# 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
× 10

^{8}B 5 × 10^{11}C 5 × 10^{14}D 5 × 10^{17}**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 3.0×10

^{8}ms^{-1}. That is, in 1 second, the light travels a distance equal to 3.0×10^{8}m. We need to find the number of wavelength in this distance.
Number of wavelength = (3.0×10

^{8}) / (600×10^{-9}) = 5×10^{14}

__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:__**(a)**

{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

**(b)**

EITHER charge and mass are not
involved in the equation in (a)

OR F

_{E}and F_{B}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
Hi, can you please provide detailed references like these for Paper 5 as well? Your blog has been very helpful. Thank you

ReplyDeleteI 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.

DeleteHi, can you please explain why the area of the force by length graph gives energy stored?

ReplyDeleteThank you