Solutions to SQA examination
1996 Higher Grade Physics
Paper I Solutions
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1. C 11. E 21. C
2. C 12. D 22. E
3. B 13. A 23. A
4. D 14. D 24. B
5. B 15. D 25. E
6. B 16. C 26. A
7. E 17. A 27. A
8. C 18. E 28. B
9. C 19. E 29. B
10.C 20. C 30. E
31. Step one : calculate the time the arrow takes to reach the target
using the fact that the horizontal velocity is constant.
t = shor/vhor
t = 30/100
t = 0.3s
Step two : calculate the vertical displacement of the ball in this time.
uver= 0m/s
t = 0.3s
a = -9.8m/s/s
sver = ?
sver = ut + 1/2(at2)
sver = 0 + 1/2(-9.8x0.32)
sver = -0.441m
Step three : calculate the radius of the target.
r = 1.5-0.9
r = 0.6m
As the arrow falls less than the radius of the target it hits
the target.
32.a.i.
a.ii. Frope = Tension(T)
As the rope is stationary the forces acting on the buoy are balanced.
Fup = w + T
Fup = 50 + 1200
Fup = 1250N
b.
The buoyancy force is not dependent on depth. Therefore, the buoyancy
force is as calculated in part a.ii.(1250N).
33.
In circuit 1 the total resistance in the circuit is equal
to the sum of the internal resistor(r) and the resistance
of the bulb (Rbulb).
Rtotal = r + Rbulb
In circuit 2 the total resistance in the circuit is equal
to the sum of the internal resistor(r) and the resistance
of the two bulbs in parallel.
Rtotal = r + Rbulb/2
The total resistance in circuit 2 is less than that in circuit 1. Thus
the current(I) flowing through the internal resistor in circuit 2 is greater.
The "lost" voltage (Vlost=Ir) across the internal resistor is
therefore greater in circuit 2. This means the voltage across the lamps
(Vlamp = E -Ir) is less in circuit 2 reducing the brightness of the
lamps.
34.a. Vrms = Vpk/21/2
Vrms = 12/1.414
Vrms = 8.5V
b. P = Irms2
Irms = Vrms/R
Irms = 8.5/4
Irms = 2.12A
P = 2.122x4
P = 18W
35. dsinq = nl
d = 1/2.5x105 = 4x10-6m
l = 600x10-9m
n = 1
q = ?
sinq = nl/d
sinq = 1x600x10-9/4x10-6
sinq = 0.15
q = 8.63o
36.a. Possible transitions : E3 -> E2
E3 -> E1
E3 -> E0
E2 -> E1
E2 -> E0
E1 -> E0
b. A total of 6 lines are in the spectrum.
Low frequency radiation will be produced when the transition
is between energy levels with a small energy gap.
The smallest energy gap is between levels E3 and E2.
c. Transitions between certain states will be more frequent than others.
The more frequent a transition, the more intense the spectral line will be.
37. Ek = Ephoton - Ework function
Ephoton = hf = hc/l
Ephoton = (6.63x10-34x3x108)/5.4x10-7
Ephoton = 3.68x10-19J
Ek = 3.7x10-19 - 2.9x10-19
Ek = 0.8x10-19
END OF QUESTION PAPER
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