# Physics 9702 Doubts | Help Page 126

__Question 634: [Deformation of Solids]__**(a)**Define

(i) stress

(ii) strain

**(b)**Explain the term

*elastic limit*.

**(c)**Explain the term

*ultimate tensile stress*.

**(d)**

(i)

**Ductile**material in the form of a wire is stretched up to its braking point. On Fig.1, sketch variation with extension x of the stretching force F.
(ii) On Fig.2, sketch variation with
x of F for a

**brittle**material up to its breaking point**(e)**

(i) Explain features of the graphs
in (d) that show characteristics of ductile and brittle materials

(ii) Force F is removed from
materials in (d) just before the breaking point is reached. Describe subsequent
change in the extension for

1. ductile material

2. brittle material

**Reference:**

*Past Exam Paper – November 2011 Paper 22 Q3*

__Solution 634:__**(a)**

(i) Stress is defined as the ratio
of force to the (cross-sectional) area.

(ii) Strain is defined as the ratio
of extension (change in length) to the

__original__length.**(b)**Elastic limit is the

__point__beyond which a material does not return to its original length / shape / size when the load / force is removed.

**(c)**Ultimate tensile stress is the (maximum stress=) maximum force /

__original__cross-sectional area a wire is able to support / before it breaks.

**(d)**

(i) A straight line from (0,0) with
correct shape in the plastic region

(ii)

__Only__a straight line from point (0,0)**(e)**

(i)

__Ductile material__: Initially, the force is proportional to the extension, then a large extension occurs for a small change in force__Brittle material__: The force is proportional to the extension until it breaks

(ii)

1. The ductile material does not
return to its original length / permanent extension (as it has entered plastic region)

2. The brittle material returns to
its original length / no extension (

**as no plastic region / still in elastic region**)

__Question 635: [Radioactivity]__
One property of Î±-particles is that
they produce high density of ionisation of air at atmospheric pressure. In this
ionisation process, a neutral atom becomes an ion pair. The ion pair is a
positively-charged particle and an electron.

**(a)**State

(i) what is meant by Î±-particle,

(ii) an approximate value for range
of Î±-particles in air at atmospheric pressure.

**(b)**Energy required to produce an ion pair in air at atmospheric pressure is 31 eV. An Î±-particle has initial kinetic energy of 8.5 × 10

^{–13}J.

(i) Show that 8.5 × 10

^{–13}J is equivalent to 5.3 MeV.
(ii) Calculate, to two significant
figures, number of ion pairs produced as the Î±-particle is stopped in air at
atmospheric pressure.

(iii) Using answer in (a)(ii),
estimate average number of ion pairs produced per unit length of the track of
the Î±-particle as it is brought to rest in air.

**Reference:**

*Past Exam Paper – June 2010 Paper 23 Q7*

__Solution 635:__**(a)**

(i) An Î±-particle is EITHER a helium
nucleus OR a particle containing two protons and two neutrons

(ii) (allow any value) between 1 cm
and 10 cm

**(b)**

(i)

{To convert Joules in eV,
we divide by the charge of an electron (1.6×10

^{-19}C).
To convert eV into MeV, we
divide by (10

^{6}).
So, to convert Joules into
MeV, we divide the value of energy (in joules) by (1.6×10

^{-19}× 10^{6}=) 1.6 × 10^{–13}}
Energy in MeV = (8.5 × 10

^{–13}) / (1.6 × 10^{–13}) = 5.3 MeV
(ii)

{The energy required to
produce an ion pair in air at atmospheric pressure is 31 eV. 1 MeV = 1×10

^{6}eV}
Number of ion pairs = (5.3 × 10

^{6}) / 31 = 1.7 × 10^{5}
(iii)

{This depends on the value
answered in part (a)(ii). Let’ say we gave the range to be 5cm.

Average number of ion-pair
produced per unit length = number of ion-pairs / range (in metre)}

Average number per unit length =
(1.7 × 10

^{5}) / (5 × 10^{-2}) = 3.4 × 10^{6}m^{-1}

__Question 636: [Waves > Superposition > Double Slit experiment]__**(a)**Apparatus used to produce interference is shown in Fig.1. Apparatus is not drawn to scale.

Laser light incident on two slits.
Laser provides light of a single wavelength. Light from the two slits produces
a fringe pattern on screen. A bright fringe is produced at C and the next
bright fringe is at B. A dark fringe is produced at P.

