2.1. Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to…..m3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to      …(mm)2

(c) A vehicle moving with a speed of 18 km h-1 covers….m in 1 s

(d) The relative density of lead is 11.3.  Its density is ….g cm-3 or ….kg m-3.              

Sol. (a) The volume of a cube of side 1cm is given by, V= 1cm3

             1 cm = (10-2)3 m3 = 10-6 m3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0cm is given by

A= Area of two caps + Curved Surface Area

  = 2Πr2 + 2Πrh

  = 2Πr(r+h)

Here, r = 2cm = 20mm, h= 10cm = 100mm

Therefore, Surface Area of cylinder = A = 2x (22/7) x20(20+100) (mm) 2

                                                                = 15085 mm2 = 1.5085 x 104 (mm) 2

                                                                = 1.5 x 104 mm2

 (c) Here, speed of the vehicle v = 18 km/hr = (18×1000)/3600 = 5m/s

Distance = Speed x Time

Thus, Distance travelled in 1s, x = vt = 5×1 = 5m

(d) Relative Density of Lead = 11.3

Density of water = 1 g/cm3

We know that relative density of lead = Density of Lead / Density of Water

Thus, Density of Lead = 11.3 x 103 kg/m3 = 1.13 x 104 kg /m3

2.2. Fill in the blanks by suitable conversion of units

(a) 1 kg m2 s-2 = ….g cm2 s-2

 (b) 1 m =…..  ly

(c) 3.0 m s-2 =…. km h-2

(d) G = 6.67 × 10–11 N m2 (kg)-2 =…. (cm) 3 / s2g-1

Sol: (a) 1kg-m2/s2 = 1(103g) (102cm2)/s2

(Since, 1kg = 1000g = 103g, 1m = 100cm =102cm)

= 103 x 104 g-cm2/s2

= 107g-cm2/s2

(b) We know that 1 light year = 9.46 x1015m

1m = 1/ (9.46 x 1015) ly

= 1.057 x 10-16 ly

(c) 3.0 m/s2 = 3 x 10-3 km x )-2              

 (Since, 1km = 1000m, 1m = 10-3km & 1hr = 60 x 60 s, 1s =  )

= 3 x 10-3 x (3600)2km/h2

= 3.88 x 104 km/h2

= 3.9 x 104 km/h2

(d) G= 6.67 x 10-11 N-m-2/kg2

         = 6.67 x 10-11 (105dyne)(102cm)2(103g)-2

= 6.67 x 10-11 x 105 x 104 x 10-6 dyne-cm2/g2

= 6.67 x 10-8 cm3/g-s2

2.3. A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s-2.  Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s.  Show that a calorie has a magnitude 4.2 α-1 β-2 γ-2 in terms of the new units.

Sol: The dimensional formulae of energy = [ML2T-2]

 Let M1, L 1, T1 and M2, L2, T2 are the units of mass, length and time in given two system.

M1 = 1kg , M2 = αkg

L1 = 1m , L2 = βm

T1 = 1s , T2 = γs

For any physical quantity , the product of its magnitude and unit is always constant .

Thus , m1u1 = m2u2

 = 4.2 x{ [M1L12T1-2]/[M2L22T2-2]}

= 4.2 x { (1/α) x (1/β)2 x (1/γ)-2 }

 m2 =  1 cal = 4.2 α-1β-2 γ2 new unit

2.4. Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:

(a) Atoms are very small objects   

 (b) A jet plane moves with great speed

 (c) The mass of Jupiter is very large

(d) The air inside this room contains a large number of molecules

 (e) A proton is much more massive than an electron        

(f) The speed of sound is much smaller than the speed of light.

Sol: (a) The above statement is true. Any dimensional physical quantity can be called large or small by specifying a standard for comparison. For example 1m is very small compared to 1 light year but very large in comparison to 1 manometer. The size of an atom is very small in comparison to salt grain or sugar cubes.

(b) A jet plane moves much faster  than a train.

(c) The mass of Jupiter is much  larger  as compared to Mercury .

(d) The air inside this room contains a large number of molecules than 1 mole of gas.

(e) The statement is already correct.

(f) The statement is already correct.

2.5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Sol:  The speed of light in vacuum according to new unit = 1 (new unit of length /s)

Time taken = (8×60 + 20)s = (480+20)s = 500s

Distance travelled in this time interval = c x t = 1 x 500 new unit length.

