TEKNIK OF ANSWERING QUESTION

7/30/2019 TEKNIK OF ANSWERING QUESTION

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60/2This question paper is CONFIDENTIAL until the examination is over.

ONFIDENTIAL*

SENTUHAN EMAS KEDUA 2011

nswer all questions [40 marks]

. The heat capacity C of a substance is defined as quantity of heat supplied to raisehe temperature of a substance by 1 K . The heat capacity C of a solid is given by thexpression

3bT aT C

here T is the absolute temperature , a and b are constant with dimension.a) Determine the dimension of heat capacity, C . [1 mark]

b) Write down the SI units of heat capacity, C [1 mark]c) Find the units for a and b. [3 marks]

nswer:a) J K -1 = Nm K -1 = (kg m s -2 )m K -1 = kg m 2 s-2 K -1

122][ K T MLT

QC [1]

b)1 JK [1]

c) Unit of C = units of aT = unit of 3bT [1]

1 JK = (unit of a) K = (unit of b) 3 K

Unit of a K JK 1

2 JK or kg m 2 s-2 K -2 [1]

Unit of b 31

K

JK

4JK or kg m

2

s-2

K -4

[1]


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ONFIDENTIAL*

. The rotating blade of a blender turns from rest with constant angular acceleration1.50 2rads .

a) How much time does it take to reach an angular velocity of 36.0 1rads ?[2 marks]

b) What is the number of revolutions in this time interval? [3 marks]

nswer:

a)t 0 [1]

s

t

0.2450.1

00.36

0

/24.00 s [1]

b)2

0 21

t t [1]

rad 432

)0.24)(50.1(210 2

Number of rev = 2432

= 68.8/68.75

1

1


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ONFIDENTIAL*

. The graph shows the variation of extension with force of a wire before reachingthe elastic limit. The original length of the wire is 2.0 m and its cross section areais 0.50 mm 2.

a) Find the work done to extend the wire to 1.20 mm. [2 marks]b) Calculate the Youngs modulus of the wire . [3 marks]nswer:

a) Work done = Fe21

=21 x 60 x 1.2 x 10 -3 J

= 0.036 J [1] b) Youngs Modulus = A

l e

F eA Fl

= Al

gradient 1

or K = 1/gradient [1]

= 63105.0

0.2

102.1

60

= 2.0 x 10 11 Pa (2, 3 or 4 sf) [1]

F(N)

e (mm)

0.20.40.60.81.01.2

0 10 20 30 40 50 60

Or area shaded correctly [1]

Or E = Stress/Strain [1]


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ONFIDENTIAL*

. The graph shows the variation of pressure with volume for 0.128 mol of an ideal gasat temperatures 600 K and 850 K.

sing the graph,(a) calculate the volume of gas at Y, and [2 marks](b) calculate the work done for the gas to change its state from W to Y. [2 marks](c) If the gas at W experiences an adiabatic expansion to reach a tempe rature of 600

, sketch the adiabatic expansion on the graph above. [1 mark]

nswer:

a. At Y, pV nRT

3232

3

101.110130.11080

85031.8128.0V

mor m[1]

b. Work done, W =2

1

1V V

d V

nRT V

= )ln( 12

V V

nRT

= )100.8103.11

ln(85031.8128.0 33

= 312.3 J atau 312 atau 310 J [1]

T2 = 850 K

T = 600 k

[1]

[1]


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ONFIDENTIAL*

c.

T2 = 850 K

T1 = 600 K

[1]


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ONFIDENTIAL*

. (a) A variable inductor is usually used to regulate the speed of an electricceiling fan instead of a variable resistor. Explain. [2marks](b) A inductor with an inductance of 0.20H is connected to a 12 V ac supply

with a frequency of 50 Hz. Calculate(i) the reactance of the inductor(ii) the rms current in the circuit [3 marks]

nswer:

a) The current varies if the self inductance is changed/ Inductor has reactance whichopposes the a.c. [1]

If a resistor is used, power is dissipated as heat loss in the resistor. /No heat isdissipated in a pure inductor [1]

b)(i)The reactance,

[1]

(ii) rms current

[1]

[1] Aor or A

X

V I

L

rmsrms

19.0191.01910.0 83.62

12

83.62

20.0502

2 fL X L


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ONFIDENTIAL*

. (a) An air wedge is formed by placing a wir e between two pieces of glassplates as shown in the diagram below. The length of the glass plate is 10cm.When monochromatic light of wavelength 475 nm is incident normally on theair wedge, interference pattern can be observed.

