# Physics 9702 Doubts | Help Page 130

__Question 651: [Amplitude modulation]__
A radio station emits an amplitude-modulated
wave for transmission of music.

**(a)**

(i) State what is meant by

*amplitude-modulated*(AM) wave.
(ii) Give two reasons why
transmitted wave is modulated, rather than transmitting the information signal
directly as a radio wave.

**(b)**Variation with frequency f of amplitude A of the transmitted wave is shown in Fig.1.

For this transmission, determine

(i) wavelength of the carrier wave,

(ii) bandwidth,

(iii) maximum frequency, in Hz, of
the transmitted audio signal.

**Reference:**

*Past Exam Paper – June 2013 Paper 41 & 43 Q11*

__Solution 651:__**(a)**

(i) For an amplitude-modulated wave,
the amplitude of the carrier wave varies (in synchrony) with the displacement
of the information signal.

(ii) Examples:

More than one radio station can
operate in same region / less interference

It enables a shorter aerial

There is an increased range / less
power required / less attenuation

There is less distortion

**(b)**

(i)

Frequency f = 909 kHz

Wavelength λ (= c / f) = (3.0 × 10

^{8}) / (909 × 10^{3}) = 330 m
(ii) Bandwidth (= 918 – 900) = 18 kHz

(iii)

{Bandwidth = 2f

_{0}where f_{0}is the maximum frequency of the transmitted audio signal. Maximum frequency = bandwidth / 2}
Maximum frequency = 9000 Hz

__Question 652: [Hooke’s law]__**(a)**Metal wire has spring constant k. Forces are applied to ends of the wire to extend it within the limit of Hooke’s law.

Show that, for extension x, strain
energy E stored in the wire is given by

E = ½ kx

^{2}**(b)**Wire in (a) now extended beyond its elastic limit. Forces causing the extension are then removed.

Variation with extension x of
tension F in wire is shown in Fig.1.

Energy E

_{s}is expended to cause a permanent extension of the wire.
(i) On Fig.1, shade area that
represents energy E

_{s}
(ii) Use Fig.1 to calculate energy E

_{s}
(iii) Suggest the change in
structure of the wire that is caused by the energy E

_{s}**Reference:**

*Past Exam Paper – November 2010 Paper 22 Q4*

__Solution 652:__**(a)**

Energy, E = average force x
extension = ½ Fx

(Hooke’s law:) extension is
proportional to the (applied) force

Hence, F = kx

So, E = ½ (kx)x = ½ kx

^{2}**(b)**

(i) Correct area shaded (between the
2 curves)

(ii) Use Fig.1 to calculate energy E

_{s}
1.0cm

^{2}represents 1.0mJ or correct units used in calculations
{On the graph, 1cm
corresponds to 5 small squares. So, 1cm × 1cm contains 25 small squares.

The x-axis represent the
extension x in mm and the y-axis represent the force F in N.

So, 25 small squares =
0.1mm × 10N = (0.1×10

^{-3}) × 10 = 1×10^{-3}J = 1mJ
By counting the number of
squares between the 2 curves, it can be found that there are about 160 small
squares (it may vary from people to people). Since 25 squares represent 1mJ,
160 would represent 160 / 25 = 6.4 mJ}

E

_{s}= 6.4 ± 0.2 mJ
(iii) The arrangement of atoms /
molecules is changed

__Question 653: [Radioactivity]__
During de-commissioning of a nuclear
reactor, a mass of 2.5 × 10

^{6}kg of steel is found to be contaminated with radioactive nickel-63 (^{63}_{28}Ni).
Total activity of the steel due to
the nickel-63 contamination is 1.7 × 10

^{14}Bq.**(a)**Calculate activity per unit mass of the steel.

**(b)**Special storage precautions need to be taken when activity per unit mass due to contamination exceeds 400 Bq kg

^{−1}.

Nickel-63 is a β-emitter with a
half-life of 92 years.

Maximum energy of an emitted
β-particle is 0.067 MeV.

(i) Use answer in (a) to calculate
the energy, in J, released per second in a mass of 1.0 kg of steel due to the
radioactive decay of the nickel.