2.6 Which of the following is the most precise device for measuring length:

(a) A Vernier callipers with 20 divisions on the sliding scale

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale

 (c) An optical instrument that can measure length to within a wavelength of light?

Sol : The instrument whose least count is minimum is called the most precise instrument .

(a) Number of divisions on Vernier scale = 20

Main Scale division (MSD) = 1mm                      

Since 20 divisions on Vernier scale will be equal to the 19 divisions on main scale

Thus Vernier Scale division (VSD) = (19/20) MSD

Least Count of Vernier Calliper’s = 1MSd – 1VSD

                                                            = 1 MSD -19/20 MSD

                                                            = (1/20) mm = (1/200) cm = .005cm

(b) Pitch of Screw gauge = 1mm

Number of divisions on circular scale = 100

Least Count of screw gauge = Pitch / Number of Divisions on circular scale

                                                   = (1/ 100) mm = (1/1000) cm

                                                   = 0.001 cm

(c) Wavelength of light (λ) = 10-7m = 10-5 cm = 0.00001 cm

Since the given optical instrument can measure length to that of wavelength of light, thus the least count of the given instrument is = 0.00001cm

The least count is minimum for given optical instrument.

Therefore, the given instrument is most precise.

2.7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

Sol :  Magnification of microscope = 100

Observed width of hair = 3.5 mm

Estimate on thickness of hair is given by

Magnification = Observed width /Real width

Real width = Observed width / Magnification

                     = 3.5/100 = 0.035 mm

2.8 Answer the following:

 (a)You are given a thread and a metre scale.  How will you estimate the diameter of the thread?

(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c)The mean diameter of a thin brass rod is to be measured by Vernier callipers.  Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Sol: (a) The diameter of a thread is so small that it cannot be measured using a metre scale. To measure the diameter of the thread, we wind a number of turns of thread on the meter scale so that the turns are closely touching one another.

Counting the number of turns (n) and measure of length (l) of the coiled thread using meter scale.

Diameter of thread = (Length of coiled thread/Number of turns)

(b) Yes , the accuracy of the gauge can be increased by increasing the number of division on circular scale as the least count of screw gauge is given by

Least Count = Pitch/Number of Division on Circular scale                                           

From the formulae it is clear that by increasing the number of divisions on circular scale. Least count can be decreased, hence the accuracy will increase.

It is correct theoretically but practically it may not be possible because its difficult to take the reading precisely as the resolution of human eye is not so high.

(c)  The mean diameter of a thin brass rod measured by a Vernier callipers from a set of 100 measurements is more reliable than the diameter obtained from a set of 5 measurements because the probability of making a positive random error is equal to the probability of a negative random error.

Therefore, in a large number of measurement these errors cancel each other and we get more reliable value.

2.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide.  The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2.  What is the linear magnification of the projector-screen arrangement?

Sol :  Area of object = 1.75cm2 =1.75×10-4 m2

Area of image = 1.55m2

Area of Magnification = {\frac {Area of image} {Area of object}} = {\frac {1.55} {1.75{\times }{{10}^{-4}}}}  = 8857

Linear Magnification = {{\sqrt {Area of magnification}}}

                                     = {{\sqrt {8857}}}   = 94.1

2.10 State the number of significant figures in the following: (a)  0.007 m2 (b)  2.64 × 1024 kg (c)  0.2370 g cm-3 (d)  6.320 J (e)  6.032 N m-2   (f)   0.0006032 m2

Sol: The number of significant figures in the given in the given quantities are given below .

(a) In 0.007 , the number of significant figure is 1.

 Because in a number less than 1 , the zero’s on the right of the decimal point but to the left of the first non-zero  are not significant.

(b) In 2.64 x 1024 , the number of significant figures are 3.

Because all non-zero digits are significant , power of 10 are not counted as significant figures.

(c) In 0.2370, the number of significant figures is 4 as well as non-zero digit left to decimal and trailing zero are significant .