Determine the diameter of the wire if 380 dark fringes are observed in theinterference pattern formed.

[3 marks]

wireglass plates

10 cm

nswer:

(a) tan2 x

tan210475

3801010 92 x x

410025.9tan x

Let the diameter of the wire be d

410025.91.0

xd

d = 9.025 x 10 -5 m

or 9.0x10 -5m

1

1

1

2 t = m (For dark fringes)

m = 380 or 379

t = 9.025x10 -5 m (m = 380)

t = 9.001x10 -5 m (m = 379)

or 9.0x10 -5 m


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ONFIDENTIAL*

b) If the wire is pushed in slightly to reduce the length of the wedge, explainny changes to the fringe width [2 marks]

The fringe width decreases. [1]

Because 380 fringes are distributed through a smaller region . [1]

Or Fringe width, tan2 x

When the wire is push in slight ly, increases, therefore the fringe widthdecrease


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ONFIDENTIAL*

(a) Explain what is meant by the half-life of a nuclide. [1 mark]

(b)A source with decay constant 6.4 x 10 -10 s-1 has an activity of 5.7 10 7 Bq.Calculate, for this source,

(i) the half-life

(ii) The time taken for the activity to fall to 6.010 5 Bq. [4 marks]

nswer:

(a) The half-life of a nuclide is the time taken for the number of radioactive nuclei to be reduced to half its original number. [1]

b) (i) [1] , [1]

(ii) [1]

[1]

Or 7.1x10 9 s

Or

s

nT t T

t n

n

A

An

o

9

2/1

2/1

101.7

570.6

21 1

1


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ONFIDENTIAL*

SECTION B

. (a) A car travelling along a flat road may be considered to have threeorces acting on it. These are represented in the Figure below.

(i) Force A has a magnitude of 8200 N and is at an angle of 28 to the vertical.Force B is horizontal and has a magnitude 1500 N.Calculate

1. The weight and the mass of the car, [2 marks]

Weight of car, C = A cos 28

= 8200 cos 28 = 7240 N [1]

[1]

2. The resultant force on the car, [2 marks]

Horizontal resultant force = 8200 sin28 1500 [1]

= 2350 N [1]

8200 N

1500

28


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ONFIDENTIAL*

3. The acceleration of the car. [1 marks]

[1]

(ii) For the car, motion is impossible with out friction. Discuss what iseant by friction and the direction in which it acts on the car. In your answer,

uggest another example where friction is useful. [3 marks]

Friction is the force between two solid surfaces which have relative motionnd it always opposes motion.

[1]

The direction of friction is forward/ in the same direction as the motion [1]

Example : 1. When we are walking or running, the friction on ourfeet is in forward direction.

2. When a load is placed on top of a conveyer belt which is moving,the friction on the load is in forward direction.

Any one or other relevant answer [1]

b) Describe a situation in which motion is produced without friction being required.[1mark]

Example : 1. A object is pulled/pushed by a boy. The object can move without thehelp of friction.

2. An object falls under gravity. [1]


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ONFIDENTIAL*

c) In a collision, when a truck of mass 12000 kg runs into the ba ck of a car of ass 1200 kg, a constant force of 72000 N acts for 0.25 s. Calculate the change in

elocity of (i) the car, [3 marks]

Change in momentum = F t [1]

=72000 0.25 = 18000 Ns [1]

[1]

(ii) the truck. [2 marks]

[1]

[1]

(d) Suggest one way in which the conditions in (c) are unrealistic. [1 mark]

The conditions in (c) ignores the presence of external forces such as the frictionalforces between the tyres of the vehicles and the road surface.

