FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Tuesday, March 24, 2015

Physics 9702 Doubts | Help Page 93

  • Physics 9702 Doubts | Help Page 93


Question 478: [Matter > Elasticity]
Energy is stored in metal wire that is extended elastically.
(a) Explain what is meant by extended elastically.

(b) Show that SІ units of energy per unit volume are kg m–1 s–2.

(c) For a wire extended elastically, elastic energy per unit volume X is given by
X = Cε2E
where C is a constant, ε is strain of the wire, and E is Young modulus of the wire.
Show that C has no units.

Reference: Past Exam Paper – June 2013 Paper 21 Q1



Solution 478:
(a) The wire returns to its original length (not ‘shape’) when the load is removed

(b)
Units of energy: Nm or kg m2 s-2
Units of volume: m3
Unit of energy / volume: kg m2 s-2 / m3 = kg m-1 s-2

(c)
Strain (= extension / original length) ε of the wire has no units
{Young modulus E = stress / strain. Stress = Force / Area}
Units of Young modulus E: kg ms-2 m-2 = kg m-1 s-2
The units of the right-hand side terms is the same as the units of the left-hand side terms. So, C has no units.










Question 479: [Work, Energy, Power]
In many old-style filament lamps, as much as 92 J of energy is emitted as thermal energy for every 8 J of energy emitted as light.
What is efficiency of the lamp, as the percentage of electrical energy converted to light energy?
A 8 %                         B 9 %                         C 91 %                                    D 92 %

Reference: Past Exam Paper – November 2011 Paper 12 Q17



Solution 479:
Answer: A.
Useful output energy (emitted as light) = 8J
Wasted energy (emitted as thermal energy) = 92J
Total energy output = 92 + 8 = 100J

Efficiency = useful output energy / total output energy = 8 / 100 = 8%








Question 480: [Waves > Interference]
Coherent waves are produced at P and at Q and travel outwards in all directions. Line RS is halfway between P and Q and perpendicular to the line joining P and Q. Distance RS is much greater than the distance PQ.

Along which line, or lines, is interference pattern observed?
A both RS and XY
B RS only
C XY only
D neither RS nor XY

Reference: Past Exam Paper – November 2011 Paper 11 Q30 & Paper 13 Q29



Solution 480:
Answer: C.
An interference pattern consists of a series of alternating constructive interference (bright spot for light waves) and destructive interference (dark spot for light waves). An interference pattern will not be observed along a line where only constructive or destructive interference occur.

Interference pattern is only observed along line XY.

Note that the waves are produced at P and Q. The phase difference between the waves from P and Q is constant along line RS (since waves are coherent and RS is midway from P and Q). Thus, the path difference along RS is always zero – that is, only constructive interference occur along this line (for constructive interference, path difference = nλ where n = 0, 1, 2, 3, …). So, there can be no interference pattern along RS.










Question 481: [Waves > Interference]
Two identical loudspeakers are connected in series to a.c. supply, as shown.

Which graph best shows variation of the intensity of the sound with distance along the line XY?


Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q26



Solution 481:
Answer: D.
For point sources, an interference pattern of alternating constructive and destructive interference would be observed along XY. However, loudspeakers are NOT point sources of sound. So, there will be no place where the sound intensity is exactly zero. [B and C are incorrect]

The intensity of the maxima is the highest at the centre on line XY, midway between the 2 loudspeakers and the maxima gradually decrease in intensity away from the central maximum because the individual amplitudes of the two sound waves decrease with distance. The intensity of waves is inversely proportional to the square of the distance apart. [A is incorrect]

The interference pattern is represented by the rise and fall of the intensity.











Question 482: [Measurements]
A micrometer screw gauge is used to measure diameter of a copper wire.
Reading with the wire in position is shown in diagram 1. Wire is removed and the jaws of the micrometer are closed. New reading is shown in diagram 2.

What is the diameter of wire?
A 1.90 mm                  B 2.45 mm                  C 2.59 mm                  D 2.73 mm

Reference: Past Exam Paper – June 2010 Paper 12 Q1



Solution 482:
Answer: B.
When the wire is removed, the reading is 0.14mm (as shown in diagram 2) instead of being zero. This means that there is a zero error in the micrometer. When the jaws are closed, instead of giving a reading of zero, the reading already has a value of 0.14mm. So, any measurement taken is greater by a value of 0.14mm.

Reading when the wire is measured = 2.59mm

So, diameter of wire = 2.59 – 0.14 = 2.45mm



17 comments:

  1. In question 481, what difference does a point source makes? Why a non-point source does not interfere completely compared to a point source? In youngs double slit, we take slits as point sources....why cant loudspeakers be chosen as point sources.

    ReplyDelete
    Replies
    1. For point sources, the wave leave the source from a single point and travels until the point where it interferes.
      For sources that are not point sources, the wave leaves the source from different points and travel until the point where they interfere. But since the points are different, the distance that the waves need to travel are also slightly different. So, there is not complete constructive or destructive interference at a single point (as is the case for point sources.) The point of interference may slightly be displaced. So, the full interference do not full at a single point. So, the intensity cannot be zero (complete destructive interference). This explains why B and C are incorrect.
      In a loudspeaker, the sound leaves it from more than one point.

      Now, for any sources, point source or not, the amplitude of the higher order points of constructive interference decreases for the reason explained in the solution. This eliminates choice A.

      Delete
    2. But there is a MCQ saying, "Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm." Here, it is possible for destructive interference to occur to make the amplitude of the sound zero, though the sources are loudspeakers not point sources.

      I was thinking, could the explanation actually be because they are connected to an a.c. supply, and hence, at a certain time, the 2 loudspeakers have different voltages flowing through them which results in the resultant waves from the loudspeakers not being coherent, and that is why destructive interference never causes the wave to be zero.

      Honestly, I'm not sure of my own explanation and it has a big chance of being completely wrong, but somehow I'm quite convinced that the a.c. supply has some role in the question. Cheers!

      Delete
    3. The 2 loudspeakers are connected to the same a.c. supply. So, that's not an issue here.

      The idea about point sources is the correct explanation here.

      Delete
    4. Admin U are the coolest person in the world.

      Delete
    5. Why the intensity of maxima is highest at the centre?

      Delete
    6. the amplitude of each wave decreases the longer they travel.
      because the 2 waves travel the shortest distance at the centre, the intensity is highest.

      Delete
  2. So there will be no point on the line where the phase difference between the waves from the two loudspeakers will be 180 degree?

    ReplyDelete
    Replies
    1. Point?
      No sound will be heard at the point where destructive interference occurs (phase difference = 180, ...)

      Delete
    2. Instead of 'no sound', I should have said a very weak sound will be heard

      Delete
  3. 9702_s06_qp_1

    qs 4
    please

    ReplyDelete
    Replies
    1. Check solution 838 at
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-167.html

      Delete
  4. may june 2012 varient 42 paper 4
    question 4 a(ii)

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/08/9702-june-2012-paper-42-worked.html

      Delete
  5. Coherent light is incident on two fine parallel slits S1 and S2 .If a dark fringe occurs at P which of the following possible phase difference for the light waves arriving at P from S1 and S2 ?

    ReplyDelete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 93