Tuesday, October 13, 2015

Physics 9702 Doubts | Help Page 210

  • Physics 9702 Doubts | Help Page 210



Question 1008: [Electromagnetism > Hall probe]
A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1.

The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off.
(a) The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved.
On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage VH.

(b) The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid.
(i) State Faraday’s law of electromagnetic induction.
(ii) On the axes of Fig. 7.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil when the current in the solenoid is switched on and then switched off.

Reference: Past Exam Paper – June 2014 Paper 41 & 43 Q7



Solution 1008:
(a) For the graph, the Hall voltage VH increases from zero when the current is switched on. VH then has a non-zero constant value. VH then returns to zero when the current is switched off.
{The Hall voltage depends on the magnetic field and the magnetic field depends on the current flowing in the solenoid. When the switch is off, no current flows, thus no magnetic field is present and VH is zero. When the switch is on, current flows and VH is not zero. VH is constant because the current is not changing (it is d.c., not a.c.).}


(b)
(i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of (magnetic) flux (linkage).

(ii) The graph consists of a pulse as the current is being switched on. There is zero e.m.f when there is current in the coil. Then, there is a pulse in the opposite direction when switching off.
{Recall Faraday’s law above. An e.m.f. is only induced when there is a ‘change’ in magnetic flux. This only occurs when the current is being switched on or off. When the current has a constant value of current (this may be zero or non-zero – as long as it is constant), no e.m.f. is induced.
The pulses are opposite because when the current is switched on, the change is from ‘off’ to ‘on’, and when the current is switched off, the change is from ‘on’ to ‘off’. This changes are opposite to each other.}












Question 1009: [Measurements > Uncertainties]
(a) State the most appropriate instrument, or instruments, for the measurement of the following.
(i) the diameter of a wire of diameter about 1 mm
(ii) the resistance of a filament lamp
(iii) the peak value of an alternating voltage

(b) The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm.
Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures.

Reference: Past Exam Paper – June 2009 Paper 21 Q1



Solution 1009:
(a)
(i) Micrometer (screw gauge) / travelling microscope

(ii) EITHER ohm-meter OR voltmeter and ammeter OR multimeter / avo on ohm setting

(iii) EITHER (calibrated) c.r.o. OR a.c. voltmeter and ×√2

(b)
Density ρ = mass m / volume V = 580 / 63 = 2.685gcm-3
% uncertainty in the mass, Δm = (10 / 580) × 100 = 1.7%
% uncertainty in the volume, ΔV = 3 × (0.1 / 6) × 100 = 5.0%
{Density ρ = m / V.
Percentage uncertainties: Δρ/ρ × 100% = [Δm/m + ΔV/V] × 100%
% uncertainty in density = 1.7 + 5.0 = 6.7%
% uncertainty in density = Δρ/ρ × 100% = 6.7% = 6.7 / 100
Density ρ = 2.685gcm-3
So, Δρ = 6.7% × ρ = (6.7 / 100) × 2.685 = 0.18}
Uncertainty in density Δρ = 0.18gcm-3
{Uncertainty is given to only 1s.f.}
So, density = 2.7 ± 0.2gcm-3











Question 1010: [Dynamics > Collisions]
Two balls X and Y are moving towards each other with speeds of 5 m s–1 and 15 m s–1 respectively.

They make a perfectly elastic head-on collision and ball Y moves to the right with a speed of 7 m s–1.
What is the speed and direction of ball X after the collision?
A 3 m s–1 to the left
B 13 m s–1 to the left
C 3 m s–1 to the right
D 13 m s–1 to the right

Reference: Past Exam Paper – June 2015 Paper 13 Q12



Solution 1010:
Answer: B.
Let the positive direction be towards the right.

For a perfectly elastic collision, both momentum and kinetic energy are conserved.

For a perfectly elastic collision,
Relative speed of approach = Relative speed of separation

‘Approach’ means getting closer to each other.
Relative speed of approach = 5 + 15 = 20 ms-1

‘Separation’ means getting away from each other.
Assume that the velocity of X is +v (assume that it continues to move to the right after the collision. If that’s not the case, the value of v would be negative, indicating that X moves to the left after the collision). Y also moves to the right with a speed of (+)7ms-1. From this assumption, for the 2 balls to be separating, Y must be faster than X, else they would be approaching. Since both are moving in the same direction, we need to take the difference between the 2 speeds to obtain the speed of separation.

Relative speed of separation = 7 – v
7 – v = 20
Speed v = 7 – 20 = – 13 ms-1
Thus, X moves at 13 m s–1 to the left.









Question 1011: [Waves > Stationary waves]
An organ pipe of length l is open at both ends. Notes are produced by the pipe when stationary waves are set up.
The speed of sound in the air column is v.
What is the lowest (fundamental) frequency of the note produced by the pipe?
A 2v / l                        B v / l                         C v / 2l                        D v / 4l

Reference: Past Exam Paper – November 2014 Paper 13 Q29



Solution 1011:
Answer: C.
Since the pipe is open at both ends, antinodes (points of maximum displacement) will be present at each end. For a note of lowest frequency, the stationary wave will consist of the 2 antinodes (mentioned above) and a single node in the middle of the pipe (at length = l / 2).

Consider a wavelength λ. This can be obtained by measuring the length between 3 consecutives antinodes (or nodes).
The distance between the 1st and the 3rd antinodes gives the wavelength.
The distance between the 1st and 2nd antinodes give half the wavelength.

Thus, in the organ pipe, l = λ / 2 for the fundamental frequency

Speed v = f λ
Fundamental frequency f0 = v / λ = v / 2l                   [since l = λ / 2]



11 comments:

  1. Assalamulaikum,

    For question 1010 why cant we use sum of momentum before collision = sum of momentum after collision?

    I have got exams on the 10th but I am struggling with momentum in past papers...
    Please help
    Jazakallah

    ReplyDelete
    Replies
    1. I can be used, but it could finally be equivalent to the relative speed formula.

      The relative speed formula is a quicker and more appropriate way to solve the problem for perfectly elastic collisions.

      Delete
  2. and even if u use relative speed of approach b4= after u dont take right or left in consideration?

    I mean b4 collision isn't Y supposed to be negative because its moving to the left?

    Jazakallah

    ReplyDelete
    Replies
    1. You need to forget the signs. As I said above, 'approach' means getting closer to each other. Since they are moving towards each other, the speeds need to be added.

      A similar logic is used for the separation.

      Delete
  3. ohk so for separation , if one is positive the other one should be negative?

    ReplyDelete
    Replies
    1. yeah, but just forget and +ve and -ve for these cases. They are getting away from each other, so they are separating. SO, we add the speeds for separation.

      A little advice: do not just memorize formulae, but try to understand what's really happening. It becomes very simple then.

      Delete
  4. Could you please explain M/J 2005 Q14 ?

    ReplyDelete
    Replies
    1. see solution 1112 at
      http://physics-ref.blogspot.com/2016/06/physics-9702-doubts-help-page-239.html

      Delete
  5. ohk,

    Thanks for the advise, Its true I do that most of the times..
    But I panic a lot during the exams, though I try not to and I can't think Straight.. I did papers from 2005 onwards, but still not quite confident for my exams..

    ReplyDelete
    Replies
    1. No problem.

      It should be enough. Try to see the mistakes you did in these papers

      Delete
    2. ok I will do so.

      Jazakallah

      Delete

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