Solutions to SQA examination

1996 Higher Grade Physics

Paper I Solutions

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



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