93 AL Physics/Essay Marking Scheme/P.4
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
93’ AL Physics: Essay
Marking Scheme
1. (a) Steady flow
- liquid elements which start at a given point always follow the same path 1
and have the same velocity at each point on the path. ½
Turbulent flow
- liquid elements which start at a given point take random paths 1
and their velocities vary in magnitude and direction ½ 3
(b) Consider a cross-section of the pipe,
the liquid layer touching the pipe wall is always stationary ½
due to adhesive force between the liquid molecules and pipe wall. ½
The velocity of liquid is greatest at the centre. ½
Internal friction exists between liquid layers with different velocities ½
because of intermolecular forces ½
so velocity falls off gradually as the pipe wall is approached ½ 3
(c) Liquid in contact with the bottom surface of the block moves at the block’s velocity v ½
Liquid in contact with the floor is stationary. ½
Thus a velocity gradient is set up in the liquid of thickness t. ½
So, force required = liquid frictional force ½
= h ´ Area ´ Velocity gradient
= ½
Assumptions :
- uniform thickness of liquid ½
- the liquid is Newtonian ½
- constant velocity gradient across the thickness of liquid ½
- only the liquid lying below the block moves (ignore edge effect) ½ 4
(d)
- The diameter of the tube is large compared with the diameters of ball-bearings 1
so that streamline conditions are satisfied. ½
- Marker A is far enough below the liquid surface 1
for the ball-bearing to have its terminal velocity at A ½
- Dip the ball-bearing in the liquid and thereby coated, before dropping so as 1
to reduce the chance of air bubbles adhering to the falling ball-bearing ½
- Avoid using ball-bearings of large radii as 1
their terminal velocities are high and vortices may form ½
- Release the ball-bearing at the centre of the tube 1
to reduce the effect of the wall of the tube on the streamlines ½
- Marker B is at a considerable distance from the bottom of the tube so as 1
to reduce the effect of the bottom of the tube on the streamlines ½ 6
2. (a) progressive wave stationary waves
- waveform advances as time goes on - waveform does not advance ½ + ½
- energy is transmitted along the direction - energy is confined within the region of the ½ + ½
of travel of the wave stationary wave
- particles within one wavelength have - all particles between two adjacent nodes ½ + ½
different phases are in phase
- all particles are vibrating - some particles (at nodes) have no vibration ½ + ½
- all vibrating particles have the same - different particles have different amplitudes, ½ + ½
amplitude in particular, amplitude is maximum at
anti-nodes
A stationary wave is formed when there is superposition of ½
two waves of nearly equal amplitude ½
and equal frequency ½
travelling in opposite directions ½ 6
(b) (i) At T1, the waves arrive in phase to produce a loud sound ½
Phase difference then increases between the waves due to different frequencies. ½
At T2, the waves are completely out of phase, little or no sound is heard. ½
Later at T3, the waves are in phase again and a loud note is heard. ½
½
1
(ii) Suppose the beat period = T, then in time T ½
number of cycles of f1 = f1T ½
number of cycles of f2 = f2T ½
Assume f1 greater than f2, then 1
f1T - f2T = 1
f1 - f2 = ½
\ beat frequency = = f1 - f2 ½ 3½
(c) Radar installed near the road sends microwaves of frequency f1 to a travelling car, then ½
the microwaves are reflected back to the radar. ½
Due to Doppler effect, ½
the observed frequency f2 of the reflected microwaves is slightly different from f1. ½
Hence by comparing the transmitted and reflected microwaves, beats are formed. ½
As the beat frequency (= f1 - f2) depends on the car speed, the car speed can be checked. ½ 3
3. (a) (i) average drift velocity: ~10-4 m/s ½
(ii) speed of electrical signal: ~ 108 m/s ½
In a current-carrying conductor, electrons tend to accelerate along the opposite
direction of the electric field inside. ½
Due to collisions between electrons and atoms (ions) of the conductor, ½
electrons move in zig-zag paths ½
and drift with small displacement in unit time. ½
1
Electric field travels at an extremely high speed in a circuit 1
so electrons at every point of the circuit are influenced by the electric field ½
nearly simultaneously as the switch is closed, ½ 5
electrical signal results at once.
