Download 1. (a) 0.1 ´ 10 = k ´ 0.05 - PLK Vicwood KT Chong Sixth Form College

97’ AL Physics/Structural Questions/Marking/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
97’ AL Physics: Structural Questions
Marking Scheme
1.
(a) T1 = k(e + x)
and
Equation of motion :
T2 = k(e - x)
1
-(T1 - T2) = m x
m x + 2kx = 0
or m x + 20x = 0


as x  -x,
 s.h.m.
1
1
1
4
1
1
(b) (i)
potential energy
0
(ii)
T
As
2.
0.2
0.4
0.6
0.8
time / s
= 0.4 s
1
2k
2 =
m
2 2
2k
(
) =
T
m
2 2
20
(
) =
0.4
m
m = 0.08 kg
1
1
1
4
(c) (i) unchanged
1
(ii) unchanged
1
2
1
1
2
2
(a) Microwaves/radiowaves
(b) (i)
plate
source
S’
97’ AL Physics/Structural Questions/Marking/P.2
(ii)
plate
source
1
Waves are diffracted at the edge of the plate and therefore reach the shadow
behind the plate.
(c) (i) Speed of S’ = 40 ms-1
u
f = f’ - f =
f
c
40
=
 1010
3  108
= 1300 Hz
(a) (i)
(ii)
1
1
T = ke
= 9.6  102 (2  0.01)
= 60.3 N
Work done = energy stored
= ½ (9.6  102) (2  0.01)2
= 1.89 J
(b) (i) Consider the wave along AB, v =
v =f
184 = f (0.36  2)
f = 256 Hz
(ii) - transverse Vs longitudinal
- stationary Vs travelling
- different speeds/wavelengths
3
1
1
2
1
1
2
1
1
2
T
m
60.3
6.4  10 4 / 0.36
= 184 ms-1
=
So
2
1
(ii) The change in frequency (i.e. the beat frequency) is very small (~10 3) compared
with the frequency of the waves (~1010), therefore if the frequency of the reflected
waves is to be measured, the instrument should be accurate at least up to the 7th
digit.
3.
1
1
1
1
3
1
1
1
3
97’ AL Physics/Structural Questions/Marking/P.3
(iii)
A
4.
B
(a) (i) The CRO of practically infinite resistance reads the e.m.f. of the cells (4.5 V).
However, the voltmeter only reads the p.d. across PQ as there is current flowing in
the circuit, some of the p.d. drops across the 10-k resistor.
(ii) 4.5 
Rv
Rv  10
= 4.1
1
1
1
1
1
3
1
Rv = 102.5 k
1
Reading: 4.5 V
1
3
1
1
1
1
4
1
1
2
(b) (i) Let r be the resistance of the coil and IC its current
Consider the shunt,
For the coil,
(ii)
0.1
IC
0.1
r
= (50  10-6 - IC)(9.8  103 + 200)
= 40 A
= 40  10-6 r
= 2.5 k
(10 - 2.5) = 50  106  R4
R4 = 150 k
(iii)
20
30
10
40
0
50
coil
R1
R2
2.5 mA
R3
50  A
S
0.1 V
R4
2.5 V
10 V
X
input terminals
+
_
2
Short circuit the input terminals and adjust the rheostat until the pointer indicates
full-scale deflection to the right.
1
3
97’ AL Physics/Structural Questions/Marking/P.4
5.
(a) The back e.m.f. induced in the inductor per unit rate of change of current.
dI
is produced by the
dt
inductor, therefore only part of the applied p.d. is used for driving the current.
dI
(V = IR + L
)
dt
(b) (i) When there is a change of current, back e.m.f. b =  L
(ii) At t = 0 s,
V = L
9 = L
dI
dt
10  10 3
1
1
1
2
1
2  10  3
L = 1.8 H
(iii) At steady state,
1
V = IR
9 = 500  10-3 R
R = 18 
1
2
1
1
2
(iv)
current/mA
B
500
B’
A’
10
C
C’
2
A
O
0
2
2500
3000
time/ms
2
(c) (i) The energy stored in the magnetic field of the inductor.
(ii)
6.
½LI2 = ½CV2
1.8(500  10-3)2 = C (350)2
C = 3.6 F
(a) (i) The electric field of the waves vibrates on certain planes; vertical planes
containing the aerial in this case.
1
1
1
1
1
3
1
1
(ii) When the antenna is vertical i.e. parallel to the aerial, the signal received is
maximum; however it decreases to a minimum when the antenna is rotated till it
is horizontal. The plane-polarized nature of the waves is demonstrated, implying
that the waves must be transverse.
