Sunday, February 14, 2016

Physics 9702 Doubts | Help Page 235

  • Physics 9702 Doubts | Help Page 235

Question 1100: [Waves > General Wave Properties]
A laser emits light of wavelength 600 nm.
What is the distance, expressed as a number of wavelengths, travelled by the light in one second?
A 5 × 108                     B 5 × 1011                    C 5 × 1014                    D 5 × 1017

Reference: Past Exam Paper – November 2008 Paper 1 Q1

Solution 1100:
Answer: C.
Wavelength λ of the light = 600 nm = 600×10-9 m

The speed of light is 3.0×108 ms-1. That is, in 1 second, the light travels a distance equal to 3.0×108 m. We need to find the number of wavelength in this distance.

Number of wavelength = (3.0×108) / (600×10-9) = 5×1014

Question 1101: [Electromagnetism]
Electrons are moving through a vacuum in a narrow beam. The electrons have speed v.
The electrons enter a region of uniform magnetic field of flux density B. Initially, the electrons are travelling at a right-angle to the magnetic field.
The path of a single electron is shown in Fig. 7.1.

Fig. 7.1
The electrons follow a curved path in the magnetic field.
A uniform electric field of field strength E is now applied in the same region as the magnetic field.
The electrons pass undeviated through the region of the two fields.
Gravitational effects may be neglected.
(a) Derive a relation between v, E and B for the electrons not to be deflected. Explain your working.      

(b) An α-particle has speed v and approaches the region of the two fields along the same path as the electron. Describe and explain the path of the α-particle as it passes through the region of the two fields.

Reference: Past Exam Paper – November 2010 Paper 43 Q7

Solution 1101:
{For the electron to pass undeviated,}The force due to E-field is equal and opposite to force due to B-field
Eq = Bqv                    
v = E / B

EITHER charge and mass are not involved in the equation in (a)
OR FE and FB are both doubled {as the charge of an α-particle is twice that of an electron}
OR E, B and v do not change           
so there is no deviation

Question 1102: [Matter > Deformation]
A spring of original length 100 mm is compressed by a force. The graph shows the variation of the length L of the spring with the compressing force F.

What is the energy stored in the spring when the length is 70 mm?
A 0.090 J                     B 0.21 J                       C 0.27 J                       D 0.63 J

Reference: Past Exam Paper – June 2008 Paper 1 Q23

Solution 1102:
Answer: A.
The original length of the spring is 100 mm. The graph also illustrates this where the compressing force F is zero.

When compressed to 70 mm, the length of the spring changes from 100 mm to 70 mm. To calculate the energy stored, we need to consider the area of the triangle formed between 70 mm and 100 mm.

At a length of 70 mm, the corresponding compressing force is 6 N.
Energy stored = ½ × ({100 – 70} × 10-3) × (6 – 0) = 0.090 J


  1. Hi, can you please provide detailed references like these for Paper 5 as well? Your blog has been very helpful. Thank you

    1. I intended to do this initially but I'm a quite busy right now. So I don;t think it's going to be available anywhere soon.

  2. Hi, can you please explain why the area of the force by length graph gives energy stored?
    Thank you


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 235