses.

What feature in Fig. 7.2 supports his theory that the second effect exists?

[1]

(d) (i) Use Fig. 7.4 to determine SF for an absorbed dose of 560 kJ kg-1.

SF = [1]

(ii) The theory proposed in (c) suggests that the resultant survival fraction (SF)R due to the two different effects, each computed individually as survival fractions (SF)1 and (SF)2 is given by the expression

(SF)R = (SF)1 x (SF)2

For the absorbed dose of 560 kJ kg-1, estimate the values of 1( )SF and

2( )SF .

1( )SF = [1]

2( )SF = [1]


19

8 When light is shone on the front of a photocell, an e.m.f. is generated in the photocell. A student realizes that the e.m.f. in photocell changes when a piece of glass is placed in front of it.

It is suggested that the e.m.f. generated is related to thickness of the glass.

Fig. 8 : Photocell Design a laboratory experiment to investigate how the e.m.f. generated varies as thickness of the glass varies. You should draw a diagram showing the arrangement of your equipment. In your account you should pay particular attention to (a) the procedure to be followed,

(b) the measurements to be taken,

(c) the control of variables,

(d) the analysis of the data,

(e) the improvements to accuracy and

(f) any safety precautions to be taken.


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Diagram


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14_15_h2_j2_ct2_p2

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