84. Q.1



PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE

84’ AL Physics: Essay

Marking Scheme

1. (a) Any reasonable examples should be marked correct

e.g. (1) a ball thrown vertically between two heights h1 &

(h1 + h) will suffer a decrease of velocity

u1 u2. Since resistance of air can be

neglected,

decrease in K.E. = increase in P.E.

½mu12 - ½mu22 = mgh, where m is mass of body

and g is free-fall acceleration. 1

e.g. (2) However if body were thrown in a viscous medium -

such as water, mechanical energy would not be

conserved. In addition work done against this

opposing force would produce heating of the medium.

In fact loss of K.E. = gain of P.E. +

gain of internal (heat) energy. 2 3

(b)

Consider two tubes of areas of cross-section A1, A2 with

pressure at ends P1 and P2. Consider movements of fluid in

time t

work done x x’

W1 = P1A1 ( (v1t)

force distance moved

Work done against P2, y y’, W2 = P2A2v2t

Net work done on fluid, W = W1 - W2

= (P1A1v1 - P2A2v2)t

If fluid incompressible A1v1t = A2v2t

W = (P1 - P2)A1v1t

Gain of P.E. of fluid = mass of moved fluid ( (h2 - h1)g

Since we have equivalently moved (v1A1t) through

height (h2 - h1)

Gain of K.E. of fluid = K.E. of yy’ - K.E. of xx’

= ½(A2v2t)v22 - ½(A1v1t)v12

Gain K.E. = ½A1v1t(v22 - v12)

If conservation of mechanical energy holds good,

(P1 - P2)A1v1t = ½A1v1t (v22 - v12)

+ A1v1t g(h2 - h1)

P1 + h1g + ½v12 = P2 + h2g + ½v22 = a constant 5

(c) (i) The condition that the fluid is incompressible is

not valid for a gas. 1

(ii) (1) In a viscous liquid conservation of mechanical

energy would not hold good - since work has

to be done against the viscous force when fluid

is moved. 1

(2) The actual velocity of the liquid would vary

from zero on the sides of the tube to a

maximum along the axis. 1 3

d) (i)

Ball experiences a sideways force and motion due to

the unequal pressures on the opposite sides of the

ball. This follows from Bernoulli's equation:

At same height level P + ½v2 = a constant.

Thus where speed of air (v) is decreased force ( P)

is increased. 2

[pic]

(ii)

Gas pipe narrows down at P - increases the rate of

flow (v) of the gas. According to Bernoulli's

equation pressure in this region will be reduced and

so air will be sucked in producing a mixture of

gas and air for the Bunsen burner. 2 4

2. (a) (i) Thermometers possess a particular physical property

which varies with temperature e.g. pressure, 1

electrical resistance.

(ii) Values of this property are measured at two

reproducible temperatures :

(1) ice point, 0 °C - P0, say.

(2) water boiling point, 100 °C - P100, say. 1

(iii) It is, then, assumed that an intermediate temperature

is measured as ( = [pic] 1 3

(b) Expression in (iii) assume a linear relation between the

physical property and temperature. Relations may vary for

different thermometers. 1

e.g. compare gas thermometer/thermocouple

(i) thermocouple - linear relation gives temperature (e

while measured property (e.m.f.) P(t gives

temperature (c.

(ii) gas thermometer property P(g would, of course, give

temperature (c. 2 3

(Note : answers should indicate ‘quantitative’ effect of

the non-linearity.)

(c) (i) Use a resistance thermometer - large thermal capacity,

and cannot react rapidly enough to follow varying 1

temperatures.

1

No current through G, [pic] 1

Dummy leads at same temperatures as leads to R(

( compensate for changes in resistance as ( varies. 1 4

(Diagram (a) not essential.)

(ii) Use a thermocouple - small thermal capacity, and can

react rapidly (establishing thermal equilibrium) so 1

as to follow varying temperatures.

1 + 1

Better arrangement

Cu connecting wires at same temperature and therefore

on connection with potentiometer thermal e.m.f.s same

and cancel.

No current in G - length AC ( e.m.f. R must be high 1

resistance to give low p.d. AB 1 5

3. (a) The light from a gas laser differs from that a gas discharge

tube :

(i) light is monchromatic (of a single discrete frequency)

while that of a discharge tube exhibits a wide

spectrum of single frequency lines or bands, 1

characteristic of gas.

