Download Assignment 5(Fall 2007)

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Assignment 5(Fall 2007)
PHYSICS (PHY101)
MARKS: 40
Due Date: 17/01/2008
Sol.
The key idea here is that current will flow until the potential of the two
spheres becomes the same. The charges are now q1 and q2 respectively with q1  q2  Q.
After being connected with the wire, we must have:
q1 q2
Q
Q
 . Solving gives q1 
and q2 
.
R1 R2
1  R2 / R1
1  R1 / R2
Thus the charge that flows in the wire is q1  q2  Q
R1  R2
.
R1  R2
Sol.
(a) Value of g at the surface of the Earth
The magnitude of the force with which Earth attracts the body towards its center is
F  mg 
GmM E
RE 2
g
GM E
RE 2
Value of g will be constant all over the surface for a perfectly spherical Earth.
The real Earth differs from our model Earth. The Earth’s crest is not uniform. There are local
density variations everywhere. The earth is approximately an ellipsoid, flattened at the poles
and bugling at the equator. That is why g varies little bit from place to place.
b) What happens to g as you go higher and higher
Variation of g with altitude
The magnitude of the force with which Earth attracts the body towards its center is
F
GmM E
 h  RE 
2
According to Newton’s second law
F  mg h 
GmM E
 h  RE 
gh 
 gh 
2
GM E
 h  RE 
2

GM E
 h  RE 
GM E

h 
R 1 

 RE 
2
2

2
E
For h = RE , or
g

h 
1 

 RE 
2
h
= 1
RE
2

 2h 
h 
g h  g 1 
 ; g 1 

 RE 
 RE 
For h  0, g h  g
c) Variation of g depth below the surface of Earth
As the particle lies inside the shell of radius d, there is no force on the particle due to the shell.
The only force exerted comes from the sphere of radius RE-d because this force has a nature of
inverse square distance that is why the entire force due to shell on this point will be zero.
3
4
Mass of the smaller sphere  M E   E    RE  d  
3

4

but  E  M E /   RE3 
3

Now the force will be
F  mg d 
 gd 
GmM E
 RE  d 
GM E
RE2
2

d 
1 

 RE 

d 
 g 1 

 RE 
For d  RE , gd  0
The value of g is zero at the center of the Earth
So acceleration due to gravity at center will completely vanish and we will feel no force of gravity
there. At the center of the earth all forces balance each other effect.
d) The maximum value of g will be at the poles. That is because earth is not a uniform sphere
as we have discussed in point (a), the Earth’s equatorial radius is greater than its polar radius
by 21km. Thus a point at the poles is closer to the dense core of the Earth than is a point on the
equator.
If we see the following equation:
g
GM E
R2
It is very much clear that g and R (the radius) are inversely proportional.
Sol.
When the source is in motion towards a stationary observer, the effect is a shortening of the wavelength
(see Fig (b) below), for the source is following after the approaching waves, and the crest therefore
come close together. If the frequency of the source is v and its speed is vs , then during each vibration it
travels a distance vs/v and each wavelength is shortened by this amount and the frequency of the
sound heard by the observer is increased.
(a) Moving Observer, Source at Rest
(b) Moving Source, Observer at Rest
The circle represents the wave fronts, spaced one wavelength apart, travelling through the medium.
Now comes to the second part of this question. The relative velocity of source and observer are not
sufficient because wave length should be known, we can not get all the information from the relative
velocity of source and observer as seen above wavelength also changes and we need initial wavelength
to find the final wavelength.
Sol.
There is no charge just below the surface. So, by Gauss's law, the electric fields there are
zero. The electric fields just above the surface are directed radially outwards and have
magnitude
Q
4 0 R
. Putting in numbers gives E=
2
8.99 109
 3.6 1012 N / C , and
4
25 10
E  0.9 10 N / C for the two cases respectively.
12
Q.5 Diagnostic ultrasound of frequency 4.50MHz is used to examine tumors in soft tissue.
a) What is the wave length in air of such a sound wave?
b) If the speed of sound in tissue is 1500m/s, what is the wavelength of this wave in
tissue?
Sol.
Wave length in air of such a sound will be:
c  f  (c  speed , f  frequency,   wavelength)

c
343ms 1

 7.62 105 m
f 4500000 Hz
b)
Wave length of this wave in tissue will be:
c  f  (c  speed , f  frequency,   wavelength)

c
1500ms 1

 3.33 104 m  0.33mm
f 4500000 Hz
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