Saturday, June 28, 2014

9702 June 2012 Paper 12 Worked Solutions | A-Level Physics

  • 9702 June 2012 Paper 12 Worked Solutions | A-Level Physics


Paper 12


1 C                  11 B                21 C                31 C
2 D                  12 D                22 C                32 C
3 D                  13 A                23 C                33 D
4 B                  14 B                24 B                34 B
5 B                  15 D                25 C                35 D
6 A                  16 D                26 D                36 D
7 A                  17 B                27 C                37 D
8 D                  18 D                28 D                38 C
9 A                  19 B                29 C                39 B
10 A                20 D                30 B                40 B






Notes for some specific questions:

3 - Ans: D. {Detailed explanations for this question is available as Solution 1117 at Physics 9702 Doubts | Help Page 240 - http://physics-ref.blogspot.com/2016/07/physics-9702-doubts-help-page-240.html}




4 – Ans: B. {Detailed explanations for this question is available as Solution 287 at Physics 9702 Doubts | Help Page 48 - http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-48.html}


5 – Ans: B. A in incorrect because the measurements are not precise since the values differ significantly and they (the measurements) are not accurate since none of the values are close to the averaged one. The measurements are not always recorded to degree of precision of measuring device (1st one). Systematic errors cannot be estimated by repeating the experiment with the same equipment.


7 - Ans: A. {Detailed explanations for this question is available as Solution 518 at Physics 9702 Doubts | Help Page 101 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-101.html}



8 – Ans: D. consider motion from point of release to M1. At time=0 velocity, u = 0ms-1. So, using the formula s = ut + ½at2, we may write x = 0(t1) + ½at12. Call this equation (1). Now, consider motion from point of release to M2. Again, using the formula s = ut + ½at2, we may write x + h = 0(t2) + ½at22. This is equation (2). Since there is no x in the answers given, we need to eliminate it from our equations. This is done by taking equation (2) – equation (1) such that x + h – x = ½at22 – ½at12. So, a = 2h / (t22 – t12).

9 – Ans: A. Brick is falling, so height decreases with time. [C and D are incorrect] Gradient of the graph gives speed of fall of brick. Due to gravity, there is an acceleration of 9.81ms-2, so speed keeps on increasing until it reaches the ground. So, gradient is not constant. [B is incorrect]


10 - Ans: A. {Detailed explanations for this question is available as Solution 947 at Physics 9702 Doubts | Help Page 195 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-195.html}


11 - Ans: B. {Detailed explanations for this question is available as Solution 513 at Physics 9702 Doubts | Help Page 100 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-100.html}


12 - Ans: D. {Detailed explanations for this question is available as Solution 932 at Physics 9702 Doubts | Help Page 192 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-192.html}


13 – Ans: A. {Detailed explanations for this question is available as Solution 425 at Physics 9702 Doubts | Help Page 80 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-80.html}


14 – Ans: B. {Detailed explanations for this question is available as Solution 974 at Physics 9702 Doubts | Help Page 201 - http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-201.html}


15 – Ans: D. {Detailed explanations for this question is available as Solution 769 at Physics 9702 Doubts | Help Page 155 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-155.html}


16 - Ans: D. {Detailed explanations for this question is available as Solution 813 at Physics 9702 Doubts | Help Page 163 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-163.html}


17 - Ans: B. {Detailed explanations for this question is available as Solution 557 at Physics 9702 Doubts | Help Page 109 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-109.html}


20 - Ans: D. {Detailed explanations for this question is available as Solution 1062 at Physics 9702 Doubts | Help Page 223 - http://physics-ref.blogspot.com/2015/11/physics-9702-doubts-help-page-223.html}



22 - Ans: C. {Detailed explanations for this question is available as Solution 332 at Physics 9702 Doubts | Help Page 58 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-58.html}


23 - Ans: C. {Detailed explanations for this question is available as Solution 361 at Physics 9702 Doubts | Help Page 64 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-64.html}


24 - Ans: B. {Detailed explanations for this question is available as Solution 476 at Physics 9702 Doubts | Help Page 92 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-92.html}


25 - Ans: C. {Detailed explanations for this question is available as Solution 363 at Physics 9702 Doubts | Help Page 65 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-65.html}


26 – Ans: D. Should allow for reflection. Total distance = 2(150) = 300m. Speed = 3x108ms-1. So, time = 1x10-6s = 1μs.


27 - Ans: C. {Detailed explanations for this question is available as Solution 640 at Physics 9702 Doubts | Help Page 127 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-127.html}



28 - Ans: D. {Detailed explanations for this question is available as Solution 166 at Physics 9702 Doubts | Help Page 28 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-28.html}


29 - Ans: C. {Detailed explanations for this question is available as Solution 366 at Physics 9702 Doubts | Help Page 66 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-66.html}


30 – Ans: B. Amplitude is the maximum displacement from equilibrium position. The first rise at position X corresponds to a maxima, the fall to a minimum corresponds to the zero position and position Y corresponds to a minimum. So, the distance XY (internodal distance) is equal to λ/2. time between X and Y =0.33/330 = 1x10-3s = 1ms.  This corresponds to 2 grids on the cro.
 