(ii) light is coherent across the laser beam (could be

broadened) i.e. waves in phase and theses could 1

produce interference.

(iii) Intensity of light is greater - concentrated in a 1 3

narrow beam, for a laser.

(b) (i) A Gas Discharge Tube

(1) Explain discrete energy levels. 1

(2) Excitation by electrical discharge (energy

transfer on collisions of ions). 1

(3) Emission on decay to ground state, frequency

given by e.g. h( = E1 - E0. 1

(4) Time taken before emission or probability of

emission depends upon atom/energy level but

also number of photons of energy h( = E1 - E0

available (to stimulate emission ) - 1

very few in discharge.

(5) The spontaneous emissions from the atoms occur

randomly and the individual waves emitted are 1 5

not in phase.

(ii) A Gas Laser

(1) Energy level excitation as in discharge tube.

(2) Difference is the large number of photons

present (light is reflected many times between

two end cavity mirrors) of energy h( = E1 - E0. 1

(3) These photons stimulate the emission of other

photons of energy h(, and these in turn 1

stimulate further emission etc. (chain-reaction).

(4) Since waves (wave-trains) are in phase there is 1 3

a large increase in output intensity.

(c) Any examples acceptable (provided they differ in field of

application)

(i) High power ruby laser - used to re-attach retina

to eye.

(1) blood-clot formed quickly by intense heat.

(2) localised due to small diameter of beam - 2

therefore no damage to surrounding eye.

(ii) Laser diodes used for telephone communications

(with optical fibres).

(1) coherent across beam - able to modulate and

(2) because of high frequency (f ~ 1015 Hz) can

accommodate a large umber of audio channels

(20 - 20 kHz).

(3) not affected by usual interference from radio 2 4

waves/mains.

4. (a) A Musical Note

(i) Propagation

Energy passed on from one region of air

(e.g. compression) to adjacent region in actual

direction of propagation (longitudinal 1

progressive wave), with actual air particles

vibrating with S.H.M. about their mean positions,

also longitudinally. N.B. for this type of wave

a medium is necessary. Distance between 1

instantaneous centres of compression or

rare-factions = wavelength (,

velocity, v = f(,

f being frequency.

In diagram C - compression

R - rarefaction

diagram/explanation 2 4

(ii) Frequency Range is 20 - 20,000 Hz (20 kHz) (audio) 1 1

(iii) Detection/Intensity Measurement

1

Microphone converts sound into a.c. electrical

signal (of same f), this signal is rectified and

produces steady meter reading indicating intensity

of sound wave. Allowance may be needed to

compensate for frequency responses of microphone

and audio amplifier. 1 2

(b) V.H.F. Radio Transmission

(i) Propagation

1

Waves are :

(1) polarised 1

(2) transverse (variations of in-phase electric and

magnetic fields ( to direction of propagation). 1

In diagram wave moves to the right with time, similar

to the sound wave given previously. N.B. for this

type of wave no medium is necessary - the changing 1 4

electric field gives rise to a changing magnetic field

and vice-versa, can take place in free space.

(ii) Frequency Range is 80 - 100 MHz, (1 MHz = 106 Hz) 1 1

(iii) Detection/Intensity Measurement

1

Voltage (a.c.) induced in an aerial (1) tuned to

V.H.F. band (particular length) and (2) orientated 1

// to E vector of wave (i.e. correct polarisation).

Tuned circuit (L/C // circuit) used to select

particular frequency from other radio waves. 1 3

N.B. do not expect discussion of use of local

oscillator/intermediate frequency amplifier.

5. (a)

1

(i) There is a difference in actual mass between a

nucleus and the individual constituent protons

and neutrons.

(ii) This mass defect (m has an energy equivalent 1

(E = (mc2) which is called the Binding Energy

(energy released if nucleus splits into

constituents.)

(iii) [pic] 1

(iv) Heat produced can be transferred to produce steam

which can be used to drive a turbine from electrical 1 4

power generation.

(b) At present only fission possible. In fusion large

temperatures (108 - 109 K) required to bring fusing nuclei 1

together against their electrostatic repulsion forces -

two difficulties :

(i) achieving high density/temperature for sufficient 1

time.