(d)In 6.320 , the number of significant figures is 4 ,( reason is same as part c)

(e)In 6.032 , the number of significant number is 4 ( reason is same as part c)

(f) In 0.0006032 , the number of significant figures is 4 ( reason is same as part a)

2.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Sol: Given, l = 4.234m, b = 1.005 m, t = 2.01 cm= 0.0201 m

Area of sheet (A) = 2(l x b + b x t + t x l)

                               = 2[(4.234 x 1.005 ) + (1.005 x 0.0201) + (0.0201 x 4.234 )

                               = 2 x 4.3604739

                               = 8.7209478 m2

As thickness has least significant figures 3, thus rounding off area up to 3 significant figures, we get

Area of Sheet = 8.72

Volume of Sheet (V) = l x b x t

                                     = 4.234 x 1.005 x 0.0201

                                     = 0.0855289

Rounding off to 3 significant figures, we get

Volume = 0.0855 m3

2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box.  What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Sol:  Given, Mass of box (m) = 2.3 kg

                      Mass of first gold piece (m1) = 20.15 g

                                                                         = 0.02015 kg

Mass of second gold piece (m2) 20.17g = 0.02017 kg

(a) Total Mass of the box (M) = m + m1 + m2

                                               = 2.3 + 0.02015 + 0.02017

                                               = 2.34032 kg

As the mass of the box has least decimal place, thus total mass of the box will have only one decimal place.

We get , Total Mass of the box = 2.3kg

(b) Difference in masses of gold pieces (ᅀm) = m2 – m1

                                                                            = 20.17 – 20.15 = 0.02 g

(the mass of gold has decimal up to two decimal places , therefore it is correct up to two decimal places.)

2.13 A physical quantity P is related to four observables a, b, c and d as follows: P = a3b2/ {{\sqrt {c}}}  d. The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P?  If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Sol : Given P={\frac {{a}^{3}{b}^{2}} {\sqrt {c}}}{\times }{\frac {1} {d}}

Maximum percentage error in physical quantity P is given by

{\frac {\Delta P} {P}}={\pm }\left [ {3\left ( {{\frac {\Delta a} {a}}} \right )} \right ]{\pm }\left [ {2\left ( {\frac {\Delta b} {b}} \right )} \right ]{\pm }\left [ {{\frac {1} {2}\left ( {{\frac {\Delta c} {c}}} \right )}} \right ]{\pm }\left [ {{\frac {\Delta d} {d}}} \right ]

Maximum percentage error in P is given by

{\frac {\Delta P} {P}{\times }100}={\pm }\left [ {3\left ( {{\frac {\Delta a} {a}}} \right ){\times }100} \right ]{\pm }\left [ {2\left ( {\frac {\Delta b} {b}{\times }100} \right )} \right ]{\pm }\left [ {{\frac {1} {2}\left ( {{\frac {\Delta c} {c}{\times }100}} \right )}} \right ]{\pm }\left [ {{\frac {\Delta d} {d}{\times }100}} \right ]

But {\frac {\Delta a} {a}}{\times }100 = 1%

{\frac {\Delta b} {b}}{\times }100 =3 %

{\frac {\Delta c} {c}}{\times }100 = 4%

{\frac {\Delta d} {d}}{\times }100 = 2%

{\frac {\Delta P} {P}}{\times }100 = {\pm }\left [ {3{\times }\left ( {1} \right )} \right ]{\pm }\left [ {2{\times }\left ( {3} \right )} \right ]{\pm }\left [ {{\frac {1} {2}{\times }\left ( {4} \right )}} \right ]{\pm }\left [ {\left ( {2} \right )} \right ]

= {\pm } [3 + 6 + 2 + 2]%

= 13 %

As the result (13%) has two significant figures, therefore the value of P=3.763 should have only two significant figures. Rounding off the value of P up to two significant figures, we get P = 3.8

2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin 2π t/T

(b) y = a sin vt

(c) y = (a/T) sin t/a

(d) y = {{\frac {a} {\sqrt {2}}}} ( sin2Πt /T + cos2Πt / T )

a = maximum displacement of the particle, v = speed of the particle.      T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Sol:  According to the principle of homogeneity of dimensions, if each term adding or subtracting in a given relation are same then it is correct, if not then it is wrong.

The dimension of LHS of each relation is [L], therefore the dimension of RHS should be [L]

(a) As 2Πt/T is dimensionless , therefore the dimension of RHS = [L]

Thus, the formulae is correct.

(b) Dimension of RHS = [L]sin [LT-1] [T] = [L]sin [L].

As the angle is not dimensionless here so the formulae is wrong. RHS is not equal to LHS.

(c) Dimension of RHS = sin  = [LT-1]sin[TL-1]

As the angle is not dimensionless thus the formulae is wrong. RHS is not equal to LHS.

(d) Dimensions of RHS = [L][ sin  + cos ] = [L]

As the angle is dimensionless and dimensions of RHS is equal to the dimensions of LHS, therefore the formulae is correct.

2.15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c.  (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c.  He writes:

m={\frac {{m}_{0}} {\sqrt {\left ( {1-{{v}^{2}}} \right )}}}

Guess where to put missing c.

