Saturday, February 7, 2015

Physics 9702 Doubts | Help Page 62

  • Physics 9702 Doubts | Help Page 62

Question 349: [Radioactive decay]
In order to trace the line of a water-pipe buried 0.4m below the surface of a field, an engineer wishes to add a radioactive isotope to the water.
Which sort of isotope should be chosen?
            Emitter                                    Half-life
A         β                                  a few hours
B         β                                  several years
C         γ                                  a few hours
D         γ                                  several years

Reference: Past Exam Paper – J80 / II / 33 & J96 / I / 30

Solution 349:
Answer: C.
The system works as follows: The isotope in the water decays by emitting either β or γ. The water itself is in the water-pipe 0.4m below the surface.  This emission is detected at the surface.

So, detection from an isotope that decays faster would allow the task of tracing the line of the water-pipe to be completed quicker. A half-life of several years would take too long for this task. [B and D are incorrect]

Now, the radiation emitted (either β or γ) needs to be able to reach the surface. It is known that these radiations have different penetration power. If the radiation cannot reach the surface, nothing will be detected.

β-particles are stopped only by a few mm thickness of aluminium.
γ-rays have the strongest penetration power (between α, β and γ). γ-rays are stopped only by a few cm thickness of lead.

So, β-particles won’t be able to reach the detector since it would long be stopped by the 0.4m of soil in the ground. γ-rays are more likely to be detected. [A is incorrect]

Note also that both of these radiations will contaminate the water. So, it can be assumed that this water is not supplied to any household. Since β-particles are charged, their interactions with the water will be more significant.

Both of these radiations are also harmful to us. But, the ground provides enough shielding for both of them.

Question 350: [Nuclear Physics]
The fission of a heavy nucleus gives, in general, two smaller nuclei, two or three neutrons, some β-particles and some γ-radiation. It is always true that the nuclei produced
A have a total rest-mass that is greater than that of the original nucleus.
B have large kinetic energies that carry off the greater part of the energy released.
C travel in exactly opposite directions.
D have neutron-to-proton ratios that are too low for stability.
E have identical neutron-to-proton ratios.

Reference: Past Exam Paper – N82 / II / 35

Solution 350:
Answer: B.
Choice B is the only choice that is always true. The nuclei, which have relatively larger masses than the other products formed, will always have large kinetic energies compared with the others. [KE = ½ mv2]

Question 351: [Current of Electricity]
Current in the circuit is 4.8 A.

What is the rate of flow and direction of flow of electrons through the resistor R?
A 3.0 × 1019 s–1           in direction X to Y
B 6.0 × 1018 s–1            in direction X to Y
C 3.0 × 1019 s–1            in direction Y to X
D 6.0 × 1018 s–1           in direction Y to X

Reference: Past Exam Paper – June 2006 Paper 1 Q31

Solution 351:
Answer: C.
Electrons (which are negatively-charged) flow from the negative terminal of the battery. The negative terminal of the battery is represented by the shorter line at the symbol of the battery. So, the direction of flow of the electrons is from Y to X. [A and B are incorrect]

In terms of current: Charge Q = It
In terms of number of electrons: Charge Q = ne       
where n and e are the number and charge of electrons respectively

So, ne = It
Rate of low of electron = n / t = I / e = 4.8 / (1.6x10-19) = 3.0x1019 s-1  

Question 352: [Electromagnetism > Moving charges]
(a) Electron is accelerated from rest in a vacuum through potential difference of 1.2 × 104 V.
Show that final speed of the electron is 6.5 × 107ms–1.

(b) The accelerated electron now enters region of uniform magnetic field acting into the plane of the paper, as illustrated.

(i) Describe path of the electron as it passes through, and beyond, the region of the magnetic field. You may draw on Fig. if you wish.

(ii) State and explain the effect on magnitude of the deflection of the electron in the magnetic field if, separately,
1. potential difference accelerating the electron is reduced,
2. magnetic field strength is increased.

Reference: Past Exam Paper – November 2005 Paper 4 Q5

Solution 352:
½ mv2 = qV    
0.5 (9.11x10-31) v2 = (1.6x10-19) (1.2x104)
Speed v = 6.49x107 ms-1  

Path within field: the path is circular in a ‘downward’ direction
Path beyond field: the path is straight, with no ‘kink’ on leaving the field

1. The speed of the electron is smaller. So, the deflection is larger.

2. {The magnetic force becomes larger when the magnetic field strength is increased. This causes the path of the electron to be circular. Since the electron is moving in circular motion, there is a centripetal force and which given by mv2 / r. This is equal to the magnetic force. So, when the force is increased, the radius r decreases. So, the deflection is greater.}
The (magnetic) force acting on the electron is larger. So, the deflection is larger.


  1. 2/O/N/03 Q.5(c)
    02/O/N/08 Q.7
    21/O/N/09 Q.7(a),(b)

    1. 1st one is explained at

    2. for nov 2008, check pat

    3. the last one is explained at

  2. Explanation for November 2008 paper 1, q16 please?

    1. The explanation has been added as solution 528 at


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