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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