VICTORIA JUNIOR COLLEGE
SUGGESTED SOLUTIONS TO
2009 PHYSICS H2 P1 PRELIM
EXAMS
1. f = Cv where C is a constant.
f
C= 2
v
kg m s -2
Units of C =
= kg m -1
-1 2
(m s )
Ans: A
2
F
IL
kg m s -2
Units of B are
= kg s -2 A -1
Am
But Q = I t. Therefore C = A s
Hence units of B are kg s-2 (C s-1)-1
= kg C-1 s-1
Ans: A
2. F = BIL ⇒ B =
3. V = πR 2 L
V
R=
πL
∆R 1 ∆V ∆L
1
1
= (
+
)= x+ y
R
2 V
L
2
2
Ans. A
4. Precision refers to the degree of
agreement of data with respect to one
another. Accuracy refers to the degree
of agreement of data with respect to an
accepted value. The accepted value of
g is taken generally to be 9.81 m s-2.
For A: g = 9.82 ± 0.02 m s-2
For B: g = 9.7 ± 0.4 m s-2
For C: g = 9.1 ± 0.3 m s-2
For D: g = 8.46 ± 0.05 m s-2
Ans. D
5. If we take the positive direction to
be downward, the acceleration of free
fall g will be positive and constant
whether the ball is moving down or up.
The breaks in the graph in (c) represent
the short time intervals when the ball
strikes the ground and is in contact
with it.
Ans: C
6. At the top of the pen’s path, the
only force acting on the pen is its
weight.
Ans: D
7. Let the force exerted by the wall on
the flagpole at P be F. The lines of
action of the 3 forces F, W and T must
pass through a single point and appear
to radiate outwards from the same
point in the form of a “Y”. When this
is satisfied, the 3 forces, when joined
head to tail, will form a closed triangle
representing equilibrium has been
achieved.
Ans: B
8. In diagram 1, torque of the couple
= 2F( 0.25l ) = 0.50 Fl = 5.0 N m
In diagram 2, torque = F (0.50 l)
= 0.50 Fl = 5.0 N m
Ans: C
9. Both iceboats travel the same
distance s, and only the horizontal
force F in the direction of motion does
work on either iceboat. Hence the total
work done between the starting line
and the finish line is the same for each
iceboat.
Ans: A
10. Loss of energy by spring = gain in
GPE + work done against resistive
force.
Let d be maximum distance traveled
by block up the inline.
1 2
kx mgd sin θ + fd
=
2
1
(1.40
=
× 103 )(0.100) 2 (0.200)(9.81)d sin 60° + 1.2d
2
d = 2.41 m
Ans: A
1
H2 Physics Papers, Notes at alevelphysics.co
Time for one revolution = T =
11.
2π
ω
= 6.2 s.
Ans: D
13. Originally, E = (1/2)mω2A2
With the frequency doubled and the
amplitude halved,
E’ = (1/2)m(2ω)2(A/2)2
=E
Ans: B
r
g
r
0
14. a = - ω2x
The acceleration is in anti-phase with
the displacement.
Ans: D
Mass does not depend on gravitational
field strength g and hence will remain
the same whether at the surface or
centre of the planet. The value of g,
however, decreases to a theoretical
zero at the centre of the planet. Hence
the weight W = mg of an object will be
zero at the planet’s centre.
Ans: D
15. ∆U = ∆Q + ∆W
12.
16. Using
3.0 m
30.00
5.0 m
Central
axis
Let T be the tension in the string.
For vertical equilibrium of mass,
Tcos30.00 = mg
(1)
For horizontal circular motion of mass,
Tsin30.00 = mrω2
(2)
(2)
gives
(1)
tan30.00 = rω2/g
tan30.00 = (3.0 + 5.0sin30.00)ω2/9.81
ω = 1.01 rad s-1
Or u = q + ∆W
∆W =u − q  by definition, this
gives the work done ON the gas.
