H
NANYANG JUNIOR COLLEGE
JC 2 PRELIMINARY EXAMINATION
Higher 2
PHYSICS
9745/01
28 September 2009
Paper 1 Multiple Choice
1 hour 15 minutes
Additional Materials:
Multiple Choice Answer Sheet
READ THESE INSTRUCTIONS FIRST
Write in soft pencil.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Write your name, class and tutor’s name on the Answer Sheet in the spaces provided unless this has been
done for you.
There are forty questions on this paper. Answer all questions. For each question there are four possible
answers A, B, C and D.
Choose the one you consider correct and record you choice in soft pencil on the separate Answer Sheet.
Read the instructions on the Answer Sheet very carefully.
Each correct answer will score one mark. A mark will not be deducted for a wrong answer.
Any rough working should be done in this booklet.
This document consists of 16 printed pages.
[Turn over
H2 Physics Papers, Notes at alevelphysics.co
2
DATA AND FORMULAE
Data
8
–1
speed of light in free space
c
=
3.00 x 10 m s
permeability of free space
µo
=
4π x 10 H m
permittivity of free space
εo
=
8.85 x 10
–7
–1
–12
–1
Fm
–9
–1
(1/(36π)) x 10 F m
e
elementary charge
h
the Planck constant
u
unified atomic mass constant
me
rest mass of electron
mp
rest mass of proton
=
=
=
=
=
–19
1.60 x 10
C
–34
6.63 x 10
Js
–27
1.66 x 10
kg
–31
9.11 x 10
kg
–27
1.67 x 10
–1
kg
–1
R
=
8.31 J K mol
NA
=
6.02 x 10 mol
the Boltzmann constant
k
=
1.38 x 10
gravitational constant
G
=
6.67 x 10
acceleration of free fall
g
=
9.81 m s
s
=
ut + ½ at
2
=
u + 2as
W
=
pΔV
hydrostatic pressure
p
=
ρgh
gravitational potential
φ
=
–Gm/r
displacement of particle in s.h.m.
x
=
xosin ωt
velocity of particle in s.h.m.
v
=
vocos ωt
=
± ω (xo − x 2 )
R
=
R1 + R2 + ....
1/R
=
1/R1 + 1/R2 + ....
electric potential
V
=
Q/4πεor
alternating current/voltage
x
=
xo sin ωt
transmission coefficient
T
=
exp(–2kd)
where k
=
radioactive decay
x
=
xo exp(–λt)
decay constant
λ
=
0.693
t 1/ 2
molar gas constant
the Avogadro constant
23
–1
–23
–1
JK
–11
2
–2
N m kg
–2
Formulae
uniformly accelerated motion
v
work done on/by a gas
resistors in series
resistors in parallel
NYJC 2009
2
2
2
8π 2 m(U − E )
h2
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H2 Physics Papers, Notes at alevelphysics.co
3
1
In physics, thermal conductivity, k, is the property of a material that indicates
its ability to conduct heat. Rate of heat flow of heat energy, dQ , can be found
dt
using the equation
(T - T )
dQ
=kA 2 1
dt
L
where k is the thermal conductivity, A is the
total cross sectional area of conducting surface, T2 – T1 is temperature
difference across the surface, and L is the thickness of conducting surface
separating the 2 temperatures. What is the SI unit for thermal conductivity?
A
W m-1 K-1
Answer: A
B
W m-2 K-2
C
W m-3 K-1
D
J m-2 K-1
 dQ 1

L
k  =  dt A (T - T ) 

2
1 
1 m

= W 2 
 m k
= W m−1 K −1
2
The velocity of a liquid in a pipe can be calculated by measuring the force on a
small disc placed in the centre of the pipe with its plane perpendicular to the
flow. The equation relating the force to the velocity
is
force =constant x (velocity)2 .
If the velocity is to be found with a maximum uncertainty of 1%, what is the
maximum permissible uncertainty in measuring the force?
