Monday, June 2, 2014

9702 November 2013 Paper 21 22 Worked Solutions | A-Level Physics

  • 9702 November 2013 Paper 21 & 22 Worked Solutions | A-Level Physics



Paper 21 and Paper 22 are similar.


Question 1
(a)        Choose any 2 :
Kelvin (K), Ampere (amp/A), Mole (mol), Candela (Cd)

(b) 
(i) Work = Force x Distance = (Mass x Acceleration) x Distance : kg x ms-2 x m
Work : kg m2 s-2
Alternatively:
Energy = ½ mv2 or mc2 : kg (ms-1)2 = kg m2 s-2

(ii) Density = mass/ volume : kgm-3
Acceleration of free fall, g : ms-2
Cross-sectional area of wire, A : m2
Original length, l0 : m

Therefore, C : kg m2 s-2 / (kg m-3)2 (ms-2)2 m2 (m)3 = kg-1 ms-2 





Question 2

(a)
Time-base setting = 0.20μscm-1
That is, 1cm (1 cell) represents 0.20μs
Speed of pulse, v = 3 x 108 ms-1
Number of cells between peaks of the 2 pulses = 4

Speed, v = distance, d / time, t

Therefore, distance travelled by pulse, d = v x t
t = 4 x 0.20 = 0.80μs
d = 3 x 108 x 0.80 x 10-6 = 240 m

Since the pulse travels from the source to the reflector, and from the reflector back to the source,
Distance between source and reflector = 240/2 = 120m  


(b)
Speed of sound = 300ms-1
Speed of light = 3 x 108 ms-1

Time taken by sound wave = 240 / 300 = 0.8s
This time corresponds to 4 cells (4cm) since the distance 240m corresponds to 4 cells.

Therefore,
4 cells represent 0.8s
1 cell represent 0.8/4 = 0.2s

Time-base setting is 0.20scm-1            [unit required]






Question 3
{Detailed explanations for this question is available as Solution 673 at Physics 9702 Doubts | Help Page 135 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-135.html}






Question 4
(a)
The torque of a couple is the product of one of the forces multiplied by the perpendicular distance between the forces.

(b)
(i)
Weight at P – vertically down
Choose any 1:
Normal reaction - vertically up
Contact force at P – at point of contact with the pin

(ii)
Torque = 35 x 0.25 x 2 = 17.5 ≈ 18 Nm

(iii)
The two 35N forces are equal and opposite and the weight and the upward / contact / reaction force are equal and opposite.

(iv)
The wheel is not in equilibrium since the (resultant) torque is not zero.







Question 5
{Detailed explanations for this question is available as Solution 540 at Physics 9702 Doubts | Help Page 106 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-106.html}







Question 6
{Detailed explanations for this question is available as Solution 389 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}







Question 7
(a)
(i)
Choose any 1:
The direction of the fields is the same.
Fields are uniform.
Constant electric field strength
E = V/d where E is the uniform electric field between the 2 parallel plates, separated by the distance d and V is the electric potential. 

(ii)
Reduce the potential difference across the plates
Increase the separation of the plates

(iii)
α-particles have opposite charge to β-particles (deflection in opposite direction)
β-particles have a range of velocities OR energies (different deflections) while α-particles all have the same velocity OR energy (constant deflection)
α-particles are more massive (do not use ‘heavier’) (deflection is less for greater field strength)

(b)
W = 234, X = 90, Y = 4 and Z = 2

(c)
A = 32, B = 16, C = 0 and D = -1

 



11 comments:

  1. hi can i have solution for paper 42 nov 2010

    ReplyDelete
    Replies
    1. solutions have already been provided.

      Delete
  2. For question 6(b)(II) to calculate the current in the wires , the resistance we use is 39.08 ohm, (which is included the resistance of heater)
    But however, to calculate the power loss in the wires , in question 6(b)(III) ,the resistance used is only the two wires. Why? And how do I know when should I use which resistance

    ReplyDelete
  3. Hi i didnt understand in question 5(a)(ii) why the distance was divided by 4? If you know please help and explain

    ReplyDelete
  4. I didnt understand question 5 (a(ii) why the distance was divided by 4? If u can explain more this will be very helpful thanks!

    ReplyDelete
  5. can i please have the solution to may june 2013 Q6 Paper 21.

    ReplyDelete
    Replies
    1. See solution 676 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html

      Delete

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