Saturday, June 13, 2015

Physics 9702 Doubts | Help Page 172

  • Physics 9702 Doubts | Help Page 172

Question 859: [Matter > Deformation]
The stress-strain graph for glass rod, up to the point at which it breaks, is shown below.

Which statement about the glass rod is correct?
A Hooke’s law is obeyed for all values of stress up to the breaking point.
B The glass is ductile.
C The glass shows plastic deformation.
D When the cross-sectional area of the rod is doubled, the ultimate tensile stress of the rod is halved.

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q20

Solution 859:
Answer: A.
Hooke’s law: F = ke
From Hooke’s law, provided the limit of proportionality has not been exceeded, the extension is linearly proportional to the force applied – that is, a graph of force against extension would be a straight line.

Stress = Force / Area
Strain = extension / Original length

Since the stress and strain are directly proportional to the force and extension respectively, a stress-strain graph would also be a straight line up to the breaking point.

Thus, for the glass rod, Hooke’s law is obeyed for all values of stress up to the breaking point. [A is correct]

Glass is brittle and it would break if enough stress is applied – it does not show plastic deformation. [B and C are incorrect]

The ultimate tensile strength is the maximum stress that a material can withstand before breaking. Usually, this is a constant value. If the cross sectional area is doubled, then the maximum force that can be applied would be twice and thus, the ultimate tensile strength is still constant. [D is incorrect]

Question 860: [Gravitation]
Mass M of a spherical planet may be assumed to be point mass at centre of planet.
(a) A stone, travelling at speed v, is in a circular orbit of radius r about the planet, as illustrated in Fig.1. 

Show that speed v is given by expression
v = (GM / r)
where G is the gravitational constant

(b) A second stone, initially at rest at infinity, travels towards the planet, as illustrated in Fig.2. 

The stone does not hit surface of the planet.
(i) Determine, in terms of gravitational constant G and mass M of planet, the speed V0 of the stone at a distance x from centre of planet. You may assume that gravitational attraction on stone is due only to the planet.

(ii) Use answer in (i) and expression in (a) to explain whether stone could enter a circular orbit about the planet

Reference: Past Exam Paper – June 2014 Paper 42 Q1

Solution 860:
The gravitational force provides / is the centripetal force
GMm / r2 = mv2 / r
So, speed, v = (GM / r)

Increase / Change in kinetic energy = Loss / Change in (gravitational) potential energy
½ mvo2 = GMm / x
vo2 = 2GM / x
vo = √(2GM / x)

{The radius of the circular orbit is r. For the 2nd stone to enter a circular orbit, its distance x from the centre of the planet should be equal to the radius of the circular orbit. That is, x = r.}
Speed, vo {of the 2nd stone} is (always) greater than the speed, v (for x = r).
So, the stone could not enter into a circular orbit

Question 861: [Measurement]
What is the ratio 10-3 THz / 103 kHz?
A 10-9                          B 10-6                          C 100                           D 103

Reference: Past Exam Paper – November 2011 Paper 12 Q2

Solution 861:
Answer: D.
10-3 THz = 10-3 × 1012 Hz = 109 Hz
103 kHz = 103 × 103 Hz = 106 Hz

Ratio = 10-3 THz / 103 kHz = 109 Hz / 106 Hz = 103

Question 862: [Waves > Interference > Double slit]
Using monochromatic light, interference fringes are produced on screen placed a distance D from a pair of slits of separation a. Separation of the fringes is x.
Both a and D are now doubled.
What is the new fringe separation?
A x / 2                         B x                              C 2x                            D 4x

Reference: Past Exam Paper – June 2010 Paper 12 Q25

Solution 862:
Answer: B.
For double slits: Wavelength λ = xa / D
where x: separation of fringes,
a: slit separation / distance between slits,
D = distance from slits to screen.

So, separation of fringes, x = λD / a

When both a and D are doubled, the fringe separation x = λD / a

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