A
0.25%
Answer D
3
B
0.5%
F = k v2
∆F
∆v
=2
F
v
C
1%
= (2) (1%)
D
2%
=2%
A boy holding a ball is standing on the floor of a lift. The lift starts to accelerate
from rest at 6.5 m s-2 upwards. After 2.0 s, he releases the ball at 1.4 m away
from the floor of the lift. At the same time, the lift stops accelerating and travels
at a constant speed. Find the time taken for the ball to reach the floor of the lift.
A
0.53 s
B
0.92 s
C
2.5 s
D
2.8 s
Ans: A
After 2.0s, speed of lift :
v = u + at
= 0 + 6.5 (2.0)
= 13 m s-2
Distance moved by ball = sball = u t + ½ a t2 = −13t + ½ (9.8) t2
(downwards +)
Distance moved by lift = slift = u t = 13t (upwards +)
sball + slift = 1.4  −13t + 4.9t2 + 13t = 1.4  t = 0.53 s
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4
An archer is firing an arrow towards a target, at a distance of 40 m away from
him. He fires horizontally, and the arrow hits a point 10.0 mm below point X. If
the bull’s eye is located 5.0 mm below point X, how far away from him should
the target be placed such that the arrow will hit the bull’s eye? Assume that he
fires with the same velocity and that air resistance can be neglected.
X
10 mm
point of impact
40 m
A
B
10 m
Let the initial velocity be u.
In y dir: y = ½ gt2
Hence, y ∝ x 2
5
;
C
20 m
In x dir: x = ut
y=
D
25 m
;
t=
28 m
x
u
g 2
x
2u 2
y2
x 2
= 22
y1
x1
x 2
5
= 22
10
40
x2 = 28m
A man is parachuting at terminal speed towards the surface of the Earth. The
force which, according to Newton’s third law, makes an action-reaction pair with
the downward force on the man is
A the tension in the harness of the parachute.
B the viscous force of the man and his parachute on the air.
C the gravitational force on the Earth due to the man .
D the viscous force of the air on the man and his parachute.
The forces acting on the falling man are the downward gravitational force on the
man by the earth, the upward air resistance or viscous force by the moving air,
the upward tension on the man by the harness of the parachute. By Newton’s 3rd
law of motion, when A exerts a force(action) on B, B exerts and equal but
opposite force(reaction) on A. The only downward force on man is exerted by
the earth. The action-reaction pair refer to the forces between the man and the
earth. Answer: C
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6
A stationary thoron nucleus (A = 220, Z =90) emits an α particle with kinetic
energy Eα. Which is the kinetic energy of the recoiling nucleus?
A
B
Eα /108
C
Eα /110
Ans: C
By cons. of momentum : M.0
KE of X = ½ mx v2 = ½ mx ( -
7
mα mx
)
mx 2
Eα /55
= mx v + mα Vα
v = -
= ½ mα Vα2 (
D
Eα /54
4
2
mα Vα
mx
Vα
220
Th
mα Vα 2
)
mx
= Eα (
He
216
X
4
) = Eα /54
216
V
A 100 g block of width 8.0 cm and height 20.0 cm rests on a rough inclined
plane. The centre of gravity of the block is located 6.0 cm above its base at the
centre of its cross-section. The angle of inclination of the plane is gradually
increased until the block topples.
8.0 cm
20.0 cm
block
8.0 cm
rough plane
Which of the following is the largest possible angle at which this occurs (such
that the block loses equilibrium by sliding)?
A
22o
B
27o
C
30o
D
45o
tan θ = 4.0 / 8.0
θ = 27o
θ
Ans: B
N
F
W
θ
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8
A mass is attached to two springs as shown (not drawn to scale). The springs
A and B have spring constants k and 3k respectively. Forces are applied to the
springs such that the free end of A is displaced x to the left and that of B is
displaced y to the right, but there is no change in the displacement of the mass.
mass
spring A
x
spring B
k
y
3k
horizontal plane
Which of the following is equal to y?
A
1/3 x
B
Tension in A = Tension in B
Ans: A
9
C
x
;
D
3/2 x
k x = 3k y
3x
Solving: y = 1/3 x
A spacecraft of mass m re-enters the Earth’s atmosphere at an angle of θ to the
horizontal. Due to air resistance, the vehicle travels at a constant speed v. The
heat-shield of the vehicle dissipates heat at a rate P, so that the mean
temperature of the vehicle remains constant. Taking g as the acceleration of
free fall, find an expression for P.
