# 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**In 1 cm, there would be (1 / 0.02 =) 50 pulses. This is too many pulses in a single cm.

^{-1}:**10ms cm**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.

^{-1}:**100**

**μs cm**{100μs = 100x10

^{-1}:^{-6}s = 1x10

^{-4}s} With this setting, a single pulse would require (0.02 / 1x10

^{-4}=) 200cm. No pulse would be seen on the screen.

**1**

**μs cm**With this setting, a single pulse would require (0.02 / 1x10

^{-1}:^{-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 ω

_{0}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 0.7ω

_{0}to about 1.3ω_{0}, where ω_{0}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 ω

_{0}(= 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 ω

_{0}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}= m

^{2}s

^{-2}

Consider the units:

**ms**

^{-1}= [m^{2}s^{-2}]^{½}Therefore, for the equation to be homogeneous, n should be ½.

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

ReplyDeleteTime for 1 revolution = period of 1 pulse = 20ms

DeleteFor 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.

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

ReplyDeleteFrom Ohm’s law: I = V / R

DeleteIncreasing 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.

Thank you .

ReplyDeletethanks

ReplyDeletefor

qn 108

I don't understand why choose your in millimeter and how you got 1/2

ReplyDeleteNote that it is NOT millimetre.

Deleteunit of h = m

Unit of g = m s^-2

So, unit of hg = m . m s^-2

For the equation to be homogeneous

the overall base unit on the left hand side should be the same as the overall unit on the right hand side.

To make the unit equal to that of speed (m s^-1), the powers of 2 should be reduced to one. This is true if then value of n is ½.