Therefore, work done BY the gas
= -u + q
Ans: D
PV
PV
1 1
= 2 2,
T1
T2
4 × 2 3× 6
=
T1
T2
9
T2 = T1
4
The temperature of the gas increases,
implying that the internal energy of the
gas increases. The graph shows an
increase in the volume of gas, implying
that work is done by the gas. From
∆U = Q + W, since ∆U is positive
while W is negative, Q will be positive.
This means heat is gained by the gas.
Ans: D
17. Phase difference, ∆φ =
x
λ
× 2π
x
× 2π
12.0
∴ x = 4.5 cm.
∴ 0.75π =
2
H2 Physics Papers, Notes at alevelphysics.co
Since the motion of M lags behind that
of N, the wave must reach N first
before it reaches M. This means that
the wave is travelling from N to M.
Ans: B
18.
Let I1 and I2 be the intensities of the
received radio signal when the angle
from vertical are θ1 and θ2
respectively.
1
Required:
---(1)
I 2 = I1
2
Since “intensity ∝ (amplitude)2”, and
the amplitude of radio signal received
by the antenna is given by
A = Ao cos θ , equation (1) gives:
1
( Ao cos θ 2 ) 2 = ( Ao cos θ1 ) 2 , where
2
Ao = amplitude of original radio wave.
1
∴ cos θ 2 =
cos θ1
2
This condition will be satisfied if
θ1 = 45o and θ2 = 60o.
Ans: C
19. To obtain a steady interference
with minimum intensity at point P,
both the sources must be coherent and
be out of phase by π radians. Both
sources must have same amplitude.
The next minimum produced at Q
corresponds to a path difference of
1 m. For two sources out of phase by
π rad, the minimum obtained at Q must
satisfy the equation
“path difference = n λ” where
n = 1. Thus, the wavelength of the
waves must be 1 m and not 2 m.
Ans. B
20. Using R = ρL/A or R ∝ ρ , the
graph will be a straight line within
each segment. Given that ρ1 > ρ2 > ρ3,
the gradient of the line graph for each
segment should decrease as we go
from P to Q.
Ans: A
21. Power dissipated in the cables
= I2R = (P/V)2R
Ans: C
22. Total resistance in the circuit
= 2r + R
Resistance across one third length of
R
the wire =
3
Using potential divider principle, emf
R
3 )E
of cell in branch circuit = (
2r + R
=
RE
3(2r + R)
Ans: D
23. The force F acting on the electron
is F = Ee where E is the electric field
strength and e is the electronic charge.
Since the electric field between two
charged parallel plates is constant, and
charge e is also constant, the force F
will also be a constant regardless of
distance x inside the space between the
charged plates.
Ans: D
24. Using W = VQ, we have
1.0
= 1.0 V
V =
1.0
Ans: A
25. The magnetic field due to the
current flowing in the wires around the
nail is proportional to the current. This
is reasonable as it is the current that
gives rise to the magnetic field. Hence,
the stronger the current, the stronger
the associated magnetic field.
Ans: B
26.
S
N
I
3
H2 Physics Papers, Notes at alevelphysics.co
When the switch is closed, a current I
flows as shown. This current flowing
through the solenoid turns the solenoid
into an electromagnet with the north
pole at the lower arm. Using Fleming’s
Left hand Rule, it can be predicted that
the rod will experience a force to the
right.
Ans: C
27. Using Fleming’s Right Hand Rule,
P is at a higher potential than Q. At
equilibrium, charge separation ceases
and no current flows in the plate.
Ans: D
28. Induced emf in section PQ has a
greater magnitude compared to that
due to section PR because PQ is longer
compared to PR.. Using Fleming’s
Right Hand Rule, Q has a higher
potential than R.
Ans: A
29. Let the power at the generating
station be P0. At the output of the first
transformer, the power available
becomes P1= 0.95P0. The power
available after the second transformer
is P2 = 0.95P1 = 0.952P0.