(i) Explain why one laser and two
slits used, instead of two lasers, to produce a visible pattern on the screen

(ii) State phase difference between
the waves that meet at

1. B

2. P

(iii)

1. State the

*principle of superposition*.
2. Use principle of superposition to
explain the dark fringe at P

**(b)**In Fig.1 the distance from the two slits to screen is 1.8 m. Distance CP is 2.3 mm and the distance between the slits is 0.25 mm.

**Reference:**

*Past Exam Paper – June 2011 Paper 22 Q6*

__Solution 636:__**(a)**

(i) To produce coherent sources or
constant phase difference

(ii)

1.

{A bright fringe is seen
due to constructive interference. So, the waves from the slits should be in
phase. The phase difference is thus 360

^{o}/ 2Ï€ or a multiple of it – that is, the 2 waves are completely in phase. A phase difference of 360^{o}/ 2Ï€ means no phase difference.}
360

^{o}/ 2Ï€ allow n × (360^{o}) or n × (2Ï€)
2.

{A dark fringe is seen due
to destructive interference. So, the waves from the slits should be out of
phase. The phase difference is thus 180

^{o}/ Ï€ or (n × 360^{o}) – 180^{o}or (n × 2Ï€) – Ï€ – that is, the 2 waves are completely out of phase.}
180

^{o}/ Ï€ allow (n × 360^{o}) – 180^{o}or (n × 2Ï€) – Ï€
(iii)

1. The principle of superposition
states that when waves overlap / meet, the (resultant) displacement is the sum
of displacements of each wave

2. At P, there is a crest on a
trough

**(b)**

Calculate wavelength of
the light provided by the laser.

{CP is the width between a
dark fringe and a bright fringe. In the equation below, the fringe width is the
distance between 2 successive bright fringe or 2 successive dark fringe. So, we
need twice the distance CP.}

Wavelength Î» = ax / D = (2× 2.3×10

^{-3}) × (0.25×10^{-3}) / 1.8 = 639nm

__Question 637: [Nucleus]__
Helium nucleus contains two protons.

In a model of helium nucleus, each
proton is considered to be a charged point mass. Separation of these point
masses is assumed to be 2.0 × 10

^{−15}m.**(a)**For the two protons in model, calculate

(i) electrostatic force

(ii) gravitational force

**(b)**Using answers in (a), suggest why

(i) there must be some other forces
between the protons in nucleus

(ii) this additional force must have
short range

**Reference:**

*Past Exam Paper – June 2014 Paper 42 Q4*

__Solution 637:__**(a)**

(i) F

_{E}= Q_{1}Q_{2}/ 4Ï€Ïµ_{o}r^{2}= (8.99x10^{9}) (1.6x10^{-19})^{2}/ (2.0x10^{-15})^{2}= 58N
(ii) F

_{G}= Gm_{1}m_{2}/ r^{2}= (6.67x10^{-11}) (1.67x10^{-27})^{2}/ (2.0x10^{-15})^{2}= 4.7x10^{-35}N**(b)**

(i) The force of

__repulsion__is (much) greater than the force of__attraction__. So, there must be some other force of__attraction__to hold the nucleus together.
(ii)

{The force must have a
short range means that the force must act only within the nucleus here [that
is, the force acts within a small, limited region, unlike the force of gravitational
of Earth for instance, which act on objects like the moon, satellite, … as well
as the objects on the surface on Earth]}

Outside the nucleus, there is
repulsion between the protons.

{The attractive force must
be short range – that is, it must act only within the nucleus because, outside,
the protons repel each other – nuclei do not attract each other. If the force
was long range, all matter would stick together.}

EITHER The attractive force must act
only in the nucleus OR If it does not have a short range, all nuclei would
stick together.

__Question 638: [Electromagnetic spectrum]__
The order of magnitude of frequency
of the longest-wavelength ultraviolet waves can be expressed as 10

^{x}Hz.
What is the value of x?

A 13 B 15 C 17 D 19

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q23 & 12 Q22*

__Solution 638:__**Answer: B.**

The order of magnitude of the frequency
of longest-wavelength ultraviolet waves is the smallest frequency for ultraviolet
light.

Range of frequencies for UV light:
7.5 x 10

^{14}– 3 x 10^{16}Hz
13, 17 and 19 are not in the range.

would I need to remember the range of frequencies for all electromagnetic waves

ReplyDeleteYes. The order of magnitudes

Deletecould u pls tell me the values of the magnitudes of frequencies of EM waves

DeleteYou may obtain this on most Physics books.

DeleteAnyway, see the table at

http://csep10.phys.utk.edu/astr162/lect/light/spectrum.html

Using your textbook would be best. It's only the powers of ten - not the complete values.

thank you sir

Delete