(ii) Use answer in (i) to suggest,
with a reason, whether the steel will be at a high temperature.

(iii) Use answer in (a) to determine
the time interval before special storage precautions for the steel are not
required.

**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q9*

__Solution 653:__**(a)**

{Activity per unit mass =
total activity / mass}

Activity per unit mass = (1.7 × 10

^{14}) / (2.5 × 10^{6}) = 6.8 × 10^{7}Bq kg^{–1}**(b)**

(i)

{To convert MeV into
joule: 1 MeV = (1.6×10

^{-19}) × 10^{6}= 1.6×10^{-13 }J
Energy released per second
in 1.0 kg of steel = Activity per unit mass × Energy of an emitted β-particle
in joules}

Energy released per second = (6.8 ×
10

^{7}) × (0.067 × 1.6 × 10^{–13}) = 7.3 × 10^{–7}J
(ii) This is a very small quantity
of energy, so the steel will not be warm.

(iii)

EITHER

Activity A = A

_{0}e^{–λt}__and__λt_{½}= ln 2
{A = 400 Bq kg

^{-1}, initial activity A_{0}= 6.8×10^{7}Bq kg^{-1}and half-life t_{½}is 92 years.}
400 = (6.8 × 10

^{7}) exp(–[ln 2 × t] / 92)
Time interval t = 1600 years

OR

Activity A = A

_{0}/ 2^{n}
Number of half-lives, n = 17.4

Time interval t = 17.4 × 92 = 1600
years

__Question 654: [Electromagnetism]__
Proton is moving with constant
velocity v. It enters uniform magnetic field that is normal to the initial
direction of motion of the proton, as shown in Fig.1.

A uniform electric field is applied
in same region as the magnetic field so that the proton passes undeviated
through the fields.

**(a)**On Fig.1, draw an arrow labelled E to show the direction of the electric field.

**(b)**Proton is replaced by other particles. The electric and magnetic fields remain unchanged.

State and explain deviation, if any,
of the following particles in the region of the fields.

(i) α-particle with initial velocity
v

(ii) electron with initial velocity
2v

**Reference:**

*Past Exam Paper – June 2006 Paper 4 Q8*

__Solution 654:__**(a)**An arrow labelled E should be drawn pointing down the page.

{From Flemming’s left hand
rule (thumb: force due to magnetic field, forefinger: field and middle finger:
current), the force on the proton due to the magnetic field only is upwards.

For the proton to pass
undeviated, the electric field should oppose the magnetic field.

The direction of an
electric field is drawn showing the direction of an electric force on a
positive charge. So, the direction of the electric field is downwards.}

**(b)**

(i)

For no deviation,
{magnetic force = electric force} Bqv
= qE.

The forces are independent of mass {thus, even if the mass of the α-particle is bigger than that
of the proton, it does not matter} and the charge ‘cancels’ {there is q on both sides of the equation}. So, there
is no deviation.

(ii)

The magnetic force (= Bq(2v) – this is greater than the magnetic force on a proton
moving with velocity v) is greater than the electric force. So, the
electrons deflects downwards.

{The magnetic field causes
a force downwards on the electron as the charge of the electron is negative,
unlike the charge of the proton used initially.

The downward electric
field causes a force upwards on the electron.

But since the magnetic
force is greater than {the magnetic field is unchanged, so B is unchanged but
the magnetic force is given by Bq(2v)} the electric force (the electric field
is unchanged, so the electric force is still = Eq and this is equal to Bqv) on
the electron.}

Here are the remaining questions for now:

ReplyDelete21/O/N/11 Q.5(b)

23/O/N/11 Q.6(c)

23/M/J/12 Q.2(c)

21/O/N/12 Q.6(c)

22/O/N/12 Q.3(b)

23/O/N/12 Q.1(e)

21/M/J/13 Q.6(c)(d)

22/O/N/13 Q.3(c)(ii)

23/O/N/13 Q.5(c)

For 21/O/N/11 Q.5(b), check solution 656 at

Deleteand for 23/O/N/12 Q.1(e), check solution 658 at

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Hi admin, thank you so much!

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