A
½ mv2
B
mgv
C
mgvsinθ
D
mgv2 sin2 θ
Ans: C
θ
Since KE remain constant, by conservation of energy,
Rate of loss in GPE = Rate of energy dissipated as heat
mgh/t = P
mgvsinθ = P (where h/t = vsinθ)
10
v
Calculate the total work done by a person when he carries a load of 2 kg and
walked for 20 m, then raises it by a vertical distance of 0.5 m, finally throwing it
with a velocity of 15 m s-1 at 40o to the horizontal. Take g = 10 m s-2 .
A
143 J
B
235 J
C
275 J
D
635 J
Ans: B
Work done in walking 20 m with the load = 0
(since support force exerted on load is vertically upwards and displacement is
horizontal, WD = Fdcosθ = 0 as θ= 900)
Work in raising the load 0.5m vertically = gain in GPE of load
= 2 (10) (0.5)
= 10 J
Work done in throwing the load = gain in KE of the load = ½ (2)(152) = 225 J
Hence, total work done = 0 + 10 + 225 = 235 J
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11
In a frictionless roller coaster ride a car of mass m, starts from rest at the top of
the first hill at point A. What will be the magnitude of the force of car on the hill
when it reaches the top of the second hill at point B? The radius of curvature of
the hill at point B is r.
A
 2(h1 − h2 ) 
mg 1 −

r


C
2mgh2
r
B
 2(h2 − h3 ) 
mg 1 −

r


D
2mg(h2 − h1 )
r
Ans : A
By COE from point A to B, ∆Ep = ∆Ek
mg(h1 – h2) = ½ m v2 ;
m v2 = 2 mg(h1 – h2)
ΣF => W – R = m v2 / r

R = W - m v2 / r = mg - m v2 / r = mg 1 −

12
2(h1 − h2 ) 

r

In the Rutherford model of the hydrogen atom, an electron of mass m and
charge -e performs uniform circular motion with angular velocity w about a
stationary proton with charge, e. If the radius of the circular orbit is r, which of
the following statement is correct?
A The force acting on the electron does not depend on r.
B The instantaneous force is always perpendicular to the instantaneous
velocity.
C The electron experiences uniform acceleration.
D The electron experiences no acceleration.
Ans : B
The electric force of attraction between nucleus and electron provides the
centripetal force.
e2
= mω2r
2
4πεor
The instantaneous force is always perpendicular to the instantaneous
velocity and causes acceleration to point towards the centre.
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13
An artificial satellite is orbiting around the Earth at a steady speed. The
astronaut in the satellite can be regarded as weightless because
A
the gravitational force acting on him is zero
B
the centripetal force he experienced is zero.
C
the satellite’s acceleration is the same as his.
D
The acceleration he experienced is zero.
answer C
When the satellite and the astronaut are both orbiting the Earth at the same
speed, they possess similar centripetal acceleration (assuming r is the same).
As such, the only net force acting on them is the gravitational force . For the
case of the astronaut, there is no reaction force acting on him by the satellite,
hence he feels weightless.
14
The table below gives the gravitational potential values at various points in the
field of a planet.
Distance from surface / km
Potential / kJ kg-1
0
- 544.6
220
- 369.2
230
- 345.8
240
Infinity
- 322.5
0
What is the gravitational acceleration at a height of 230 km on this planet?
A
1.50 m s-2
2.34 m s-2
C
3.46 m s-2
D
9.81 m s-2
dV
−322.5 − ( −369.2) 46.7
g=
2.34 m s −2
−
=
−
= =
dr
240 − 220
20
answer B
15
B
A mass is attached to the end of a vertical spring. It is given a downward
displacement of 3.0 cm from its equilibrium position and then released. The
mass is found to oscillate with period of 0.50s. What is the maximum speed of
the oscillation?
