# 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 V

_{0}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:__**(a)**

The gravitational force provides /
is the centripetal force

GMm / r

^{2}= mv^{2}/ r
So, speed, v = √(GM
/ r)

**(b)**

(i)

Increase / Change in kinetic energy
= Loss / Change in (gravitational) potential energy

½ mv

_{o}^{2}= GMm / x
v

_{o}^{2}= 2GM / x
v

_{o}= √(2GM / x)
(ii)

{The radius of the
circular orbit is r. For the 2

^{nd}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, v

_{o}{of the 2^{nd}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 / 10^{3}kHz?
A 10

^{-9}B 10^{-6}C 10^{0}D 10^{3}**Reference:**

*Past Exam Paper – November 2011 Paper 12 Q2*

__Solution 861:__**Answer: D.**

10

^{-3 }THz = 10^{-3}× 10^{12 }Hz = 10^{9 }Hz
10

^{3 }kHz = 10^{3}× 10^{3 }Hz = 10^{6 }Hz
Ratio = 10

^{-3}THz / 10^{3}kHz = 10^{9 }Hz / 10^{6 }Hz = 10^{3}

__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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