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Physics Practice Paper Paper 2 Suggested Answers (12-13)
Section A: Astronomy and Space Science
MC
1-8
CACBCDCB
Explanations to selected mc
1.2 The angle between the north celestial pole and the zenith is 900 – 22.50.
The celestial sphere completes one revolution every day.
1.5 Luminosity is proportional radius2 and (temperature)4.
Luminosity ratio = (1200/6)2 x (10000/25000)4 = 1024
1.7 Ratio of intensity = 2.512(11.05-9.54) = 4.02
1.8 T2 = a3
28.62 = a3 => a = 9.35 AU
a = d + d’ => (4.24 + d)/2 = 9.35
A9
(a)
(b)
=> d’ = 14.5 AU
Main sequence stars are similar to blackbodies.
[1M]
Therefore, the spectrum of a typical O type star with surface temperature
30 000 K is well approximated by Fig. 1.
From Fig. 1, most of the emitted energy in the visible band comes from the blue
end (~400 nm) of the visible band,
[1A]
so O type stars are blue.
The spectrum of a typical M type star with surface temperature 3 000 K is well
approximated by Fig. 2.
From Fig. 2, most of the emitted energy in the visible band comes from the red
end (~ 700 nm) of the visible band,
[1A]
so M type stars are red.
By doppler effect:
v
v

= r => Δλ = r (λ0)
[1M]
c
c
0
(5  10 4 )
(434.0) = 0.07 nm
(3  10 8 )
It will look redder than it actually is,
because the whole spectrum is red shifted due to doppler effect.
Stefan's Law:
L = 4R2T4
For the Hλ line: Δλ =
(c)
(d)
[1A]
[1A]
[1A]
L⊙ = 4R⊙ 2 T⊙ 4
For the O star :
L  R 


L  R 
2
4
T 
L
2  30000 
  [1M] =>
 2.5 
  4474
L
 5800 
 T 
4
[1A]
1
1
=
= 1.3 pc
[1A]
p
0.76 pc
Section B: Atomic World
MC
1-8 C B B B B B B C
2.2 Kmax = hf -  = 6.63 x10-34 x 3 x 6.20 x 1014 - 6.63 x10-34 x 6.20 x 1014 = 8.22 x 10-19 J
(e)
d=
3  6.63  10 34
nh
=> 9.11  10–31  v  0.212  10–9 =
=> v = 1.64  106 m s–1
2
2
2.5 To ionize a hydrogen atom from ground state, 13.6 eV is required. If this amount of energy comes
from a photon, the photon energy hf will be equal to 13.6 eV. hf = 13.6 x 1.6 x 10-19 => f = 3.28 x
1015 Hz
2.6 eV = (1/2) mv2
1.6 x 10-19 x 50 = (1/2) (9.11 x 10-31) v2
=> v = 4.191 x 106 m/s
2.4 L = mevr =
 = h/mv = 6.63 x 10-34 / (9.11 x 10-31 x 4.191 x 106) = 0.174 nm
180
180



= 1.22 

=>  = 429 nm
3
2r


2  5 10
2.8 Self-cleaning glasses make use of water-attracting property.
Q2
2.7 0.003 = 1.22 
Section C: Energy and Use of Energy
MC 1-8 C B C C B C A B
3.1 End-use efficiency = [3 x 2600 x (90-20) / (5 x 60)] / 2500 = 72.8%
3.2 Binding energy per nucleon = 1.88482 x 931 / 238 = 7.37 MeV
3.3 Electrical power output = 1000 x 20 x cos 30o x 0.12 = 2078 W
3.4 (1) is incorrect. Unit of OTTV is Wm-2.
(2) is correct because the material used in a building can affect the rates of heat transfer by
conduction and radiation. As a result, the OTTV value will be affected as well.
(3) is correct as the current OTTV limits for podiums and tower buildings are 70 Wm-2 and 30
Wm-2.
3.5
A(TH  TC )
Q
 UA(TH  TC )
=k
t
d
Decrease in heat transfer = Q / t = U A (TH – TC) = (0.5-0.2) x 600 x (32-24.5) = 1350 J/s
3.6 E 

200
 1  10 7 
=> r = 12.6 km
A
4r 2
3.8 The construction of a wind power station or hydroelectric power station may cause pollution to
the environment.
Q3
(a) (i) Power input to the motor = 12/ 0.8 = 15 kW
Current delivered by battery = P/V = 15000 / 80 = 187.5 A
(ii) Time for the battery to deliver all the energy t = E/P = 28 kWh / 15kW = 1.867 h
Maximum travelling distance = vt = 60 x 1.867 = 112 km
(b) Increase in KE = (1/2) mv2 = 125000 J
Power output by the car = E/t = 125000 / 10 = 12500 W
Power output by battery = 12500 / 0.8 = 15625 W
(c) The regenerative braking system converts the kinetic energy of the car to electrical energy during
braking. This energy is stored in the battery.
(d) 1) Electric vehicles have a higher end-use energy efficiency.
2) Electric cars produce less air pollutants
3) The cost of operation is much cheaper for electric cars.
4) Electric cars produce less noise.