A
0.21 m s-1
answer B
NYJC 2009
B
vmax = ω A
0.38 m s-1
= (2π/T) A
C
0.53 m s-1
D
0.88 m s-1
= (2π/0.50) x 0.03 = 0.38 m s-1
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9
16
A pendulum is driven by a periodic driving force of frequency f. The variation of
pendulum’s amplitude with driving frequency was determined for experiments
done in a container of fixed volume of gas, and at two temperatures T1 and T2.
When T2 > T1, determine which of the following graphs best illustrates how
amplitude of the pendulum changes with the driving frequency.
Amplitude
A
Amplitude
B
T2
A
T1
B
T2
T1
Driving frequency
C
Amplitude
Driving frequency
Amplitude
D
T2
T1
C
D
T1
T2
Driving frequency
Driving frequency
Ans: B
At higher temperature and with a fixed volume of gas, the randomly
distributed KE of the gas molecules will be higher and will increase the
damping effect on the pendulum’s motion. B shows the correct graph of how
amplitude varies with driving frequency when higher damping occurs at
temperature T2.
17
Which statement about internal energy is correct?
A
B
C
D
The internal energy of a system can be increased without transfer of
energy by heating.
The internal energy of a system depends only on its temperature.
When two ideal gas systems have equal internal energies, they are in
thermal equilibrium.
When work is done on a gas, its internal energy always rises.
answer A
A - Without transfer of heat i.e. Q = 0 , ∆U = W + Q = W + 0 . If W is ‘+’ i.e.
work done on the gas eg in compression, then ∆U is ‘+’ and internal energy
increases.
B – Internal energy depends on temperature only if the system is an ideal gas;
otherwise internal energy depends also on the volume i.e the separation of
the molecules which affects its potential energy.
C – U = 3NkT/2 . U depends on N and T. Two systems having the same U
have the same product of NT. Thus their temperatures are equal (systems
are in thermal equilibrium) if they have the same number of molecules.
D- ∆U = W + Q . ∆U may not be ‘+’ when W is ‘+’ as Q can be numerically
more ‘-’ than W is ‘+’ such that the net value is ‘-’ .
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18
An ideal gas undergoes an expansion at a constant pressure of 1.3 x 105 Pa
from a volume of 1.3 x 10-4 m3 to 3.6 x 10-4 m3 . During this expansion, 75 J of
heat is supplied to the gas. Its pressure is then increased to 3.6 x 105 Pa at
constant volume. After which, it is compressed to its original volume at
constant pressure. Which statement about the gas is correct?
A
The work done by the gas is 83 J.
B
Its internal energy increases by 6 J during the expansion.
C
The work done by the gas is 30 J during the expansion.
D
The net work done on the gas is 30 J.
p
Answer C
Work done = p ∆V
A – The work done ON the gas is 83 J.
B – During expansion, the work done by gas is 30 J and the
heat supplied is 75 J , hence ∆U = 75 – 30 = 45 J.
C – The work done by the gas is 30 J during the expansion.
D - The net work done on the gas = work done on the gas
- work done by the gas = 83 J - 30 J = 53 J
E
F
H
G
V
19
A point source of sound emits energy equally in all directions at a constant rate
and a person 8 m from the source listens. After a while, the intensity of the
source is doubled. If the person wishes the sound to seem as loud as before,
how far should he be standing now?
B 2 2 m
A 2m
C 8 2 m
Initially, I = k
Answer : C
D 2 8 m
a2
r2
( a 2 )2
r2
a2
( a 2 )2
For intensity to be same at distance r1 from source; k
=k 2
r
r12
When intensity is doubled at the same distance, I1 = k
r1 = 8 2
20
The distance between 2 points of a progressive transverse wave having a
phase difference of π is 40 cm. If the frequency of the wave is 300 Hz,
3
what is the speed of the wave?
