 # Solutions to SQA examination

### Paper I Solutions

```1. D		11. D			21. B
2. A		12. C			22. A
3. D		13. D			23. B
4. D		14. B			24. C
5. B		15. B			25. E
6. D		16. A			26. A
7. B		17. C			27. E
8. E		18. D			28. B
9. C		19. B			29. E
10.B		20. C			30. E

31.	Calculate the magnitude of the displacement vector (s) using Pythagoras' theorem
and calculate the corresponding bearing using basic trigonometry.

s2= AB2+BC2				Let angle BAC = q
s2= 502+1502				tanq=BC/BA
s2= 2,500+22,500			tanq=150/50
s2= 25,000				tanq=3
s= (25000)1/2				q=TAN-13
s= 158.11m				q=71.6O

S=158.11m at a bearing of 071.6O

32.	To solve this problem use conservation of momentum.

Pbefore=Pafter

Pbefore = mballuball+mstudentustudent
Pbefore = 0.42xuball+50x0
Pbefore = 0.42xuball

Pafter = (mball+mstudent)x v
Pafter = (0.42+50)0.10
Pafter = 50.42x0.1
Pafter = 5.042kgm/s

Equating Pbefore to Pafter leads to:

0.42xuball	 = 5.042
uball 		= 5.042/0.42 (m/s)
uball 		= 12.00m/s

33.	Step 1 :Calculate the effective resistance of the parallel network
containing the 6W and 3W resistors.

1/Rp=1/R1+1/R2
1/Rp=1/61+1/32
1/Rp=3/6
Rp=6/3
Rp=2W

Step 2 :Add the 2W resistor to the value calculated above.
R=2W+2W
R=4W

Step 3 :The 4W resistor is in parallel with the resistance calculated in step 2.
The total resistance is therefore given by:

1/Rtotal= 1/4 + 1/4
1/Rtotal= 2/4
Rtotal = 2W

34.a.	When the wheatstone bridge is balanced:

R1/R2 = Rv/Rt
Rt = RvxR2/R1
Rt = 0.225x2.2/3.3
Rt = 0.15kW
Rt = 150W

b.	At 80mV the temperature of the thermistor has increased by 4oC.
The temperature of the thermistor is therefore 24oC.

35.a.	E = hf		f=v/l
E = hv/l
l = hv/E
l = 6.63x10-34x3x108/6.9x10-19
l = 2.88x10-7m

b.	As 4.0x10-7m is greater than the maximum wavelength that will cause an
electron to be emitted, no electrons will be emitted from the zinc surface.

36.a.	When an electron in a high energy level is made to fall to a lower energy level
in an atom as a result of a passing photon, with an energy equal to the energy gap
between the high energy level and the lower energy level, a photon produced.
This is called stimulated emission of radiation where the passing photon is called
the stimulating photon.

b.	The stimulating and the stimulated radiation have the same frequency and wavelength. They
are also travelling in the same direction and are in phase.

37.	To solve this problem use the grating equation dsinq = nl.

d = 1/300 mm = (1/300)x10-3m = 3.33x10-6m
q = 24.5o
n = 2
l = ?

l = dsinq/n
l = 3.33x10-6sin24.5o/2
l = 6.91x10-7m = 691nm

END OF QUESTION PAPER