Solution for Sec 4 Higher Physics Sample Questions
1. (a)
(b)
Parallel – High [1]
Series – Low [1] Single element – Medium [1]
Total energy dissipated is given by (V2/R)t and is related to power dissipated. [1]
For the parallel combination, total power dissipated = 2(2402/R) = 115200/R [1]
For the series combination, total power dissipated = 2(1202/R) = 28800/R [1]
For the single element, total power dissipated = 2402/R = 57600/R [1]
[Accept other possible explanation]
(c)
2. (a)
[1] for each correct answer
Heat setting
Switches to close
High
S1, S3 and S4
Medium
S1 and S4
Low
S1 and S2
When the wheels are rotating in the external magnetic field, it cuts the magnetic field lines,
which results in an emf induced. [1]
As a result, induced current in the form of eddy currents are generated within the wheel.
Based on Lenz’s Law, this induced current is in a direction which opposes the change causing
it, thus generating a force that opposes the original rotation of the wheel. [1]
Since a retarding force acts on the wheel, the train undergoes deceleration. [1]
(b)
Kinetic energy  Electrical energy  Thermal energy
[1] for each correct conversion.
(c)
When the wheels rotate slower, the rate of cutting of field lines decreases [1], resulting in
smaller emf and thus current induced based on Faraday’s Law of EM Induction. [1]
A smaller current induced in turn results in smaller braking force. [1]
(d)
It does not require large amount of physical contact, thus there is little wearing of the wheel
material. [1]
The braking force decreases while the train slows down, thus ensuring smooth braking. [1]
(e)
It does not require large amount of physical contact, thus there is little wearing of the wheel
material. [1]
3
(a) (i) l = 25.0 - 2.0
=23.0 m
(ii)
E = ½ k (l – Lo)2 [1]
(iii)
Total E = initial P.E
(b) (i)
= 25mg
[1]
Assumption: Air resistance is negligible.
[1]
l = 25.0 – 8.0 - 2.0
= 15.0 m
[1]
(ii)
(c)
[1]
F = k (15.0 – Lo) [1]
(i)
At the equilibrium position.
[1]
(ii)
His velocity increases at a constant rate [1] till the rope starts to stretch,
as he experiences acceleration due to gravity. [1]
When the rope stretches, his velocity increases at a decreasing rate [1]
as the upward elastic force increases while his weight remains constant. [1]
I = V/ R = 2.0 / 10.0 =0.20 A
[2]
4 (a)
(b)
(i)
RPJ =
 (0.250  x)
S

x
0.5S
[1]
(ii)
(iii)
RPQ =
 (0.500  x)
[1]
[1]
E2 = VPJ = IPJ . RPJ
=
(c)
x
S
0.5S
 (0.500  x) 2 x

10.0 =
S
S

10.0

S 0.500  x
= 0.20[


(0.250  x)
S

x
]
0.5S [1]
0.500  2 x
0.500  x [1]
Potential difference across the length of wire PJ decreases as the jockey is moved from J to P.
[1]
Hence the e.m.f. of E2 is now greater than p.d across wire PJ.
[1]
Hence current will flow from the positive terminal of E2 through PJ to the galvanometer causing
it to deflect to one side.
[1]
5.(a)
(i)
r1  8.00 2  1.15 2  8.08m
r2  8.00 2  1.85 2  8.21m
Path difference = r2 – r1 = 0.13 m
‫ג‬/2 = 0.13 m
‫ = ג‬0.26 m
v = f‫ג‬
[1]
[1]
343 = f (0.26)
F = 1.3 kHz
(ii)
[1]
He hears a sound as he walks away from P. [1]
Constructive interference occurs just after point P. [1]
5.(b)
(i)
1) Different colour of light through the fiber will travel at different speed /wavelength and
reach the end point at different time resulting in chromatic dispersion. [1]
2) Different path taken by light of the same colour will cause the light to reach the end
point at different time resulting is distortion of the information it carries.[1]
(ii)
Overlapping can occur when the first pulse travels at the lower speed
limit and takes
10000
s to reach the end
1.95  10 8
And the second pulse travels at the max, speed limit and takes
10000
s
2.05  10 8
to reach the end. [1] for time for both pulse
Thus if we are to avoid overlapping, the minimum time lag between the
emission of the first pulse and second pulse is the difference =
10000
=2.5 x 10-6 s.
8
2.05  10
10000
1.95  10 8
[1] for diff. in time
The intensity vs time diagram of the light pulse is shown below,
Intensity
2.5x10-6
time
As such, we can deduce that the period = 2 x 2.5 x 10-6s
And thus the maximum frequency = 1/T = 200 kHz.
[1] for finding frequency
6.
(a)
Temp = (26 + 50) = 76 oC
[1]
drawing tangent in graph at y coordinate of 76oC
6.7 oC per min
(b)
(i)
Straight line through origin [1]
OR
Smooth curve from origin [1]
(See diagram below)
[1]
[1]
(ii)
The oil has cooled to the same temperature as the surroundings
OR
oil is in thermal equilibrium with surroundings. [1]
The rate of decrease of temperature will be zero at this point.
OR
No net transfer of thermal energy between surroundings and oil.
(c)
Allowing for uncertainties in readings, points lie on straight line and
the line passes through the origin.
(d)
[1]
[1]
**Award either 2 or 0 marks.
Yes.
At higher excess temperature, the points do not lie on the straight line. [2]
OR
No.
At higher excess temperature, the points still lie close to the straight line. [2]
OR
No conclusion.
There are too few points to draw a valid conclusion. [2]
7(a)
Direction of electron flow
[1] – arrow downwards
[1]
Using Fleming’s right-hand rule, [1] applied force on the antenna is due north, and the
component of the Earth’s magnetic field is due East, hence the conventional current is into the
plane of the paper.
Thus the electron current is of an opposite direction from conventional current. [1]
(b)
On the figure below, indicate which end of the antenna is at a higher potential. [1]
(c)
Equating electric and magnetic forces, eq (i) = eq(iii)
 qv Bsin θ = k
qq1
r2
 vB sin θ = E
E = 75 (6.0 x 10-5) sin55o

E = 0.0037 V/m
(d)
[1]
[2]
[1]
[1] – arrow perpendicular to B-field
For maximum induced current, the airplane should be flying perpendicular to the B-field. [1]
The rate of change of magnetic flux linkage will be maximum at this angle since θ = 90o and sin θ
will be the maximum value. [1]