Sunday, May 3, 2015

Physics 9702 Doubts | Help Page 130

  • Physics 9702 Doubts | Help Page 130

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

Frequency f = 909 kHz
Wavelength λ (= c / f) = (3.0 × 108) / (909 × 103) = 330 m

(ii) Bandwidth (= 918 – 900) = 18 kHz

{Bandwidth = 2f0 where f0 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 kx2

(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 Es is expended to cause a permanent extension of the wire.
(i) On Fig.1, shade area that represents energy Es
(ii) Use Fig.1 to calculate energy Es
(iii) Suggest the change in structure of the wire that is caused by the energy Es

Reference: Past Exam Paper – November 2010 Paper 22 Q4

Solution 652:
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 kx2

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

(ii) Use Fig.1 to calculate energy Es
1.0cm2 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 = 1×10-3J = 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}
Es = 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 × 106 kg of steel is found to be contaminated with radioactive nickel-63 (6328Ni).
Total activity of the steel due to the nickel-63 contamination is 1.7 × 1014 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:
{Activity per unit mass = total activity / mass}
Activity per unit mass = (1.7 × 1014) / (2.5 × 106) = 6.8 × 107 Bq kg–1

{To convert MeV into joule: 1 MeV = (1.6×10-19) × 106 = 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 × 107) × (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.

Activity A = A0 e–λt    and      λt½ = ln 2
{A = 400 Bq kg-1, initial activity A0 = 6.8×107 Bq kg-1 and half-life t½ is 92 years.}
400 = (6.8 × 107) exp(–[ln 2 × t] / 92)
Time interval t = 1600 years

Activity A = A0 / 2n
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.}

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.

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.}


  1. Here are the remaining questions for now:

    21/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)

    1. For 21/O/N/11 Q.5(b), check solution 656 at
      and for 23/O/N/12 Q.1(e), check solution 658 at

  2. Hi admin, thank you so much!
    I owe you! :)

  3. This blog is amazing and has helped me a lot to get my queries solved. Please keep posting more for the recent papers as well.

  4. appreciate your help! Thank youuu!!


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