(b) v = average drift velocity
e = electronic charge
A = area of cross-section of the wire 1
n = no. of conduction electrons per unit volume
Suppose electrons drift a distance l along a wire in time t, the charge q flows through a cross-section
of the wire is
q = nlAe ½
Drift velocity v = l/t ½
current in the wire i = q/t
= (nlAe)/(l/v) ½
= nAve 3
(c)
2½
2½
When a current flows through the meter, the coil experiences a couple, ½
the coil then turns until it is stopped by the increasing tension in the springs. ½
Thus the larger the current through the meter, the greater the forces on the coils, ½
and a greater angular deflection results. ½
To achieve a linear scale:
Set up a radial magnetic field ½
then flux density B of the field is nearly constant at the coil 1 The coil stops rotating when G coil = G spring ½ NBAI = kq (q = angular deflection) ½
hence I µ q ½
(N, B, A & k are constants) ½ 8
4. (a) : maximum kinetic energy of photoelectrons ½
hn : energy of incident photon ½
f : work function of metal ½
- the work required to remove an electron from the metal surface ½ 2
(b) (i)
½ + ½
Photoelectrons emerge with different speeds (or K.E.). ½
When V is positive, current is constant
because all photoelectrons can reach electrode D ½
Current falls when V is negative
because the less energetic photoelectrons cannot overcome the potential barrier. ½
When V reaches the stopping potential (V = -Vs) ½
even the most energetic photoelectrons are repelled back so no current flows. ½ 3½
(ii)
(I) increased light intensity
constant frequency ® same maximum K.E. ½
® Vs remains unchanged
intensity increased ® no. of photons increases ½
® no. of photoelectrons ejected increases ½
® current increases 3½
(II) increased light frequency
frequency increased ® energy of each photon increases ½
® maximum K.E. of photoelectrons increases ½
® magnitude of Vs increases
For the same intensity, if energy of each photon increases
® no. of photons decreases ½
® no. of photoelectrons ejected decreases ½
® current decreases 2
(c)
Light is focused on the ‘soundtrack’ of a moving film. ½
The ‘soundtrack’ varies the intensity of light falling on a photocell. ½
The photocell creates a varying current, ½
thus produces a voltage which is amplified for driving a loudspeaker. ½ 2
(d)
1
A beam of electrons strikes a thin film of graphite just beyond the anode. ½
A diffraction pattern, 1
consisting of concentric rings, is observed on the fluorescent screen ½
showing the wave-like behaviour of electrons. 3
5. (a) (i)
LRC in series ½
a.c. supply & milliammeter ½
Set up the above circuit,
set the signal generator output to a value, say 3 V, with a measurable current, ½
and increase the frequency stepwise from a low value, say 10 Hz, ½
check whether the output is constant at the previous setting, 3 V, ½
then record the corresponding current readings on the a.c. milliammeter, ½
when frequency increases, the current reading rises and then drops.
(ii) (iii)
f0 = resonant frequency ½
current I and voltage across resistor VR are in phase, ½
I = VR/R ½
½ mark For frequency < f0,
for V0 lags behind VR0 V lags behind VR,
½ mark so VR0 < V0 ½
for VC0 > VL0 \ current I0 = = Imax
½ mark For frequency = f0
for V0 , VR0 in phase V and VR in phase,
½ mark so VR0 = V0 ½
for VC0 = VL0 \ current I0 = Imax = ½
½ mark For frequency > f0,
for V0 leads VR0 V leads VR,
½ mark so VR0 < V0 ½ 11
for VC0 < VL0 \ current I0 = = Imax
(b) (i) Radio signals from different transmitting stations induce e.m.f.s ½
of various frequencies in the aerial, ½
which cause currents flowing in the aerial coil. ½
Then currents of the same frequencies are induced in coil L by mutual induction. ½
If C is adjusted so that ½
the resonant frequency of the LCR circuit equals the frequency of the wanted station, ½
a large current and p.d. at that frequency only will develop across C ½
This selected and amplified p.d. is then applied to the next stage of the receiver.
(ii) Use an inductor of high L/R ratio 1
so increases the resonant current and hence the voltage across C. ½ 5
6. (a) (i) (ii)
P.E. Curve
- shape correct ½
- max. at xc - A and xc + A ½
K.E. Curve
- shape correct ½
- zero at xc - A and xc + A ½
xc marked ½
In S.H.M., the total mechanical energy E is conserved ½
At maximum displacement, K.E. = 0, P.E. = Umax ½
\Total mechanical energy of the system
E = K.E. + P.E. = Umax (constant)
so K.E. = Umax - P.E. for xc - A £ x £ xc + A ½ 4
(b) (i) mass of each spring element, dM = 1
speed of spring element at a distance l from the fixed end, V1 = 1
\ kinetic energy of this spring element = ½ (dM) (V1)2
= ½
= ½
\ Total kinetic energy of the spring = ½
= ½
= 4
(ii) Extension of spring, y = x - xc ½
Total energy = K.E. of spring + K.E. of block + P.E. of spring
\ E = 1
½
; ½
and = 0 ½
\ 1
or
For S.H.M. ½
\ ½
and T = ½
= ½ 6
(c) Damped oscillation takes place, ½
energy of the system decreases, ½
amplitude of oscillation gradually decreases to zero, ½
the frequency is smaller and the period is longer ½ 2