A
360 m s-1
B
720 m s-1
Answer : B
Since phase difference of
π
3
C
36000 m s-1
D
72000 m s-1
= 40cm, the wavelength = 40 x 6 = 2.40 m
Thus, speed = ƒ λ = 300 x 2.40 = 720 ms-1
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21
Two loudspeakers L1 and L2, driven by a common oscillator and amplifier, are
set up as shown. As the frequency of the oscillator increases from zero, the
detector at D recorded a series of maximum and minimum signals. At what
frequency is the first maximum observed? (Speed of sound = 330 m s-1)
D
A
165 Hz
B
330 Hz
C
495 Hz
D
660 Hz
Answer B
Path difference = 1m
For maximum (constructive interference) to occur, path diff = nλ
n
330
=1
f
The frequency increases from zero so the first maximum occurs at the when
n=1
(1)
330
=1
f
f = 330 Hz
22
A diffraction grating is placed at the centre of a circular 0 – 360o scale. A beam
of laser light is incident normally on the grating. The zero order maximum
occurs at a scale reading of 180o and a first order maximum occurs at a scale
reading of 155o. A second order maximum would be observed at
A
205 o
B
213 o
C
223 o
D
238 o
Answer D
d sin θ = n λ
d sin (180-155) = 1 λ
d sin θ2 = 2 λ
solving simultaneously, θ2 = 58o
Second order maxima would be observed at (180 + 58)0 = 2380
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23
The table shows the electric potential V measured
from various distances x from P along the line PQ.
V/V
65
The electric field strength along PQ for x = 5.0 m
is approximately
A
12 V m
-1
x/m
3.0
56
4.0
39
5.0
22
6.0
17
7.0
towards P.
-1
B
12 V m towards Q.
C
17 V m-1 towards P.
D
17 V m-1 towards Q.
dV
22 − 39
E=
−
=
−
=
17 V m −1
dx
6.0 − 5.0
Since E field direction is in direction on decreasing potential, direction is towards Q.
Ans: D
24
Two charges, - 9.0 × 10-6 C and + 2.5 × 10-3 C are placed 10.0 m apart.
When their separation is changed to 5.0 m, the electrostatic potential energy of
the system
A
decreases by 20 J.
B
decreases by 6.0 J.
C
increases by 6.0 J.
D
increases by 20 J.
Ans: A
Change in Electrostatic PE
= Final Electrostatic PE – Initial Electrostatic PE
QQ  1 1
1 Q1Q2
1 Q1Q2
= 1 2 − 
=
−
4πε 0  rf ri 
4πε 0 rf
4πε 0 ri
1 
 1
= 9.0 × 109 ( −9.0 × 10 −6 )(2.5 × 10 −3 ) 
−

 5.0 10.0 
= - 20 J
25
The electron beam current in a cathode ray oscilloscope is 40 µA. The electron
beam strikes the screen perpendicular to its path and sweeps horizontally
across the screen at a speed of 0.80 cm s-1.
What is the number of electrons arriving on the screen in 2.0 cm length of the
horizontal trace?
A 6.25 x 1014
Answer : A
NYJC 2009
B 1.00 x 1014
C 3.13 x 1013
D 3.13 x 1010
I = ne /t
n=It/e
= 40 x 10-6 x 2.0 / ( 0.80 x 1.6 x 10-19 )
= 6.25 x 1014
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26
A potential difference of 12 V is applied across a resistor for a time interval of
10 s. The current flowing through the resistor is 1 A. Which of the following
statements is incorrect?
A
B
C
D
The resistance is 12 Ω.
The potential difference across the resistor is 12 J C-1.
The charge passing through the resistor is 10 C.
The energy dissipated is 12 J.
Answer : D
V = IR
R = 12 / 1 = 12
I=Q/t
Q = 1.10 = 10 C
E = P t = IVt = 1x12x10 = 120 J
27
In the circuit diagram below, X is an ideal diode. The power supply has
negligible internal resistance and the voltmeter reads 5.0 V.
4Ω
4Ω
V
4Ω
X
If the connections to the terminals of the voltage supply are reversed, the
voltmeter reading would be
A
2.0 V
B
3.3 V
C
5.0 V
D
7.5 V
Answer : B
Initially, current is unable to flow thru X, thus, if the voltmeter reads 5.0V, the
power supply is 10.0V,
When the power supply is reversed, the current will flow thru X, thus, now,
effective resistance across the voltmeter is 2.0 Ω.