 Sol : The relation written by the boy is m={\frac {{m}_{0}} {\sqrt {\left ( {1-{{v}^{2}}} \right )}}}

According to the principle of homogeneity, the dimension of LHS should be equal to the dimension of RHS.

The LHS has dimension of mass, thus RHS should also have dimension of mass.

So, it is only possible when the denominator term is dimensionless.

In order to make the denominator dimensionless, we have to make the term v2 dimensionless, which is possible only by dividing v2 by c2. Thus the relation should be rewritten as

m={\frac {{m}_{0}} {\sqrt {\left ( {1-{\frac {{v}^{2}} {{c}^{2}}}} \right )}}}

Thus LHS = RHS

2.16 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10-10m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Sol: Radius of hydrogen atom (r) = 0.5 Å = 0.5 x 10-10 m

Volume of each hydrogen atom = 4/3(­Π r3)

                                                           = (4/3) x 3.14 x (0.5 x 10-10)3

                                                           = 5.234 x 10-31 m3

Number of atoms in one mole of hydrogen atom = 6.022 x 1023 = Avogadro No.

Atomic Volume of 1 mole of hydrogen = number of hydrogen atoms x volume of one hydrogen atom.

V = 5.234 x 10-31 x 6.022 x 1023 m3     = 3.152 x 10-7 m3

2.17 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen?   (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Sol: Given molar volume of a gas = 22.4 L = 22.4 x 10-3 m3

Diameter of hydrogen molecule (d) = 1 Å = 1 x 10-10 m

Radius of hydrogen molecule = 0.5 x 10-10 m

Volume of each hydrogen atom = 4/3(­Π r3)

                                                           = (4/3) x 3.14 x (0.5 x 10-10)3

                                                           = 5.234 x 10-31 m3

Number of atoms in one mole of hydrogen atom = 6.022 x 1023 = Avogadro No.

Atomic Volume of 1 mole of hydrogen = number of hydrogen atoms x volume of one hydrogen atom.

V = 5.234 x 10-31 x 6.022 x 1023 m3

    = 3.152 x 10-7 m3

The ratio of molar volume to atomic volume = (22.4 x 10-3)/ (3.154 x 10-7)

                                                                                 = 7.1 x 104

This ratio is very large which shows that intermolecular separation in a gas is much larger than the size of a molecule.

2.18 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Sol:  A line joining the object to the eye is called the line of sight. The relative speed of any object outside the fast moving train is decided by angular speed of its line of sight. The line of sight of nearby objects i.e. trees , houses etc. cross large angles in short interval of time and therefore appears to run fast in opposite direction .

The line of sight of very far away objects i.e. hill tops , the moon , the stars crosses at very short angle in the same time  interval and hence appears to be stationary i.e. moving in the direction of train.

2.19 The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun.  That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale.  It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Sol : The parallax (θ) of a star is the angle subtended by the semi- major axis of the earth’s orbit on that star.

 b = AB = Base line

    = Diameter of earths orbit

     = 3 x 1011m

Parallax angle (θ) = 1s = 1/60 min

                                 = 1/ (60 x 60 )

                                 ={ 1/(60 x 60 ) } x (Π/180) rad = 4.85 x 10-6 rad

From parallax method (l) = b/2θ

                                               = ( 3 x 1011) / ( 2 x 4.85 x 10-6)

                                                = 3. 08 x 1016  m

1 parsec = 3.08 x 1016  m

2.20 The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs?  How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Sol: distance of the nearest star = 4.29 ly

                                                           = (4.29 x 9.46 x 1015 ) m

                                                           =  (4.29 x 9.46 x 1015 )/ (3.08 x 1016 ) parsec

                                                           = 1.318 parsec

                                                           = 1.32 parsec

2.21 Precise measurements of physical quantities are a need of science.  For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Sol: Precise measurement of physical quantity such as length, time, mass etc. is a basic requirement of development of astronomy, nuclear physics, medical sciences, and crystallography.

  • In the measurement of astronomical distances such as distance of moon from earth by laser beam, an accurate measurement of time is required. A very small mistake in the measurement of that time can produce a large mistake in the accurate distance of moon from the earth which can cause of failure to reach moon. This time is of order 10-9.
  • X-ray spectroscopy is used to find the interatomic separation.
  • To measure the mass of atoms, the mass spectrometer is developed.
  • In case of medical science also precise measurement of length is necessary to find location , size and mass of tumour or any infected area of body. Any mistake could cause fatal situations.