Some forces is used to do work to change the shape of the car or truck.

Any one above [1]


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ONFIDENTIAL*

Use the Figure to

(i) Explain why the motion of particle P is simple harmonic. [2 marks]

The acceleration is directly proportional to the displacement and [1]always acts towards the equilibrium position/against the displacement . [1]

(ii) Show that the frequency of oscillation of particle P is 464 Hz. [2 marks]

Gradient of graph = - w2

w2 = 3400/0.4x10 -3

2 f = 8.5x10 6 ( w = 2 f )

f = 464 Hz

(iii) The magnitude of the gradient of the line in the Fig. is G . Show that,for a particle of mass m oscillating with amplitude A, its maximumkinetic energy E max is given by

[3 marks]Speed of particle is

1

1

2max

2

222

2

22

21

0max..

)(21

21

(

mGA E

xat imumis E K

graphof Gradient

x Awm

mv E

x Av

k

1

1

1


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ONFIDENTIAL*

11. (a) In the kinetic theory of gasses, sever al assumptions are made concerningan ideal gas. State four of the assumption s. [4 marks]

Assumption of kinetic theory of gases: ( any four ) [4]

1. Gas consists of a large number of molecules which move freely andrandomly.

2. Gas molecules collide elastically with each other and with the walls of thecontainer.

3. The volume of the gas molecule is very small and can be neglected compareto the volume of the gas.

4. The intermolecular forces are very small and can be neglected.

5. The time of collision between molecules can be neglected compare to the timeinterval between two consecutive collisions of a molecule.

(b) The equation for the kinetic theory of gas is given by

p = 231

c

Where p and are pressure and density of the gas respectively and < c2 > is

the mean square speed for the molecules in the gas.By comparing this equation with the equation pV m= RT , derive an equationfor the average kinetic energy for one molecule of a monatomic gas in termsof the temperature T, gas constant R and the Boltzmann constant k .

[4 marks]

p = 231

c = 231

cV M

m; ( =

mV

M ) [1]

pV m = 231 c M ; M = mass of 1 mole of gas

But pV m = RT

2

31 c M = RT [1]


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ONFIDENTIAL*

2

21

c M = 23

RT

2

21

cm N A =23 RT ; ( M = N A m ) [1]

So the average kinetic energy for a molecule of the gas is

2

21

cm = T N R

A23

= kT 23

where k = A

N

R[1]

c) Calculate the total kinetic energy for a molecule of monatomic gas attemperature 400K [2 marks]

Ek = 3/2 kT

= 3/2 (1.38x10 -23)(400)

= 8.28x10 -21 J

1

1


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ONFIDENTIAL*

Sketch on the same axes two graphs to show the distribution of molecularspeed for a gas at two different temperatures T 1 and T 2 where T 2 > T 1.

Label the position of the mean speed vm , the root mean square speed v andthe most probable speed vo on each graph. [3 marks]

Compare the two graphs and explain the physical characteristic of the twospeed distributions by giving reasons for your answer [2 marks]

The area under the graphs T 1 and T 2 are the same because the number of

molecules remains the same.The value of vo, vm and v are higher for higher temperature because thekinetic energy of the gas molecules increases as the temperature increases.Both graphs start from origin because the minimum speed at anytemperature is zero and no upper limit. Any two [2]

T1

T2

0 vovm v vovm v S eed

Graph T 1 and T 2 correct and labeled [1]

All the speeds are correctly labeled for T 1 [1]

All the speed are correctly labeled for T 2 [1]

Origin not labeled -1

Not starting from origin -1 axes not labeled -1


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ONFIDENTIAL*

2. (a) Two thin conducting plates have an area of 0.50 m 2 each. They arelaced parallel to each other and 20 mm apart. One plate is maintained at +75 Vhile the other at - 75 V by a d.c. supply.

