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A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances.
The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?


Radius of the lunar orbit is the distance between the earth and the moon. 

Laser beam takes 2.56 second to reflect back, therefore distance of the moon from the earth is given by, 

straight d space equals space 1 half space ct space
space space space equals space 1 half cross times 3 cross times 10 to the power of 8 cross times 2.56 space straight m

space space space equals space 3.84 space cross times 10 to the power of 8 space straight m


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The unit of length convenient on the nuclear scale is a fermi : 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3 

where,
 r is the radius of the nucleus, A its mass number, and
ro is a constant equal to about, 1.2 f.
Show that the rule implies that nuclear mass density is nearly constant for different nuclei.
Estimate the mass density of sodium nucleus. Compare it with the average mass density of a
sodium atom obtained in Exercise. 2.27.


Radius of nucleus r is given by the relation, 

                 r = r0 A1/3

r0 = 1.2 f = 1.2 × 10-15 m 

Volume of nucleus, V = (4 / 3) π r

                                = (4 / 3) π (rA1/3)3 

                                = (4 / 3) π rA       ... (i) 

Now, the mass of a nuclei M is equal to its mass number.

That is, 

M = A amu = A × 1.66 × 10–27 kg

Density of nucleus, ρ = Mass of nucleus / Volume of nucleus

                            = A X 1.66 × 10-27 / (4/3) π r03 A

                            = 3 X 1.66 × 10-27 / 4 π r03  Kg m-3

Density of sodium nucleus is given by,

ρSodium = 3 × 1.66 × 10-27 / 4 × 3.14 × (1.2 × 10-15)3

          = 4.98 × 1018 / 21.71

          = 2.29 × 10
17 Kg m-3 




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It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Distance of the Moon from the Earth = 3.84 × 108 m 

Distance of the Sun from the Earth = 1.496 × 1011 m 


Diameter of the Sun = 1.39 × 109 m 

During a lunar eclipse, the position of the sun, moon and earth is as shown below:





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A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate
objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?
(Speed of sound in water = 1450 m s –1).


Let the distance between the ship and the enemy submarine be ‘S’.


Speed space of space sound space in space water space equals space 1450 space straight m divided by straight s space

Time space lag space between space transmssion space and
reception space of space Sonar space waves space equals space 77 space straight s

In space this space time space lag comma space
sound space waves space travel space straight a space distance space which space is space twice
space the space distance space between space the space ship space and space the space submarine space left parenthesis 2 straight S right parenthesis.

Time space taken space for space the space sound space to space
reach space the space submarine space equals space 1 divided by 2 space cross times space 77 space equals space 38.5 space straight s space

Therefore comma space

Distance space between space the space ship space and space the space submarine comma space

left parenthesis straight S right parenthesis space equals space 1450 space cross times space 38.5 space

space space space space space space equals space 55825 space straight m space equals space 55.8 space km
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The farthest objects in our Universe discovered by modern astronomers are so distant  that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

We have here, 

Time taken by quasar light to reach Earth = 3 billion years 

straight t space equals space 3.0 space cross times space 10 to the power of 9 space years space
space space equals 3.0 space cross times space 10 to the power of 9 space cross times 365 cross times 24 cross times 60 cross times 60 space straight s
space space equals space 9.46 space cross times space 10 to the power of 16 space straight s space

Speed space of space light space equals space 3 space cross times space 10 to the power of 8 space straight m divided by straight s

Distance space between space Earth space and space quasar space is comma space

equals space left parenthesis 3 space cross times space 10 to the power of 8 right parenthesis space cross times space 9.46 space cross times space 10 to the power of 16 space straight s space

equals space 283824 space cross times space 1020 space straight m space

equals space 2.8 space cross times space 10 to the power of 22 space km

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