2.22 Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) The total mass of rain-bearing clouds over India during the Monsoon

(b) The mass of an elephant

(c) The wind speed during a storm

 (d) The number of strands of hair on your head

 (e) The number of air molecules in your classroom.

Sol : (a) The total mass of rain-bearing clouds over India during the Monsoon

 The total mass of rain bearing clouds over India during the monsoon :

If meteorological record 150 cm of average rainfall during monsoon then

Height of average rainfall (h) = 150 cm = 1.5 m

Area of India (A) = 3.3 million sq. km = 3.3 x 1012 m2

Volume of rain water = A x h = 3.3 x 1012 x 1.5 = 4.95 x 1012

Density of water = р= 1 x 103 kg/m3

Mass of water = Density x volume = 4.95 x 1012 x 103 = 4.95 x 1015 kg

(b) The mass of an elephant

We can measure the mass of an elephant using law of flotation which states that the mass of an object dipped in a liquid is equal to the mass of liquid displaced by the submerged part of the body in liquid.

Let A known base area of a plank   floating in the sea. Let the depth of sea be d1.

Volume of water displaced = A d1

 Measuring the depth of the plank with an elephant on-board.

 Volume of water displaced = A d2

From the above equations, the volume of water displaced by the elephant          =Р d– Р d2

Water density = Р

Elephant’s mass = AР (d2 – d1)

( c ) The wind speed during storm

For measuring the speed of wind, anemometer is used.

When wind blows anemometer rotates and it gives rotation per second which measures the speed of wind.

(d ) The number of strands on hair.

Assuming uniform distribution of strands of hair on head ,

The number of hair = {\frac {areaofhead} {areaofcross-sectionofhair}}

The thickness of a strand of hair is measured by appropriate instrument , if it is obtained

 d = 4 x 10 -5 m = 4 x 10-3 cm

The area of cross- section of human hair = Π(d/2)2

                                                                           = (3.14 x 4 x 4 x 10-6 )/4

                                                                           = 3.14 x 4 x 10-6  cm2

Average radius of human head (r ) = 10 cm

Area of human head = Πr2 = 3.14 x 100 = 314 cm2

The number of strands of hair = 314/(3.14 x 4 x 10-6 )

                                                       = 25 x 106

( e) The number of air molecules in your classroom

We measure the length (l) , breadth (b) and height (h) of our classroom .

Thus volume of classroom = l x b x h

We know that at NTP, one mole of air molecules occupies a volume of 22.4 x 10-3 m3.

Number of air molecules = Avogadro no./ volume of 1 mole of air molecules

                                              = ( 6.022 x 1023 / 22.4 x 10-3) x l x b x h

2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ?  Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m.

Sol :   Given mass of sun =  2.0 ×1030 kg

          Radius of sun (r) =   7.0 × 108 m

Density of the sun = Mass of the sun (M) / Volume of the sun (V)

{{\rho }={\frac {M} {\frac {4{\pi }{{r}^{3}}} {3}}}}

= (¾) x (M/4Πr3)

 = 3 x 2 x 1030 / {4 x 3.14 x (7 x 108)3}

= 3 x 1030 / (6.28 x 343 x 1024)

= 1.392 x 103

= 1.4 x 103 kg/m3

The density is of the order of density of solids and liquids and not of gases.

The temperature of inner core of the sun is 107K while the temperature of the outer layers is nearly 6000K.

At so high temperature no matter can exist in its solid or liquid state. Every matter is highly ionised and present as a mixture of nucleus, free electrons and ions which is plasma.

The density of plasma is so high due to inward gravitational attraction on outer layers due to inner layers.

2.24 When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter .

Sol : Distance of Jupiter from earth (d) = 8.247 x 106 km

Angular diameter of Jupiter (θ) = 35.72’’

But 1  = 60’ = (60 x 60)’’

1’’ = 1/(60 x60)

     = Π/(60 x 60 x 180) rad

1’’ = 4.85 x 10-6 rad

Angular diameter of Jupiter (θ) = 35.72 x 4.85 x 10-6) rad

If D is the diameter of Jupiter then ,

Angular diameter (θ) = D/d

D= θd

    = 35.72 x 4.85 x 10-6 x 824.7 x106

    = 142873 km

     = 1.429 x 105 km

Thus diameter of Jupiter = 1.429 x 105 km

For more clear understanding download the pdf given above.

Categories: General


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