Thursday, June 11, 2015

Physics 9702 Doubts | Help Page 170

  • Physics 9702 Doubts | Help Page 170

Question 851: [Waves > Electromagnetic spectrum]
The number of wavelengths of visible light in one metre is of the order of
A 104.                          B 106.                          C 108.                          D 1010.

Reference: Past Exam Paper – June 2005 Paper 1 Q24

Solution 851:
Answer: B.
The range of wavelength of visible light for a typical is can be considered to be around 400 – 700 nm.

Consider the mean value: 550 nm = 550 × 10-9 m

Number of wavelengths in one metre = 1 / (550 × 10-9) ≈ 2 × 106

Question 852: [Waves > Polarisation]
If a wave can be polarised, it must be
A a longitudinal wave.
B an electromagnetic wave.
C a sound wave.
D a transverse wave.

Reference: Past Exam Paper – June 2011 Paper 12 Q27

Solution 852:
Answer: D.
Polarisation is the restriction of the vibrations in a wave so that the vibrations occur in a single plane. Polarisation is a property exhibited by transverse waves only (since the direction of vibration is perpendicular to the direction of propagation in a transverse wave – that is, there are 2 planes).
This applies to all transverse waves, not only an EM wave.

Polarisation does not occur for longitudinal waves such as sound waves (since the direction of vibration is parallel to the direction of propagation in a longitudinal wave – that is, there is only 1 plane).

Question 853: [Dynamics > Newton’s laws of motion]
A shot-put champion accelerates a 7.0 kg metal ball in straight line. The ball moves from rest to speed of 12 m s–1 in a distance of 1.2 m.
What is the average resultant force on the metal ball?
A 70 N                        B 210 N                      C 420 N                      D 840 N

Reference: Past Exam Paper – June 2014 Paper 13 Q17

Solution 853:
Answer: C.
Resultant force F = ma

Equation of uniformly acceleration motion: v2 = u2 + 2as
Initial speed, u = 0 and final speed, v = 12ms-1. Distance s = 1.2m

122 = 0 + 2a(1.2)
Acceleration, a = 60ms-2

Resultant force = ma = 7.0 (60) = 420N

Question 854: [Nuclear Physics]
Two α-particles with equal energies are deflected by a large nucleus.
Which diagram best represents their paths?

Reference: Past Exam Paper – November 2013 Paper 13 Q40

Solution 854:
Answer: A.
To correctly answer this question, we need to consider the paths of the α-particles independently – that is, one at a time. Alternatively, it is correct to assume that the effect of the repulsion between the α-particles is much smaller compared to the effect of the large nucleus, and thus can be neglected.

The large nucleus may cause an attractive force (if it is negatively charged) or a repulsive force (if it is positively charged).

If it is positively charged, the large nucleus would deflect the α-particles away from it. The closer the α-particle, the stronger is the repulsive force and thus, the deflection is larger. [A is correct and B is incorrect]

If it is negatively charged, the large nucleus would deflect the α-particles towards it. However, attraction occurs TOWARDS the large nucleus – that is, the path would be more circular than just a deflection as is the case when the large nucleus is positively charged. [C is incorrect]
Again, the closer the α-particle, the greater the attraction. [D is incorrect]


  1. Hi,could you please answer 9702/01/O/N/05 Q24?

    1. Check solution 1055 at

  2. Can you please answer 9702/01/O/N/06 Q25?

    1. See solution 1109 at


If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 170