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Monday, November 17, 2014

Physics 9702 Doubts | Help Page 13

  • Physics 9702 Doubts | Help Page 13



Question 70: [Matter > Young modulus]
Steel wire and brass wire are joined end to end and are hung vertically with steel wire attached to point on the ceiling. Steel wire is twice as long as brass wire and has half the diameter.
Large mass is hung from end of brass wire so that both wires are stretched elastically.
Young modulus for steel is 2.0 × 1011Pa and for brass is 1.0 × 1011Pa.
What is the ratio of extension of steel to extension of brass?
A 2                  B 4                  C 8                  D 16

Reference: Past Exam Paper – November 2012 Paper 11 Q25



Solution 70:
Answer: B.
Young modulus, E = stress / strain
Stress = Force, F / Area, A
Strain = extension, e / original length, L
E = (F / A) / (e / L) = FL / Ae
Extension, e = FL / AE

Let length of brass wire be L and let diameter of brass wire = d
Length of steel wire = 2L and diameter of steel wire = d/2
The weight, F is the same for both.  Area is proportional to d2.

For steel, extension, es (2L) / (d2/4) Es = 8L / (2.0x1011 d2)
For brass, extension, eb L / d2 Eb = L / (1.0x011 d2)

Ratio = es / eb = [8 / 2.0x1011] / [1 / 1.0x1011] = 4








Question 71: [Forces > Moments]
Spindle is attached at one end to centre of lever of length 1.20 m and at its other end to centre of disc of radius 0.20 m. String is wrapped round disc, passes over a pulley and is attached to 900 N weight.

What is the minimum force F, applied to each end of lever, that could lift the weight?
A 75 N                        B 150 N                      C 300 N                      D 950 N

Reference: Past Exam Paper – June 2009 Paper 1 Q13 & June 2014 Paper 11 Q13



Solution 71:
Answer: B.
Distance of Force F from the centre of the spindle = 1.20 / 2 = 0.60m.
The forces F cause a clockwise moment.
The 900N weight acts at a distance 0.20m (equal to the radius) from the spindle, causing an anticlockwise moment.

So for minimum force, the clockwise moment should be equal to the anticlockwise moment.
F (0.60) + F (0.60) = 900 (0.20).
1.20F = 180N
Minimum Force F = 180 / 1.20 = 150N









Question 72: [Forces > Resultant force]
Tractor of mass 1000 kg is connected by tow-bar to trailer of mass 1000 kg. Total resistance to motion has constant value of 4000 N. One quarter of this resistance acts on trailer.
When tractor and trailer are moving along horizontal ground at constant speed of 6 m s–1, what is the force exerted on tractor by the tow-bar?


A 0 N                         B 1000 N                                C 3000 N                    D 4000 N

Reference: Past Exam Paper – June 2014 Paper 12 Q10



Solution 72:
Answer: B.
The tractor {where the driver is} and trailer {found at the back} are moving. {The question involves equilibrium of an object moving with constant velocity}

Resultant force on the trailer is zero during motion {since the speed is constant, resultant acceleration, and hence resultant force is zero}, so the (forward) force exerted by tractor on trailer is (= 4000/4 =) 1000 N. (since one quarter of the resistance acts on the trailer)

This force is equal and opposite to the force exerted on tractor by the tow-bar.








Question 73: [Energy > Centre of mass > Potential energy]
Uniform solid cuboid of concrete of dimensions 0.50 m × 1.20 m × 0.40 m and weight 4000 N rests on flat surface with the 1.20 m edge vertical as shown in diagram 1.

What is the minimum energy required to roll cuboid through 90° to position shown in diagram 2 with the 0.50 m edge vertical?


A 200 J                        B 400 J                        C 1400 J                      D 2600 J

Reference: Past Exam Paper – June 2014 Paper 13 Q14



Solution 73:
Go to
 
 

49 comments:

  1. 43/m/j/13 no. 4(c) pls

    ReplyDelete
    Replies
    1. Explained as solution 237 at
      http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-38.html

      Delete
  2. salam! in q73, in the diagram on the right-hand side, the cuboid should not technically fall as the point of contact is in line(blue-dotted line) with the centre of mass, so why does it fall?

    ReplyDelete
    Replies
    1. Wslm. Well, it may be in some kind of equilibrium, but that equilibrium is not stable. In reality, you would expect it to fall, right?

      + the question asks for the minimum energy required to the cube to roll as shown. This minimum energy is the energy required to raise the centre of gravity as shown. Afterwards, it may fall to the right (that is, roll) or regain its original position.

