Q43 of 44 Page 9

(a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° fromthe earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.


(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.


(Comment : This is to illustrate why a telescope can magnify planets but not stars.)


(a) Parsec is defined as the distance at which 1 A.U. long arc subtends an angle of 1s. The angle subtended by an arc of length L and radius r is given as:


Here L = 1 A.U., θ = 1s and r = 1 parsec. Therefore





(b) Given:


Sun’s diameter as seen from earth =


Star’s distance from earth = 2 parsecs


Magnification of the telescope = 100


Minimum resolution eye due to atmosphere = 1’


The size of the star as seen from the earth can be considered by placing the sun at the same distance from earth as the star since it is sun like. Thus if sun appears to be in diameter from earth at 1 A.U. then for 2 parsecs or as calculated in (a) from 2 x 2.06 x 105 A.U. its angular diameter will be,



When seen from the telescope, the star will be 100 times magnified and therefore,



Converting to arc minutes the angular diameter will be 7.25 x 10-5 arc minutes. Since due to atmospheric fluctuations, the eye cannot resolve below 1 arc minute, the sun like star appears to be 1 arc minute in angular diameter.


(c) Given:


Ratio between mar’s diameter and earth’s diameter =


Let De, Dma and Ds be the diameters of earth, mars and the sun. Then we have,



Also, we know that



Therefore, we have,



To find the size of mars as seen from earth, we first place mars at distance equal to the earth-sun distance form earth i.e. 1 A.U. At 1 A.U. the sun appears to be in diameter, therefore for mars,



So, at mars’ original distance of A.U. mars will have a diameter of



With the telescope used in the previous problem,



Therefore, after magnification mars appears to be 30 arc minutes which is more than 1 arc minute and hence is not influenced by atmospheric fluctuations and is magnified. Thus, the distance of stars are so large in comparison to planets that stars cannot be magnified on earth while planets can be. To study stars, the stars are made brighter by allowing more light to enter the telescopes by using different designs.


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40

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

 41

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T =


where k is a dimensionless constant and g is acceleration due to gravity.


 42

In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:


(a) Why do we dissolve oleic acid in alcohol?


(b) What is the role of lycopodium powder?


(c) What would be the volume of oleic acid in each mL of solution prepared?


(d) How will you calculate the volume of n drops of this solution of oleic acid?


(e) What will be the volume of oleic acid in one drop of this solution?


 44

Einstein’s mass - energy relation emerging out of his famous theory of relativity relates mass (m ) to energy (E ) as E = mc2, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV= 1.6×10-13J; the masses are measured in unified atomic mass unit (u) where

1u = 1.67 × 10-27 kg.


(a) Show that the energy equivalent of 1 u is 931.5 MeV.


(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.