31 - Ans: C. {Detailed explanations for this question is available as Solution 678 at Physics 9702 Doubts | Help Page 137 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-137.html}


32 - Ans: C. {Detailed explanations for this question is available as Solution 1114 at Physics 9702 Doubts | Help Page 239 - http://physics-ref.blogspot.com/2016/06/physics-9702-doubts-help-page-239.html}



33 – Ans: D. R=ρL/A.  A is incorrect since the charged particles (ions and electrons) are already present in the wire. B is incorrect because the wire with the smaller diameter, and hence, smaller cross-sectional area resulting in a higher resistance. So, for same current to flow in both of them [I=V/R], a larger potential difference should be applied to the smaller wire, causing the average velocity of the charged particles to be faster in the smaller wire. C incorrect because larger diameter > smaller resistance > smaller pd required > less energy provided to charged particles.



35 - Ans: D. {Detailed explanations for this question is available as Solution 364 at Physics 9702 Doubts | Help Page 65 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-65.html}


36 - Ans: D. {Detailed explanations for this question is available as Solution 426 at Physics 9702 Doubts | Help Page 81 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-81.html}


37 – Ans: D.{Detailed explanations for this question is available as Solution 789 at Physics 9702 Doubts | Help Page 159 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-159.html}


38 - Ans: C. {Detailed explanations for this question is available as Solution 182 at Physics 9702 Doubts | Help Page 29 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-29.html}




Explanations have been provided for questions that were found to be more difficult. If you are having other doubts that have not been explained here, you may ask your questions through comments.

49 comments:

  1. How about Q8 and 18 ?
    Thank you

    ReplyDelete
  2. Explanation for Qu8 is already available above.

    18 – Ans: D. Power produced by generator = VI = 230 x 32 = 7360W. Power = Work done / time. Taking KE = PE. Power available in water = mgh / t = (m/t)gh = (200)x9.81x8 = 15696W. Efficiency = 7360/15696 = 0.468 = 47%.

    ReplyDelete
  3. please explain 38

    ReplyDelete
  4. Replies
    1. Explanations for Q23 has been added

      Delete
    2. Q25 has now been explained

      Delete
    3. Explanations for Q29 is now available

      Delete
  5. 9702/13/M/J/12?
    I want worked Solution for this paper.

    ReplyDelete
    Replies
    1. the paper would be publish later.
      in the meantime, you may ask for the specific questions from that paper

      Delete
  6. May June 2013 varient 13.
    Question# 12?

    ReplyDelete
    Replies
    1. check at
      http://physics-ref.blogspot.com/2015/01/9702-june-2013-paper-13-worked.html

      Delete
  7. Q36 solution please?

    ReplyDelete
    Replies
    1. Question 36 solution has been added

      Delete
  8. For question 13, why are we subtracting the fricition from the 2kg box's weight?whereas the friction is acting on the 8kg box?

    ReplyDelete
    Replies
    1. The 8kg box and the 2kbg box form part of a single system. If the friction on the 8kg box is greater, this also causes the 2kg box to fall with a smaller acceleration, and if it is large enough, the 2kg may even remain still.

      + it is NOT being subtracted from the 2kg box. The weight of the 2kg box causes an acceleration downwards, and on the horizontal surface, this acceleration is to the right. Friction, which opposes motion, is to the left.

      So, both are forces that area acting on the whole system

      Delete
  9. Can you provide a diagram from q 15. It's confusing, where exactly is the torque acting?
    I mean, how does the clockwise direction maintain equilibrium? I don't get it.

    ReplyDelete
    Replies
    1. The explanations have been updated

      Delete
    2. Thank you. You are doing a wonderful job!

      Delete
  10. May June 2012 variant 11 questions - 36, 34, 24 please thanks!

    ReplyDelete
    Replies
    1. For Q36, see solution 762 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-154.html

      Delete
  11. please explain questions 8,12,15,23,31 of may 2012 variant 11
    thanks :)

    ReplyDelete
    Replies
    1. For Q8, see solution 784 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-158.html

      For Q12, see solution 178 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-29.html

      For Q15, see solution 777 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

      For Q23, see solution 334 at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-58.html

      Delete
  12. thank you for the above explanations. explain q31 of the same variant aswell

    ReplyDelete
    Replies
    1. See solution 793 at
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-160.html

      Delete
  13. Question 16, why is it not C? Initial acceleration should be 9.81ms^-2 right? And then this decreases as the speed increases since resultant forces (viscous drag) increase so the gradient should fall. Also, when do we know if value of g is positive or negative?

    ReplyDelete
    Replies
    1. Details have been added for Q16

      Delete
  14. October/November 2012 paper13 qu25. Could you please explain this question? I found D as the answer by finding the spring constant of each wire and then multiplying it by 2 since the lift is supported by two cables in parallel and then substituted my values in E=kL/A, where A is the total cross sectional area of both wires. It is the wrong answer, however.

    ReplyDelete
    Replies
    1. See solution 824 at
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-164.html

      Delete
  15. can you explain Q.32 ?

    ReplyDelete
  16. your question 33 explanation is incorrect. isn't it ?

    ReplyDelete
    Replies
    1. Let me know what exactly you believe to be incorrect.

      Delete

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