1
1
1
(iii) The waves reflected from the plate to the antenna suffers a phase change of ,
destructive interference occurs between the direct and reflected waves reaching the
antenna. Therefore the microammeter reading decreases.
1
1
1
3
1
1
2
(b) (i) When alternating currents (of various frequencies) flow in coil L’, e.m.f.’s of the
same frequencies can be induced in L by mutual induction.
3
97’ AL Physics/Structural Questions/Marking/P.5
7.
(ii) Although the currents in the aerial coil induce currents of various frequencies in
coil L, only the current with frequency equals the resonant frequency of the LC
circuit can develop a large p.d. (at that frequency) across C.
1
1
2
(iii) Resistance of coil L can be minimized so that the reception is better as the
resonant current at the wanted frequency increases.
1
1
2
(a) (i) As the capacitor charges up (Q ), voltage across it increases (VC =
the e.m.f. for driving the current decreases, so
V  VC
=I
R
Q
) therefore
C
(ii) The charges stored in the capacitor.
1
1
2
1
1
4
4
(iii)
I / mA
80
70
60
50
40
30
20
I
II
10
0
1
2
3
4
5
t /s
(b) (i)
potential / kV
1.0
0.5
0
3
P
Q
R
S
T
position
-0.5
-1.0
3
(ii) The strip deflects to the right and the extent of deflection remains unchanged
when moving the strip between the plates.
This is because the electric field is constant between the plates.
1
1
1
3
(iii) unchanged.
1
1
97’ AL Physics/Structural Questions/Marking/P.6
8.
(a) (i)
High frequency
alternating p.d.
+V
-V
narrow gap of
negligible width
0V
1
1
(ii)
mv 2
r
qBr
v =
m
As the force on the proton is perpendicular to its motion, no work is done on the
proton.
qvB =
(iii) Time
=
r
v
= (
1
1
3
1
m
)
qB
which is constant
(b) (i) 2 (e  10 kV) = 20 keV
1 MeV
= 50 rev
20 keV
m
= 50 (2 
)
qB
(ii) No. of revolutions required =
Time required
= 100
1
2
1
1
1
(by (a)(iii))
1
  166
.  10 27
160
.  10 19  15
.
= 2.17  10-6 s
9.
1
(a) Vout = A0(V+ - V-)
15
V+ - V- =
= 1.5  10-4 V or 150 V
10 5
1
3
1
1
(b) (i)
Vin
Vout
2
0V
2
97’ AL Physics/Structural Questions/Marking/P.7
(ii) As the open-loop gain is very large (infinite for an ideal op amp), the two input
terminals are nearly at the same potential
i.e.
Vin = V+  V- = Vout
- used as a buffer between a high impedance (low current) circuit and a low
impedance (high current) circuit (e.g. electrometer)
(c) (i) When it is dark, most of the p.d. drops across the LDR as its resistance is higher
than that of the 50-k rheostat, so VY < VX and LED is off as Vout is negative.
When it is bright, resistance of LDR decreases, p.d. across the 50-k rheostat
increases until VY > VX, LED is on as Vout becomes positive.
(ii) 5 V
resistance of LDR is 10 k
Rrheostat / 10 k
= 30 k / 15 k
Rrheostat = 20 k
1
1
1
1
1
1
3
1
1
1
10. (a) (i) When d  7 mm, absorption of  rays increases with d.
For d > 7 mm, nearly all  rays are absorbed but the aluminium plates have little or
no effect on  rays.
2
1
(iii) The penetrating power of -particles is very weak, most are stopped by the
absorber plates.
1
(b) (i) The ‘corrected’ count rates (N’) are obtained by deducting the background count
rate from the count rates (N) taken.
2
24.8
3.21
3
22.0
3.09
4
20.0
3.00
5
18.2
2.90
6
16.8
2.82
1
1
1
(ii)
1
28.0
3.33
3
1
1
(ii) 1.5  0.1 mm
d/mm
N’ / s-1
ln N’
3
7
15.7
2.75
1
1
97’ AL Physics/Structural Questions/Marking/P.8
ln N’
3.4
3.3
3.2
3.1
3
3.0
2.9
2.8
2.7
0
1
2
3
4
5
6
7
d/mm
4
(iii)
2.86  314
.
56
.  2.8
= -0.1 mm-1
Slope =
or
ln N’ = -0.1 d + 3.42
N’ = 31 e 0.1d
1
1
2