(ii) preventing hot plasma from touching container. 1 3

(c)

[pic]

3

(i) To increase probability of neutron capture, they

are slowed down by the moderator - water/graphite. 1

(ii) To stabilise reaction rate control rods of neutron

absorbing rods of boron coated steel are used i.e. 1

on average only one neutron (slightly more) produced

at each fission.

(iii) Heat produced (K.E. of products) conducted away by

coolant - e.g. water.

Steam is produced by heat exchange with circulating 1 6

coolant - drives turbine to produce electrical power.

(d) Precautions against hazards (any 2)

(i) slowing down reactor by further insertion of control

rods to prevent ‘run-away’.

(ii) concrete shield to give protection from neutrons

and (-rays.

(iii) water heated to steam indirectly through heat

exchange system (less chance of radio-active leakage).

(iv) all system remotely controlled - no direct handling

of controls by personnel. 2 2

6. (a)

On closing key K

(i) back e.m.f., [pic] 1

(ii) p.d. across R, IR both act in opposite directions to e.m.f. E 1

Applying Kirchoff Law to circuit,

E - [pic]- IR = 0 1 3

(b)

Since there are N turns in series, induced e.m.f. at any

inclination of coil to B, θ, is Eind = [pic] 2

(normal component of magnetic flux through coil.)

Since θ = (t, Eind = BAN( sin (t 1 3

(c) Following the procedure of (a) Eind replaces E,

i.e. BAN( sin (t - [pic]- IR = 0 1

Substituting I = P cos (t + Q sin (t, we obtain

BAN( sin (t - L(-(P sin (t + (Q cos (t)

- R(P cos (t + Q sin (t) = 0

i.e. sin (t (BAN( + (LP - RQ) + cos (t (-(LQ - RP) = 0

This equation must be satisfied for all values of t. 1

(i) t = 0, (LQ + RP = 0

(ii) t = π/(2(), BAN( + (LP - RQ = 0 1

Solving for Q, equation (ii) becomes Q(R + (2L2/R) = BAN(

and Q = BAN (R /(R2 + (2L2) 1

Also P = - (LQ/R = -BAN (2L/(R2 + (2L2) 1 5

Expression for current, I = [pic][pic]

(d) Instantaneous power loss due to Joule heating = I2R,

say Pdissipated.

Substituting, Pdissipated = [pic]((2L2cos2(t - 2(LR cos (t sin (t+ R2sin2(t) 1

Considering the time average (over a complete cycle)

term by term, cos2(t = ½(1 + cos 2(t) = ½

cos (t sin (t = ½ sin 2(t = 0 1

sin2(t = ½(1 - cos 2(t) = ½

Thus Pdissipated = [pic] 1

If (L >> R, Pdissipated B2A2N2R/ 2L2 i.e. independent of (. 1 4

-----------------------

emission

excitation

ground state

(atom most stable)

E0

E1

E2

E(

microphone

audio amplifier

rectifier

meter

sound wave

correct polarised aerial

tuned circuit

amplifier

rectifier

meter

E.M. wave

K

E

I

back e.m.f.

p.d.

( = (t

(

B cos (

B

B

E

E

v

B

v

(

[pic]

[pic]

[pic]

[pic]

Binding energy per nucleon / MeV

FISSION

FUSION

Mass number

9

8

7

6

5

4

3

2

1

240

220

200

180

160

140

120

100

80

60

40

20

0

( (C

0 (C

E1

R2

R1

B

C

A

R

G

G

Cu

Cu

Pt-Rh

Pt

unknown temperature t

ice water

ice water

to potentiometer

R(

(b)

(a)

G

Dummy leads

S

Q

P

Mica spacers

Silica tube

mica

Platinum wire

Dummy leads

Leads to platinum wire

[pic]

[pic]

[pic]

(ii)

(i)

qðe

qðc

θe

θc

θ

100

0

Motion of ball

Force on ball

Speed of air flow increases pressure reduced

Spin

Speed of air flow decreases pressure increased

h2

p2A2

Area A2

v2(t

v2

h1

p1A1

Area A1

v1(t

v1

x’

x

y’

y

velocity v

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