Therefore, the p.d across the voltmeter reading is now 2.0/6.0 x 10.0 = 3.3 V
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28
In the circuit shown, X can be connected to A, B, C or D.
18 Ω
30Ω
X
G
10 Ω
6Ω
A
7Ω
B
2Ω
C
15 Ω
D
In order to obtain a zero reading on the galvanometer G, which point, A, B, C or
D should X be connected to?
Answer : D
There will be no current through the galvanometer G if the p.d across is 0 V.
Thus, for the first path, the resistance ratio is 30 : 18 ⇒ 5:3
Hence, the second path, resistance ratio should be 5 : 3 ⇒ 25:15
29
A and B are two parallel wires carrying currents of magnitudes 1.0 A and 2.0 A
respectively. Which of the following statements is definitely wrong?
A
The external magnetic field experienced by A is stronger than that
experienced by B.
B
The external magnetic force experienced by A is stronger than that
experienced by B.
C
A and B attract each other.
D
A and B repel each other.
Q29 Ans: B
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30
An electron enters a region at point O at an angle of 45o to the y-axis as shown .
The region contains a magnetic field which is directed into the plane of the page.
An electric field can be applied such that the electron is able to pass through
the region undeflected. Which of the following correctly describes the direction
of this field from O ?
y
Magnetic
field
y
A
D
B
x
x
45o
O
O
C
Magnetic
field
y
current
FE
45o
FB
31
Q30
x
Ans: C
O
Two coils X and Y are arranged as shown.
Coil X is connected to a current supply
and coil Y is connected to a voltmeter.
X
Which of the following will NOT produce a
reading in the voltmeter?
Y
voltmeter
current supply
A
Using a high direct current supply.
B
Using a low alternating current supply.
C
Quickly increasing the direct current supply.
D
Slowly decreasing the direct current supply.
V
Using a high direct current supply, the flux linkage in Y will be high but not
changing. Thus no e.m.f. will be induced in Y to produce a reading.
Ans: A
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32
Two coils P and Q enter a magnetic field at the same time, as shown. P has the
same length but twice the width of Q and is moving with half the speed of Q.
P
vP
Q
vQ
Magnetic
field
Which of the following correctly shows the e.m.f. induced in the coils as they
enter the field?
e.m.f.
e.m.f.
P
P
Q
Q
time
time
A
C
e.m.f.
e.m.f.
Q
P
Q
time
B
P
time
D
E.m.f. induced = Rate of change of flux linkage = B w v (where w is width of coil)
Q is half the width of P but moving at twice the speed of P. Thus e.m.f. induced in P and
Q are the same.
Because of its higher speed, Q will enter the field fully in half the time P takes.
Ans: D
NYJC 2009
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co
17
33
The graph shows a rectangular waveform with a peak current of 3.0 A.
Current / A
3.0
time
0
-3.0
In order to obtain the same r.m.s. current with the same frequency as the
waveform shown, a sinusoidal current should have a peak value of
A
B
2.1 A
C
3.0 A
D
4.2 A
6.0 A
Ans: C
For the circuit shown, the r.m.s. current will be 3.0 A. To achieve an equivalent r.m.s.
current in a sinusoidal current, the peak value have to be 3.0 ( √2) = 4.2 A
34
An a.c. supply is connected across a resistor R as shown in Fig. 34.1. The
voltage across AB varies sinusoidally with a peak voltage of 5.0 V as shown in
Fig. 34.2.
Voltage / VAB
B
A
5.0
0
0.02
0.04
time /s
-5.0
R
Fig. 34.2
Fig. 34.1
Which is the following statements about the setup is not correct?
A
The frequency of the supply is 50 Hz.
B
An electron will travel from A through R to B and reverses its direction of
travel every 0.02 s.
C
A steady d.c. supply of
5
V will transfer the same power as the a.c.
2
D
supply.
A p-n semiconductor diode connected to the circuit will be able to rectify
the a.c. source to obtain a d.c. output at R.
Ans: B
From Fig. 34.2, we can see that one period of the alternating voltage AB is 0.02 s.