(i) Determine the amount of charge stored on each plate. [3 marks]

[1]

Q = CV

= 2.212 x 10 -10 x [75-(-75)] ( V = 150 V) [1]

= 3.318 x10 -8 C /3.3x10 -8 C [1]

(ii) Calculate the energy per unit volume stored in the electric field betweenhe plates. [2 marks]

W = CV 2 or W = QV or W = Q 2/C [1]

Energy per unit volume =

=

= 2.489 x or 2.5x10 -4 J m -3 [1]


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ONFIDENTIAL*

(b) A parallel-plate capacitor of capacitance 3.0 and a resistor of largeesistance 1.0M are connected in series with a 12 V battery having negligiblenternal resistance.

E

(i) Sketch graphs to show the time variation of th e voltage across thecapacitor and the current through it when switch S is close. Labelclearly the maximum voltage and current on the respective graphs

[4 marks]

V /V

12

t /s I/ A

1.2x10 -5

t /s

0

0

Graph correct [1]

Value correct [1]

No label for axes -1

No origin -1

Graph correct [1]

Value correct [1]

No label for axes -1

No origin -1


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ONFIDENTIAL*

(ii)What is the time constant of the circuit. [1mark]

= 3.0 s [1]

(iii) Calculate the time taken for the capacitor to be 50% charged. [2marks]

2ln21

RC t

eQQ

eQQ

RC t

oo

RC t

o

t = 2.08 s or 2.1 s

(iv) With the switch S closed, the plates of the capacitor are pulled slightlyapart. Describe the possible changes in the potential difference and theenergy store in the capacitor.

[3marks]The capacitance of the capacitor decre ases because d increases [1]

The charges on the capacitor decrease s /the potential difference remainconstant. [1]

The energy stored decreases. [1]

1

1


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ONFIDENTIAL*

3 (a) The X-rays production process is the inverse of that of thephotoelectric effect. Explain this statement. [2 marks]

In the photoelectric effect, electrons are emitted from the metal surface when it isilluminated by light, that is phot on. [1]

In an X-rays tube, electromagnetic radiation, that is photons, are emitted from themetal surface when the metal is hit by electrons. [1]

(b) The graph shows the X-rays spectrum produced by a X-ray tube.

(i) What is the effect of increasing the anode voltage to the characteristicX-rays and continuous X-rays spectra? [2marks]

Wavelength of the characteristic lines remains unchanged . [1]

Minimum wavelength of the continuous spectrum decreases but theintensity increases. [1]


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ONFIDENTIAL*

(ii) What is the effect of increasing the filament curre nt to the continuousand characteristic X-ray spectra? [2marks]

The positions of the characteristics peaks remains unchanged [1]

min remains unchanged, but intensity increases for all wavelength. [1]

or show in graph :


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ONFIDENTIAL*

(d) A beam of X-rays of wavelength 0.154 nm is incident on a crystal of interplanar spacing 0.288 nm.

(i) Calculate the maximum order of diffracted X -ray beams that can bedetected.

2 d sin = n

Max order = 3

(ii) Calculate the Braggs angle for the highest order of diffraction.[5 marks]

From Braggs equation , 2dsin = n , n = 3

[1] [1]

1

1

1


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ONFIDENTIAL*

ii. carbon-12 nucleus [2 marks]

6 M p + 6 M n M c + 6 M e [1]

= 0.10223 u

Total binding energy = 0.10223x1.66x10 -27x(3x10 8)2

= 95.46 MeV

Binding energy per nucleon = 7.955 MeV [1]

Why is the uranium-238 nucleus less stable than carbon -12 nucleus?[1 mark]

Since the binding energy per nucleon of uranium-238 is less than the bindingenergy per nucleon of carbon-12. Therefore uranium-238 is less stable thancarbon-12. [1]


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(c) An element of unknown mass is mixed with in a mass spectrometer.The radii of curvature of (the path of the ions) of the element and ionsof are 26.2 cm and 22.4 cm respectively. What is the element? Stateany assumption made. [6 marks]

[1]

Assuming that the ions of the element have the same charge andsame velocity as the ions of , [1] , [1]

[1]

A = 14 [1]

The element is nitrogen [1]

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TEKNIK OF ANSWERING QUESTION