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 α-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 α-particle,
(ii) an approximate value for range
of α-particles in air at atmospheric pressure.
(b) Energy required to produce an ion pair in air at atmospheric
pressure is 31 eV. An α-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 α-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 α-particle as it is brought to rest in air.
Reference: Past Exam Paper – June 2010 Paper 23 Q7
Solution 635:
(a)
(i) An α-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 (106).
So, to convert Joules into
MeV, we divide the value of energy (in joules) by (1.6×10-19 × 106
=) 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 = 1×106
eV}
Number of ion pairs = (5.3 × 106)
/ 31 = 1.7 × 105
(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 × 105) / (5 × 10-2) = 3.4 × 106 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:
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) FE = Q1Q2
/ 4πϵor2 = (8.99x109) (1.6x10-19)2
/ (2.0x10-15)2 = 58N
(ii) FG = Gm1m2
/ r2 = (6.67x10-11) (1.67x10-27)2 /
(2.0x10-15)2 = 4.7x10-35N
(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 10x 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 1014 – 3 x 1016 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
DeleteThe webpage is no longer available. Where is it in my physics textbook...its not in Cambridge physics coursebook
ReplyDelete???
DeleteGood day. In solution 636 II 1. you stated that the phase difference is 360 or multiples of it. May I know why the phase difference cannot be written as 0 degree since there is no phase difference. How is 360 degree phase difference different from 0 degree? Thank you.
ReplyDeleteit's basically the same. it means that in each case, the waves are in phase
Delete