Therefore, the charge will flow in one direction and reverses every 0.01 s.
NYJC 2009
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co
18
35
An electron with kinetic energy E has a de Broglie wavelength of λ . Which of
the following graphs correctly represents the relationship between λ and E?
λ
λ
1
1
E
0
E
0
A
B
λ
λ
1
1
E
0
E
0
C
D
Ans: C
E = ½ m v2 ; (mv )2 = p2 = 2mE
de Broglie wavelength, λ = h/p = h / √ (2mE) ;
λ ∝ 1/√ E
The graph of λ vs 1/√ E is a positive gradient straight line through the origin .
36
Light quanta each of energy 3.5 x 10-19 J fall on the cathode of a photocell. The
current through the cell is just reduced to zero by applying a reverse voltage to
make the cathode 0.25 V positive with respect to the anode. The minimum
energy (in J) required to remove an electron from the cathode is
A
B
2.9 x 10-19
3.1 x 10-19
C
3.5 x 10-19
D
3.9 x 10-19
Ans: B
φ = hf - eVs = 3.5 x 10-19 - 1.6 x10-19 x 0.25 = 3.1 x 10 -19 J
NYJC 2009
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co
19
37
Consider the simple circuit as shown in Fig. 37.1. The input signal is a sine
wave as shown in Fig. 37.2 .
VAB
C
A
t
B
D
Fig.37.2
Fig.37.1
The output signal would be
VCD
VCD
t
t
A
C
VCD
VCD
t
t
B
D
Answer B
C
C
A
+
A
-
B
-
B
+
D
VAB = positive
Diode is forward biased
VCD = VAB
NYJC 2009
D
VAB = negative
Diode is reverse biased
VCD = 0
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co
20
38
Below are energy level schemes of possible laser materials. The “pump” is
where excitation takes place. The “laser transition” indicates the transition
where lasing should occur. The transition “fast decay” is fast compared to the
pumping or excitation process as well as the lasing transition.
Which of these level schemes is best for facilitating population inversion?
A
B
fast decay
pump
fast decay
laser
transition
pump
fast decay
C
laser
transition
D
fast decay
pump
laser
transition
pump
laser
transition
Answer A
The fast decay transitions in scheme A facilitates population inversion best
because it causes the upper lasing level to be populated and the lower lasing level
to be emptied quickly.
NYJC 2009
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co
21
39
A deuteron (2H nucleus) consists of a proton and a neutron. If the binding
energy of a deuteron is 1.12 MeV per nucleon, what is the mass of the deuteron?
Given that 1eV = 1.60 x 10 -19 J ; mproton = 1.007825 u ; mneutron = 1.008665 u
A
2.014091 u
B
2.015291 u
C
2.016490 u
D
2.018889 u
Ans: A
Mass of nucleon = mproton + mneutron = 1.007825 u + 1.008665 u = 2.01649 u
Energy equivalent = m c2 = 2.01649 x 1.66 x 10-27 x (3.00 x 108)2
= 3.0126 x 10-10 J
Binding energy = 1.12 Mev = 2 x 1.12 x 106 x 1.60 x l0-19 = 3.584 x 10-13 J
Mass of the deuteron = (3.0126 x 10-10 - 3.584 x 10-13) / (3.00 x 108)2
= 3.3434 x 10-27
Mass of the deuteron = 3.3434 x 10-27/ 1.66 x 10-27 = 2.014091 u
40
The 14C : 12C ratio of living matter has a constant value during life but the ratio
decreases after death because 14C is unstable and decays by β-emission with
a half-life of 5740 years. Two archeological samples of organic matter were
found at a site. The 14C : 12C ratio of one sample is greater than the other by a
factor of 8. The difference in age (in years) between the two samples is
A
45920
B
17220
C
1913
D
718
Ans: B
The 14C : 12C ratio of one sample is greater than the other by a factor of 8 (ie 23 ).
=> one sample has decayed 3 more half-lifes
=>
The difference in age between the two samples is 3 x 5740 = 17220 yrs
NYJC 2009
9745/01/PRE/09
H2 Physics Papers, Notes at alevelphysics.co