      Delete
  3. there are 3 types of equilibrium: stable, unstable and neutral. the cuboid should be in stable equilibrium in position 2 as the centre of gravity is in line with point of contact of cuboid and surface. so in that position(position2), the cuboid wont fall

    ReplyDelete
    Replies
    1. consider it to be as in diagram 1 first - its vertical. A force is required to move it to the position we are discussing.

      By applying the force, the cuboid is moving. From its motion, it reaches the position being discussed.

      But if you want to consider more laws of physics, the cuboid has been made to be in a state of motion. From Newton's 1st law, the cuboid would tend to continue to be in a state of motion unless a force is applied to stop it. But no force is being applied. So, after reaching the position I drawn, the cuboid would tend to continue to move and it will fall. That's its inertia.


      Well, it's actually a simple case. Try to do it on your own in reality and you'll observe the cuboid would fall

      Delete
    2. yeah u r ryt! this is definitely an everyday example and i really wanted to know the physics behind it. JazakAllah!

      Delete
  4. is there a website like this for chemistry???

    ReplyDelete
  5. The tension in a spring of natural length lo is first increased from zero to T1, causing the length to increase to l1. The tension is then reduced to T2, causing the length to decrease to l2.
    Which area of the graph represents the work done by the spring during this reduction in length?
    May/June 2002, Question#24.

    ReplyDelete
    Replies
    1. check q488 at
      http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-94.html

      Delete
  6. can you please do question 10 b i and ii from w10qp43 ?

    ReplyDelete
    Replies
    1. See solution 603 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-119.html

      Delete
  7. Replies
    1. Go to
      http://physics-ref.blogspot.com/2015/02/9702-june-2013-paper-11-worked.html

      Delete
  8. Hi, for question 71, why isn't the two forces F considered as a couple? Why are they not treated with the torque formula of force * distance?
    Thanks

    ReplyDelete
    Replies
    1. It can be done this way. The magnitude of only one force is considered but the distance is taken as 1.20m.
      The answer would be the same.

      Delete
    2. How Does 900N Weight Produces ACW moment ?

      Delete
    3. Hey sorry I don't get why 900N weight produces ACW moment. Could someone pls explain to me? Tqvm

      Delete
    4. The 900 N force appears as the tension in the string.

      Consider the spindle as the pivot. The 900 N in the string causes an anticlockwise moment on the disc since the pivot is the spindle. the disc moves in an anticlockwise direction

      Delete
  9. Salam,Can you explain me how did you get 4 in Q72

    ReplyDelete
    Replies
    1. Wslm.
      The question says "One quarter of this resistance acts on trailer."

      Delete
  10. could you please direct me to the answers for june 2015 paper 11,12,and 13 if they're available
    thanks!

    ReplyDelete
    Replies
    1. you need to search for them. some of them are already solved, but not all.
      To do this, go to Google search and type '9702 physics reference' followed by the fist lines of the question.

      Delete
  11. 9702_s14_qp_11 question no. 11. Thanks in advance

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/11/9702-june-2014-paper-11-worked.html

      Delete
  12. can you explain banking part in circular motion

    ReplyDelete
  13. can u explain this:
    A bucket of water can be swung in a vertical circle so that the water remains in the bucket even though it is upside down at the top of the circle?

    ReplyDelete
  14. can u please answer my two qs that I send before

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2015/12/circular-motion-in-vertical-plane.html

      Delete
  15. please help me with q5b 23/mj/2013

    ReplyDelete
    Replies
    1. See solution 691 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-140.html

      Delete
  16. for solution 73 what if the question was for maximum energy ? hen we will find the difference in the potential energy of the two positions given the question itself?

    ReplyDelete
    Replies
    1. it would be irrelevant though, as any value above this minimum would cause it to roll

      Delete
  17. Replies
    1. which variant? try to see if it's available at
      http://physics-ref.blogspot.com/2014/05/physics-9702-notes-worked-solutions-for.html

      Delete
  18. Q13 speciman paper 2008 ownwards

    ReplyDelete
  19. I get almost all of your solutions.I just don't understand one thing why do you do all of this for us and how do you get all of the tough questions while we need to come here to try and understand.

    ReplyDelete
  20. For Q17, where is the pivot, is it the whole spindle is the pivot?

    ReplyDelete
    Replies
    1. The (vertical section of the) spindle may taken as the pivot.

      Delete

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