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Sunday, July 13, 2014

9702 June 2012 Paper 22 Worked Solutions | A-Level Physics

  • 9702 June 2012 Paper 22 Worked Solutions | A-Level Physics



Question 1
{Detailed explanations for this question is available as Solution 1022 at Physics 9702 Doubts | Help Page 213 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html}





Question 2
A ball is thrown vertically down towards the ground and rebounds as illustrated in Fig. 2.1.





Question 3
{Detailed explanations for this question is available as Solution 614 at Physics 9702 Doubts | Help Page 121 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html}






Question 4
(a)
Pd across XY:
Total resistance in circuit = 8 + 12 = 20kΩ
Current = 12 / (20x103)           or potential divider formula
p.d = [12/(20x103)] x 12x103 = 7.2V

(b)
Resistance LDR = 4.0kΩ. Current in ammeter:
Parallel resistance (across XY) = [(1/4) + (1/12)]-1 = 3kΩ
Total resistance in circuit = 3 + 8 = 113kΩ
Current = 12 / (11x103) = 1.09x10-3A or 1.1x10-3A

(c)
Intensity of light on LDR increased.
(i)
Change to ammeter reading:
The resistance of the LDR decreases. So, the total resistance (of the circuit) is less. Hence, current increases.

(ii)
Change to p.d across XY:
The resistance across XY is now less. So, there is less proportion of the 12V across XY. Hence, p.d. is less.






Question 5
{Detailed explanations for this question is available as Solution 626 at Physics 9702 Doubts | Help Page 124 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html}





Question 6
{Detailed explanations for this question is available as Solution 631 at Physics 9702 Doubts | Help Page 125 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-125.html}





Question 7
{Detailed explanations for this question is available as Solution 641 at Physics 9702 Doubts | Help Page 127 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-127.html}

19 comments:

  1. could you explain 3(B2) in detail please.It would be really appreciated! ^_^ in desperate need of help

    ReplyDelete
    Replies
    1. Details have been added for the question. See again

      Delete
  2. In question 5(c), why must the line pass through 1.5 on the y axis? i get that it has to pass through 0.25 on the x axis but how did you find that it has to pass through 1.5 on the y axis?

    ReplyDelete
    Replies
    1. Details have been updated for question 5

      Delete
    2. thank you :)
      can you also upload the diagram for question 6(b)?

      Delete
  3. can you please draw question 6(b)?

    ReplyDelete
  4. Hi, I am just wondering if the uncertainty is 0.09 x 10^-3,shouldnt the actual measurement be multiples of the uncertainty? or follow the decimal place of uncertainty?Confused about this part...I would really appreciate if you can answer me in time.Thanks!!!

    ReplyDelete
    Replies
    1. If you are referring to a specific question, give me the reference of the question. It's difficult to explain otherwise.

      Delete
    2. Please tell me the difference between 9702 June 2012 Paper 22 Q 1c and 9702 October 14 P23 Q2b),your help is much appreciated ^^

      Delete
    3. The explanation has been updated. See if it helps.

      Delete
  5. Can you please solve 0ctober November 2001 Paper 2 Question 7?
    Anyways great blog. Keep up the good work!

    ReplyDelete
    Replies
    1. Check solution 1093 at
      http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-232.html

      Delete
  6. these have been extremely helpful,by any chance do you have one for chemistry papers 9701?

    ReplyDelete
    Replies
    1. tried to make one at
      http://chemistry-ref.blogspot.com/

      but it requires too time work/time. if anyone's willing to contribute it would help

      Delete
  7. Dear Admin, thanks for sharing the workings. Referring to Question 2(b), the velocity vs. time plot, shouldn't the graph start at v = -8.4 m/s at t = 0 s instead, since the velocity is downward (negative)? Also, shouldn't the gradient of the graph be -9.8 m/s^2, since the gravitational acceleration is downward? Looking forward your kind advice. Thank you.

    ReplyDelete
    Replies
    1. it depends. if we take the downward direction as positive, then the graph is as above.

      Delete
  8. HOW DID YOU FIND THE MAXIMUM HEIGHT IN QUESTION 2?

    ReplyDelete

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