Extrapolating, the power available
after the 8th transformer is P8 = 0.958P0
≈ 0.66P0. Power loss is therefore
0.34P0. Percentage of energy lost is
 P8 
1 −  x100 ≈ 34 %
 P0 
Ans: C
30. Given V = V0 sin (ω t )
If the frequency is doubled while the
voltage amplitude is reduced to onethird the original value, then
1 
V =  V0  sin (2ω t )
3 
Ans: B
31. The root-mean-square value of an
alternating sinusoidal e.m.f. is
E
6.0
≈ 4.2 V
Vrms = 0 =
2
2
Ans: C
32. K = hf − φ
⇒=
KE1 h(2 f ) − φ when frequency is
doubled.
⇒ KE1 = 2( K + φ ) − φ
⇒ KE1 = 2 K + φ
Intensity does not affect the maximum
kinetic energy of the electrons.
Ans: C
33. The shortest wavelength X-ray
photon is produced in Bremstrahlung
radiation when the full kinetic energy
of the electron impinging on the target
material is practically lost in a single
deceleration process. Thus,
hc/λmin = eV, where V is the potential
difference used to accelerate the
electrons. Hence, λmin ∝ 1/V. To half
the minimum wavelength, the
accelerating potential difference has to
be doubled.
Ans. D
34. The photon energy absorbed by the
metal vapour atoms will be re-radiated
uniformly in all directions, resulting in
lower intensities (dark lines) observed
in the transmitted or emergent light.
Ans. C
35. By the Energy-Time Uncertainty

Principle, ∆E∆t ≥
2
1 6.63 × 10 −34
∴ ∆E (2.2 × 10 −6 ) ≥ ×
2
2π
∴ uncertainty in particle’s energy, ∆E
≥ 2.4 × 10-29 J
By the mass-energy equivalence
equation, E= mc2
∴∆E= ∆mc2
4
H2 Physics Papers, Notes at alevelphysics.co
∴ uncertainty in the mass,
∆E
2.4 × 10 −29
∆m ≥ 2 =
c
(3.0 × 10 8 ) 2
= 2.7 × 10-46 kg.
∴ minimum uncertainty in the mass
= 2.7 × 10-46 kg.
Ans: B
36. Since the square of the amplitude
2
of the wave function Ψ at any point
is a measure of the probability density
function of the particle at the point, the
fact that the amplitude ofΨ decreases
upon passing through the barrier
implies that the probability of finding
the particle on the right side of the
barrier is less than that on the left side
of the barrier.
Since the energy of the particle is
1 2 p2
,
given by KE = E = mv =
2
2m
where the momentum is given by
h
p = (de Broglie’s relation), we
λ
conclude that both the energy E as well
as the momentum p of the particle are
unchanged after passing through the
barrier, since the de Broglie
wavelength λ is unchanged.
Ans: D
37. For the most energetic electron in
the valence band (with energy E2) to
reach the conduction band, it must
absorb a minimum of E5 − E2 of
energy
Ans: C
it becomes positively charged (gains a
positive potential). Since the p-type
material gains electrons from the ntype material, it becomes negativelycharged (gains a negative potential).
Hence, the p-type material has a lower
electrical potential compared to the ntype material. Hence, statement A is
incorrect.
Ans: A
B: Since the n-type material has a
positive potential, the electrons in this
side will have a negative potential
energy, by U = qV, where q is the
negative electron charge and V is the
positive potential. Statement B is
correct.
C: If the p-n junction is forwardbiased, current flows from the p-side to
n-side. Statement C is correct
D: The electrical voltage described
above amounts to about 0.6 V.
Statement D is correct.
39. The net energy released = ∆mc2
= (mL + mp – 2mH)c2
Ans. B
40. N = N 0 e − λt or ln N = ln N 0 − λt
From the graph,
-λ = gradient of graph
4 −1
=
λ = 0.050 s-1
60
ln 2 ln 2
t 1 = = 14 s
=
0.05
λ
2
Ans: C
********** END *********
38. A: When a p-type semiconductor is
joined to an n-type semiconductor,
mobile electrons from the n-side will
diffuse to the p-side where they
recombine and neutralise each other,
creating a region with no free charge
carriers, called the depletion region.
Since the n-type material loses
electrons in this neutralisation process,
5
H2 Physics Papers, Notes at alevelphysics.co