Wednesday, April 29, 2015

Physics 9702 Doubts | Help Page 126

  • 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:
(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 360o / 2π or a multiple of it – that is, the 2 waves are completely in phase. A phase difference of 360o / 2π means no phase difference.}
360o / 2π          allow n × (360o)          or n × (2π)

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 180o / π or (n × 360o) – 180o or (n × 2π) – π – that is, the 2 waves are completely out of phase.}
180o / π            allow (n × 360o) – 180o           or (n × 2π) – π

(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 λ = 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) 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.



5 comments:

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

    ReplyDelete
    Replies
    1. Yes. The order of magnitudes

      Delete
    2. could u pls tell me the values of the magnitudes of frequencies of EM waves

      Delete
    3. You may obtain this on most Physics books.

      Anyway, 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.

      Delete

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