Saturday, November 22, 2014

Physics 9702 Doubts | Help Page 20

  • Physics 9702 Doubts | Help Page 20


Question 105: [C.R.O > Time-base setting]
Diagram shows cathode-ray oscilloscope (c.r.o.) being used to measure rate of rotation of a flywheel.

Flywheel has small magnet M mounted on it. Each time magnet passes the coil, a voltage pulse is generated, which is passed to c.r.o. Display of c.r.o. is 10 cm wide.
Flywheel is rotating at rate of about 3000 revolutions per minute.
Which time-base setting will display clearly separate pulses on screen?
A 1 s cm–1                   B 10 ms cm–1               C 100 μs cm–1            D 1 μs cm–1

Reference: Past Exam Paper – November 2010 Paper 11 Q5



Solution 105:
Answer: B.
The flywheel rotates 3000 revolutions in 1 minute (60 seconds).
Time (in seconds) for 1 revolution = 60 / 3000 = 0.02s = 20ms

Consider the time-base settings:
1s cm-1: In 1 cm, there would be (1 / 0.02 =) 50 pulses. This is too many pulses in a single cm.

10ms cm-1: In this case, in every (20 / 10 =) 2 cm, one pulse is seen. So, on the screen a maximum of 5 pulses may be seen. This is a correct time-base setting.

100μs cm-1: {100μs = 100x10-6s = 1x10-4s} With this setting, a single pulse would require (0.02 / 1x10-4 =) 200cm. No pulse would be seen on the screen.

1μs cm-1: With this setting, a single pulse would require (0.02 / 1x10-6 =) 20000cm. No pulse would be seen on the screen.









Question 106: [Oscillations & Electromagnetism]
Magnet is suspended vertically from fixed point by means of spring, as shown.

One end of magnet hangs inside a coil of wire. Coil is connected in series with a resistor R.
(a) Magnet is displaced vertically a small distance D and then released.
Fig. shows variation with time t of vertical displacement d of magnet from its equilibrium position.


(i) State and explain, by reference to electromagnetic induction, the nature of oscillations of the magnet.
(ii) Calculate angular frequency ω0 of oscillations.

(b) Resistance of resistor R is increased.
Magnet is again displaced a vertical distance D and released.
On Fig, sketch variation with time t of displacement d of magnet.

(c) Resistor R in Fig. is replaced by variable-frequency signal generator of constant r.m.s. output voltage.
Angular frequency ω of generator is gradually increased from about 0.7ω0 to about 1.3ω0, where ω0 is angular frequency calculated in (a)(ii).
(i) On axes of Fig, sketch graph to show variation with ω of amplitude A of oscillations of magnet.
(ii) State name of phenomenon illustrated in graph.
(iii) Briefly describe one situation where phenomenon named in (ii) is useful and one situation where it should be avoided.

Reference: Past Exam Paper – June 2007 Paper 4 Q7



Solution 106:
(a)
(i)
The oscillations are damped / amplitude decreases. As the magnet moves, flux is cut by the coil and an e.m.f / current is induced in the coil, causing energy loss in the load OR a force on the magnet. This energy is derived from the oscillations of the magnet OR the force opposes the motion of the magnet.

(ii)
(From graph,) Period T = 0.60s
Angular frequency ω0 (= 2π / T) = 10.5 rad s-1

(b)
The sketch should consist of a sinusoidal wave with the period unchanged {same as the previous one – they intercept the x-axis at the same points} or slightly smaller. It has the same initial displacement, but there is less damping {since there is less damping, the displacements are greater than the previous graph at all the other points – other than the initial displacement. This implies for both the +ve and –ve displacements}.
{A sample of the required is drawn below. Note that the graph you draw SHOULD DEFINITELY LOOK BETTER THAN THIS. I could not draw the graph properly on the computer. Sorry}




(c)
(i)
The general shape of the sketch is a peaked curve with the peak at ω0 and the amplitude is never zero.




(ii)
Resonance

(iii)
Useful: e.g. a child on swing, microwave oven heating
Avoid: e.g. vibrating panels, vibrating bridges









Question 107: [Resistance > Filament Lamp]
Graph shows variation with potential difference (p.d.) of current in a lamp filament.

Which statement explains shape of this graph?
A As filament temperature rises, electrons can pass more easily through the filament.
B It takes time for filament to reach its working temperature.
C The power output of filament is proportional to the square of the current in it.
D The resistance of filament increases with a rise in temperature.

Reference: Past Exam Paper – November 2011 Paper 12 Q33



Solution 107:
Answer: D.
Ohm’s law: V = IR

At any point on the graph, the ratio of (I / V) gives the value of (1 / R) where R is the resistance of the filament lamp.
It can be seen from the graph that this ratio decreases as the p.d. increases [the gradient decreases]. So, this means that the resistance R of the filament lamp increases.

When resistance increases, less current can flow. [A is incorrect]

Note that when current passes through the filament, its temperature increases. Therefore, from the graph, it can be concluded that the resistance of the filament increases with a rise in temperature.
[B and C do not explain the shape of the graph]








Question 108 : [Units]
Speed v of liquid leaving a tube depends on the change in pressure ΔP and density ρ of the liquid. The speed is given by equation
v = k (ΔP / ρ)n
where k is a constant that has no units.
What is value of n ?
A ½                 B 1                  C 3/2               D 2

Reference: Past Exam Paper – June 2014 Paper 11 Q3


Solution 108:
Answer: A.
Units of velocity = ms-1
The overall units of the right-hand side of the equation should also be ms-1 for the equation to be homogeneous.

k has no units and be neglected for dimension analysis.

Change in Pressure, ΔP = Δhρg
So, ΔP / ρ = Δhρg / ρ = Δhg

Units of ΔP / ρ:
[ΔP / ρ] = [Δhg] = m ms-2 = m2s-2

Consider the units: ms-1 = [m2s-2]½   
Therefore, for the equation to be homogeneous, n should be ½.




5 comments:

  1. In question 105 why didn't you calculate in centimetres? how did you do A and B?

    ReplyDelete
    Replies
    1. Time for 1 revolution = period of 1 pulse = 20ms

      For A,
      the setting is chosen such that 1cm (1 square on the screen) represents a time of 1s. But 1 pulse takes 20ms. In 1 second, there are 1/0.02 = 50 pulses. 20ms = 0.02s

      For B,
      1cm represents 10ms. The period is 20ms. So, in every 2cm (2 squares), a pulse is seen. Since the screen is 10cm, 5 pulses will be seen. This is good enough to analyse the pulses.



      Calculating in seconds seemed easier. And actually, I just started solving without even thinking about whether it's better to calculate in cm or second. But, now I think that it's really better to calculate in second.

      Delete
  2. For question 106 b), why is there less damping as resistance of resistor R is increased ?

    ReplyDelete
    Replies
    1. From Ohm’s law: I = V / R
      Increasing the resistance R causes the current induced in the circuit to decrease.

      Power dissipated = (I^2)R.

      The power dissipated depends on the square of current I. So, as I decreases, the power dissipated also decreases, even if the resistance R increases. The dependence of current I is higher than on the resistance.

      So, less energy is loss. Since this energy is derived from the oscillation of the magnet, less energy is now taken from the oscillation. So, there is less damping.

      Delete

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